Lab manual for various experiments in chemical engineering

Chemical Engineering Department, Birla Institute of Technology & Science Pilani, Pilani Campus

A LABORATORY MANUAL FOR Chemical Engineering Laboratory –I Edited By Dr Suresh Gupta & Dr Hare Krishna Mohanta

Department of Chemical Engineering, Birla Institute of Technology & Science (BITS), Pilani September 2014

1


CONTENTS CYCLE - I S. No.

Experiment

Page No.

1.

a. b.

Flow through Fluidized Bed (Gas and Water) Flow through Packed Bed (Gas and Water)

3

2.

a. b. c. a. b. c. a. b.

Losses due to pipe fittings Losses due to friction in pipes Drag Coefficient determination Bernoulli’s Theorem verification Discharge through venturi, orifice and rotameter Flow through tubular pipe Pitot tube experiment (Air and Water) Reynolds Apparatus

10

5.

a. b.

Centrifugal pump characteristics Reciprocating pump characteristics

42

6.

a. b. c. a. b. a. b. a. b. a. b.

Heat Pipe demonstrator Thermal Conductivity of solids Thermal conductivity of liquids Drop wise and film wise condensation Unsteady state heat transfer unit Heat Transfer in agitated vessel Fluidized bed heat transfer unit Parallel flow & Counter flow heat exchanger Shell and Tube heat exchanger Plate type Heat Exchanger Finned tube heat exchanger

52

3.

4.

7. 8. 9. 10.

22

34

61 76 83 92

CYCLE - II S. No.

Experiment

Page No.

11.

Stefan-Boltzmann Apparatus

151

12.

Cross-circulation drying apparatus

110

Vapour in air diffusion Open pan evaporator

114

14.

Simple/Differential distillation setup

120

15.

Batch crystallizer

125 128

17.

Steam distillation setup Vapor liquid equilibrium setup Two phase flow

18.

Mass transfer with chemical reaction

141

19.

Adsorption in packed bed

144

20.

Humidification in wetted wall column

148

13.

16.

a. b.

a. b.

2

136


EXPERIMENT NO. 1(a)

FLOW THROUGH FLUIDIZED BED (AIR & WATER) 1. Aim Study of hydrodynamic and bed characteristics for flow through fluidized beds. 2. Objective  To determine the minimum fluidization velocity experimentally as well as theoretically.  To study the bed expansion characteristics of the fluidized bed (plot log NRe vs. porosity and pressure drop vs. NRe). 3. Apparatus Stop watch, graduated cylinders, beakers 4. Theory When a liquid or a gas is passed at very low velocity up through a bed of solid particles, the particles do not move, and the pressure drop is given by the Ergun equation. If the fluid velocity is steadily increased, the pressure drop and the drag on individual particles increase, and eventually the particles start to move and become suspended in the fluid. The terms ‘fluidization’ and ‘fluidized bed’ are used to describe the condition of fully suspended particles, since the suspension behaves like a dense fluid. Fluidized beds are used extensively in the chemical process industries, particularly for the cracking of high-molecular-weight petroleum fractions. Such beds inherently possess excellent heat transfer and mixing characteristics. These are devices in which a large surface area of contact between a liquid and a gas, or a solid and a gas or liquid is obtained for achieving rapid mass and heat transfer and for chemical reactions. The fluidized bed is one of the best known contacting methods used in the processing industry, for instance in oil refinery plants. Among its chief advantages are that the particles are well mixed leading to low temperature gradients, they are suitable for both small and large scale operations and they allow continuous processing. There are many well established operations that utilize this technology, including cracking and reforming of hydrocarbons, coal carbonization and gasification, ore roasting, Fisher-Tropsch synthesis, coking, aluminium production, melamine production, and coating preparations. Nowadays, you will find fluidized beds used in catalyst regeneration, solid-gas reactors, combustion of coal, roasting of ores, drying, and gas adsorption operations. The application of fluidization is also well recognized in nuclear engineering as a unit operation for example, in uranium extraction, nuclear fuel fabrication, reprocessing of fuel and waste disposal. When a fluid is admitted at the bottom of a packed bed of solids at a low flow rate, it passes upward through bed without causing any particle motion. If the particles are quite small, flow in the channels between the particles will be laminar and the pressure drop across the bed will be proportional to the superficial velocity (Vo) and for turbulent situations, pressure drop across the bed increase nonlinearly with the increase in the superficial velocity. As the velocity is gradually increased, the pressure drop increases, but particles do not move and the bed height remains the same. At a certain velocity, the pressure drop across the bed counterbalances the force of gravity on the particles or the weight of the bed, and any other further increase in velocity causes the particles to move and the true fluidization begins. For a 3


high enough fluid velocity, the friction force is large enough to lift the particles. This represents the onset of fluidization once the bed is fluidized pressure drop across the bed remains constant, but the bed height continues to increase with increasing flow.

Figure 1 Fluidization regimes In order to determine the pressure drop through a fixed bed for any flow condition, the Ergun equation (1952) can be used:

dp is the size of particles (µm) L is the height of the bed (m) µ is the viscosity of sir (N/m2.s) U is the average superficial velocity (m/s) ϵ is the bed voidage or porosity ρ is the density of air/water (kg/m3) ∆P is the pressure drop across the bed (N/m2) The average Reynolds number based on superficial velocity which is given by,

If the Reynolds number is less than 10 then it is laminar flow and is greater than 2000 it is turbulent flow. The rest of the values lie in the transition regime. If the flow rate of air/water Q is measured in litres, A is the bed cross-sectional area and U is the superficial velocity in m/s, then

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Theoretically at incipient fluidization (the stage in the fluidized bed where the force on the solid is enough to balance the weight of the solid material),

∆P is in mm of manometer The pressure drop at fluidization can also be predicted by the equation,

ρp is the particle density (kg/m3) ρ is the fluid density (kg/m3) g is the gravitational force (m/s2) Minimum fluidization velocity Umf, the minimum fluidizing velocity, is frequently used in fluid-bed calculations and in quantifying one of the particle properties. This parameter is best measured in small-scale equipment at ambient conditions. The correlation given below can then be used to back calculate dp. This gives a particle size that takes into account effects of size distribution and sphericity. The correlation can then be used to estimate Umf at process conditions. If Umf cannot be determined experimentally, use the expression below directly. Assumption: - Consider the uniform particle size. Remf = (1135.7 +0.0408Ar) 0.5 - 33.7

(Wen and Yu correlation for particles dp> 100 µm)

Re mf  d p  f U mf / 

Ar  d 3 p  f  s   f g /  2

(Ar is the Archimedes Number)

For particles of dp < 100 µm, Baeyens and Geldart (1977) can be used,

5. Experimental procedure 1. The height of the static bed Z1 i.e. when there is no flow of water/air (porosity 1) was noted. 2. The flow of air/water in the column is started and the flow rates from the rotameter were noted. 3. The corresponding bed heights and pressure drop values were noted. 4. The flow rates were increased steadily and similar data were collected at different intervals. 5. 6 to 8 readings of flow rates were varied and reading were taken. 6. Steady state flow rate of water was ensured at each point.

5


6. Observations and calculations Particle size Porosity Inside diameter of the column (D) Cross-sectional area of fluidized column Density of water ρ (at T0C) Viscosity of water µ (at T0C) Density of air Viscosity of air

= = = = = = = =

8 mm 0.6 0.055m 0.002375 m² 1000 kg/m3 0.798*10-3 kg/m-s 1.1687 kg/m3 1.8633x10-5 Pa.s

Velocity V (m/s) = volumetric flow rate of water / cross section area of column Initial height of static bed in the column (Z1) Porosity of the fluidized Bed: If Z1 and 1 are the height and porosity of the static bed and Z2 and 2 are the height and porosity of fluidized bed, the (1 1 ) (Total volume of column equal to total volume of solid) Z 2  Z1 (1 2 ) Z2 = 2 Z2 = Z1 (1- 1) Z2 - Z1 (1- 1) = 2 Z2 

2 = Z2 - Z1 (1- 1) / Z2

Note: Keep the observation tables for air and water separately. The sample calculations for the same should be shown separately.

Obs No.

Volumetric Height of Flow rate of Velocity(v) Porosity Bed (Z2) water/air in m/s () m³/s

Pressure drop (P)

NRe

7. Results & Discussions References: 1. Warren, L McCabe, Smith, J C , and Harriott, P, Unit Operations of Chemical Engineering, 6th edition, McGraw Hill, New Delhi, India, 2000. 2. Kunii, Levenspeil O, Fluidized Engineering, Robert E Krieger Publishing Company, Huntington, New York, 1977

6


EXPERIMENT NO. 1(b)

FLOW THROUGH PACKED BED (AIR & WATER) 1.

Aim a) To plot Modified Reynolds’s No. (NRe,m) vs. Modified friction factor (f) on a log-log graph b) To verify Ergun’s equation

2. Apparatus Stop watch, Graduated cylinder, packing as Hollow rings of Glass, Manometer, beakers etc. 3. Theory A packed bed is a hollow tube, pipe, or other vessel that is filled with a packing material. The packing can be randomly filled with small objects like Raschig rings, Pall rings or else it can be a specifically designed structured packing. Packed column is a pressure vessel that has a packed section. In general, packed towers are used for bringing two phases in contact with one another and there will be strong interaction between the fluids in this case between solid and fluid. As the fluid passes through the bed, it does so through the voids presents in the bed. The voids form continuous channels throughout the bed. The flow may be laminar through some channels and turbulent in other channels. Just as with straight pipes, Ergun relates the flows and pressure drops to a Reynolds Number and friction factor respectively. The Reynolds number for packed beds, Rep , depends upon the controlled variable i.e. Superficial velocity Vo and the system parameters ρ, ε, μ, and Dp and is defined as (Bird et. al., 1996),

Dp is the equivalent spherical diameter of the particle, V0 is the superficial velocity defined as the volumetric flow rate divided by the cross-sectional area of the column, ρ is the fluid density, ε is the dimensionless void fraction defined as the volume of void space over the total volume of packing, and μ is the fluid viscosity. The friction factor, fp, depends upon V0 and the pressure drop, ΔP, and system parameters, and is defined as (Bird et al., 1996)

P = Pressure drop in kg/cm2  = porosity L = effectively length of the bed, cm V = Superficial-fluid velocity, based on empty cross-section of pipe, m/sec  = density of the fluid  = viscosity of fluid

7


4.

Working principle

Figure 1 Schematic of Packed Bed Column The packed column shown above is filled with packings, in our case the packing is Raschig glass rings. Now, in this the water is passed through the bottom of column and the flow happens through the packing thus creating friction , this friction leads to pressure drop in the column which is measured by the manometer attached to the two ends of packed column. Also, the flow here is regulated by the rotameter attached to the column. An increase in flow increases the friction and thus leads to an increase in pressure drop. 5. Procedure a. Water/air was allowed to flow from bottom to top in a packed bed. b. Flow of water/air is regulated to vary the flow rate. c. The corresponding pressure drop values were noted by means of the manometer for 6 to 8 different rates of flow of fluid. d. The flow rate values were noted and it was ensured that the steady state flow rates have been achieved. 6. Observations Inside diameter of the tube Volume of each particle Surface area of the particle Porosity of the bed Effective length of packing Temperature of fluid Manometer fluid,  Hg Density of water ρ (at T0C) Viscosity of water µ (at T0C) Density of air Viscosity of air Manometer Reading h (metres)

= = = = = = = = = = =

8

50 mm 1.57x10-6 m3 0.000628 m2 0.3 750 mm ambient 13,534 kg/m³ 1000 kg/m3 0.798*10-3 kg/m-s 1.1687 kg/m3 1.8633x10-5 Pa.s


S. No.

Volumetric Manometer flow rate of reading. water. Q  h (m) of Hg 3 m /sec

Superficial Velocity of Pressure drop/length water = Vo (Q /Cross  P h g ( m   )  N / m2 sectional area of the L L column m/s)

Note: Calculate friction factor and modified Reynolds No. and plot on a log-log graph separately for air and water. The observation columns and sample calculations for both of them should be shown separately. Calculations Certain assumptions are carried out before calculating the Friction factor and Reynolds number. First, we assume that there is no channeling in the packed bed. Channeling occurs when the fluid flowing through the packed bed finds a “preferred path” through the bed. We also assume that the diameter of the packing is much smaller than the diameter of the column as well. The maximum recommended particle diameter is one-fifth of the column diameter. We assume that velocity, particle diameter and void fraction behaves as a bulk behavior and hence we can use an average values. Initially, we calculate diameter of the particle, DP = Calculate friction factor as,

and modified Reynolds number as,

7. Results and Discussion Modified Reynold’s number, friction factor are calculated accordingly to the changes in variables affecting and Ergun’s equation is verified. References: a) Bird, R. Byron, Transport Phenomena. Madison, Wisconsin: John Wiley & Sons, 1996. b) Geankoplis, Christie J., Transport Process and Unit Operations. 4th ed., New Jersey: Prentice Hall, 2003. c) Ergun, Sabri, “Fluid Flow through Packed Columns.” Chemical Engineering Progress, Vol. 48, No 2. , 89-94 , 1952.

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EXPERIMENT NO. 2(a)

LOSSES DUE TO PIPE FITTINGS, SUDDEN ENLARGEMENT & CONTRACTION 1 Aim To determine the loss co-efficient for following sections (a) Sudden expansion (b) Sudden contraction (c) Tee – junction (d) Pipe bend (e) Elbow 2. Apparatus FHP (Fractional Horse Power) pump driven pipe system consisting of elbow bend, Teejunction, sudden enlargement and contraction, manometer with tapping into pipe for head readings, sump and discharge tanks, stopwatch. 3. Theory In Hydraulic Engineering practice, it is frequently necessary to estimate the head loss incurred by a fluid as it flows along a pipeline. For example, it may be desired to predict the rate of flow along a proposed pipe connecting two reservoirs at different levels. Or it may be necessary to calculate what additional head would be required to double the rate of flow along an existing pipeline. Loss of head is incurred by fluid mixing, which occurs at fittings such as bends or valves, and by frictional resistance at the pipe wall. Where there are numerous fittings and the pipes are short, the major part of the head loss will be due to the local mixing near the fittings. For a long pipeline, on the other hand, skin friction at the pipe wall will predominate. Loss of head due to change in cross-section, bends, elbows, valves and fittings of all types fall into the category of minor losses in pipe lines. In long pipe lines the friction losses are much larger than these minor losses and hence the latter are often neglected. But, in shorter pipelines their consideration is necessary for the correct estimate of losses. When there is any type of bend in pipe, the velocity of flow changes, due to which the separation of the flow from the boundary and also formation of eddies, takes place. Thus the energy is lost. The losses of head due to bend in pipe: V2 hL  K L 2g The minor losses in contraction can be expressed as: V12 hL  K L 2g The minor losses in enlargement can be expressed as: (V  V2 ) 2 hL  K L 1 2g Where, hL = minor loss or head loss KL = Loss coefficient V = velocity of fluid. 10


V1 V2

= velocity of fluid in pipe of small Diameter. = velocity of fluid in pipe of large Diameter.

4.Experimental set-up The apparatus consist of a ½” smooth and sharp bends and elbow, a sudden expansion from ½” to 1”, sudden contraction from 1” to ½” and ½” ball valve, gate valve, tee fitting and pipe union joint. Pressure tapings are provided at inlet and outlet of these fittings at suitable distance. A differential manometer fitted in the line gives pressure loss due to fittings. Supply to the pipeline is made through centrifugal pump, which deliver water from sump tank. The flow of water in pipeline is regulated by means of Control valve & By-Pass valve. Discharge is measured with the help of measuring tank and stop watch. 4.1 Utilities Required: a) Power supply: Single Phase, 220 Volts, 50 Hz, 5 Amp with Earth. b) De-mineralized Water. c) Drain. d) Space required: 1.6m x 0.6m

Schematic Diagram for Losses due Pipe Fittings, Sudden Enlargement & Contraction Apparatus 5

Experimental procedure: a) Close the drain valves provided. b) Fill Sump tank ¾ with clean water and ensure that no foreign particles are there. c) Close all flow control valves given on the water line, pressure taps of manometer connected to different pipe -fittings and open by-pass valve. d) Now switch on the main power supply (220 Volts AC, 50 Hz) and start the pump. e) Operate the flow control valve to regulate the flow of water in the desired test section. f) Open the pressure taps of manometer of related test section, very slowly to avoid the blow of water on manometer fluid. 11


g) Now open the air release valve provided on the manometer, slowly to release the air in manometer. When there is no air in the manometer, close the air release valves. h) Adjust water flow rate in desired section with the help of control valve and record the manometer reading. i) Measure the flow of water, discharged through desired test section, using stop watch and measuring tank. j) Repeat same procedure for different flow rates of water, operating control valve and by-pass valve. k) When experiment is over for one desired test section, open the By-Pass Valve fully. Then close the flow control valve of running test section and open the Control valve of secondly desired test section. l) Repeat same procedure for the 9 selected test sections of the two bends, elbow, contraction, expansion, tee and union joint and both gate and ball valves. m) When experiment is over, close all manometers pressure taps first, switch off pump, and switch off power supply to panel. 6 Observation & Calculations 6.1 Specification: Sudden Enlargement : From 16.5mm to 27mm Sudden Contraction : From 27mm to 16.5mm. Bend : 1/2" Elbow : 1/2" Ball valve : 1/2" Gate valve : 1/2" Tee joint : 1/2" Union joint : 1/2" Water Circulation : FHP Pump. Flow Measurement : Using Measuring Tank with Piezometer, Capacity 40 Ltrs. Sump Tank : Capacity 80 Ltrs. Stop Watch : Electronic. Control Panel Comprises of: Standard make On/Off Switch, Mains Indicator, etc. 6.2 Formulae: (a) Discharge: A* R Q t *100 (b) Velocity: Q (Velocity in ½” pipe) V1  a1 Q (Velocity in 1” pipe) V2  a2 (c) Loss of Head (for Contraction): V2 hL  K L 1 2g (d) Loss of Head (for Expansion): (V  V2 ) 2 hL  K L 1  K LC 2g

---------------- (1)

---------------- (2) ---------------- (2a)

---------------- (3)

---------------- (4) 12


(e) Experimental loss of head h *13.6 hL  100 (f) Loss Co-efficient: 2g K L  hL 2 V1 6.3 Data: A = Area of measuring tank s = Specific gravity of Hg g = Acceleration due to gravity d1 = Dia. of smaller pipe d2 = Dia. of larger pipe a1 = Cross-sectional area of Small Dia. pipe a2 = Cross-sectional area of Large Dia. pipe 6.4 Observation Table: S. Type of Fitting Pressure No. Difference, (cm) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 6.5 Calculation Table: S. Discharge, Velocity, No Q V1

--------------- (5)

---------------- (6) = = = = = = =

0.1m2 13.6 9.81 m/sec2 0.0165 m 0.027m. 2.14 x 10-4 m2 5.73 x 10-4 m2

Rise of water level in Time taken for R h measuring tank R (cm) (sec.)

Velocity, V2

Loss of Kinetic Energy, C

6.6 Nomenclature: Q = discharge, m3/s V1 = velocity of fluid in pipe of Small Diameter (m). V2 = velocity of fluid in pipe of Large Diameter (m). a1 = Cross-sectional area of Small Dia. pipe a2 = Cross-sectional area of Large Dia. pipe KL = loss coefficient. 13

Loss of head hL

Loss coefficient K L


h hL R t

= = = =

7.

Result &discussion

8.

Conclusion

9.

Precaution & maintenance instructions: Do not run the pump at low voltage i.e. less than 180 Volts. Never fully close, the delivery line and by-pass line valves simultaneously. Always keep apparatus free from dust. To prevent clogging of moving parts, run pump at least once in a fortnight. Frequently Grease/Oil the rotating parts, once in three months. Always use clean water. It apparatus will not in use for more than one month, drain the apparatus completely, and fill pump with cutting oil.

a) b) c) d) e)

pressure difference, m head loss, m Rise of water level in measuring tank (m). Time taken for R (sec.).

f) 9.1 Troubleshooting: a) If pump gets jam, open the back cover of pump and rotate the shaft manually. b) If pump gets heat up, switch off the main power for 15 minutes, avoid closing the flow control valve, and by pass valve simultaneously during operation. References: a) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”, 4th ed., McGraw Hill, NY, 1985. b) Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering Vol. - 1”, 5th ed., Asian Books ltd., ND, 1996. c) Brown, G.G., “Unit Operations”, 1st ed., CBS, ND, 1995. d) Foust, A.S., et. al., “Principles of Unit Operations”, 2nd ed., John Wiley, NY, 1980.

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EXPERIMENT NO. 2(b) LOSSES DUE TO PIPE FRICTION 1. Aim To determine the losses due to friction in pipes and friction factor for Darcy - Weisbach equation 2. Apparatus FHP (Fractional Horse Power) pump driven pipe system consisting of Galvanized Iron (GI) pipe of ½”, 1” and Stainless Steel (SS) pipe of ½” diameter, manometer with tapping into pipe for head readings, sump and discharge tanks, stopwatch. 3. Theory When a fluid is flowing through a pipe, it experiences some resistance due to which some of the energy is lost. This loss of energy in the pipelines comes under major energy losses and minor energy losses. In long pipelines, the friction losses are much larger than the minor losses and hence, the latter are often neglected. The losses due to friction in the pipelines are known as major energy losses. The friction in the pipeline is due to a viscous drag between the stream bands of fluid. The stream bands adjacent to the solid surface are always at rest relative to the wetted surface. The viscous drag is due to the molecular attractions between the molecular of the fluid. It is found that the total friction resistance to fluid flow depends on the following: 1. The area of the wetted surface 2. The density of the fluid 3. The surface roughness 4. It is independent of the fluid pressure 5. It increase with the square of the velocity It depends on a frictional co-efficient. The loss of head in pipe due to friction is calculated from Darcy-Weisbach equation which has been given by: 4 f LV 2 hf  2g d Where: hf = f = L = V = d = g =

loss of head due to friction Co-efficient of friction distance between pressure point mean velocity of fluid diameter of pipe Acceleration due to Gravity

4. Experimental setup The apparatus consist of three pipes of varying diameters and materials for which common inlet connection is provided with control valve to regulate the flow, near the down stream end of the pipe. Pressure tapings are taken at suitable distance apart between which a manometer is provided to study the pressure loss due to the friction. Discharge is measured with the help of measuring tank and stopwatch.

15


Schematic Diagram for Friction in Pipe Lines Apparatus 4.1 Utilities Required a) Power supply: Single Phase, 220 Volts, 50 Hz, 5 Amp with Earth. b) De-mineralized Water Supply. c) Drain. d) Space required : 1.6m x 0.6m 5. Experimental procedure a) Close the drain valves provided. b) Fill Sump tank ¾ with clean water and ensure that no foreign particles are there. c) Close all flow control valves given on the water line, pressure taps of manometer connected to different pipe -fittings and open by-pass valve. d) Now switch on the main power supply (220 Volts AC, 50 Hz) and start the pump. e) Operate the flow control valve to regulate the flow of water in the desired test section. f) Open the pressure taps of manometer of related test section, very slowly to avoid the blow of water on manometer fluid. g) Now open the valve provided on the manometer, slowly to release the air in manometer. When there is no air in the manometer, close the valves. h) Adjust water flow rate in desired section with the help of control valve and record the manometer reading. i) Measure the flow of water, discharged through desired test section, using stop watch and measuring tank. j) Repeat same procedure for different flow rates of water, operating control valve and by-pass valve. k) When experiment is over for one desired test section, open the By-Pass Valve fully. Then close the flow control valve of running test section and open the Control valve of secondly desired test section. l) Repeat same procedure for different flow rates of water, operating Control Valve and By-Pass valve.

16


m) When experiment is over, close all manometers pressure taps first, switch off pump, and switch off power supply to panel. 6. Observation & calculation 6.1 Specification Pipes (3 Nos.) : Pipe Test Section Water Circulation Flow Measurement

: : :

Material GI of ½” & 1” diameter. SS of ½” diameter Length 1.1 m for each Pipe FHP Pump. Using Measuring Tank with Piezometer, Capacity, 40 Ltrs. Capacity 80 Ltrs. Electronic.

Sump Tank : Stop Watch : Control Panel Comprises of: Standard make On/Off Switch, Mains Indicator, etc. Tanks will be made of Stainless Steel. The whole set-up is well designed and arranged in a good quality painted structure. 6.2 Formulae 1. Head losses,  h  m  hf   1 , m of water. 100  W  2. Co-efficient of Friction: h 2 gd f  f 4 LV 2 3. Discharge (Q): AxR 3 Q m /s t *100 4. Velocity of Fluid: Q V  m/ s a 6.3 Data: Area of measuring tank, A Sp. gravity of Hg Acceleration due to gravity, g Inside Diameter of Pipe, d For GI pipe (1”) For GI pipe (1/2”) For SS pipe (1/2”)

Cross-section area of pipe, a For GI pipe (1”) For GI pipe (1/2”) For SS pipe (1/2”) Distance between pressure points, L

= = =

0.1m2 13.6 9.81m/sec2

= = =

0.027 m 0.016.5 m 0.016 m

= = = =

5.70 x 10-4 m2 2.14 x 10-4 m2 2.01 x 10-4 m2 1.1 m each for all pipes

6.4 Observation Table: (for For GI pipe 1”, GI pipe 1/2”, and SS pipe 1/2”) 17


Extent of Pressure difference Rise of water level Valve open h (cm) measuring Tank R (cm) ½ Open Full Open

in Time taken, t (sec.)

6.5 Calculation Table: Pressure Extent of Valve head, hf open (m)

Discharge, Q (m3/s)

Velocity Fluid, V (m/s)

of

Fanning factor, f

Friction

½ Open Full Open 6.6 Nomenclature: A = Area of measuring tank a = Cross-section area of pipe d = Inside Diameter of Pipe g = Acceleration due to gravity h = manometer reading, cm hf = Pressure head of water (m) R = Rise of water level in measuring Tank (m) t = Time taken for R (sec) L = distance between two points 7.

Result & discussion

8. Precautions 1. Do not run the pump at low voltage i.e. less than 180 Volts. 2. Never fully close the Delivery line and By-Pass line Valves simultaneously. 3. Always keep apparatus free from dust. 4. To prevent clogging of moving parts, Run Pump at least once in a fortnight. 5. Frequently Grease/Oil the rotating parts, once in three months. 6. Always use clean water. 7. If apparatus will not in use for more than one month, drain the apparatus completely, and fill pump with cutting oil. 8.1 Troubleshooting: 1. If pump gets jam, open the back cover of pump and rotate the shaft manually. 2. If pump gets heat up, switch off the main power for 15 minutes and avoid closing the flow control valve and by pass valve simultaneously during operation. References e) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”, 4th ed, McGraw Hill, NY, 1985. f) Foust, A.S., et. al., “Principles of Unit Operations”, 2nd ed., John Wiley, NY, 1980. g) Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering Vol. - 1”, 5th ed., Asian Books ltd., ND, 1996.

18


EXPERIMENT NO. 2(c)

DRAG COEFFICIENT 1. Aim & Apparatus To verify Stoke’s law and to study the variation of the drag coefficient with Reynolds number (Re) for sphere. Experimental setup of 3 cylinders filled with three different liquids – demineralised water, glycerine and hydraulic oil, spherical balls, measuring scales and stopwatch. 2. Theory When a body moves through any fluid, it experiences a resistance, which acts in a direction opposite to that of the motion of the body. This resistance is called the drag force (FD) and it depends on the size of the body, velocity with which it moves and the viscosity of the fluid. According to stoke, the drag force acting on a sphere moving through a fluid due to its weight is given by the following expression: FD = 3πµU₀Dp (i) Where Dp is the diameter of sphere, µ is the viscosity of fluid, and U₀ is called the terminal fall velocity. Terminal velocity is defined as the velocity attained by a body in falling through a fluid at rest, when the drag force on the body is equal to the submerged weight of the body. For equilibrium condition in the case of a freely falling sphere through a liquid, drag force plus the buoyant force must be equal to the weight of the sphere and is given by the expression (ii) 2 Where Fg is the drag force, g is acceleration due to gravity (9.8 m/sec ), ρp is density of spherical particle, and ρf is the density of fluid stream. The drag coefficient predicted by Stoke’s law is given by the expression , where (iii) where ν is the kinematic viscosity of fluid stream (m2/sec). Thus, coefficient of drag CD varies with Reynolds number. Experiments have shown that Eq. (ii) holds well for Re < 0.2, and the sphere is falling in an infinite fluid. If the fluid is not infinite in extent but is confined within a container (finite dimensions), then the resistance to motion is increased, and in such a case the modified value of drag coefficient, as given by the following expression, should be used: CD = , (1+2.1 ) (vi) Where D is the smallest lateral dimensions of the container and D is the diameter of sphere. Also the observed fall velocity U is corrected in Eq. (iii) by using the following expression in order to get the fall velocity corresponding to infinite fluid medium: Corrected velocity, U₀ = U (1+2.4 ) Where, D is the diameter of the container. 3. Experimental set-up The set-up consists of three transparent vertical cylinders of diameter 100 mm and height 1 m. A hopper with a valve is provided at the bottom of the cylinder to collect the spheres. The cylinder is supported by four vertical posts and fixed to a MS table. A vertical scale is fixed

19


on the surface of a cylinder. The cylinders are filled with fluids of varying viscosities such as glycerine, hydraulic oil and water. 4. Experimental procedure 1. Measure the diameters of spheres and note down their materials. 2. Determine the mass of the spheres on electronic balance. 3. Mark two lines on the cylinder for measurement of the vertical distance (L) for the determination of terminal velocity. The upper line should be at a depth of 100 mm or more from the free surface so that the terminal velocity is achieved. 4. Hold the sphere with a finger and a thumb and bring it up to the fluid level. Leave the sphere gently (wet the sphere before dropping it). 5. Note down the time taken by the sphere in falling through distance L. 6. Repeat steps (4) and (5) for other diameters of spheres (Set 1). 7. Repeat steps (4) and (6) for spheres of other material (Set 2).

Liquid 1

Liquid 2

Liquid 3

Fig 1: Stoke’s Apparatus 5. Observations and calculations 5.1 Specification: Liquid in the cylinder Mass density of liquid, ρf

Mass density of sphere Kinematic viscosity of liquid at T 20°C, v

Diameter of cylinder, D

= = 1261kg/m³ (glycerol), 1080kg/m³ (hydraulic oil), 1000kg/m³ (water) = 2070 kg/m3 = 1.006x10-6 m²/s (water), 1.180x10-6 m²/s (glycerol) 1.25*10-4 m2/s (hydraulic oil) = 87mm 20


Distance of fall, L = Diameter of sphere = 0.819cm Time of falling for sphere (t) = ____ sec Terminal Velocity (U) (Length/time for fall) = ______ (m/sec) 5.2 Observation Table: S.No Fluid type 1.

Water

2.

Glycerol

3.

Hydraulic Oil

Length of liquid column (m)

5.3 Calculation Table: Fluid Terminal Corrected type velocity terminal velocity

Reynolds number

Discharge coefficient

Sphere falling time (s)

Corrected Discharge coefficient

Fg

FD

6. Result & discussions Plot a log-log graph of CD (corrected discharge coefficient) vs. NRe (Reynolds number) 7. Conclusion 8. Precautions   

The sphere needs to be dropped from the centre of the cylinder. Extreme care should be taken while taking out the sphere from the cylinder to avoid loss of excess fluid. Wet the sphere with the specific fluid before dropping it in the cylinder.

References: h) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”, 7th ed., McGraw Hill, NY, 2005. i) W.L Badger & J.T. Banchero., “Introduction to Chemical Engineering”, 26th reprint, McGraw Hill, NY, 2011.

21


EXPERIMENT NO. 3(a)

BERNOULLI’S THEOREM 1. Aim & Objective To experimentally verify Bernoulli equation for fluid flow in horizontal pipe. 2. Apparatus FHP pump run setup consisting of 4 rectangular tanks, horizontal pipe section, 8 piezometric tubes with measuring scales, stopwatch. 3. Theory Energy of a Liquid in Motion The energy, in general, may be defined as the capacity to do work. Though the energy exists in many forms, yet the following are important from the subject point of view: a) Potential Energy b) Kinetic Energy, and c) Pressure Energy. Potential Energy of a Liquid in Motion It is the energy possessed by a liquid particle, by virtue of its position. If a liquid particle is Z meters above the horizontal datum (arbitrarily chosen), the potential energy of the particle will be Z meter-kilogram per kg of liquid. Potential head of the liquid, at that point, will be Z meters of the liquid. Kinetic Energy of a Liquid Particle in Motion: It is the energy, possessed by a liquid particle, by virtue of its motion or velocity. If a liquid particle is flowing with a mean velocity of v meter per second, then the kinetic energy of the particle will be v2/ 2g m-kg per kg of liquid. Velocity head of the liquid, at that velocity, will be v2/ 2g meter of the liquid. Pressure Energy of a Liquid Particle in Motion: It is the energy, possessed by a liquid particle, by virtue of its existing pressure. If a liquid particle is under a pressure of p kg / m2, then the pressure energy of the particle will be p/w m-kg per kg of liquid, where w is the specific weight of the liquid

Total Energy of a Liquid Particle in Motion: The total energy of a liquid particle, in motion, is the sum of its potential energy, kinetic energy and pressure energy. Mathematically, Total Energy,

EZ

v2 p  m-kg/ kg of liquid 2g w

BERNOULLI’S EQUATION: It states, “For a perfect incompressible liquid, flowing in a continuous stream, the total energy of a particle remains the same; while the particle moves from one point to another.” This statement is based on the assumption that there are no losses due to friction in pipe. Mathematically,

v2 p E  Z   = Constant 2g w

--------- (1)

Bernoulli Equation: 22


---------- (2) This is the basic from of Bernoulli equation for steady incompressible inviscid flows. It may be written for any two points 1 and 2 on the same streamline as

--------- (3) The constant of Bernoulli equation, can be named as total head (ho) has different values on different streamlines.

-------- (4) The total head may be regarded as the sum of the piezometric head h = p/(ρg) + z and the kinetic head v2/2g. Bernoulli equation is arrived from the following assumptions: a) Steady flow - common assumption applicable to many flows. b) Incompressible flow - acceptable if the flow Mach number is less than 0.3. c) Frictionless flow - very restrictive; solid walls introduce friction effects. d) Valid for flow along a single streamline; i.e., different streamlines may have different ho. e) No shaft work - no pump or turbines on the streamline. f) No transfer of heat - either added or removed. Range of validity of the Bernoulli Equation: Bernoulli equation is valid along any streamline in any steady, inviscid, incompressible flow. There are no restrictions on the shape of the streamline or on the geometry of the overall flow. The equation is valid for flow in one, two or three dimensions. Modifications on Bernoulli equation: Bernoulli equation can be corrected and used in the following form for real cases. *

Where 'q' is the work done by pump and 'w' is the work done by the fluid and h is the head loss by friction. 4. Experimental Set-up The apparatus is made from transparent acrylic and has both the convergent and divergent sections. Water is supplied from the constant head tank attached to the test section. Constant level is maintained in the supply tank. Piezometric tubes are attached at different distance on the test section. Water discharges to the discharge tank attached at the far end of the test section and from there it goes to the measuring tank through valve. The entire setup is mounted on a stand.

23


Figure 1: Schematic diagram of experimental setup 5.

Experimental procedure a) Fill the sump tank with water. b) Before starting the pump, make sure that the bypass valve is fully open and the discharge valve is closed. c) Switch ON the Pump. d) Flow control valve to the supply tank is opened and bypass valve is closed slowly and simultaneously. Keep drain valve of the discharge tank fully open. e) After we get steady height of liquid in the supply tank adjust the drain valve on the discharge tank so as to get steady level there also. f) Collect the predetermined quantity of water in the measuring tank and measure the time required for the same. g) Also record the height of liquid in each of the piezometric tube.

piezome tric tube position

6. Observations Sr. No. H1 Height. diff. in measuring tank H2 (m) ΔH 3 Volume (m ) V Time (sec) T Vol. Flow rate Q (m3/sec) H1=P1/w(m) 1 v1 = Q/A1 v12/ 2g (m) 24


2

3

4

5

6

7

8

Total Head H2=P2/w(m) v2 = Q/A2 v22/ 2g (m) Total Head H3=P3/w(m) v3 = Q/A3 v3 2/ 2g (m) Total Head H4=P4/w(m) v4 = Q/A4 v42/ 2g (m) Total Head H5=P5/w(m) v5 = Q/A5 v5 2/ 2g (m) Total Head H6=P6/w(m) v6 = Q/A6 v6 2/ 2g (m) Total Head H7=P7/w(m) v7 = Q/ A7 v7 2/ 2g (m) Total Head H8=P8/w(m) v8 = Q/ A8 v82/ 2g (m) Total Head

Calculations: Length of the test section = 0.4 m Piezo tube 1 at 0.05 m c/s area A1 = 0.000961 Piezo tube 2 at 0.085 m c/s area A2 = 0.00053 Piezo tube 3 at 0.115 m c/s area A3 = 0.000314 Piezo tube 4 at 0.18 m c/s area A4 = 0.000572 Piezo tube 5 at 0.240 m c/s area A5 = 0.000854 Piezo tube 6 at 0.300 m c/s area A6 = 0.001256 Piezo tube 7 at 0.355 m c/s area A7 = 0.00166 Piezo tube 8 at 0.40 m c/s area A8 = 0.00215 Cross Sectional Area of Measuring Tank (Am) = 0.09 m2 Height Difference in measuring tank (H) = = Volume Collected (V) = = Volumetric Flow rate (Q) = = Linear Point Velocity (vi) = = 25

m2 m2 m2 m2 m2 m2 m2 m2 H1- H2 m H * Am m3 V/T m3/s Q/ Ai m/ sec


= [vi2 / (2*g)] = m Total Head at point –1 = Static Head + Velocity Head = Hi + vi2 / (2*g) = m Check for all the points that sum of velocity head and pressure head is almost constant. Here potential energy is not being considered as the test section lies in one horizontal plane only. Velocity head at point – 1

7. Result: Draw graph of static head, velocity head, and total head Vs distance of the piezometric tubes. 8.

Conclusions

9.

Precautions

References

(a) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”, 7th ed., McGraw Hill, NY, 2005. (b) W.L Badger & J.T. Banchero., “Introduction to Chemical Engineering”, 26th reprint, McGraw Hill, NY, 2011. (c) C.J. Geankoplis., “Transport Processes and Separation Process Principles”, 4th ed., PHI, 2009.

26


EXPERIMENT NO. 3(b)

VENTURIMETER, ORIFICEMETER & ROTAMETER CALIBRATION 1.

Aim & Objective a) To determine the co-efficient of discharge through Venturimeter & Orificemeter. b) To calibrate the Rotameter

2.

Theory

VENTURIMETER:

A Venturimeter consists of; 1. An inlet section followed by a convergent cone. 2. A Cylindrical Throat. 3. A gradually divergent cone. The inlet section of the Venturimeter is of the same diameter as that of the pipe, which is followed by a convergent cone. The convergent cone is a short pipe, which tapers from the original size of the pipe to that of the Throat of the Venturimeter. The Throat of the Venturimeter is a short parallel side tube having its cross-sectional area smaller than that of the pipe. The divergent cone of the Venturimeter is gradually diverging pipe with its crosssectional area increasing from that of the Throat to the original size of the pipe. At inlet section & Throat of the Venturimeter, pressure taps are provided. ORIFICEMETER:

An Orificemeter consists of a flat circular plate with a circular hole called Orifice, which is concentric with the pipe axis. An orifice plate is a thin plate with a hole in the middle. It is usually placed in a pipe in which fluid flows. When the fluid reaches the orifice plate, the fluid is forced to converge to go through the small hole; the point of maximum convergence actually occurs shortly downstream of the physical orifice, at the so-called vena contracta point. As it does so, the velocity and the pressure changes. Beyond the vena contracta, the fluid expands and the velocity and pressure change once again. By measuring the difference in fluid pressure between the normal pipe section and at the vena contracta, the volumetric and mass flow rates can be obtained from Bernoulli's equation. ROTAMETER:

The Rotameter is a variable – area meter that consists of an enlarging transparent tube and a metering “float” (actually heavier than the liquid) that is displaced upward by the upward flow fluid through the tube. The tube is graduated to read the flow directly. Notches in the float cause it to rotate and thus maintain a central position in the tube. The float is an indicating element, and greater the flow rate, the higher the float rides in the tube.

3.

Experimental set up

27


Schematic Diagram for Venturimeter, Orificemeter & Rotameter Apparatus

Flow Diagram of Venturimeter

4.

Flow Diagram of Orificemeter

Experimental procedure

VENTURIMETER & ORIFICEMETER:

Starting Procedure: a) Clean the apparatus and make All Tanks free from Dust. b) Close the drain valves provided. c) Fill Sump tank ¾ with Clean Water and ensure that no foreign particles are there. d) Close all Flow Control Valves given on the water line and open By-Pass Valve. e) Close all Pressure Taps of Manometer connected to Venturimeter & Orificemeter. f) Ensure that On/Off Switch given on the Panel is at OFF position. g) Now switch on the Main Power Supply (220 Volts AC, 50 Hz). h) Switch on the Pump. i) Operate the Flow Control Valve to regulate the flow of water in the desired Test Section. j) Open the Pressure Taps of Manometer of related Test section, very slowly to avoid the blow of water on manometer fluid. 28


k) Now open the Air release Valve provided on the Manometer, slowly to release the air in manometer. l) When there is no air in the manometer, close the Air release valves. m) Adjust water flow rate in desired section with the help of Control Valve. n) Record the Manometer reading. o) Measure the flow of water, discharged through desired test section, using Stop Watch and Measuring Tank. p) Repeat Steps the same procedure for different flow rates of water, operating Control Valve and By-Pass valve. q) When experiment is over for one desired test section, open the By-Pass Valve fully. Then close the flow control valve of running test section and open the Control valve of secondly desired test section. CALIBRATION OF ROTAMETER:

a) b) c) d)

Close the ball valves provided in the Venturimeter and Orificemeter pipelines. Open the ball valve provided in the Rotameter pipeline. Now switch on the main power supply and switch on the pump. Set the flow rate with the help of by pass and flow control valves provided in Rotameter pipeline. e) Measure the discharge with the help of measuring tank and stopwatch. f) The actual discharge, verify the set value of Rotameter. g) Repeat the same procedure for different flow rates. Closing Procedure: a) When experiment is over, close all Manometers Pressure Taps first. b) Switch off Pump. c) Switch off Power Supply to Panel. 5.

Observation

For venturimeter & orificemeter OBSERVATION TABLE: S.No. Pressure Rise of Water level in difference, h Measuring Tank, R (cm) (cm)

Time taken for R, t (sec.)

Average Co-efficient of Discharge: For rotameter

OBSERVATION TABLE: S.No.

Rotameter Reading in LPH, Qth

Rise of Water level in Measuring Tank, R (cm)

29

Time taken for R, t (sec.)


FORMULAE: For both Venturimeter & Orificemeter: 1. Head losses,  h  m  H  1 , m of water 100  W  2. Theoretical discharge (Qt): a a 2 gH Qt  1 2 , m3/s 2 2 a1  a 2 3.

------------------- (1)

------------------- (2)

Actual discharge (Qa): AxR Qa  , m3/s t *100 Co-efficient of discharge (Cd): Cd = Qa/Qt

4.

------------------- (3) ------------------- (4)

For Rotameter: 5. Actual discharge:

Qa 

AxR x3600 x1000 , m3/s t *100

------------------- (5)

DATA:H= Head loss h = Pressure difference (m) A = 0.1 m2 ρm = Density of manometric fluid = 13600 kg/ m3 ρW = Density of working fluid = 1000 kg/ m3 g = Acceleration due to Gravity = 9.81 m/sec2 For Venturimeter: d1 = Dia. at inlet of the venturimeter = 28mm = d2 = Dia. at throat of the Venturimeter = 14mm = 2 2 a1 = d1 /4 cm Area at inlet of Venturimeter = 2 2 a2 = d2 /4 cm Area at throat of Venturimeter =

For Orificemeter: d1 = Dia. at inlet of Orificemeter = 28mm. = d2 = Dia. of Orifice Plate = 14mm. = a1 = d12/4 Area at inlet of Orificemeter = a2 =  d22/4 m2 Area of Orifice Plate = 6.

0.028m 0.014m 6.157 * 10-4 m2 1.539* 10-4 m2

0.028m 0.014m 6.157 * 10-4m2 1.539* 10-4 m2

Result and discussion

For Venturimeter and Orificemeter S. No. Actual discharge Qa 3 , m /s

Theo. Discharge Qt , m3/s 30

Cd = Qa/Qt


For Rotameter S. No. Rotameter Reading in Qth, m3/s

Theo. Discharge Qt, m3/s

7.

Conclusion

8.

Precaution 1. Do not run the pump at low voltage i.e. less than 180 Volts. 2. Never fully close the Delivery line and By-Pass line Valves simultaneously. 3. Always keep apparatus free from dust. 4. To prevent clogging of moving parts, Run Pump at least once in a fortnight.

31


EXPERIMENT NO. 3(c)

FLOW THROUGH HELICAL COIL 1. Aim  To determine the critical Reynolds number of a fluid flowing through the helical coil.  To determine the friction factor for flow of water through helical coil. 2. Apparatus Helical coil experiment set up, Stop watch, Vernier Caliper 3. Theory Helical coils are used widely in processing industries for cooling and heating applications since the centrifugal forces experienced by the fluid acts to promote contact with the channel wall, thereby tending to insure good contact with the wall, and an enhanced heat transfer Some of their main advantages over straight tubes are high heat and mass transfer coefficients, and space economy in terms of area per unit volume. Helical coils are used for heating or cooling in process tanks. When a fluid flows through a curved tube, centrifugal force acting upon the various elements of fluid moving with different velocities causes secondary circulation. Secondary flow results in higher heat transfer characteristics. Further, secondary flow stabilizes the laminar flow leading of a higher critical Reynolds number for transition from laminar to turbulent flow. The use of helically coiled exchangers continues to increase. Applications include liquid heating/cooling, steam heaters, vaporizers, cryogenic cooling and vent condensing. Listed below are the details for standard services in which helical exchangers warrant consideration. Seal Coolers, Condensers, Cryogenic Vaporizers, Compressor Inter- and After-Coolers. 4. Experimental Procedure a) With the discharge valve closed, switch ON the pump. b) Slowly open the discharge valve and note down the manometer reading. c) Note down the time taken for 1 cm raise in the discharge tank. d) Repeat the same procedure for different flow rates of water. 5. Observation d = Inner diameter of helical pipe = 0.932×10-2 m v = velocity of water in pipe (m/sec) µ = dynamic viscosity of water at 25 oC (0.890 ×10-3 Ns/m2) A = cross-sectional area of coil = m2 H1 = one side manometer reading. H2=other side manometer reading. H2 – H1 = difference in manometer reading. V= volume of water collected. T= time in seconds

OBSERVATION TABLE: S. No

H1

H2

V(m3)

H1-H2

32

T(sec)

Q (m3/s)


1 2 3 4 6. Calculation Volume collected is calculated by measuring tank, V = l * b * height (Units in meter) = m3 Flow rate Q = V/T = m3 /sec Velocity v = Q/A = m/sec On the basis of velocity we can find Reynolds number, Re = ρ * v * d ρ = density of water = 1000 kg/m3 v= velocity d = inner diameter of coil = dynamic viscosity of water Calculation for fanning friction factor F = (∆p/ ρ g) * (D/L) * (2g/v2) ∆p=pressure difference in manometer g =9.81 m/s2 D = overall diameter of coil = 0.235 m H = overall height of coil = 0.35 m N = No. of turns = 14 R =D/2 LENGTH OF COIL

N2 * R2 9R + 10 H Put the value of L in the above equation and find out value of friction factor of helical coil. L=

7. Result and Conclusion 

The critical Reynolds number of the fluid flowing through the coil was found to be _______.



The friction factor for flow of water through helical coil was found to be ______

8. Precaution

33


EXPERIMENT NO. 4(a)

STUDY OF PITOT TUBE – AIR 1. Aim To determine the velocity profile for the flow of air through a circular pipe using Pitot Tube. 2. Objective To measure the velocity at different points in the flow. 3. Theory A pitot tube is a device to measure the point velocities in the flow field of a fluid. It generally consists two concentric tubes arranged parallel to the flow .The outer tube is perforated at one end with small holes in its wall, which are perpendicular to the flow direction and lead to the annular space. The annular space is otherwise sealed except for a manometer lead on the other end .The inner tube has a small opening at one end (the same end in which holes are there in the outer tube) parallel to the flow and pointed in the direction opposite to the flow direction. The other end of the tube has a manometer lead. The two tubes are connected to two limbs of a manometer through the manometer leads. At equilibrium there is no fluid within the pitot tube. The annular space serves to transmit the static pressure. The flowing fluid is brought to rest at the entrance of the inner tube, and this tube transmits the impact pressure (or stagnation pressure) equivalent to the kinetic energy of the flowing fluid at that position. If and are the stagnation pressure measured by the inner and outer tubes respectively, the point velocity, , at a given position can be obtained from the Bernoulli equation, by neglecting the frictional losses, as u0 = √ {2(Ps-P0)/ℓf} = √ {2gh(ℓm-ℓf)/ℓf} Where ℓf and ℓm are the densities of the process fluid and the manometric fluid respectively, h is the difference in the level of the manometric fluid in the two limbs of the manometer, and g is the acceleration due to gravity. Thus by changing the position of the pitot tube the local velocities in the flow field can be obtained. The velocity profile at a given cross section of the pipe is obtained by moving the pitot along the diameter of the pipe. By knowing the point velocities at a given cross section, the volumetric flow rate, Q, can be determined Q = 0∫D/2 2ᴫru0dr Where D is the pipe diameter, and r is the radius at which u0 is measured. 4. Requirements Pipe line with provision for supply of air, pitot tube, inclined manometer. 5.

6.

Procedure 1. Start the flow of air by switching on the blower. 2. Divide the pipe diameter in 9 equal parts to fix the radial position at which the point velocities are to be determined. 3. Keep the pitot tube at the radial position. 4. Once the flow steadies (indicated by unchanging or slightly fluctuating level difference in the manometer), record the level difference in the manometer. 5. Record the ambient temperature (taken same as the air temperature). 6. Repeat step 4 by moving the pitot tube at the radial determined in step 2. Observations 34


Ambient temperature (T) = Air density at T = Density of manometric liquid at T = S.No.

7. Formula 1)Point velocity, u0

kg/m³ kg/ m³ r,mm

h,mm

u0 = √ {2gh(ℓm-ℓf)/ℓf}

2)Volumetric flow rate,Q Q = 0∫D/2 2ᴫru0dr

8.

9.

Results and discussion 1. Calculate the point velocities at different radial positions. 2. Plot the point velocity versus the radial position. 3. Calculate the volumetric flow rate from the point velocities . Conclusions

35


STUDY OF PITOT TUBE - WATER 1. Aim To determine the velocity profile across the cross section of pipe for the flow of water using Pitot tube and thereby determine the co-efficient of Pitot tube for different flow rates. 2. Objective To measure the velocity at different points in a pipe for different flow rates. 3. Theory It is a device used for measuring the velocity of flow at any point in a pipe. It is based on the principle that if the velocity of flow at a point becomes zero, there is increase in pressure due to the conversion of the kinetic energy into pressure energy. The Pitot tube consists of a capillary tube, bent at right angle. The lower end, which is bent through 90, is directed in the upstream direction. The liquid rises up in the tube due to conversion of kinetic energy into pressure energy. The velocity is determined by measuring the rise of liquid in the tube.

When a Pitot tube is used for measuring the velocity of flow in a pipe or other closed conduit the Pitot tube may be inserted in the pipe as shown in figure. Since a Pitot tube measures the stagnation pressure head (or the total head) at its dipped end. The pressure head may be determined directly by using piezometeric readings between the Pitot tube and pressure taping at the pipe surface. Consider two points (1) and (2) at the same level in such a way that point (2) is just at the inlet of the Pitot -tube and point (1) is far away from the tube. At point (1) the pressure is p1 and the velocity of the stream is v1. However, at point (2), is called stagnation point, the fluid is brought to rest and the energy has been converted to pressure energy. Therefore the pressure at (2) is p2, the velocity v2 is zero and since (1) and (2) are in the same horizontal plane, so z1 = z2. Applying Bernoulli’s equation at points (1) and (2) 2 p1 v1 p2 v 2 2    w 2g w 2g v2 = 0 v 21 p2 p1   2g w w 36


p  p v1  2 g  2  1  w w

v1  2 gH This is theoretical velocity. Actual velocity v1 act  Cv 2gH 4. Pitot tube

Requirements :

Material SS/copper of compatible size fitted with scale. Test Section : Material Clear Acrylic, compatible to 1” dia Pipe. Water Circulation : FHP Pump. Flow Measurement : Using Measuring Tank, Capacity 40 Ltrs. Sump Tank : Capacity 70 Ltrs. Stop Watch : Electronic. Control Panel Comprises of: Standard make On/Off Switch, Mains Indicator, etc.

5. Experimental set-up The apparatus consists of a Pitot tube made of SS and fixed below a pointer gauge. The pointer gauge is capable to measure the position of Pitot tube in transparent test section. The pipe has a flow control valve to regulate the flow. Piezometric tubes are provided to determine the velocity head. A pump is provided to circulate the water. Discharge is measured with the help of measuring tank and stopwatch 6. Experimental procedure Starting Procedure: 1. Clean the apparatus and make Tank free from Dust. 2. Close the drain valves provided. 3. Fill Sump tank ¾ with Clean Water and ensure that no foreign particles are there. 4. Close all Flow Control Valves given on the water line and open By-Pass Valve. 5. Ensure that On/Off Switch given on the Panel is at OFF position. 6. Now switch on the Main Power Supply (220 Volts AC, 50 Hz). 7. Switch on the Pump. 8. Operate the Flow Control Valve to regulate the flow of water through orifice. 9. Adjust water flow rate to desired rate with the help of flow Control Valve. 10. Set the Pitot tube at the center of test section 11. Record the piezometric reading and measure the discharge with the help of measuring tank and stop watch. 12. Now move the Pitot tube up and down at the same flow rate and note the piezometric readings to find out the velocity at different points in pipe. 13. Calculate the co – efficient of Pitot tube from actual and theoretical velocities and plot the velocities at different points inside the pipe. 14. Repeat the same procedure for different flow rates of water, operating Control Valve, and By-Pass valve.

37


Closing Procedure: 1. When experiment is over Switch off Pump. 2. Switch off Power Supply to Panel. 3. Drain water from all tanks with the help of given drain valves. FORMULAE 1. Discharge, Q

A* R t

----------------------- (1)

Q a

----------------------- (2)

2. Actual Velocity, v act 

3. Theoretical velocity v th  2 gh

------------------------ (3)

4. Co – efficient of Pitot tube, v C v  act v th

------------------------- (4)

5. Velocity at any point, v  Cv 2 gh 7. Observation table S.No. Pressure head at different points on up side 8 mm 6 mm 4 mm 1.

Pressure head at center 0

Pressure head at R(cm) different points on down side 4 mm 6 mm 8 mm

t(sec)

2. 3.

CALCULATION TABLE S. Cv v8 No. 1.

v6

v4

v0

2. 3. DATA 38

v4

v6

v8


A a g

= = =

0.1 m2 0.0006157 m2 9.81 m/ s2

NOMENCLATURE A = Area of measuring tank. a = Cross section area of test section R = Rise of water level in measuring tank. h = Piezometric difference CV = Co efficient of Pitot tube g = Acceleration due to gravity va = actual velocity of fluid. Q = discharge at outlet. t = time for R.

8.   

Result & discussions Calculate the point velocities at different radial positions. Calculate the co-efficient of Pitot Tube for different flow rates. Plot the point velocity versus the radial position.

9.

Conclusion

10.

Precautions Do not run the pump at low voltage i.e. less than 180 Volts. Never fully close the Delivery line and By-Pass line Valves simultaneously. Always keep apparatus free from dust. To prevent clogging of moving parts, Run Pump at least once in a fortnight. Frequently Grease/Oil the rotating parts, once in three months. Always use clean water. If apparatus will not in use for more than one month, drain the apparatus completely, and fill pump with cutting oil.

1. 2. 3. 4. 5. 6. 7.

39


EXPERIMENT NO. 4(b)

STUDY OF REYNOLDS APPRATUS 1. Aim To study the flow regime for liquid flow in a pipe, and to find the critical Reynold's number. 2. Objective To find the volumetric flow rate corresponding to various flow regimes. 3. Theory Osborne Reynolds (1883) demonstrated different flow regimes when a fluid flows through a pipe. These regimes are: laminar, turbulent and transition or intermediate regimes. For a given pippe and fluid, laminar flow exists at low velocity characterized by ordered sliding past of liquid layer over one another without lateral mixing; while turbulent flow is found at high velocity in which the flow pattern is random and there is intense lateral mixing in the fluid. In between these two extremes, a transition flow regime is observed in which the flow alters between laminar and turbulent flows. Reynolds’s found that the nature of flow depends on the diameter of the pipe, and viscosity, density and average velocity of the fluid; and these factors can be combined into one dimensionless group, called the Reynolds’s number, the magnitude of which would indicate the flow regime. The Reynolds number, Re is defined as Re = Du/µ = Du/v Where D = pipe diameter, u= fluid velocity, ρ= fluid density, µ = fluid viscosity, and v = kinematic viscosity. Physically, Reynolds number is interpreted as the ratio of the inertial force to viscous force acting on the fluid. In general, D and v represent some characteristic dimension of the flow domain and characteristic velocity of the fluid. Thus laminar flow exists at small Reynolds number, and turbulent flow at large Reynolds number. The minimum Reynolds number at which laminar flow disappears is called the critical Reynolds number. The value of critical Reynolds number depends on the geometry of the flow domain (circular or rectangular pipe, open or closed channel, flat plate etc.) and flow configuration (flow around a bluff body, in packed bed etc.) 4. Requirements Reynolds apparatus, water source, dye, measuring cylinder, stop watch. 5.

PROCEDURE 1. Fill in the tank with water, and the dye- chamber with dye. 2. Note the water temperature. 3. Start the water flow and maintain a small flow rate, enough to fill the whole pipe cross section. 4. Once the flow stabilizes, start the dye injection. The injection rate should be just enough to give a clear visible streak of the dye. 5. Observe the pattern of the dye streak. The dye should flow in a straight line. 6. Increase the water in small and equal increments, and observe the dye streak. 7. Repeat step 6 until some undulations commence in the streak. Note the corresponding volumetric flow rate of water, which is the critical Reynolds number. Appearance of the undulations signifies the initiation of the intermediate or transition flow.

40


Note At this point the undulations will be unstable so that there will be some portion of the dye streak which will be undulating and some portion which will not. 8. Keep increasing the flow rate of the liquid further until at one point there is found a complete dispersion of the dye (indicated by the liquid getting colour through the cross section) just as it comes out of the injection needle. This point shows the conversion to a fully turbulent regime. 9. Note the corresponding the volumetric flow rate. 6.

Observations Temperature of the liquid Pipe diameter

S.No 1 2 3

Flow Regime Laminar Transition Fully Turbulent

= = Volume of water collected

Volumetric flow rate corresponding to start of the Laminar flow Volumetric flow rate corresponding to start of the transition flow Volumetric flow rate corresponding to fully turbulent flow Liquid density at observed temperature Liquid viscosity at the observed temperature

m

Time taken

= = = = =

m³/s m³/s m³/s kg/m³ kg/m.

7. Results and discussion 1. Calculate the Reynolds number corresponding to the transition flow and turbulent flow. The liquid velocity is calculate as u = 4Q/ π D², Where Q is the volumetric flow rate. 2. Compare the critical Reynolds number and the Reynolds number for transition to fully turbulent flow observed with those reported in the literature. Discuss the possible sources of discrepancy, if any. 8.

Conclusion

41


EXPERIMENT NO. 5 (a)

CENTRIFUGAL PUMP CHARACTERISTICS 1. AIM Study and analysis of centrifugal pump characteristics. 2. Objectives 1. To study the operating characteristics of a centrifugal pump. 2. To study the operating characteristics of two centrifugal pumps operated in series and in parallel. 3. Theory Pumps are the fluid moving machineries which increase the mechanical energy of the fluids to be displaced. The energy increase may be used to increase the velocity, the pressure or the elevation of the fluids. A large number of pumps, differing widely in principle and mechanical construction, have been developed to meet a wide variety of operating conditions. For selection of pumps for a specific application requires the knowledge of operating conditions of the system and applicability of different available pumps. The mechanical energy of the liquid is increased by centrifugal action. Centrifugal pumps are classified as single suction and double suction pumps depending upon the suction from either one side or from both sides respectively. In a single suction centrifugal pump the liquid enters through a suction connection concentric with the axis of a high-speed rotary element called the impeller, which carries radial vanes integrally cast in it. Liquid flows outward in the spaces between the vanes and leaves the impeller at a considerably greater velocity with respect to the ground than at the entrance to the impeller. In a properly functioning pump the space between the vanes is completely filled with liquid flowing without cavitation. The liquid leaving the outer periphery of the impeller is collected in a spiral casing called the volute and leaves the pump through a tangential discharge connection. In the volute, the velocity head of the liquid from the impeller is converted into pressure head. The power is supplied to the fluid by the impeller and is transmitted to the impeller by the torque of the drive shaft, which usually is driven by a direct-connected motor at constant speed, commonly at 1750 rpm. Another common type uses a double-suction impeller, which accepts liquid from both sides. Also, the impeller itself may be a simple open spider, or it may be enclosed or shrouded. 4. Experimental Set up 4.1. Requirements Centrifugal pump test rig, bucket, watch. 4.2. Schematic Diagram of Experimental Setup The schematic diagram of the experimental setup is shown below:

42


P2

P4

V2

V8

V4

V5

P1

V1

C1

V3

C2

P3 V-Notch

Measuring Tank

Sump Tank

OPERATION SINGLE PUMP SERIES PARALLEL

: : :

OPEN

CLOSE

V1, V4, V5 V1, V2, V5 V1, V3, V4, V5

V2, V3, V8, V9 V3, V4, V8, V9 V2,V8,V9

5. Experimental Procedure First, prime the pump by pouring water through valve V8. Subsequently, the following sequence of operations are to be carried out: 5.1. Calibration of V-Notch 1. Measure the width of the V-notch at the top and depth of the V-notch. 2. Fill the storage tank with water. 3. Open valves V1, V2 and V5, and close valves V3, V4 & V9. 4. Start running the pumps. 5. Fill the channel with water until water starts spilling over the notch to the outlet. 6. Stop the water supply by closing the bench supply valve (V5) with the pump in running condition. 7. Allow the water above the crest height to spill over the notch. 8. When the water level is at the level of the crest of the notch, bring the point gauge exactly to the water surface, and note the reading (say R1) on the longer scale which coincides with the “zero” of the shorter scale. (This is the datum level for subsequent heightmeasurements when the water flows in the channel.) 9. Open valve V5 completely and wait for the level (or head) of water in the channel to stabilize. 10. Bring the point gauge to the surface of the water by rotating the knob attached to the scale, and note the reading (say R2) on the longer scale that coincides with the “zero” of the shorter scale.

43


11. Collect the water in the bucket for an arbitrary but known duration, and determine the flow rate of water by dividing the volume (or mass) of water collected by the time of collection. 12. The difference (R2-R1) is called the head which is related to the volumetric flow rate by the following equation: 5

8  Q   C d tan  2 g h 2 ; Where, h = R2-R1  15 

13. Adjust the water flow rate at seven different values from the maximum to zero flow rate in the channel by manipulating valve V5 and determine the water flow rate and the corresponding head. 14. Calculate the discharge coefficient (Cd) in each case, and take the arithmetic average of these values as the Cd of the notch. 5.2. Pump Characteristics (Single Pump) 1. After carrying out the steps 1 to 3 mentioned in “Calibration of V-Notch”, open the valve V5 fully. 2. Note the RPM of the pump from the control panel. 3. Note the gage pressure on the outlet line (P4 in the figure) and on the inlet line (P1 in the figure). The difference of the readings in P4 and P1 is the head delivered by the pump. 4. Note the power supplied to the pump from the display on the control panel by pressing the appropriate button (labeled W). 5. Note the head of the V-notch. 6. Repeat steps 2 to 5 for seven different flow rates of water spanning the whole range of flow rate up to zero flow by manipulating V5. 5.3. Pump Characteristics (Pump in Series) 1. Fully open the valves V1, V2 and V5, and close valves V3 and V4. 2. Note the RPM of the pump from the control panel. 3. Note the gage pressure on the outlet line (P4 in the figure) and on the inlet line (P1 in the figure). 4. Note the power supplied to the pump from the display on the control panel by pressing the appropriate button (labeled W). 5. Note the head of the V-notch. 6. Repeat steps 2 to 5 for seven different flow rates of water spanning the whole range of flow rate up to zero flow by manipulating V5. 7. Repeat steps 1 to 6 for three different RPMs. 5.4. Pump Characteristics (Pump in Parallel) 1. Fully open the valves V1, V3, V4 and V5, and close valve V2. 2. Note the RPM of the pump from the control panel. 3. Note the gage pressure on the outlet line (P4 in the figure) and on the inlet line (P1 in the figure). 4. Note the power supplied to the pump from the display on the control panel by pressing the appropriate button (labeled W). 5. Note the head of the V-notch. 6. Repeat steps 2 to 5 for seven different flow rates of water spanning the whole range of flow rate up to zero flow by manipulating V5. 7. Repeat steps 1 to 6 for three different RPMs. 6. Observations 44


6.1. Calibration of V-Notch Width across the top of the V-notch Depth of the V-notch Initial point gauge reading at zero flow rate (R1)

S No

Head of the V-notch (R2 – R1)

Point gauge reading, R2

= = = Amount of Water collected

Duration of collecting water

6.2 Pump Characteristics* 1. RPM = 2. Initial point gauge reading at zero flow rate (R1) = 3. Length of Pipe in single pump operation = 4. Length of Pipe in series operation = 5. Length of Pipe in parallel operation = 6. Elevation = 7. Diameter of the suction pipe = 8. Diameter of the discharge pipe = S No

Point gauge Reading, R2

Gauge pressure reading P1 P4

Head of The V-notch

Power to the pump

* Make separate tables for different configurations (single/series/parallel) of the pumps and for different RPMs.

7. Model Calculations For steady incompressible flow of a liquid of density , the developed head is given as

P2  P1 V22  V12 H   ( Z 2  Z1 )  h f g 2g

(1)

where subscripts 1 and 2 signify the values at the suction and delivery ports of the pump; P is the pressure; V is the velocity; Z is the elevation; and hf is the friction head (that is, the frictional losses) in the line between the suction and delivery ports. hf is given by

L  hf  4 f  D 



V Kf   2g

2

(2)

where f is the friction factor, L is length of the pipe, D is the diameter of the pipe, Kf is the loss factor for fittings. Friction factor (f) depends on roughness and type of flow

45


(laminar/turbulent) and read from friction factor chart. Alternatively, it can be calculated from the equations: f 

16 ; N Re

for laminar flow

The friction factor for turbulent flow can be calculated by using Colebrook equations (either implicit form or explicit form) given below: 7.1. Implicit Forms of Colebrook Equation There are at least three forms of the Colebrook Equation that can be found in current literature on hydraulics. These are:   2.51  2 log10    3.7 D f N Re f 

1

   

 2 18.7  1.74  2 log10   D f N Re f 

1

(3)    

  9.3 D  1.14  2 log10   2 log101     f   N Re  D 

1

(4)     f  

(5)

where, f is the Friction Factor and is dimensionless ε is the Absolute Roughness and is in units of length D is the Inside Diameter and, as these formulas are written, is in the same units as ε. NRe is the Reynolds Number and is dimensionless. Note that ε/D is the Relative Roughness and is dimensionless. These three equations are referred to as “Implicit” Equations. “Implicit” means that “f”, the Friction Factor, is “Implied or understood though not directly expressed”2. Simply stated, the equations ARE NOT in the form of “f = ………”. These are sometimes referred to as “equivalent” but the results will vary when calculated to the fourth significant digit. These equations can be solved for “f” given the Relative Roughness (ε/D) and the Reynolds Number, (NRe), by iteration. Such iterations can be performed using an electronic spreadsheet. A spreadsheet, “Friction Factor Formulas for Cheresource.xls” is available at http://www.cheresources.com/colebrook1.shtml presented for demonstration. The spreadsheet contains four worksheets. The first “Tab” is labeled “Iterations”. The Iterative solutions are generated by breaking the formulas in two parts, that which is left of the equal sign and that which is right of the equal sign (See row 20 as an example). The Iteration then tests values of “f” that will result in the difference between the two sides to be zero or very close to zero. (A complete explanation was published in the ASHRAE Journal of September, 2002: see Reference 4 for details) 7.2. Explicit Forms of Colebrook Equation 46


There are four Explicit forms of Colebrook Equations reported in literature. (Ref: available on internet at http://www.cheresources.com/colebrook1.shtml). Out of these the Serghide’s solution is the most accurate one, which is given below: Serghide’s Solution (see Reference 3 for details). f  ( A  ( B  A)2 /(C  (2B)  A))2

(6)

where

A  2 log10[( / 3.7D)  (12 / N Re )] B  2 log10[( / 3.7D)  (2.51A / N Re )] C  2 log10[( / 3.7D)  (2.51B / NRe )] Serghide can be used across the entire range of the Moody Diagram. Its accuracy is unparalleled amongst the Explicit Equations evaluated here. It appears to be based on Eqn. 3, as do all the Explicit Equations presented. There is less deviation between Serghide and Eqn. 3 then there is between Eqn. 3 and either Eqn. 4 or Eqn. 5.

1 f



 4.07 log N Re



f  0.6 ;

(7)

Eqn. 7 is the von Karman equation for turbulent flow which can be used only if the pipe is smooth. In our case, the pipe is not smooth and hence this equation cannot be used. Colebrook equation has to be used. Power developed by the pump (or the power delivered to the fluid) Pf is given as Pf  QHg

(8)

where Q is the volumetric flow rate of the liquid. The mechanical efficiency of the pump  is given as



Pf

(9)

PB

where PB is the total power supplied to the pump drive from an external source. Operating (or performance) characteristics of a pump is commonly illustrated by plots of actual head developed H, power consumption Pf, and efficiency  versus the volumetric flow rate. 8. Results & Discussion 8.1. Calibration of V-notch S No Volumetric flow rate

Head of the V-notch

47

Cd


Average Cd = 8.2. Pump Characteristics (RPM = ) Volumetric flow Head developed by S No rate the pump

Power delivered by the pump

Efficiency

1. Plot head, power delivered to the pump, power delivered to the liquid, and efficiency versus the capacity of the pump (i.e., flow rate of water). 2. Suggest the operating point from the efficiency versus flow rate plot. 9. Conclusions 10. Precautions Do not run the pump with the delivery valve closed for a long time to avoid damage of the pump. References 1. Warren, L McCabe, Smith, J C , and Harriott, P, Unit Operations of Chemical Engineering, 6th edition, McGraw Hill, New Delhi, India, 2000. 2. Babu, B.V.,"Pumps: Selection & Trouble Shooting", IPT (Indian Plumbing Today), Vol. 2005 (No. 6), pp. 41-49, November-December, 2005. 3. T.K.Serghide’s implementation of Steffenson’s accelerated convergence technique, reportedly to have appeared in Chemical Engineering March 5, 1984. 4. Lester, T. “Calculating Pressure Drops in Piping Systems.” ASHRAE Journal Sept. 2002.

EXPERIMENT NO. 5 (b)

RECIPROCATING PUMP CHARACTERISTICS 1. AIM Study and analysis of operating characteristics of double acting reciprocating pump. 2. Objectives To study the operating characteristics of a double acting piston-type reciprocating pump. 3. Theory A piston pump consists of a cylinder with a reciprocating piston. The fluid is drawn through an inlet check valve into the cylinder by the withdrawal of a piston and is then forced out through a discharge check valve on the return stroke. Most piston pumps are double acting with the liquid admitted alternately on each side of the piston so that one part of the cylinder is being filled while the other is being emptied. The energy added per unit mass of liquid by 48


the pump is termed as the head developed by the pump, which is expressed in length dimension. 4. Experimental Set up 4.1. Requirements Reciprocating pump test rig, watch. 4.2. Experimental Setup

V1 P2 V2

Measuring Tank P1 LL V3

Sump Tank

V4 To Drain

5. Experimental Procedure 1. Fill the storage tank with water and ensure that the measuring tank is empty. 2. Open the valve on the delivery line, and allow the flow from the delivery line to drain directly into the sump tank with the help of the drain valve. 3. Adjust the RPM of the motor to an arbitrary but known value. 4. Note the gauge pressures at the suction and delivery line of the pump. 5. Note the power to the pump from the display on the control panel by pressing appropriate button. 6. Collect water in the measuring tank for a given duration of time and note the change in water level in this tank during this period. 7. Repeat steps 2 to 6 for seven different flow rates of water spanning the whole range of flow rate (excluding zero flow rate) by manipulating the delivery valve. 8. Repeat steps 4 to 7 for three different RPMs.

i. ii. iii. iv.

6. Observations Technical specifications of the pump: Type: Double acting Power: Connected to 1 HP, 3 phase, 440V AC motor. Pump bore (diameter): 44.5 mm Pump stroke: 35 mm 49


Inside diameter of the suction pipe Inside diameter of the delivery pipe RPM Length of the suction pipe Length of the delivery pipe Elevation

S No

Time of collection of water in measuring tank

= = = = = =

Change in water level in the measuring tank

Gauge pressure reading Suction Delivery (P1) (P2)

Power to the pump

Note: Make separate tables for different RPMs.

7. Model Calculations For steady incompressible flow of a liquid of density , the developed head is given as

H

P2  P1 V22  V12   ( Z 2  Z1 )  h f g 2g

(1)

where subscripts 1 and 2 signify the values at the delivery and suction ports of the pump; P is the pressure; V is the velocity; Z is the elevation; and hf is the friction head (that is, the frictional losses) in the line between the suction and delivery ports. hf is to be calculated as described in previous experiment, i.e., centrifugal pump characteristics. Power developed by the pump (or the power delivered to the fluid) Pf is given as Pf  QHg

(2)

where Q is the volumetric flow rate of the liquid. The mechanical efficiency of the pump  is given as



Pf

(3)

PB

where PB is the total power supplied to the pump drive from an external source. Volumetric efficiency (V) of a reciprocating pump is defined as the ratio of the actual discharge (Q) to the theoretical discharge (Qth). The theoretical discharge is the volume swept by the piston of the pump per unit time. The difference between the theoretical and actual discharge of the pump is called the slip. Q (4) V  Qth

50


where Qth for a double acting pump is given by     2  Qth  2 d B2  L N  4   60 

(5)

where dB and L are the bore diameter and length of the stroke of the pump, and N is the speed of rotation of the crank (attached to the motor) in RPM. Operating (or performance) characteristics of a pump is commonly illustrated by plots of actual head developed H, power consumption PB and Pf and efficiency  versus the capacity of the pump (i.e., the volumetric flow rate of liquid). 8. Results & Discussion RPM Theoretical discharge of the pump S No

Volumetric flow rate

= =

Head developed by the pump

Power delivered by the pump

Η

ηV

1. Plot head, power delivered to pump, power delivered to the liquid, and mechanical efficiency versus the capacity of the pump (flow rate of water). 2. Suggest the operating point from the efficiency versus flow rate plot. 9. Conclusions 10. Precautions Never run the pump with the delivery valve closed. References 3. Warren, L McCabe, Smith, J C , and Harriott, P, Unit Operations of Chemical Engineering, 6th edition, McGraw Hill, New Delhi, India, 2000. 4. Babu, B.V.,"Pumps: Selection & Trouble Shooting", IPT (Indian Plumbing Today), Vol. 2005 (No. 6), pp. 41-49, November-December, 2005.

51



HEAT PIPE DEMONSTRATOR 1. Aim To study the performance of heating of heat pipe and compare its working with the best conductor. 2. Objective To calculate the thermal conductivity of the given heat pipes. 3. Theory A heat pipe or heat pin is a heat-transfer device that combines the principles of both thermal conductivity and phase transition to efficiently manage the transfer of heat between two solid interfaces. At the hot interface of a heat pipe a liquid in contact with a thermally conductive solid surface turns into a vapor by absorbing heat from that surface. The vapor then travels along the heat pipe to the cold interface and condenses back into a liquid - releasing the latent heat. The liquid then returns to the hot interface through either as capillary action, centrifugal force, or gravity, and the cycle repeats. Heat pipes are thermal superconductors, due to the very high heat transfer coefficients for boiling and condensation. 3.1 Working Principle The heat pipe is a heat-transfer element with extremely high thermal conductivity and large heat transfer capacity, which accomplishes the heat transfer through the latent heat of phase change of the working fluid inside the sealed vacuum tube. For the typical gravity heat pipe, the up end of the heat pipe releases heat, while the working media is condensed into the liquid. The condensed liquid, with the pull of gravity, will return to the hot side along the inner wall of the heat pipe and will be heated and vaporized again. In this way, the heat will be transferred from one end to the other successively by the heat transfer of phase change, the thermal resistance inside the heat pipe is very small, which results in a larger heat transfer capacity at a smaller temperature difference. In terms of the gravity heat pipe, due to its simple structure, single-direction heat conductivity, and special heat transfer mechanism (the heat exchange between the hot fluid and the cold one is done outside the heat pipe), the heat transfer can be easily enhanced, the working fluid is readily evaporated and boiled and the heat pipe is quick to start up the heattransfer capability of the heat pipe can rival and even exceed that of stainless steel and copper. The heat pipe can be used individually or as combinedly. The heat pipe heat exchanger consisting of heat pipes is characterized by such merits as high heat transfer efficiency, low flow resistance, compact structure, high reliability, and excellent maintenance economy, and is widely applied in the space technology, electronic, metallurgical, motive power, and petroleum and chemical trades.

52


Fig. Liquid and Vapor pressure distribution along the heat pipe 4. Experimental Set up 4.1. Requirements The apparatus consist of three different test specimen viz. stainless steel, copper & heat pipe (Al) which is banded with water cut at the top and heaters at the bottom. 4.2. Schematic Diagram of Experimental Setup The schematic diagram of the experimental setup is shown below

:

5. Procedure 1. Plug-in socket of control panel & set up instrument in proper position. 2. Fill up the cup provided on each of the pipe with water. 3. Start the main switch of control panel. 53


4. Increase slowly the input to heater by the dimmerstat starting from 0 volts position and ampere I changes accordingly. 5. Adjust input so that it correspond out to certain Watts to be calculated (i.e. Q = V * I) out as maximum by help of dimmerstat by varying Voltage. 6. See that both inputs remain constant throughout the experiment. 7. The steady state condition can be checked by rectal temperature of thermocouple of 1 to 8 zone temperature indicator provided on Control panel out of which 1 to 6 channel temperature Indicator are actually working with 1,2 – Stainless Steel , 3,4 – Heat Pipe (Al) and 5,6 – Copper material heat pipe. 8. Note down the reading in the all the tube with Time up to which it reaches before and after steady state as given in observation table. 6. Observations & Calculations Data given: 1. Length of SS, Copper and Heat pipe (L) :- 300 mm 2. Diameter of pipe (Dp) :- 250 mm 3. Area of pipe (Ap) :- 0.04908 m2 6.1 Observation table for S.S. pipe. Thermocouple No.

Thermocouple Position (mm) (Before Steady State)

1

50

2

200

Temperature (K) (Before Steady State)

Temperature (K) (At Steady State)





6.2 Observation table for Heat pipe. Thermocouple No.


3

50

4

200

6.3 Observation table for Copper pipe. Thermocouple No.


5

50

6

200

6.4 Formulae & Calculations QA = V * I

54


QA = - KA * AP * QA – Power, Watts V – Voltage Potential, Volts I – Ampere Current, Ampere QA - Heat Transfer Rate, Watts Ap - Area of Pipe, m2 T - Change in temperature from Start to Steady state ,K x - Change in position of Thermocouple from Start to Steady state (i.e. 50 & 200 mm) KA - Thermal Conductivity of Material A, W / m . K For Copper, Heat pipe and Stainless Steel pipe : KA

=

=

W / m .K

7. Results Thermocouple Position (mm)

Thermal Conductivity ( W/m K) SS pipe

Heat pipe

Copper pipe

50 200 8. Precautions 1. Ensure that the Dimmerstat is at 0 position at before switching on and off the equipment. 2. Increase the energy input gradually to the heater during initial set-up experimentation. 3. Never use the heater at full wattage for longer period of time.


THERMAL CONDUCTIVITY OF SOLIDS 1. Aim To measure the thermal conductivity of solids. 2. Objective (i) To Determination the Overall Thermal Conductivity of Composite wall. (ii) To check that the Thermal Resistances in Composite wall are connected in series.

55


3. Theory Thermal conductivity is defined as the fundamental property of material which gives the measure of the effectivity of material in transmitting heat through it. For the measurement of the thermal conductivity K what is required is to have one dimensional heat flow through the flat specimen, an arrangement for maintaining its faces at the constant temperature and metering method to measure the heat flow through a known area. Knowing the heat input to the central plate heater, the temperature difference across the each specimen, its thickness and the area, one can calculate the K by the following formula. K = q * L / [2 * A* ( - )] Where, K Thermal Conductivity of the sample, W / m q Heat flow rate in the specimen, W A Area of the specimen, m² Hot side average temperature, Cold side average temperature L Thickness of the specimen, m 4. Experimental Set up 4.1. Requirements Thermal Conductivity Apparatus with composite wall (Mild steel, Hylam (Paper Grade) & Wood), C clamps 4.2. Schematic Diagram of Experimental Setup

Two sections of composite walls are positioned on either side of the plate heater (Ni-Cr wire packed in upper and lower mica sheets, 1000 W) Two thermocouples (2&3) are used to measure the hot face temperature at the upper and lower heater plate, 4 & 5 are used to measure the temperature at the other end of mild steel plate (25mm), 6 & 7 to measure temperature at far end of hylam sheet (20 mm) and 8 & 9 for far end of the wooden plate (12 mm). (see figure). Diameter of the plates is 300 mm Specimens are held in position by the help of C clamps. The whole assembly is enclosed in wooden box with one side transparent of visualization. Voltmeter and Ammeter are used to measure the energy input to the heater. This energy input to the heater can be varied using Dimmer stat. Digital temperature indicator with selector switch on the control panel indicates the temperature at different positions in the composite 56


wall. Indicator lamp Indicates ON/OFF position of the heater. MCB has been provided to switch ON/OFF the power to the equipment. 5. Procedure The specimens are placed on either side of the heating plate assembly uniformly touching each other. Then predetermined heat input is given to the heater using the Dimmer stat. The input of the heater (current & voltage) and the thermocouple readings are observed every 5 minutes till a reasonably steady state condition is reached (steady temperature gradient). The readings are recorded in the observation table. The final steady state values are taken for calculations. 6.Observations and calculations: 6.1 Observation Table S.No. Volt. Amp. T2 (V) (A) ( )

T3 T4 ( ) ( )

T5 ( )

T6 ( )

T7 ( )

T8 ( )

T9 ( )

6.2 Calculations Heat transfer Area Perpendicular to Heat Flow, A = (π /4) * D² = ________ m² Heat Input Q = V * I = ________ W Thermal Conductivity for individual specimen, = (Q * ) / [2 * * ( – ], W/m0C i = 1, 2 and 3 for MS, Hylam and Wooden Plate respectively. Where, = ( + ) / 2 for MS plate, ( + ) / 2 for Hylam Plate, ( + ) / 2 for Wooden Plate, = ( + ) / 2 for MS Plate, ( + ) / 2 for Hylam Plate, ( + ) / 2 for Wooden Plate, Overall Thermal Conductivity, = (Q * L) / [2 * A *((( + ) / 2) – (( + ) / 2)))], W / m Indiviual Thermal Resistance, Ri= L/KiAi, 0C/W In Series, Thermal Resistance Rt of the composite wall is given by RT= ∑ Ri , i = 1, 2 and 3 for MS, Hylam and Wooden Plate respectively. 7. Results Overall thermal conductivity of the material is……. W/m0C 8. Precautions 1. Ensure that the Dimmer state is at 0 position at before switching on and off the equipment. 2. Increase the energy input gradually to the heater during initial set-up experimentation. 57


3. Never use the heater at full wattage for longer period of time.

EXPERIMENT NO. 6 (c)

THERMAL CONDUCTIVITY OF LIQUIDS 1. Aim To measure of the effectivity of the liquid in transmitting heat through it. 2. Objective To determine the Thermal conductivity of liquid. 3. Theory Thermal conductivity is the property of a material to conduct heat. Heat transfers occurs across materials of high thermal conductivity than across materials of low thermal conductivity. Thermal conductivity of materials is temperature dependent. For the measurement of the thermal conductivity K what is required is to have one dimensional heat flow through the flat specimen, an arrangement for maintaining its faces at the constant temperature and metering method to measure the heat flow through a known area.

58


Knowing the heat input to the central plate heater, the temperature difference across the each specimen, its thickness and the area, one can calculate the K by the following formula. K = q * L / [A * ( Theater avg – Tsample liquid temp ) ] where, K Thermal Conductivity of the sample, W / m C q Heat flow rate in the specimen, W A Area of the specimen, m2 TH,av Hot side average temperature, C TC,av Cold side average temperature, C L Thickness of the test section, m Thermal conductivity is important in building insulation and related fields. However, materials used in such trades are rarely subjected to chemical purity standards.Several construction materials' ''k'' values are listed below. These should be considered approximate due to the uncertainties related to material definitions. The following table is meant as a small sample of data to illustrate the thermal conductivity of various types of substances. Some typical thermal conductivity (k values) Material

Ice Water Glycerol Rubber (92%) Alcohols OR Oils Air Oxygen (O2) Nitrogen (N2)

Thermal Conductivity (W·m−1·K−1) 1.6 - 2.2 0.32 0.29 0.16a 0.1- 0.21 0.024 - 0.0262 0.0238 0.0234 - 0.026

Temperature (K) 293 293 293 303 293 273-300 293 293 - 300

4. Experimental Set up 4.1. Requirements Thermal Conductivity Apparatus with composite wall (Mild steel, Hylam (Paper Grade) & Wood), C clamps . 4.2 Description Test Section (SS 304, 5 mm width, 150 mm diameter) is resting on a heater (500 W, Ni-Cr) .The heat passing across the test section is removed by the cooling water jacket. The housing made of Mild Steel has been filled with glass wool to ensure minimum heat loss to the surroundings. The total assembly of test section, , main heater and cooling jacket are held in position by the help of bolts. Voltmeter and Ammeter are used to measure the energy input to the heater. This energy input to the heater can be varied using Dimmerstat. Digital Temperature indicator with selector switch on the control panel indicates the temperature at different positions. Indicator Lamp indicates ON/ OFF position of the heater. MCB has been provided to switch ON/ OFF the power to the equipment. 59


5. Procedure 1. Fill the test section with the liquid whose thermal conductivity is to be measured. 2. Switch ON the Main heater and set the desired heat input through the test section using Dimmerstat. 3. Start the cooling water supply to the cooling water jacket. 4. Now observe the temperatures T1 to T6 after 15 minutes and note down their values once they become reasonably constant . 6. Observations and calculations 6.1 Observation Table Sr. Volt Amp. T1 T2 No (V) (A) (C) (C) 1 2 3 4 5

T3 (C)

T4 (C)

T5 (C)

T6 (C)

T1= heater temperature T2= cold water inlet temp to upper plate T3= hot water outlet temp of upper plate T4= cold water inlet to lower plate T5= hot water outlet of lower plate T6= sample liquid inner temp T Heater = Heater temperature= T1 Tsample = sample liquid temperature Diameter of test section = 150 mm Thickness of test section = 20 mm 6.2 Calculations Heat Transfer Area Perpendicular to Heat Flow, A = (2 *  * r) = ------------------ m2 Heat Input Q = V * I = ------------ Watt Thermal Conductivity , K = (Q * L) / [A * (Theater – T 6)] = -----------

W/ m C.

7. Results Thermal conductivity of the material is……. W/m0C 8. Precautions 1. Ensure that the Dimmerstat is at 0 position at before switching on and off the equipment 2. Increase the energy input gradually to the heater during initial set-up experimentation 3. Never use the heater at full wattage for longer period of time

60



DROPWISE AND FILWISE CONDENSATION 1. Aim:  To determine the inside and outside heat transfer coefficient of Filmwise and Dropwise condenser.  To study the Dropwise and Filmwise condensation phenomena  2. Apparatus: Filmwise and Dropwise Condensers enclosed in a Borosilicate Glass Tube with flow control valves, Steam generator with heating elements, Digital Temperature Indicator with selector switch, Rotameter. 3. Theory: Condensation Heat Transfer: The process of condensation is the reverse of boiling. Whenever a saturated vapor comes in contact with a surface at a lower temperature, condensation occurs. There are two modes of condensation; filmwise, in which the condensate wets the surface forming a continuous film which covers the entire surface and dropwise in which the vapor condenses into small liquid droplets of various sizes which fall down the surface in a random fashion. Filmwise condensation generally occurs on clean uncontaminated surfaces. In this type of condensation the film covering the entire surface grows in thickness as it moves down the surface by gravity. There exists a thermal gradient in the film and so it acts as a resistance to heat transfer. In dropwise condensation a large portion of the area of the plate is directly exposed to the vapor, making heat transfer rates much larger (5 to10 times) than those in filmwise condensation.

61


4. Experimental Setup DROPWISE – FILMWISE CONDENSATION APPARATUS STEAM INLET

T1 WATER INLET

T2-3

T6

T7

T4

T5

Apparatus description: The apparatus consist of: Steam Generator: (8 liter capacity) equipped with 2 kW heater and Pressure Gauge, Manual Release Valve, Feed Line and Steam Line. Dropwise Condenser: MOC: Copper with chrome plating

Dimensions: ID (di) = 16 mm OD (do) = 19 mm Length (L) = 170 mm Filmwise Condenser: MOC: Copper with Natural finish Dimensions: ID (di) = 16 mm OD (do) = 19 mm Length (L) = 170 mm Temperature Indicator with Selector Switch measures the Temperature of : T1 Steam Chamber Temp. T2 Cooling Water Inet to dropwise condenser T3 Cooling Water Inet to filmwise condenser T4 Dropwise Condenser outlet Temp T5 Filmwise Condenser outlet Temp T6 Dropwise condensation surface Temp T7 Filmwise condensation surface Temp 62


5. EXPEIRMENTAL PROCEDURE Fill the steam generator with about 10 – 15 liter of water (preferably soft). Connect supply socket to the mains and switch on the heater. Switch on the heater while keeping the steam line and feed line valves in closed position. Adjust the temperature of the steam near to the 105 – 110 OC. Allow the steam generation to take place. This may take 30 – 40 minutes depending on the initial temperature of the feed water. The pressure of the generated steam will be indicated on the pressure gauge. Note down the Pressure reading of the steam inlet. h) Now select the condenser to be tested first and open the ball valve of the same for cooling water supply. i) Now start the supply of cooling water in the selected condenser (Dropwise or Filmwise) j) Depending upon the type of condenser under test Dropwise or Filmwise condensation can be visualized. k) If water flow rate is low than steam pressure in chamber will rise and pressure gauge will read the pressure. l) If the water flow rate is matched than condensation will occur at more or less at atmospheric pressure. m) Process of Dropwise and Filmwise condensation can be easily viewed through the front glass window of main unit. n) Note down the inlet temperature of the cooling water, Outlet Cooling water Temperature as indicated by the DTI. o) Slowly open the steam line valve and allow the steam to enter the steam chamber. p) Observe the condensation phenomena and also note down the condenser temperature, steam inlet temperature. q) Measure and note the cooling water flow rate through Rotameter provided. r) Repeat the above procedure for the second type of condenser. s) Use the wiper provided if the fog/ mist restricts the visualization of the glass vessel t) At the time of steady state ,the outlet temperature of the water flow inside the tube becomes constant. PRECAUTIONS: a) Do not start heater supply unless water is filled in the test unit. b) Operate gently the selector switch of temperature indicator to read various temperatures. c) Increase the temperature gradually of the heater during initial set-up experimentation. d) Never use the heater at full wattage for longer period of time. e) Use the proper range of Rotameter. f) Operate the change over switch of temperature indicator gently from one position to other, i.e. from 1 to 4 position a) b) c) d) e) f) g)

6. Observations: Filmwise Condensation: Cooling Water Flow Rate (mw): LPM Temperature: T1 Steam Temp: T3 Cooling Water In Filmwise Condenser: T7 Filmwise Condenser Surface Temp: T5 Cooling water outlet from film wise condensation

63


Dropwise Condensation: Cooling Water Flow Rate (mw):

LPM

Temperature: T1 Steam Temp: T2 Cooling Water in Drop wise Condenser: T6 Drop wise Condenser Surface Temp: T4 Cooling water outlet from drop wise condensation surface Calculations: Normally steam will not be pressurized, but the pressure gauge reads some pressure than properties of steam should be taken at that pressure or otherwise atmospheric pressure will be taken. First calculate the heat transfer coefficient inside the condenser under test. For this properties of water are taken at bulk mean temperature of water i.e. for dropwise condensation mean temp = (T2 + T4) / 2. Film-wise condensation mean temp = (T3+T5)/2 Following properties are required:  Density of water  kg / m3  Kinematic Viscosity  m2/ sec  Thermal Conductivity ‘k’ kcal / hr m C  Prandtl Number Pr  Reynolds Number NRe =  v di /  Where v = (mw*4/  * di2) = ( Flow rate/Area) di = Inner Diameter of Condenser If this value of NRe > 2100 then flow is turbulent, below this value flow is laminar. Normally flow will be turbulent in the tube.  Nusselt Number NuD = 0.023 (ReD)0.8 (Pr)0.4  Inside heat transfer coefficient ( hi ) = NuD * k/ di kcal / hr m2 C Calculate heat transfer coefficient on outer surface of the condenser HO: For this properties of water are taken at bulk mean temperature of condensate (Ts + Tw) / 2.  Density of water  kg / m3  Kinematic Viscosity  m2/ sec  Thermal Conductivity ‘k’ kcal / hr m C  Prandtl Number Pr  Reynolds Number NRe =  v di /  Where, Ts Temperature of steam, T6-7 Temperature of condenser wall (DROPWISE AND FILMWISE) do outside diameter of condenser 1. ho = 0.725 * 2 * g * k3 / ( Ts - Tw) * do From these values overall heat transfer coefficient (U) can be calculated, 1/ U = 1 / hi + (di / do) (1 / ho) U= kcal / hr m2 C The same procedure can be repeated for another condenser. 64


Except for some exceptional cases overall heat transfer coefficient for dropwise condensation will be higher than that of filmwise condensation. Results may vary from theory to some degree due to unavoidable heat losses from the glass tube walls. 7. Result:

GIVEN DATA: First calculate the heat transfer coefficient inside the condenser under test. For this properties of water are taken at bulk mean temperature of water i.e. (T2 + T4) / 2. Following properties are required: PHYSICAL PROPERTIES AT WATER MEAN TEMPERATURE

Physical Properties

Dropwise Condenser

Filmwise Condenser 99.4

Steam Temperature OC

Ts

99.3

Mean Temperature oC

Tm(T2+T4 / 2)

35

Density of Water kg/m3



993.95

µ

0.00073

0.00073



0.732*10-6

0.732*10-6

Q

2

2

m

0.000033

0.000033

V

0.164

0.164

K

0.537

0.537

CP

0.997

0.997

Viscosity Kg/ m s Kinematics Viscosity m2 / s Cooling Water Flowrate LPM Vol. Flowrate m3/s Velocity m/s Thermal Conductivity kcal / hr m C Specific Heat

35 993.95

Now calculate the heat transfer coefficient outside the condenser under test. For this properties of condensate are taken at bulk mean temperature of Condenser wall and the Steam Inlet Temperature i.e. (Tw + Ts) / 2. Following properties are required: PHYSICAL PROPERTIES AT CONDENSATE MEAN TEMPERATURE Physical Properties

Dropwise Condenser 65

Filmwise Condenser


Steam Temperature OC

T1

99.3

99.4

Mean Tamp 0 C

Tm

78.45

82.2

Density of Water kg/m3



971.8

971.8

Viscosity Kg/ m s

µ

0.000355

0.000355



0.365*10-6

0.365*10-6

Q

2

Vol. Flowrate m3/s

m

0.000033

Velocity m/s

V

0.164

Thermal Conductivity kcal / hr m C

K

0.579

CP

0.997

Kinematics m2 / s Cooling Flowrate LPM

Specific Heat

Viscosity Water

66



UNSTEADY STATE HEAT TRANSFER UNIT AIM (1) To study the unsteady state temperature response of finite geometric shapes. (2) To calculate the value of surface conductance (h). 1.

THEORY:

When the temperature at any given point in a system changes with time, the heat transfer is at unsteady state. The phenomenon of unsteady state heat transfer is widely utilized in industrial processes, such as the cooling of metal ball bearings. The calculation of heat transfer coefficients makes it possible, among other things, to predict the amount of time a system takes to reach steady state conditions. In this experiment, we want to use a lumped capacity analysis to simplify our calculations, in which only the effects of convection are significant. This occurs when the internal resistance is negligible, and the Biot Number, Bi, is less than 0.1 as described in Equation (1).1 Bi 

hx k

(1)

where k is the thermal conductivity and x is a characteristic dimension of the body obtained from the volume to area ratio.

Derivation of Lumped-Sum Analysis: ln

T  T T0  T



h t xC p 

(1*)

where x is the relative radius, described by: x

V A

(2*)

V is the volume of the cylinder and A is the surface area. For a long cylinder this becomes: x

r 2

(3*)

Plugging into equation (1*)

ln

T  T 2h  t rC p  T0  T

(4*)

67


ln

T  T 2h  t rC p  T0  T

(2)

where Cp is the heat capacity and  is the density of the material. T is the temperature of a large bath, T is the center temperature of the cylinder at time, t, and T0 is the initial center temperature. This lumped sum model makes several approximations, including that of an infinite tank. As discussed in previous reports, this approximation is not valid for the tank in the Rothfus lab using materials with high internal resistance, such as Plexiglas. However, using materials with low internal resistance such as those in this experiment, the lumped sum model is a close approximation. In addition to the approximation of an infinite bath, the h value is the average value over the entire surface of the cylinder. This introduces a problem of conduction through the ends of the cylinder, which must be accounted for by using an appropriate length (L) to diameter (D) ratio2: L (3)  2 D Using Plexiglas end caps helps to eliminate heat flow through the ends of the cylinders, making the infinite cylinder approximation more accurate. Using various L/D ratios, it is possible to test the accuracy of this assumption, which is addressed in this experiment.

In general, the average heat-transfer coefficient on immersed bodies is correlated in terms of the Nusselt number1: m 1/ 3 N Nu  CN Re N Pr

(4)

where NPr is the Prandtl number which is constant for this experiment. The variables C and m are constants based on the Reynolds number. From all previous reports using similar cylinders, the Reynolds number was found to be within the range of 4 x 103 to 4 x 104 corresponding to C=0.683 and m=0.6183.

N Re 

Dv

(5)



where v, , and  are properties of water which are constant throughout the experiment.

N Nu 

hD k

(6)

68


where k is the thermal conductivity of water, and D is the diameter of the cylinder. From these equations it can be seen that h will decrease with increased diameter. h

C1 D 0.382

(7)

where C1 is a proportionality constant. This relationship will hold true for infinite cylinders, but with increasing diameter and constant, finite length, the end effects will become more apparent, and therefore the Biot number will increase. While equation (6) contains a relation between the k value of the surrounding water and the convective heat transfer coefficient, it does not indicate any relationship between the h value and the thermal conductivity of the material itself. In fact, this equation would imply that the material has no affect on the heat transfer coefficient, as long as the lumped sum approximation has been justified, meaning the internal resistance is relatively low. However, in previous experiments it has been noted that the heat transfer coefficient varies with changing material properties.4 When the effects of internal resistance are not negligible, the lumped-sum analysis described above is invalid, and Heisler charts must be used to calculate the convective heat transfer coefficients. To use these charts, dimensionless numbers are calculated and applied to a chart with data for objects of varying geometry. In this case, we will use the following described values and apply them to a Heisler chart (see figure A 4.1 below) to determine the heat transfer coefficient of a long cylinder.5

X   t x2 1 Y

T  T T  T0

m n



(8)

(9)

k hx

(10)

x x1

(11) k

 Cp

(12)

X is a dimensionless number which accounts for the effects of the shape and material, where k is the solid thermal conductivity, and x1 is the radius of the cylinder. Y is a dimensionless temperature constant, m is the inverse of the Biot number, relating conductive effects and convective effects, and n is a relative radius, in which x is the location of the thermocouple. The results gathered from Heisler charts will be accurate even with significant internal resistance and are recommended anytime the Biot number is greater than 0.1.

69


Figure A4.1: Heisler chart. 2.

Experimental setup:

In this experiment, the effects of material selection and radius on the heat transfer coefficient, h, of a solid cylinder were measured. For this experiment the materials were ordered and were prepared by cutting the appropriate material into its desired length. Then Sphere & Cylinder were drilled in the centre to a depth that corresponded to half its length (See Figure A2.1 for an example of how this was done). A Type J, iron/ constantan, thermocouple was glued into the centre of each cylinder using an epoxy with a high thermal conductivity. This particular epoxy was chosen because it would not have any affect on the temperature readings that were obtained from the thermocouple. In essence, the presence of the epoxy could be ignored. (For more accurate results, the ends of the cylinders would be sealed with Plexiglas end caps and a two-part epoxy. The epoxy and Plexiglas acted as an insulator to prevent heat loss from the ends of the cylinders.) The apparatus consist of a 20 litre SS 304 tank equipped with 1.5 kW heater and a drain valve. The tank is well insulated from outside to prevent heat losses to the surroundings. The temperature of the fluid inside the tank can be controlled by the regulator provided on the tank and is measured using the digital temperature indicator. Two shapes of known geometry and metals are provided with thermocouples to study the unsteady state heat transfer. T1 Hot Oil/ Water Bath Temperature T2 Mild Steel Sphere T3 Copper Cylinder Test Body Shape 1: Geometry MOC

= Cylinder = Copper 70


Diameter = 50 mm Length = 150 mm Density () = 8900 kg/m3 Specific Heat (Cp) = 0.38 kJ/ kg oC Test Body Shape 2: Geometry MOC Diameter Density () Specific Heat (Cp)

= Sphere = Mild Steel = 50 mm = =

71


3.

Schematic diagram: Stirrer Reference Thermocouple

Hot Water Bath: Constantly stirred. Kept at 65 C 4. Procedure: A hot water bath is used . This water bath provides a constant temperature environment for the experiment to be conducted .The test body was placed in the ambient temperature and is allowed to achive this temperature . Once this temperature is achieved ,as shown by the reference thermocouple , the test body to be tested is then transferred to the hot water bath at particular temperature (about 65 degree c), which was kept constant using thermostat provided. The cylinder is allowed to equilibrate to this temperature .digital temperature indicator and stopwatch is used to record the temperature read by thermocouple every 5 seconds. I. Fill the heating tank with fluid upto 75% level so that test specimen can be dipped into the fluid. II. Switch on the heater and maintain the desired temperature of the water /oil using the regulator provided on the tank . III. Note down this temperature (Ta) and intial temperature of the body as indicated by the DTI. IV. Now dip the selected body into this hot fluid with the help of stand provided. V. With selector switch of DTI at proper thermocouple number and start noting down the unsteady state temperature response (T) for a regular time interval (about 5 or 10 secs interval).

72


VI.

After the temperature reaches the steady state take the body out of the heating tank and note down the unsteady state temperature response of the body while cooling .Or dip the body into the cold water bath and note down the unsteady state temperature response (T) for a regular time interval for cooling.

Repeat the above procedure with second body. 5. OBSERVATION TABLE: Body Shape : Copper Cylinder Diameter : 50 MM Length :150 mm Volume : 2.9437 * 10-4 MM2 Surface Area : 0.0235 M Thermal Conductivity : 206 w/m.k Density : Specific Heat : Cycle : Heating / Cooling Surrounding Temperature (Ta) : C Initial body Temperature (Ti) : C Sr. No.

Time (sec) (t)

Temperature Response (C) T = T3

Unit Surface (T - Ta/ Biot Conductance Ti – Ta) Number 2 o (w/ m * C) h Bi

Fourier Number Fo

GRAPHS: Plot [(T-Ta)/ (Ti-Ta)] vs. t for heating as well as cooling cycle. OBSERVATION TABLE: Body Shape Diameter Volume Surface Area Thermal Conductivity Density Specific Heat Cycle Surrounding Temperature (Ta) Initial body Temperature (Ti) Sr. No.

Time (sec) (t)

Temperature Response (C) T = T3

: MS Sphere : 50 MM :2.9437 * 10-4 MM2 :0.0235 M : 206 w/m.k :8900 kg/m3 : 0.38 kJ/ kg oC : Heating / Cooling : C : C Unit Surface (T - Ta/ Biot Conductance Ti – Ta) Number (w/ m2* oC) h Bi

73

Fourier Number Fo


GRAPHS: Plot [(T-Ta)/ (Ti-Ta)] vs. t for heating as well as cooling cycle.

SAMPLE CALCULATION: Sample calculation of h using lumped-sum model (For Aluminium D=0.05 M, L=0.150 M) First, obtain the equation of the line.where x=(v/As),X= Next, use the slope of that equation in the following equation:

h  slope 

 r Cp   2



SAMPLE CALCULATION OF H FROM HEISLER CHART First, choose a point on the graph of Y vs. t. Then substitute those values into the following equations:

X Y

 t

 x2 1 T  T

T  T0 k m h  x1



Using the Heisler chart, find the point, which corresponds to a value for m. Rearranging the equation for m yields the following:

h

k  m  x1

Sample calculation of C1 value: For 3.0-inch diameter aluminum cylinder, from Table 2 the average experimental h is 1335.

h = 1335

W  in m2  K

D = 3.0 in. 74


h

C1 D0.382

Therefore, C1 = 2031

(equation (9)) W  in m2  K

h values for all trials Material trials:

L/D trials:

D=1.5 in. L=6 in. h(W/m^2*K) h(heisler) steel 1 974.771 945 steel 2 1039.984 945 steel 3 985.068 859 AVERAGE 999.941 916 stdev 35.05835762 49.65212315

aluminum D=2 in. h(W/m^2*K) h(heisler) L=4 in 1 1801.954 1622 L=4 in 2 1823.516 1352 L=4 in. 3 1829.677 1352 AVERAGE 1818.382 1442 stdev 14.55702862 155.8845727

aluminum L=6 in. h(W/m^2*K) h(heisler) D=1 in 1 2086.937 D=1 in 2 2083.856 D=1 in 3 2048.432 AVERAGE 2073.075 stdev 21.39699107

Radius Trials:

copper1 copper2 copper3 AVERAGE stdev

2136.796 2042.045 1979.967 2052.936 78.97970765

L=6 in. 1 1943.6467 1474 L=6 in. 2 1919.0046 1622 L=6 in. 3 1885.1217 1622 AVERAGE 1915.924333 1573 stdev 29.38383805 85.44783984

D=1.5 in 1 D=1.5 in 2 D=1.5 in 3 AVERAGE stdev

brass1 brass2 brass3 AVERAGE stdev

1809.447 1365 1753.195 1365 1906.326 1560 1822.989333 1430 77.45851718 112.5833025

L=7.5 in. 1 2008.3322 1622 L=7.5 in. 2 1959.048 1707 L=7.5 in. 3 1949.8072 1622 AVERAGE 1972.3958 1650 stdev 31.46294217 49.07477288

D=2 in 1 1697.273 D=2 in 2 1152.051 D=2 in 3 1497.05 AVERAGE 1448.791333 stdev 275.7959962

aluminum1 aluminum2 aluminum3 AVERAGE stdev

1266.024 1520.152 1240.691 1342.289 154.5537942

L=9 in. 1 1774.232 L=9 in. 2 1811.1953 L=9 in. 3 1801.9545 AVERAGE 1795.793933 stdev 19.23631521

D=2.5 in 1 1478.568 1298 D=2.5 in 2 1259.093 1442 D=2.5 in 3 1517.072 1442 AVERAGE 1418.244333 1394 stdev 139.1671645 83.13843876

1622 1622 1622 1622 0

1266.024 1520.152 1240.691 1342.289 154.5537942

D=3 in 1 1270.644 1298 D=3 in 2 1270.644 1298 D=3 in 3 1464.706 1442 AVERAGE 1335.331333 1346 stdev 112.0417479 83.13843876

6.

Results

7.

Discussion

8.

Conclusion

9.

PRECAUTIONS 1. Wait till steady state is achieved. 2. Use stabilized power supply.

75



HEAT TRANSFER IN AGITATED VESSEL (JACKET & COIL)

1. Aim & Objective To study heat transfer phenomena in agitated vessel with jacket & helical coil system and comparison experimental overall heat transfer coefficients in both jacket & coil. 2. Apparatus 2 nos. pump driven setup consisting of agitated vessel with helical coil, 2 sump tanks, PID temperature controller, temperature indicators, rotameters. 3. Theory Whenever the fluid motion is varied by external means the heat is transferred by forced convection. Most of the time fluid is agitated by circulating the hot and cold fluids at rapid rates on the opposite sides of pipes or tubes. The rate of heat transfer by forced convection to an incompressible fluid traveling in turbulent flow in a pipe of uniform diameter at constant mass rate has been found to be influenced by the velocity, density, specific heat, thermal conductivity, viscosity of fluid as well as the inside diameter of the pipe. The velocity, viscosity, density and diameter affect the thickness of fluid film at the pipe wall through which the heat must first be conducted and they also influence the extent of fluid mixing. The thermal conductivity of the fluid and the specific heat reflects the variation of the average fluid temperature as a result of unit heat absorption. A simple jacketed pan or kettle is very commonly used in the chemical industries as a reaction vessel. In many cases, such as in nitration or sulphonation reactions, heat has to be removed or added to the mixture in order either to control the rate of reaction or to bring it to completion. The addition or removal of heat is conveniently arranged by passing steam or water through a jacket fitted to the outside of the vessel or through a helical coil fitted to the inside. In either case some form of agitator is used to obtain even distribution in the vessel. This may be of the anchor type for very thick mixes or a propeller or turbine if the contents are too viscous. 4. Experimental Setup: Agitated vessel consist impeller mounted on the shaft, which is driven by a motor. The formation of swirling water in an agitated vessel can be prevented by introduction of baffles. The schematic is shown in below figure

76


Fig 1: Schematic diagram of Heat Transfer in Agitated Vessel experiment 5. Experimental Procedure:  Open the drain cock of the vessel and remove all the water from inside.  Fill the agitated vessel with about 4-5 liter of water (about 70-75% height of the vessel) for a batch operation.  Switch on the heater and heat the water to the desired degree of hotness (about 70-80 C). Intermittently switch ON the pump with bypass line valve fully open and supply valve fully closed to ensure thorough mixing of water in the tank to ensure uniform temperature.  Measure the temperature of bulk water periodically with the help of the thermocouple placed in the thermowell.  Now start the hot water supply in the coil or jacket depending upon the experiment. And measure its inlet and outlet temperatures periodically with the thermocouples placed in the thermowells.  Note down the values of cooling water inlet and outlet temperatures and cooling water flow rate, hot water inlet and outlet temperatures and hot water flow rate. 6. Observations & Calculations Thermal conductivity of SS (K) Diameter of agitated vessel (Da) Height of agitated vessel (h) Diameter of coil (Dc) Length of coil (Lc)

= 16.86 W/ m.K = 0.73 m = 0.265 m = 0.11m = 5.115 m

6.1 Properties of water at 331 K (58 0C) Density of water

984.1 Kg/cm3 77


Specific heat of water

4187 J/Kg.K

Viscosity of water

485 * 10-6 Pa.sec

6.2 Observation table For hot water in coil: S.No Hot water Cold water T1 flow rate flow rate (° C) (litres/min) (litres/min)

For hot water in jacket: S.No Hot water Cold water T1 flow rate flow rate (° C) (litres/min) (litres/min)

T2 (° C)

T2 (° C)

T3 (° C)

T5 (° C)

T4 (° C)

T6 (° C)

6.3 Calculations: 6.3.1 Experimental Value calculations: For hot water in coil: Heat Gained by the Cooling Water (Q)

= mw Cp (T4– T3)/60 = .................J/ sec = Uexpt Acoil [ T2 - ( T3 + T4 ) / 2 ]

For hot water in jacket: Heat Gained by the Cooling Water (Q)

= mw Cp (T6 – T5)/60 = .................. J/ sec = Uexpt Ajacket [ T2 - ( T5 + T6) / 2 ] Calculate the value of Uexpt from the above equations for both jacket and coil. Where, QC = Cold water flow rate, LPM. QH = Hot water flow rate, LPM. T1 = Bulk water inlet temperature of vessel. T2 = Bulk water outlet temperature of vessel. T3 = Coil inlet temperature, oC. T4 = Coil outlet temperature, oC. T5 = Jacket inlet temperature, oC. T6 = Jacket outlet temperature, oC. Ajacket = 0.11 m2 7. Results & Conclusions Experimental heat transfer coefficients for hot water flow through coil are _______ W/m2 °C

78


Experimental heat transfer coefficients for hot water flow through jacket are _______ W/m2 °C 8. PRECAUTIONS: Make sure during the test period Hot Water Tank should not be emptied totally and the heater must not be exposed to air if the Heater is ON, otherwise it will be damaged.

79



FLUIDIZED BED HEAT TRANSFER UNIT 1. AIM  To study the performance of the heat transfer in Fluidized bed  To study the heat transfer coefficient for heat transfer in fluidized bed. 2. Apparatus: 2 nos. pump driven setup consisting of fluidized bed, 2 sump tanks, PID temperature controller, temperature indicators, rotameters. 3. THEORY: When a liquid or a gas is passed at very low velocity up through a bed of solid particles, the particles do not move ,and the pressure drop is given by the Ergun equation. If the fluid velocity is steadily increased ,the pressure drop and the drag on individual particles increase,and eventually the particles start to move and become suspended in the fluid. The terms “fluidization” and “fluidized bed” are used to describe the condition of fully suspended particles,since the suspension behaves like a dense fluid. Fluidized beds are used extensively in the chemical process industries, particularly for the cracking of high-molecular-weight petroleum fractions.Such beds inherently possess excellent heat transfer and mixing characteristics.These are devices in which a large surface area of contact between a liquid and a gas ,or a solid and a gas or liquid is obtained for achieving rapid mass and heat transfer and for chemical reactions. The fluidized bed is one of the best known contacting methods used in the processing industry, for instance in oil refinery plants. Among its chief advantages are that the particles are well mixed leading to low temperature gradients, they are suitable for both small and large scale operations and they allow continuous processing. There are many well established operations that utilize this technology, including cracking and reforming of hydrocarbons, coal carbonization and gasification, ore roasting, Fisher-Tropsch synthesis, coking, aluminum production, melamine production, and coating preparations. Nowadays, you will find fluidized beds used in catalyst regeneration, solid-gas reactors, combustion of coal, roasting of ores, drying, and gas adsorption operations The application of fluidization is also well recognized in nuclear engineering as a unit operation for example, in uranium extraction, nuclear fuel fabrication, reprocessing of fuel and waste disposal. 4. EXPERIMENTAL PROCEDURE:  Fill the water in cold water supply tank with about 100 litre of clean tap water. After filling tank upto 50% start water pump and fill the overhead heater tank by controlling bypass and supply valve.  As the heater tank fills with water to a level of heater, Switch on the immersion type heater provided in the hot water tank and heat the water to the desired temperature (about 60-70 C). Intermittently switch ON the pump with bypass line valve fully open and supply valve fully closed to ensure thorough mixing of water in the tank to ensure uniform temperature.  After achieving the desired temperature of water in the hot water tank, allow the hot water to flow through inner pipe side and adjust the flow rate to the desired value using the valve of rotameter. Drain the exit of the hot water to the drainage.  Start the cold water supply on the outer pipe side and adjust the flow rate to the desired value-using valve of rotameter. Now place the outlet of outer tube side in to the drain line. 80


  

Observe the inlet and outlet temperature of both cold and hot water streams and note down them after they achieve steady state. Also note down the Flow rates of hot water and cold water with the help of Rotameters. Repeat the above procedure either by changing the Flow rates or by changing the inlet temperature of the hot water.

5. OBSERVATION TABLE AND CALCULATION : Density of water  = 1 kg/l Diameter of bed do = 2 inch Length of bed L = 12 inch Specific heat of water = CpH = CpC = 4187 J/Kg.K Hot Water Inner Tube Side

Cold Water Outer Tube Side

Inlet Flow Rate Temp. mh T1 LPM (C)

Flow Rate mc LPM

Outlet Temp. T2 (C)

Flow rate of hot water in kg/ s

Inlet Temp. T3 (C)

mH = mh *  / 60 =____________ kg/ s

Heat Transferred by the Hot Water to the Cold Water QH = mH * CpH * (T1 – T2) =____________ W Flow rate of cold water in kg/ s

mC = mc *  / 60 = ___________ kg/ s

Heat Gained by the Cold Water from the Hot Water QC = mC * CpC * (T4 – T3) = ____________ W Average Heat

Area for the Heat Transfer,

Q = (QH + QC) / 2 = ____________ W A =  * do * L = ____________ m2

For Parallel flow ( T1 – T3 ) - ( T2 – T4 )

LMTD

=

--------------------------------

ln ( ( T1 – T3 ) / ( T2 – T4 ) )

= _____________ 0C Overall Heat transfer coefficient can be calculated using, 81



U = Q / (A * LMTD) = ____________ W/ m2 C 6. RESULT AND DISCUSSIONS: FOR PARALLEL FLOW OVERALL HEAT TRANSFER COEFFICIENT = ---------------- W/ m2.C 7. PRECAUTIONS: Make sure during the test period Hot Water Tank should not be emptied totally and the heater must not be exposed to air if the Heater is ON, otherwise it will be damaged.

82



PARALLEL FLOW AND COUNTER FLOW HEAT EXCHANGER 1. Aim a) To calculate the overall Heat transfer coefficient of the heat exchanger. b) To calculate the local heat transfer coefficient of the heat exchanger using Seider Tate Equation and Dittus-Boelter Equation. 2. Objective To study the heat transfer phenomena in parallel and counter flow arrangements. 3. Theory The double-pipe heat exchanger is one of the simplest types of heat exchangers. It is called a double-pipe exchanger because one fluid flows inside a pipe and the other fluid flows between that pipe and another pipe that surrounds the first. This is a concentric tube construction. Flow in a double-pipe heat exchanger can be co-current or counter-current. There are two flow configurations: co-current is when the flow of the two streams is in the same direction, counter current is when the flow of the streams is in opposite directions. As conditions in the pipes change: inlet temperatures, flow rates, fluid properties, fluid composition, etc., the amount of heat transferred also changes. This transient behaviour leads to change in process temperatures, which will lead to a point where the temperature distribution becomes steady. When heat is beginning to be transferred, this changes the temperature of the fluids. Until these temperatures reach a steady state their behaviour is dependent on time. In this double-pipe heat exchanger a hot process fluid flowing through the inner pipe transfers its heat to cooling water flowing in the outer pipe. The system is in steady state until conditions change, such as flow rate or inlet temperature. These changes in conditions cause the temperature distribution to change with time until a new steady state is reached. The new steady state will be observed once the inlet and outlet temperatures for the process and coolant fluid become stable. In reality, the temperatures will never be completely stable, but with large enough changes in inlet temperatures or flow rates a relative steady state can be experimentally observed. As with any process the analysis of a heat exchanger begins with an energy and material balance. Before doing a complete energy balance a few assumptions can be made. The first assumption is that the energy lost to the surroundings from the cooling water or from the Ubends in the inner pipe to the surroundings is negligible. We also assume negligible potential or kinetic energy changes and constant physical properties such as specific heats and density. These assumptions also simplify the basic heat-exchanger equations.

83


84


4. Calculation Assumption  Steady state conditions exist.  Fluid properties remain constant and are evaluated at a temperature of fluid. Nomenclature T refers to the temperature of the warmer fluid. t refers to the temperature of the cooler fluid. w subscript refers to the warmer fluid. h subscript refers to hydraulic diameter c subscript refers to the cooler fluid. a subscript refers to the annular flow area or dimension. p subscript refers to the tubular flow area or dimension. 1 subscript refers to an inlet condition. 2 subscript refers to an outlet condition. e subscript refers to equivalent diameter. Fluid Properties

Counter current: Flow rate Flow rate mh S.no mc (Kg/m) (Kg/m)

Hot water in (ºC)

Hot water out (ºC)

85

Cold water in (ºC)

Cold water out (ºC)


Co-current:

Tubing Sizes IDa = IDp =

ODp =

Flow Area Flow rate Flow rate mh S.no mc (Kg/m) (Kg/m)

Hot water in (ºC)

Hot water out (ºC)

Cold water in (ºC)

Area of pipe A= π*D*L = Area of annulus A = π*De*L = Fluid Velocity 5. Results The Over All heat transfer coefficient = The local heat transfer coefficient =

E EXPERIMENT NO. 9 (b)

SHELL AND TUBE HEAT EXCHANGER 1. Aim Study of Shell and Tube Heat Exchanger 2. Objective To calculate overall heat transfer coefficient for shell & tube heat exchanger. 86

Cold water out (ºC)


3. Theory Heat Exchanger is device in which heat is transferred from one fluid to another. The necessity for doing this arises in a multitude of industrial applications. Common examples of heat exchangers are the radiator of a car, the condenser at the back of a domestic refrigerator and the steam boiler of a thermal power plant. Heat Exchangers are classified in three categories: 1) Transfer Type. 2) Storage Type. 3) Direct Contact Type. A transfer type of heat exchanger is one on which both fluids pass simultaneously through the device and heat is transferred through separating walls. In practice, most of the heat exchangers used are transfer type ones. Shell and tube heat exchangers are the most widely used in chemical process industries. A shell and tube exchanger consists of a bundle of tubes enclosed in a cylindrical shell. The ends of the tubes are fitted into tube sheets, which separate the shell-side and tube-side fluids. Baffles are provided in the shell to direct the fluid flow and support the tubes. The tubes are arranged in triangular or square pitch. One fluid flows inside the tube and is called the tube side fluid, the other fluid flows outside the tubes and is called the shell side fluid. The shell and tube heat exchangers are further classified according to flow arrangement as 1. Single Pass 2. Multiple Pass The simplest shell and tube heat exchanger is 1-1 heat exchanger (one shell pass and one tube pass), the other types are 1-2 heat exchangers and 2-4 heat exchanger. The transfer of heat from the hot fluid to the wall or tube surface is accompanied by convection through the tube wall or plate by conduction, and then by convection to the cold fluid. The flow arrangement in a shell and tube heat exchanger could either be co-current or countercurrent. The heat transfer coefficients outside the tube bundles are referred as shell side coefficients. Baffles also increase the convection coefficient of the shell side fluid by inducing turbulence and a cross flow velocity component. Any heat exchanger design requires rigorous analysis. One of the most essential parts of the heat exchanger analysis is determination of the overall heat transfer coefficient. The overall heat transfer coefficient is defined in terms of the total thermal resistance to heat transfer between two fluids. The heat lost by the hot fluid can be calculated = Heat Transfer rate to the hot water. qh q h  m h C Ph Thi  Tho  KCal/hr Heat taken by the cold fluid can also be calculated qc = Heat Transfer rate to the cold water. q c  m c CPc Tco  Tci  KCal/hr

Qavg  Uo 

qc  q h 2 Qavg

Ao Tm

4. Experimental setup 4.1. DESCRIPTION

87


The apparatus consists of 1-2 Pass Shell and Tube heat exchanger. The hot fluid is hot water, which is attained from an insulating water bath using a magnetic drive pump and it flow through the inner tube while the cold water flowing through the annuals. For flow measurement rotameters are provided at inlet of cold water and outlet of hot water line. The hot water bath is of recycled type with Digital Temperature Controller 0 to 100 oC. 4.2. REQUIREMENTS Water supply 20 lit/min (approx.), Drain, Electricity Supply: 1 Phase, 220 V AC, and 4 kW, Floor area of 1.5 m x 0.75 m 4.3. SPECIFICATIONS 1. Shell Material = S.S. Dia. = 220 mm Length = 500 mm 25% cut baffles at 100 mm distance 4 Nos. 2. Tube Material = S.S OD = 16 mm ID = 13 mm Length of tubes = 500 mm Nos. of tubes = 24 3. Temperature Controller = Digital 0 – 199.9oC 4. Temperature Sensors = RTD PT-100 type (5 nos.) 5. Temperature Indicator = Digital 0 to 200oC with multi-channel switch. 6. Electric Heater = 230 V AC 2 kW (2 Nos.) 7. Flow measurement = Rotameter (2 No.) 8. Water Bath = Material: SS insulated with ceramic wool and powder coated MS outer Shell fitted with heating elements. 9. Pump = FHP magnetic drive pump (max. operating temp 85oC).

5. Experimental procedure: Starting Procedure: 1. Clean the apparatus and make Water Bath free from Dust. 2. Close all the drain valves provided. 3. Fill Water Bath ¾ with Clean Water and ensure that no foreign particles are there. 4. Connect Cold water supply to the inlet of Cold water Rotameter Line. 5. Connect Outlet of Cold water from Shell to Drain. 6. Ensure that all On/Off Switches given on the Panel are at OFF position. 7. Now switch on the Main Power Supply (220 V AC, 50 Hz). 8. Switch on Heater by operating Rotary Switch given on the Panel. 9. Set Temperature of the Water Bath with the help of Digital Temperature Controller. 10. Open Flow Control Valve and By-Pass Valve for Hot Water Supply. 11. Switch on Magnetic Pump for Hot Water supply. 12. Adjust Hot water flow rate with the help of Flow Control Valve and Rotameter. 13. Record the temperatures of Hot and Cold water Inlet & Outlet when steady state is achieved. Closing Procedure: 1. When experiment is over, Switch off heater first. 2. Switch of Magnetic Pump for Hot Water supply. 88


3. 4. 5. 6. 7.

Switch off Power Supply to Panel. Stop cold water supply with the help of Flow Control Valve. Stop Hot water supply with the help of Flow Control Valve. Drain Cold and Hot water from the shell with the help of given drain valves. Drain Water Bath with the help of Drain valve.

6. Observation & calculation: DATA: Inside heat transfer area, Ai Outside heat transfer area, Ao OBSERVATION TABLE: S. Hot water side No. Flow rate mh Kg/hr 1 2 3 4

= =

o

Thi C

o

Tho C

3.187 x 10-3 m2 4.827 x 10-3 m2

Cold water side Flow rate mc kg/hr

Tci oC

Tco oC

CALCULATIONS: 1. Rate of heat transfer from hot water, Qh  M h C ph Thi  Tho  , Watt 2. Rate of heat transfer to cold water, Qc  M c C pc Tco  Tci  , Watt 3. Average heat transfer, Q  Qc Q h , Watt 2 4. LMTD, T  T1 Tm  2 T ln 2 T1 Where: T1 = Thi - TCi (for parallel flow) = Thi - TCo (for counter flow) and T2 = Tho - TCo (for parallel flow) = Tho- TCi (for counter flow)  Note that in a special case of Counter Flow Exchanger exists when the heat capacity rates Cc & Ch are equal, then Th i - Tc o = T h o - T c i thereby making  Ti =  To . In this case  LMTD is of the form 0/0 and so undefined. But it is obvious that since T is constant throughout the exchanger, hence  Tm =  Ti =  To (acc. to ref. Fundamental of Engineering Heat & Mass Transfer by R.C. Sachdeva, Pg. 499) 5. Overall heat transfer coefficient, Q , W/m2-OC Ui  Ai FTm

89


Q , W/m2-OC AO FTm Heat Transfer rate, is calculated as Qh = -------- W Qc = -------- W Qh  Qc Q = W 2 (Assume Cph = Cpc = 4.176 J /kg- ºC) LMTD - logarithmic mean temperature difference that can be calculated as per the following formula: Ti  To LMTD = Tm = lnTi To  UO 

Where

Ti

= T h i - T c i (for parallel flow) = T h i - Tc o (for counter flow) and To = T ho -T c o (for parallel flow) = T ho - T c i (for counter flow) Note that in a special case of Counter Flow Exchanger exists when the heat capacity rates C c & Ch are equal, then Th i - Tc o = T h o - T c i thereby making  Ti =  To. In this case, LMTD is of the form 0/0 and so undefined. But it is obvious that since T is constant throughout the exchanger, hence  Tm =  Ti = To Overall heat transfer coefficient can be calculated by using. Q = U ATm  Ui = Q/(Ai Tm) W/m² ºC Uo = Q/(Ao Tm) W/m² ºC NOMENCLATURE: Mc = Cold water flow rate, Kg/s Mh = Hot water flow rate, Kg/s Tci = Cold water inlet temp. Thi = Hot water inlet temp. Tco = Cold water outlet temp. Tho = Hot water outlet temp. Tc = Mean temp. of cold water Th = Mean temp. of hot water , c = Density of Cold fluid, kg/m3 Cpc = Specific heat of cold fluid, J/kg-C pc = Thermal Conductivity of cold fluid, W/mC h = Density of Hot fluid, kg/m3 Cph = Specific heat of Hot fluid, J/kg oC Ph = Thermal Conductivity of Hot fluid, W/mC Qh = Heat lost by hot water, W Qc = Heat gained by cold water, W Q = Average heat transfer, W LMTD = Logarithm mean temp. difference A = Area of Heat Transfer, m2 Do = Outer dia of S.S tube, m L = length of the tube, m 90


U

=

Overall heat transfer coefficient, W / m2 oC

7. Results & discussion 8. Conclusion 9. Precautions & maintenance instructions References: 1. Holman, J.P., “Heat Transfer”, 8th ed.,McGraw Hill, NY, 1976. 2. Kern, D.Q., “Process Heat Transfer”, 1st ed.,McGraw Hill, NY, 1965. 3. McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”, 4th ed.McGraw Hill, NY, 1985. 4. Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering Vol. - 1”, 5th ed.Asian Books ltd., ND, 1996.

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PLATE TYPE HEAT EXCHANGE 1. Aim: To analyze the performance of an existing plate type Heat Exchanger. 2. Objectives: 1. To calculate Overall Heat Transfer Coefficient & effectiveness for plate type Heat Exchanger. 2. To analyze effects of changing the flow rate for hot water & cold water fluids. 3. Theory: A heat exchanger is a device that facilitates the transfer of heat between two fluids at different temperatures. The main mechanisms for heat transfer in heat exchangers are conduction and convection. There are many types of heat exchangers. An example of one is a plate and frame heat exchanger. A plate and frame heat exchanger is a device that utilizes corrugated pressed plates for heat by counter-current flow. Other types of heat exchangers include the miniature, brazed heat exchanger, traditional gasket plate heat exchangers, welded heat exchangers, plate and shell heat exchanger, and the spiral heat exchanger. Plate and frame heat exchangers, are used in many different processes at a broad range of temperatures and with a wide variety of substances. Plate and frame heat exchangers have been used in industry since 1930. During recent years, research into Plate and frame heat exchangers has increased considerably and there is now a state-of-the-art compilation of knowledge on this topic. During liquid-liquid heat exchange, it is important to maintain relatively low temperatures and pressures. The heat exchange process should not exceed temperatures and pressures of 250 degrees Celsius and 25 atmospheres respectively. Plate and frame heat exchangers are not exclusive for liquid-liquid heat exchange, but also have some usage in liquid-gas and liquid-condensing-vapor combination heat exchange. Plate and frame heat exchangers are devices that are made up of many plates stacked together on a frame, and either bolted or welded together. Each plate, consists of tiny little passageways, through which, the fluids are passed. The cold and hot fluids travel through two different passage ways in a counter current flow, which maximizes the heat transfer by creating a temperature gradient that is constantly from the hot fluid to the cold fluid, as illustrated below.

92


FIG 1: Single-pass countercurrent flow. In Fig 1, it can be seen how two fluids move within the plates. The black arrows, for example, represent the hot fluid and the gray arrows represent the cold fluid. As can be seen in the above diagram, the fluids move in opposing directions as they cross the plates.

FIG 2: Parts of a typical plate and frame heat exchanger FIG 2 above shows the different parts of a Plate and frame heat exchangers. The fixed head is where the two fluids enter. One fluid enters through the bottom, and the other fluid enters through the top. The movable follower is the back of Plate and frame heat exchangers. It can be adjusted back and forth for the addition or removal of plates in the plate pack. The guiding

93


and carrying bar provide support for the plate pack and the follower runs along these bars. The clamping bolts secure the stacked structure.

FIG 3(a) Intermitting

FIG 3(b) Chevron

Fig 3 above shows the two main types of plates. The first employs the intermitting corrugations. These corrugations are at right angles with the fluid flows. The fluid goes down the plate and hits the corrugation line, runs to the other end of the plate, and goes down through the slit at the end of the line, after which, the fluid hits another corrugation plate and continues this pattern throughout the plate. The maximum gap size between the corrugations is 3 to 5 mm maximum. With this style of plate, turbulence in the flow is increased by constantly hitting the plate. The size of the corrugation gaps is directly proportional to the amount of turbulent increase in the fluid. The chevron corrugations are slanted downward at an angle of beta. The fluid flows steadily down the plate crisscrossing back forth. The plate in FIG 3(b) has the troughs going in the opposite direction for the other fluids. This maximizes the crossing points of the two fluids, thus, increasing the heat transfer. If the angle is about 80 degrees, the heat transfer increases due to the swirling flow of the two fluids in opposite directions in opposing plates.

FIG 4: Flow patern of fluids in plate type heat exchanger 3.1. Advantages:

94

Chemical Engineering Department, Birla Institute of Technology & Science Pilani, Pilani Campus         

Flexibility in the efficiency: The number of plates can be adjusted for more or less heat transfer Compactness: The numerous amounts of little corrugations increase the surface area for heat transfer in a small volume. Low Cost: The plates are inexpensive and easy to produce Less Corrosion: the plates are typically made of stainless steel Low Maintenance: very easy to clean Temperature Control: can maintain a relatively low temperature difference in fluid because of the small diameters inside the corrugations. High thermal efficiency Reduced Liquid Volume: due to narrow flow channels Due to their efficiency, they can be selected to use less coolant

3.2. Disadvantages:  Temperature and Pressure Limitations: Temperature cannot go above 250 degrees Celsius and pressure cannot go exceed 25 atmospheres.  Leakage: Corrosive materials can cause leakage  Pressure Increase: The small diameter of the corrugations can cause a heavy increase in the pressure, so pumping costs must to taken into account. 3.3. Applications: Some Plate Heat Exchangers are designed specifically for use in the food, dairy and brewing industries. A major feature of Plate and frame heat exchangers used for this process is a sanitary plate designed to achieve optimum distribution of the product over the entire plate surface. These liquid food plants are designed for the concentration of malt, beer, yeast, fruit juices, pulp and other liquid food products to remove water and stabilize enzymes, prior to product storage. The Plate and Frame Heat Exchanger, can also be used in the cooling and heating of fibrous materials, such as, fruit juices and fluids containing pulp, fruit purees, turbid fruit juices, dairy mixes, citrus pulp, and highly viscous liquids. The unique capabilities of the PHE makes it suitable for a wide range of applications that extend beyond refrigeration such as: refrigerant evaporating & condensing, heat pumps, steam heating, engine or hydraulic oil cooling, swimming pool heating, and heat recovery for industrial applications e.g. waste water, dye works, paper manufacture etc. When choosing a heat exchanger, the design engineer has to take into account many factors. Some of these factors include, fluid characteristics, operating pressure, operating temperature and the range of possible flow rates. 4. Experimental Setup: The miniature exchanger supplied for the Lab project consists of a pack of 11 plates, 9 of which have water on both sides that contributes to heat transfer. The plates have sealing gaskets and are held together in a frame between a fixed end plate and moving end plate. Two nuts/bolts passing through the end plates compress the plates and gaskets together. Hot and cold fluids flow between channels on alternate sides of the plates to promote heat transfer. The plate heat exchanger with 11 plates is configured for 5 passes in series. Although the overall flow arrangement may be either countercurrent or Co-current, the flow arrangements on either side of each individual plate alternates between countercurrent and Co-current patterns. 95


Each end plate of the heat exchanger incorporates tapings for the hot and cold fluids to enter/leave the exchanger, and thermocouples in each of the tapings allow the temperatures of the fluids to be measured. The four type K thermocouple temperature sensors are labeled T1 to T4 for identification and each lead is terminated with a miniature thermocouple plug for connection to the appropriate socket on the service unit. Flexible tubing attached to each fluid inlet/outlet is terminated with a ferrule to allow rapid connection to the appropriate quick release fittings on the HT30X service unit. In normal countercurrent operation the flexible connections are hot water inlet adjacent to temperature sensor T1, hot water outlet adjacent to temperature sensor T2, cold-water inlet adjacent to temperature sensor T3, and cold-water outlet adjacent to temperature sensor T4. The pattern of holes in the plates and the shape of the gaskets determine the direction of low through the exchanger. The pates are made of SS-316 and incorporate a locating groove for the gasket. Each plate has a pressed chevron pattern to promote turbulence and provide multiple support points. As a result of the turbulence promoters, which cause flow separation around protuberances, turbulent-like behavior can begin at Reynolds numbers as low as several hundred, thereby enhancing heat transfer performance. Silicone rubber gasket on each plate ensures that the adjacent flow channels are sealed from each other. 5. Experimental Procedure: 1. Fill the thermic fluid tank with about 75 liter of thermic fluid (say water - here). 2. Switch on the immersion type heater (6 kW) provided in the thermic fluid tank and heat the thermic fluid to the desired temperature (about 50-60 C). Intermittently switch ON the pump with bypass line valve fully open and supply valve fully closed to ensure through mixing of thermic fluid in the tank to ensure uniform temperature. 3. After achieving the desired temperature of thermic fluid in the thermic fluid tank, switch ON the pump (0.5 HP) and allow the hot thermic fluid to flow through plates and adjust the flow rate to the desired value using the valve for about five minutes. Recycle the exit of the hot thermic fluid to the thermic fluid tank. 4. Start the cold water supply into plates and adjust the flow rate to the desired value. 5. Monitor the hot thermic fluid inlet temperature and maintain it at the constant value by switching the heater either on/ off with the help of thermostat provided on the control panel of the tank. 6. Observe the inlet and outlet temperature of both cold water and hot thermic fluid streams and note down them after they achieve steady state. 7. Also note down the flow rates of hot thermic fluid and cold water with the help of Rotameters. 8. Repeat the above procedure either by changing the flow rates or by changing the inlet temperature of the hot thermic fluid. 6. Observations: S. No.

Hot Water Flow rate mh (LPM)

Cold Water Inlet Temp T1 (C)

Outlet Temp T2 (C)

96

Flow rate mc (LPM)

Inlet Temp. T3 (C)



7. Calculations: The actual area of heat transfer is computed using the following equation: A = N*a = N * L * W Where, N= number of plates a= projected area of a single plate L= the length component of the projected area of the plate W= the width component of the projected area of the plate Flow rate of hot water in kg/ s mH = mh *  / 60 = __________ kg/ s Heat Transferred by the Hot Water to the Cold Water QH = mH * CpH * (T1 – T2) = ___________ W Flow rate of cold water in kg/ s mC = mc *  / 60 = __________ kg/ s Heat Gained by the Cold Water from the Hot Water QC = mC * CpC * (T4 – T3) = ___________ W True Temperature Difference T 

T11  T3 3  (T 22  T4 4) T11  T3 3 ln T 22  T 444 

= ______________0C Now, average heat transfer Q = (QH + QC )/ 2 = ______________ W  Designed Overall Heat Transfer Co-efficient Q A * Tm = __________ W/m2 0C

Uc =

97


E= EFFECTIVENESS OF THE EXCHANGER.

=MCp)cold/(MCp)min * ∆ Tcold / ∆ Thot = ∆ Tcold / ∆ Thot where, ∆ Tcold= the change in temperature of the cold fluid ∆ Thot= the change in temperature of the hot fluid 7.1. Model Calculations: S. No. Hot Water Flow rate mh

1 2 3 4 5

(LPM) 2 4 6 8 4

Cold Water Inlet Temp T1 (C) 41 42 47 48 48

Outlet Temp T2 (C) 32 36 40 44 43

Flow rate mc (LPM) 3 6 6 4 8

Inlet Temp. T3 (C) 32 34 35 38 42

The actual area of heat transfer is computed using the following equation: A = N*a = N * L * W Where, N= number of plates =8 L= the length component of the projected area of the plate = 550 mm W= the width component of the projected area of the plate = 125 mm A = N*a = N * L * W = 8 * 0.550 * 0.125 = 0.55 m2 Sample calculations for reading no -2:

Flow rate of hot water in kg/ s mH = mh *  / 60 = 4 * 1000/(1000*60) = 0.066 kg/sec Heat Transferred by the Hot Water to the Cold Water QH = mH * CpH * (T1 – T2) = 0.066 * 4.187 * (42 – 36) = 1.6580 W(J/SEC) Flow rate of cold water in kg/ s

98

Outlet Temp. T4 (C) 36 37 42 46 44


mC = mc *  / 60 = 6 * 1000/(1000*60) = 0.1 kg/sec Heat Gained by the Cold Water from the Hot Water QC = mC * CpC * (T4 – T3) = 0.1 * 4.187 * (37 – 34) = 1.2561 W(J/SEC) True Temperature Difference Tlm 

T1  T 4  (T 2  T 3) T1  T 4 ln T 2  T 3

=3/(ln 2.5) = 3.27 0C Now, average heat transfer Q = (QH + QC )/ 2 = 2.9141 W  Designed Overall Heat Transfer Co-efficient Q A * Tm = 2.9141/(0.55 * 3.27) = 1.6202 W/m2 0C Effectiveness of heat exchanger: =MCp)cold/(MCp)min * ∆ Tcold / ∆ Thot = ∆ Tcold / ∆ Thot = (37–34 )/ (42 – 36) = 3/6 * 100 =50 % where, ∆ Tcold= the change in temperature of the cold fluid ∆ Thot= the change in temperature of the hot fluid

8. Results & Discussion: Hence, the performance of the plate type Heat Exchanger has been analysed and the Overall Heat Transfer Coefficient & effectiveness for the plate type Heat Exchanger is calculated. 1. Overall Heat Transfer Coefficient of the plate type Heat Exchanger = 2. Effectiveness of the plate type Heat Exchanger =

99


9. Conclusions: 10. Precautions: 1. Make sure during the test period Hot Thermic Tank should not be emptied totally and the heater must not be exposed to air if the Heater is ON, otherwise it will be damaged. 2. Take all observations within 5 to 10 minitues before cold water temperaturrises significantly. 11. References: 1. http://en.wikipedia.org/wiki/Plate_heat_exchanger 2. Lawry, F.J., “Plate type heat exchangers”, chem. Eng., June 29, 1959, 89-94. 3. www.tv-me.com/download.php?a=apv.pdf


FINNED TUBE HEAT EXCHANGER 1. Aim: To analyze the performance of an existing finned tube Heat Exchanger. 2. Objective: To calculate overall heat transfer coefficient, logarithmic mean temperature difference and effectiveness of finned tube heat exchanger. 3.Theory The heat which is conducted through a body must frequently be removed by some convection process. For example, the heat lost by conduction through a furnace wall must be dissipated to the surroundings through convection. In heat exchanger application a finned tube arrangement might be used to remove heat from a hot fluid. Consider the one-dimensional fin exposed to a surrounding fluid at a temperature T. The temperature of the base of the fin To. Equivalent diameter of annular, de = 4 x Net free flow area / wetted or heated perimeter Total Wetted Perimeter, Pw =  (Ds + do) + 2YNF Heat Transfer Perimeter, PHT = do + 2YNF Net free area in the annular of a longitudinally finned exchanger is ANF = [(/4) (Ds2 + do2)] - (XYNF)

100


Fig1: Crossection of pipe and finned tube Equivalent Diameter Based on wetted perimeter (hydraulic diameter) (de)w = 4ANF / Pw Based on heat transfer perimeter (de)HT = 4 ANF / PHT Use (de)w for the calculation of Re-No, Gz-No, D/L for Annular fluid and (de)HT for the calculation of HT coefficient on annular side and also in evaluating the Gr-No. Total outside heat transfer surface: (At) Ao = 2 ( do L - NF L X) Af = 2 NF L (2Y + X) At = Ao + Af = 2L (do + 2NFY) For bare tube Af = At = Ao Heat Transfer co-relations For turbulent flow (Re>10,000, L/D>60, 0-671) For laminar flow : (Re < 2100) Nu = 1.86 (Re Pr D/L )0.33  For Gz > 100 0.33 Nu = 1.86 (Re Pr D/L )  + 0.87 [1 + 0.015 Gr0.33] 0.14 For liquids  = (b / w ) For gases  = (Tb / Tw)n For temperature ratio of 0.5 - 2, n=0, =1 For transition flow: Nu = 0.116 (Re0.67 - 125) Pr0..33 [1+ (D/L)0.67]  Fin Performance Fin Effectiveness (f) is defined as the actual heat transferred by the fin divided by the heat that ideally would be transferred if the entire fin were at base temperature. tanh(mY) f = mY m = (hfPf / hfaf)1/2 = (2hf / kf X)0.5 for rectangular fin 101


af

= 2(L + X)  2L

Fin effectiveness f is defined as the ratio of heat transferred through the fin to heat transferred through the same bare surface having no fins. f = f (surface area of fin / cross sectional area of fin) Efficiency of the total heat transfer surface is expressed as weighted for efficiency: (f)wtd = f (Af /At) + Ao /At Over all heat transfer coefficient - U, Q = U A (LMTD) 4. Experimental Setup: Fins increase heat transfer surface per unit length and reduce the size of heat exchanger required for a given service. Fins are usually welded to the tube but may also be integrally formed. Heat exchangers with fins are made up of copper. Consider a horizontal finned tube (Longitudinal) double pipe heat exchanger (water to air). Water flow rate can be measured by rotameter and air flow rate can be measured by using orifice provided.

5. Procedure: a) Clean the apparatus and make water bath free from dust. b) Close all the drain valves provided. c) Fill water bath ¾ with clean water and ensure that no foreign particles are there. d) Ensure that all ON/OFF switches given on the panel are at OFF position. e) Select the parallel or counter flow of air and operate the ball valves given on the air stream line. f) Now switch on the main power supply (220 V AC, 50 Hz). g) Switch on heater by operating switch given on the panel. h) Set temperature of the water bath with the help of digital temperature controller. i) Open flow control valve for hot water supply. j) Switch on magnetic pump for hot water supply. k) Adjust hot water flow rate with the help of flow control valve provided before rotameter. l) Record the hot water flow rate with the help of rotameter and cold air flow rate with the help of orifice-meter provided. m) Record the temperatures of hot and cold fluid inlet & outlet when steady state is achieved. n) Take readings at different flow rates of hot water and cold air. o) For second run again fix the valve position for parallel or counter flow of air stream. p) Repeat the steps 8 to 14 of experiment for different flow rates of fluids. q) Repeat the steps 5 to 16 for another flow of air stream into the heat exchanger. 6. Observations: Flow arrangement System given Tube side fluid Annular fluid Inner Tube Longitudinal Fins Outer tube Water Flow Measurement

: : : : : : : :

Counter flow / Parallel flow. copper pipe with fins Hot water. Cold air stream. Dia 9 mm, Length 1 m. Material Copper. Width 12mm, Length 1m, 4 Nos. Material: Copper. Dia. 42 mm. Length 1 m. Material G.I. Rotameter. 102


Air stream Flow Measurement: Hot Water Tank : Hot Water Circulation

:

Heaters

:

Orifice-meter made of Stainless steel, Double Wall, insulated with ceramic wool Magnetic Pump made of Polypropylene to circulate Hot Water. Maximum working temperature is 85°C. 2 kW Nichrome wire heater (1 Nos.)

Digital Temperature Controller : 0–199.9°C. (For Hot Water Tank) Digital Temperature Indicator: 0-199.9oC, with multi-channel switch Temperature sensors : RTD PT-100 type 5 Nos. With Standard make On/Off switch, Mains Indicator etc. A good quality painted rigid MS Structure is provided to support all the parts

Data: Copper tube outside dia ,do Copper tube inner dia, di Fin length or tube length, L Fin height, Y Fin thickness, X No. of fins, Nf I D of outer shell, Ds I D of orificemeter

= = = = = = = =

Observation Table: Water flow rate Ww Sr no. Thi oC (kg/h)

Hot water inlet, (T1) = Hot water outlet, (T2) = Cold air stream inlet, (T3) = Cold air stream outlet (parallel), (T4) = Cold air stream outlet (counter), (T5) = Cold fluid i.e. air mean temp, (Tc)

=

Hot fluid i.e water mean temp, (Th) =

12mm 9 mm 1000 mm 12 mm 2 mm 04 42 mm 6.25mm

ThooC

TcioC

Thi Tho Tci TCo TCo TCO  TCi

O

2  Th i

O

ThO

2

103

C C

TcooC


From data hand book: Thermal conductivity of fin material (Copper), Kf Properties of water at temp T Density  Specific heat Cp Thermal Conductivity k

= 386.0 W/m-oC = ------- oC = 1000 kg/m3 = -------J/kgoC = ------- W/m -oC

7. Calculations: Total heat transfer area At

= At = 2L (do + 2NFY) = ---------- m2 Thi Qc = Wc Cpc ( Tco – Tci) Qh = Wh Cph (Tho - Thi) Tco  Qavg = ( Qh + Qc) /2 Logarithmic Mean Temp Difference (LMTD) T2  T1 LMTD = T ln( 2 ) 0 T1 Over all Heat Transfer Coefficient Q U = AxLMTD Use NTU method to determine the effectiveness of the heat exchanger. For shell side fluid (air) Cs = Wc Cpc For tube side fluid (hot water) CT = Wh Cph Identify the minimum either Cmin = Cs & Cmax = CT Or Cmin = CT & Cmax = Cs Cmin = -------Cmax = ----------C  CR = min C max UA NTU = C min % For counter flow: Effectiveness, 

=

Tci

L

CR CR

1  exp[NTU(1  C R )] 1  C R exp[NTU(1  C R )] NTU = UA/Cmin

For parallel flow: 

Tho

=

1  exp[NTU(1  C R )] 1  CR

For each run calculate NTU &  (Try to keep CR fixed) Plot %  vs NTU for counter flow and for co-counter flow. Nomenclature: 104


Ds do de NF X Y L ANF Qf Cp G h k W U Q

= = = = = = = = = = = = = = = =

inner diameter of shell (outer pipe), m the outer diameter of inner pipe (bare tube), m Equivalent dia. of annular No. of fins per tube Fin thickness, m Fin height, m Nominal length of heat exchanger and length of longitudinal fin (in meters) Net free area, m2 Cross Sectional area of fin, m2 Specific heat, KJ/kg-OC Mean mass velocity, kg/m2-s Heat transfer coefficient, w/m2-OC Thermal conductivity, w/mOC Water flow rate Kg / s. Over all heat transfer co-efficient; w/m2k Heat load, w

8. Result & Discussions: Hence, the performance of the finned tube Heat Exchanger has been analysed and the Overall Heat Transfer Coefficient, logarithmic mean temperature difference, & effectiveness for the finned tube Heat Exchanger is calculated. 1. Logarithmic mean temperature difference of the finned heat exchanger is= 2. Overall Heat Transfer Coefficient of the finned tube Heat Exchanger = 3. Effectiveness of the finned tube Heat Exchanger = 9. Conclusion: 10. Precautions: 1. Never switch on main power supply before ensuring that all the on/off switches given on the panel are at off position. 2. Never Switch on Heaters before filling water bath ¾ with clean water. It may damage heaters. 3. Never run the Pump at low Voltage i.e. less than 180 Volts. 4. Never fully close the Delivery and By-pass line Valves simultaneously. 5. Always keep apparatus free from dust. 6. To prevent clogging of moving parts, run Pump at least once in a fortnight. 7. Frequently Grease/Oil the rotating parts, once in three months. Always use clean water. References: 1. Sinnott, R.K., Coulson & Richardson’s Chemical Engineering, Vol. 6, ButterworthHienemann, Oxford, 1996. 2. http://www.enggcyclopedia.com/2012/03/finned-tube-type-heat-exchangers/

105


EXPERIMENT NO. 11

STEFAN-BOLTZMANN APPARATUS 1. AIM To find out the Stefan-Boltzmann constant. 2. Objective To study radiation heat transfer of black body. 3. Theory All substances at all temperature (above absolute Kelvin) emit thermal radiation. Thermal radiation is an electromagnetic wave and does not require any material medium for propagation. All bodies can emit radiation and have also the capacity to absorb all of a part of the radiation coming from the surrounding towards it. The most commonly used law of thermal radiation is the Stefan Boltzmann law which states that thermal radiation heat flux or emissive power of a black surface is proportional to the fourth power of absolute temperature of the surface and is given by,

The constant of proportionally is called the Stefan Boltzmann constant and has the value of 5.67 x 10-8 W/m² K4. The Stefan Boltzmann law can be derived by integrating the Planck’s law over the entire spectrum of wavelength from 0 to infinity. The objective of this experimental set up is to measure the value of this constant fairly closely, by an easy experimental arrangement. 4. Experimental Description The apparatus is centered on a flanged copper hemisphere B fixed on a flat non-conducting plate A. The outer surface of B is enclosed in a metal water jacket used to heat B to some suitable constant temperature. One RTD PT-100 type temperature sensor is attached to the inner wall of hemisphere B to measure its temperature and to be read by a temperature indicator. The disc D, which is mounted in an insulating Bakelite sleeve is fitted in a hole drilled in the center of the base plate A. An RTD PT-100 temperature sensor is used to measure the temperature of D i.e. TD. The Temperature Sensor is mounted on the disc to study the rise of its temperature. 5. Utilities Required Electricity Supply: 1 Phase, 220 V AC, 2 kW. Table for set-up support 6. Experimental Procedure a) Heat the water in the tank by the immersion heater up to a temperature of about below 90 C. b) The disc D is removed before pouring the hot water in the jacket. c) The hot water is poured in the water jacket. 106


d) The hemispherical enclosure B and A will come to some Uniform temperature in a short time after filling the hot water in the jacket. The thermal inertia of hot water is quite adequate to prevent significant cooling in the time required to conduct the experiment. e) The disc D is now inserted in A at a time when its temperature is TD. f) Start noting the temperature change for every five second for a minute. 7. Specifications Hemispherical enclosure diameter Base plate, Bakelite diameter No. of Temperature Sensor mounted on B No. of Temperature Sensor mounted on D Temperature indicator digital (RTD PT-100) Immersion water heater of suitable capacity and tank for hot water

= = = = = =

200 mm 250 mm. 1 1 0-199.90C 1.5 KW

The surface of B and A forming the enclosure are black to make their absorptivity to be approximately unity. The copper surface of the disc D is also blacked. 8. Observations & Calculations 8.1 Data Mass of water in the hemispherical tank Specific heat of water Area of the disc, AD 8.2 Observation Table Time t (second)

: 5.1x10-3 kg : 4180 J/kg K : 3.14x10-3 m2

Temperature TD in ºC

5 10 15 20 25 30 8.3 Calculation The radiation energy falling on D from the enclosure is given by

The emissivity of the disc D is taken as unity, (assuming black disc). The radiant energy disc D is emitting into enclosure will be

Net heat input to disc D per unit time is given by subtracting both the equations,

If the disc D has a mass m and specific heat s then a short time after D is inserted in A, 107


In this equation

denotes the rate of rise of temperature of the disc D at the instant

when its temperature is TD and will vary with time. It is clearly measured at time t = 0 before heat conducted from A to D begins to have any significant effect. This is obtained from plot of temperature rise of D with respect to time and obtaining its slope at t = 0 when temperature is TD. The temperature sensor mounted on disc is to be used for this purpose. Note that the disc D with its sleeve is placed quickly in position and start recording the temperature at fixed time intervals. The whole process must be completed in 30 seconds of time. Longer disc D is left in position the greater is the probability of errors due to heat convection from A to D. Temperature of water Temperature of hemispherical enclosure at A, T Temperature of disc at the instant when it is inserted at D, TD

: ____0C : ____ 0C : ____ 0C

Temperature time response of the disc. Note down the temperature TD at the time interval of 5 second. Plot the graph of TD and time (in seconds). Obtain from the graph, Value of  can be obtained by using,

Nomenclature: AD T TD m s

= = = = =

Area of disc D. Temperature of enclosure Temperature of disc mass of disc specific heat of the disc material.

9. Precautions & Maintenance Instructions a) Always use clean water and heater should be completely dipped in the water before switch ON the heater. b) Always take the reading for the first minute after fixing the disc. c) Use the stabilize AC single phase supply. d) Never switch on mains power supply before ensuring that all the ON/OFF switches given on the panel are at OFF position. e) Voltage to heater should be constant. f) Keep all the assembly undisturbed. g) Never run the apparatus if power supply is less than 180 V and above than 240 V. h) Operate selector switch of temperature indicator gently. i) Always keep the apparatus free from dust. 108


j) Don’t switch ON the heater before filling the water into the bath. 10. Troubleshooting 1. If electric panel is not showing the input on the mains light. Check the fuse and also check the main supply. 2. If digital temperature indicator displays “1” on the screen check the computer socket if loose tight it. 3. If temperature of any sensor is not displayed in digital temperature indicator check the connection and rectify that. References 5. Holman, J.P., “Heat Transfer”, 8th ed., pp. 396, McGraw Hill, NY, 1976. 6. Kern, D.Q., “Process Heat Transfer”, 1st ed., pp. 69, 74, McGraw Hill, NY, 1965. 7. Perry, R.H., Green, D.(editors), “Perry’s Chemical Engineers’ Handbook”, 6th ed., pp. 10/53, McGraw Hill, NY, 1985. 8. Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering Vol. - 1”, 5th ed., pp. 390, Asian Books ltd., ND, 1996.

109


EXPERIMENT NO. 12

CROSS-CIRCULATION DRYING 1. Aim Study of drying characteristics of porous and non-porous solids under forced draft condition with cross flow of air. 2. Objectives 1. To determine the critical moisture content. 2. To calculate the total drying time. 3. To calculate the humidity of inlet and outlet air with respect to drying time. 3. Theory Drying of solid generally means the removal of relatively small amounts of a liquid from the solid material to reduce the liquid content to an acceptably low value. There are various modes of drying a material in various types of dryers. In the present experiment, we study the cross-circulation drying in an adiabatic or direct dryer. In this case, hot gas (air) is blown over a bed of wet solids (brick particles wetted with water) under constant drying conditions by maintaining the temperature, humidity and the velocity of the air across the drying surface constant. A drying curve is then obtained by plotting drying rate against moisture content of the solid. In drying, it is necessary to remove free moisture from the surface and also moisture from the interior of the material. Thus there is interplay of the surface evaporation, inter-particle and intra-particle diffusion processes during the drying of a material. This gives rise to several distinct periods in the drying curve as the moisture content of the solid is reduced from the high initial value to its final value, as explained below in terms of different periods: I0 (Initial Period): An initial period during which the drying rate may increase or decrease rapidly from an initial value and the drying conditions are adjusting themselves to the steady state condition, which is the next period. This period is of relatively short duration and in some experiments may be unobservable. Therefore, this period is neglected in the mathematical analysis of the process. I (Constant Rate Period): An early stage of drying during which the drying rate remains at a constant value, that is, is independent of the moisture content. During the constant rate period, it is assumed that drying takes place from a saturated surface of the material by diffusion of the water vapor through a stationary air film into the air stream. This period may be absent if the initial moisture content of the solid is less than a certain minimum. II (Falling Rate Period 1): During this period, there is insufficient water on the surface to maintain a continuous film of water. The entire surface is no longer wetted, and wetted area continually decreases in this first falling-rate period until the surface is completely dry. During this period, the drying rate decreases more or less linearly with continued decrease of water content. III (Falling Rate Period 2): This second falling-rate period begins when it may be assumed that the surface is dry. In this period evaporation will be taking place from within the solid and the vapor reaches the surface by molecular diffusion through the material. The drying rate in this zone decreases further, but generally in a non-linear fashion with the moisture content. The moisture content at which the drying rate falls for the first time is the first critical moisture content, and the moisture content at which the drying rate falls for the second time is the second critical moisture content. In case of non-porous solids, due to the absence of intra-particle moisture and hence intra-particle diffusion, the constant-rate period is followed 110


by only one falling rate-period so that there is only one critical moisture content (see McCabe et al., 2001, for detail). Drying continues until equilibrium moisture content in the solid is attained. Nature of the drying curve depends on the nature of the solid (porous or non-porous), and the temperature, humidity and flow rate of the drying medium. The total time of drying, tT, is determined from the following equation (the derivation is given in McCabe et al., 2001)

tT 

ms  Xc  ( X 1  X c )  X c ln  ARc  X2 

(1)

Where, ms A Rc X Xc X1 X2

= mass of bone-dry solid = area of drying = rate at first critical point = free-moisture content, mass of water per unit mass of dry solid = free-moisture content at first critical point = initial free-moisture content = final free-moisture content

Free moisture content, X, is given by

X  XT  X 

(2)

Where, XT = total free-moisture content X* = equilibrium free-moisture content The equilibrium moisture content of a non-porous insoluble material is practically zero so that the whole moisture content is free moisture. 4. Experimental Setup 4.1. Requirements Dryer assembly, blower, beaker, water, brick particles (porous), glass beads (non-porous), heater, physical balance, digital anemometer, thermometers (4 in number). 4.2. Schematic Diagram of Experimental Setup

111


Anemometer Measurement space

Dryer

Wet bulb temperature thermometer

Dry bulb temperature thermometer Inlet of air

Outlet of air

Blower

Valve

Heater

5. Experimental Procedure 1. Weigh the empty pan of the dryer. 2. Take out the pan from the dryer, and after filling it with about 250 gm of brick particles place it back in the drying chamber. Note the weight of the pan with its contents. 3. Take out the water-cups for wet-bulb temperature (WBT) thermometer from the dryer and keep them aside. Inject water in the water-cups using a 5 ml-pipette after every 30 minute. 4. Switch on the heater and the blower. Adjust the valve on the inlet line to give a constant air velocity at about 4 m/s. Measure the air velocity with the digital anemometer. 5. Keep blowing the air for about 30 minutes for the system to reach a steady state, which is indicated by an insignificant change in the weight of the filled pan. Note the dry-bulb temperature and the weight of the pan at this state. 6. Take out the pan from the dryer and soak the brick particles in water for about 10 minutes in a beaker. Then spread the wet brick particles over the pan uniformly. Fill the cups of the WBT thermometers with water and wrap wet cotton around the bulb of these thermometers. 7. Put one cup in the cup-holder and one WBT thermometer each at the inlet and outlet ports of the dryer. Ensure that the bulb of the thermometer remains dipped in the cup. 8. Place the pan back in the drying chamber. 9. Note: Step 8 should follow step 7 with very little time gap. 10. Record the dry-bulb and wet-bulb temperatures at the inlet and the outlet of the dryer, and weight of the pan. These are the reading at time t = 0. 11. Record the dry-bulb and wet-bulb temperatures at the inlet and the outlet of the dryer and the weight of the pan, initially at intervals of about 2 minutes, and later at intervals of about 5 minutes as the rate of drying (indicated by the rate of change of the weight of the pan) decreases. 12. Continue the run until there is no significant change in the weight of the pan (and so the temperatures; also the pan plus solid would weigh almost the same as the one noted in step 5 before wetting the particles). 13. Repeat the experiment with the glass beads at the two air velocities of about 2 and 4 m/s. 6. Observations 1. Surface area of pan 2. Air velocity 3. Weight of pan, w1

= 0.0245 m2 = = 112


4. Weight of (pan + dry brick particles), w2 (at step 5)

=

Material (glass beads/brick particles): S No

Time

Weight of pan and wet solid, w3

Inlet temperature Dry-bulb Wet-bulb

Outlet Temperature Dry-bulb Wet-bulb

7. Model Calculations 1. Calculate the total moisture content (XT) as (w3 – w2)/(w2 - w1), and plot it with time. 2. Determine the rate of drying by finding the slope of the XT vs. t curve at different times. (Truly speaking, in case of porous brick particles, one should find the equilibrium moisture content and then determine the rate of drying using the free moisture content; since it is difficult to find the equilibrium moisture content [why?], we are working with the total moisture content). 3. Make the drying-curve by plotting rate of drying vs. moisture content. 4. Find the critical moisture content(s) and the corresponding drying rate(s) from the drying curve. 5. Check the equilibrium moisture content of glass beads. 6. Using Eq. 1 calculate the total time of drying. 7. Compute the humidity of the air at the inlet and the outlet of the dryer at 5 minutes interval and plot it against time at inlet and outlet. 8. Results & Discussion (Tables, Graphs, Comparison with those reported in literature) 9. Conclusions 10. Precautions Reference 1. Warren, L McCabe, Smith, J C , and Harriott, P, Unit Operations of Chemical Engineering, 6th edition, McGraw Hill, New Delhi, India, 2000. 2. Richardson, J F, Harker, J H and Backhurst, J R, Coulson and Richardson's Chemical Engineering, 5th edition, Vol-2, Asian Books Private Limited, New Delhi, India, 2002.

113


EXPERIMENT NO. 13(a)

OPEN PAN EVAPORATOR 1. Aim To determine the overall heat transfer coefficient and the economy of Open Pan Evaporator when evaporating saturated sodium chloride brine. 2. Theory Evaporation is a process for concentrating a solution by vaporizing part or all of the solventusually water. The objective of the evaporation is to concentrate a solution consisting of a nonvolatile solute and a volatile solvent. Essentially, evaporation may be considered to be a special case of heat transfer in which heat is transferred from condensing vapors, from hot gases, or directly by radiation to a liquid at constant temperature, usually its boiling point at the operating pressure. By far the most common evaporators are those in which heat is transferred through a metal wall from condensing steam to a boiling liquid. Evaporation differs from the drying in that the residue is a liquid sometimes a highly viscous one rather than a solid. It differs from the distillation, in that the vapor usually is a mixture, no attempt is made in the evaporation step to separate the vapor into fractions. It differs from the crystallization in that emphasis placed on concentrating a solution, e.g. in the evaporation of brine to produce common salt, the line between evaporation and crystallization is far from the sharp. Evaporation sometimes produces slurry of crystals in saturated mother liquor. Normally, in evaporation the thick liquor is the valuable product and the vapor is condensed and discarded. In one specific situation, however, the reverse is true. Mineral – bearing water often is evaporated to give a solid – free product for boiler feed, for special process requirements, or for human consumption. This technique is often called water distillation, but technically it is evaporation. The vapor coming out of evaporator can be used as heating media for another evaporator which will be operating at pressure lower than the pressure in the evaporator from which vapors are issuing so as to provide sufficient temperature gradient for heat transfer in that evaporator. When single evaporator is put into service and vapors leaving the evaporator are condensed and discarded, the method is known as single effect evaporation. The economy of single effect evaporator is always less than one. Generally for evaporation of one kg of water from a solution, 1 to 1.3 kg of steam is required. 3. Procedure a) Prepare approximately 11.6 kg of saturated sodium chloride brine (20 % w/w). b) Charge this material to the open pan evaporator with bottom drain valve in closed position. c) Open the steam supply valve and also the steam to flow to the steam jacket, collect the condensate in a separate vessel. Also note down the pressure indicated by the pressure gauge. d) After an hour stop the steam supply and measure the volume of condensate collected as well as temperature of the boiling liquor. e) Open the bottom drain valve of an evaporator and weigh the concentrate of sodium chloride. f) Wash the open pan evaporator to remove any traces of sodium chloride. 4.

Observation and Calculation

4.1 Observations: 114


        

Initial weight of saturated sodium chloride brine (20% w/w) taken = 11.6 kg Final weight of saturated sodium chloride brine = ------kg Time of operation () = 1hr Pressure of the steam supplied to the steam jacket (P) = 0.5 kg/ cm2 Temperature of boiling liquid (T2) = -------- C Temp. of steam inlet (T1) = --------C Temp. of condensate (T3) = --------C Amount of condensate collected (w) = ------kg Latent heat of vaporization of steam at the given pressure from steam table () =-------- kcal/ kg

4.2 Calculations 1) Amount of water evaporated (m) = (Initial wt. of satd. NaCl brine (20% w/w) taken – Final weight of satd. NaCl brine) = -------- kg 2) Total energy given up by the steam to the brine solution (Q) =*w = -------kcal 3) Overall Heat Transfer Coefficient (U): Q/  = U * A * (T1 – T2) Where, Q/  = heat transferred, kcal/ hr A= area of heat transfer surface, m2 t2 = temperature of condensing steam, C t1 = temperature of boiling liquor, C U = ----------- Kcal / Hr m2 OC 4) Economy, kg of water evaporated per kg steam = (m/ w)* 100 5.

Result 1) Overall Heat Transfer Coefficient (U) =--------- Kcal / Hr m2 OC 2) Economy, kg of water evaporated per kg steam =---------

6.

Conclusion

7.

Precaution    

Keep partially open the steam trap valve during the performance of an experiment for keeping the flow of steam continuous. Do not increase the pressure above 1.5 kg/cm2 in steam generator. Circulate the water in pipes after completion of an experiment to avoid the chocking of solids inside the pipeline. Drain the water from steam generator and condensate collector after completion of an experiment.

115


 

Operate gently the selector switch of temperature indicator to read various temperatures. Increase the temperature gradually of the heater during initial set-up experimentation.

116


EXPERIMENT NO. 13(b) VAPOR IN AIR DIFFUSION 1. Aim To determine the diffusion coefficient of acetone (liquid) in air by using Arnold's cell. 2. Theory Diffusion describes the spread of particles through random motion from regions of higher concentration to regions of lower concentration. The time dependence of the statistical distribution in space is given by the diffusion equation. The concept of diffusion is tied to that of mass transfer driven by a concentration gradient, but diffusion can still occur when there is no concentration gradient (but there will be no net flux). If two gases are inter diffusing with continual supply of fresh gas and removal of the products of diffusion, this diffusion reaches an equilibrium state with constant concentration gradients. This is known as steady state diffusion. If also there is no total flow in either direction the rates of diffusion of A and B, N A and NB are equal but have opposite sign. The diffusivity or diffusion coefficient, D is a property of the system dependent upon temperature, pressure and nature of the components. The dimension is length2/time. This is the case of pseudo-steady-state diffusion in which one of the boundaries shift with time with the effect that the length of the diffusion path changes, though only by a small amount over a long period of exposure. When this condition exists:

N

A, Z



D

AB

Z



P 1  RT P

PA1  PA2 

B ,lm

---------- (1) The molar flux at any instant in the gas phase Where Z=z1-z2 = the length of the diffusion path in time t.

 P D AB   RT

  1 PA1  PA2   t  A,L 1 zt2  z12  M 2  PB ,lm





3. Experimental Setup 3.1 Apparatus: Arnold's cell, Thermometer, Scale 3.2 Chemicals: Acetone. 4. Experimental Procedure a) Acetone is filled in capillary tube and air bubble, if present, is removed from the tube carefully. b) Note down the initial height (H0) of the acetone level in the tube. c) The tube is placed in a water bath to maintain the constant temperature. Note down the temperature of the water by thermometer. d) Switch on blower, which blown air across the opening of capillary tube continuously to remove the vapors evaporated that rises from the surface of the liquid. 117


e) At regular interval of time note down the drop in the level of acetone. f) Do necessary calculations and find out the diffusivity (D) from the equation. g) Also calculate the diffusivity from the GILLILAND’S equation (as shown below) at the same temperature using standard data and compare the result with the experimental one. h) Repeat the same procedure for two-three different temperatures by increasing and maintaining constant temperature of the water bath and calculate the diffusivity by both the methods. State your conclusion precisely at the end. 9.

Observation Table

S.No

Temp c

Initial ht. of liquid (mm)

Final ht. of liquid(mm)

Time (sec.)

1 2 3

6. Calculation a) Vapor pressure of acetone at given temp. (Calculate by Antoine Equation) b) PA1 = ______ mm hg (Convert to N/m2, 1mmhg=133.2 N/m2 ) c) PB1 = P-PA1 d) PA2 = 0 ( Pure B(air) is flowing) e) PB2 = P-PA2

PB , lm 

PB 2  PB 1

P  ln  B 2   PB1  f) Z1 = ____________ m , Zt =_______ m g) Diffusivity is given by,

  P  1 PA1  PA2   t  A,L 1 zt2  z12 D AB   M 2  RT  PB ,lm 2 = __________ m /sec



DAB

7. Diffusivity by Gilliland’s method

D

4.3 *10 T  1 / M 4

3/ 2



P * VA

1/ 3

 1/ M B 

1/ 2

A

 VB



1/ 3 2

VA = Molar volume of acetone = 0.0074 m3/kmol VB = Molar volume of air = 0.01386 m3/kmol MA = Molecular weight of acetone = 58 kg/kmol 118




MB = Molecular weight of air = 18 kg/kmol T = Absolute temperature = __________ 0 K So, D = __________ m 2/sec from Gilliland’s eqn.

8. Data Given a) P= 1 Std.atm. = 760 mm hg = 101325 N/m2 b) R = 8314 N.m/Kmol. K c) Room Temp = ______ C d) Density of acetone = 1540 Kg/m3 e) Mol. of acetone = 58 Kg/Kmol. Antoine Constant For acetone f) g) h)

A = 16.6513 B = 2940.46 C = - 35.93

9. Result Temperature (K)

DAB

DTheo.

10. Conclusion Observe and state the effect of Temperature on Diffusion.

119


EXPERIMENT NO. 14

DIFFERENTIAL DISTILLATION 1. Aim To verify the Raleigh’s equation for a differential distillation in a binary system and to estimate relative volatility of the binary system 2. Introduction Differential distillation is also known as simple distillation. Usually it is operated in a batch mode. It is used to separate a liquid mixture whose components have fairly large difference in their boiling points. In this mode of distillation a charge of liquid is placed in the still and it is distilled off until the liquid charged is all gone or until the operation is no longer doing any beneficial separating. In batch distillation operations, the liquid in the original charge is constantly being depleted of its more volatile components. This mode of distillation may also be used for analytical evaluation of boiling ranges and vaporization characteristics if mixtures in laboratories. 3. Theory Differential distillation refers to a batch distillation in which only one vaporization sate is involved. Lard Rayleigh developed a mathematical equation which represents differential distillation. The following are the major assumptions in deriving Rayleigh’s equation 1. Still liquid composition is uniform. 2. The distillation process is carried out slowly so that the vapour leaving the still is in equilibrium with the liquid in still. 3. There is no entrainment of the liquid during the distillation 4. There is no condensation of the vapour evolved before it reaches the condenser. 3.1 Rayleigh’s equation derivation Let L be the number of moles of the mixture at a given instant. Let x be the mole fraction of more volatile component in liquid mixture. Let y  be the mole fraction of more volatile component in vapour that is in equilibrium with liquid that is in still. If dL is the differential amount of liquid vaporized then by applying the material balance with respect to the more volatile component we get Input = Lx (1.1)  Output = y dL (1.2) Accumulation = L  dLx  dx  (1.3) Therefore Lx  L  dLx  dx   y  dL Neglecting dLdx , we get y  dL  xdL  Ldx dL dx Therefore,   L y  x  The boundary conditions are: Initially, L = F and x  x F ; F : moles of feed mixture Finally, L = W and x  x W ; W : moles of with drawn or bottom mixture W

dL  L F



xW

 y

xF

dx x





(1.4) (1.5) (1.6)

(1.7)

120

Chemical Engineering Department, Birla Institute of Technology & Science Pilani, Pilani Campus x

T dx  F  ln       W  xW y  x





(1.8)

Using the vapour liquid equilibrium data for the binary mixture system, the plot is made of 1 vs x .Area under the curve between the ordinate at x F and x W gives the values of  y  x  the integral on the R.H.S of Eq.(1.8). From this we obtain the value of W, the amount of with drawn (or) bottoms, to be left to attain the given composition x W of the liquid. Alternatively, if we want to calculate the composition of with drawn, when W is known, then L.H.S of Eq.(1.8) can be found and value of x W adjusted to obtain the area under the curve equal to the

 W .

value of ln F

3.2 Rayleigh’s equation for an ideal system For an ideal system, average relative volatility,  ave , is constant. Therefore for an ideal system we can write the following equilibrium relation α ave x y  (1.9) 1  α ave  1x Substituting Eq.(1.9) in Eq.(1.8) and upon manipulation we get the following relation  Fx F   F1  x F     α aveln   ln  (1.10)  Wx W   W1  x W   Eqs.(1.8) and (1.10) are known as Rayleigh’s equations for differential distillation. 4. Experimental setup The experimental setup to study differential distillation is shown in Figure P1.1. The distillation still may be fabricated from steel with substantial thickness recommended. The vapors of the chemical distilled are collected as distillate is shown in the Figure P1.1.

121


Figure P1.1. Differential distillation experimental setup 5. Experimental procedure The following procedure may be followed while conducting the experiment 1. The still is charged with known composition binary system 2. Ensure water flow in the condenser 3. Heating mantle is switched on and heating is carried out sufficiently and slowly. When the mixture starts boiling, collect the distillate in a flask. 4. This procedure is continued for sufficient length of time (say 2hours) until approximately 2/3 of the feed distilled. 5. Meantime prepare a calibration chart using known concentrations for the binary system. For calibration purpose use specific gravity technique (or) refractive index (RI) method. 6. Measure the specific gravity of the distillate (or) RI of the distillate. Similarly cool the bottoms to room temperature and measure specific gravity (or) RI of bottoms 6. System used and calculations The Following binary Chemical may be taken for simple distillation Ethanol (A)-water (B) Data required for the calculations from the literature 1. Molecular weights of compounds A and B 2. Densities of components A and B (or) refractive indices of components A and B 3. Vapour- liquid equilibrium (VLE) of components A and B at atmospheric pressure 4. Plot calibration chart 7. Calculations 1. Calculate the amounts and compositions of distillate for the sample i.e D and x D . While doing make use of calibration plot for composition calculations.

122


2. Using total material balance and more volatile component balances to calculate amount of residue (bottom),W and composition of the residue, x W 1 3. Plot  vs. and measure the area between the curve and x-axis to the left of y  x  x the ordinate x  x F ,when the area = ln F/W  , final x W

ln F/W   4. Calculate

 Fx F 5. Using ln   Wx W

xF

 dx y

 W xW

ln F



x



 100

  F1  x F     α aveln   estimate α ave   W1  x W  

8. Presentation of results 1. Mole fraction of mole volatile component in the feed; mole fraction of more volatile component in the distillate; mole fraction of more volatile component in the bottoms 2. Amounts of distillate and bottoms and their compositions 3. α ave = average relative volatility of the system between x W and x F 4. Report % errors 5. One can also compare estimated α ave value with the value obtain with the help of y   1 y  equilibrium data. Make use the definition of α as α    and then calculate α ave  x1  x    .  Fx F   F1  x F     α aveln   estimate α ave 6. Using ln   Wx W   W1  x W  

VLE DATA ETHANOL WATER SYSTEM

x

y*

x

y*

0.0 1.9 7.21 9.66 12.38 16.61 23.37 26.08

0.0 17 38.91 43.75 47.04 50.89 54.45 55.8

32.73 39.65 50.79 51.98 57.32 67.63 74.72 89.43

58.26 61.22 65.64 65.99 68.41 73.85 78.15 89.43

123


CALIBRATION CHART

1.36

1.355

Refractive Index

1.35

1.345

1.34

1.335

1.33 0

0.1

0.2

0.3

0.4

0.5

Mole fraction of Ethanol in Ethanol-Water System

References: 1. Max Peters, “Elementary Chemical Engineering”, 2nd ed, McGraw-Hill Book Company, New York, 1984, pp. 203-204. 2. V.G.Pangarkar and R.R.bhave, “Mass transfer Operations: A laboratory manual in Chemical Engineering”, Chemical Engineering Education Development Centre, Indian Institute of Technology Madras, Madras, 1981, pp.20-26. 3. R..E.Treybal, Mass Transfer Operations, McGraw-Hill Book Company, 3rd ed, New York, 1984, pp. 367-370. 4. F.Molyneux, Laboratory exercises in Chemical Engineering, Leonard Hill Books, London, 1967, p.137. 5. ‘Chemical Engineering Laboratory Manual’ published by IIT Bombay, obtained on the Webpage http://www.che.iitb.ac.in/courses/uglab/uglabs.html accessed on 25.08.2011

124

0.6


EXPERIMENT NO. 15

BATCH CRYSTALLIZATION 1. Aim Determine the yield of crystals using agitated batch crystallizer. 2. Objective To study the yield of crystals of Sodium Sulfate (Na2SO4) or Sodium Chloride (NaCl) from its saturated solution using open tank type agitated batch crystallizer. 3. Theory Crystallization is an important operation in processing as a method of both purification and of providing crystalline materials in a desired size range. In a crystal, the constituent molecules, ions or atoms are arranged in a regular manner with the result that the crystal shape is independent of size and, if a crystal grows, each of the faces develops in regular manner. The presence of impurities will, however, usually result in the formation of an irregular crystal. Generally large regular crystals are a guarantee of the purity of the material, though a number of pairs of materials form “mixed crystals”. In recent years, techniques have been developed for growing perfect crystals, which are used in the production of semiconductor devices, laser beams, and artificial gems. The crystallization process consists essentially of two stages which generally proceed simultaneously, but which can to some extent be independently controlled. The first stage is the formation of small particles or nuclei, which must exist in the solution before crystallization can start, and the second stage is the growth of nuclei. If the number of nuclei can be controlled, the size of the crystals ultimately formed may be regulated, and this forms one of the most important features of the crystallization process. Yield of crystals: The yield of crystals produced by a given cooling may be estimated from the concentration of the initial solution and the solubility at the final temperature, allowing for any evaporation, by making solvent and solute balances as follows: For the solvent, usually water, the initial solvent present is equal to the sum of the final solvent in the mother liquor, the water of crystallization within the crystals and any water evaporated or: w1 = w2 + y(R-1)/ R + w1E ------------------- (i) Where, w1 and w2 are the initial and final masses of solvent in the liquor y is the yield of crystals R is the ratio (molecular weight of hydrate/ molecular weight of anhydrous salt) And E is the ratio (mass of solvent evaporated/ mass of solvent in the initial solution) For the solute: w1c1 = w2c2 + y/ R ---------------------- (ii) Where, c1 and c2 are the initial and final concentrations of the solution in the terms of mass of anhydrous salt per unit mass of solvent. From equation (1): 125


w2 = w1(1 – E) – y(R - 1)/ R And in equation (2) w1c1 = c2 [w1(1 – E) – y(R - 1)/ R] + y/ R From which the yield, y = Rw1[c1 – c2 (1 – E)]/ [1 – c2(R-1)]

---------------------- (iii)

----------------------- (iv) ------------------------ (v)

Broadly, crystallizers may be classified according to whether they are batch or continuous in operation, and continuous crystallizers may be divided into linear and stirred types which are referred to later. Crystallizers may also be classified according to which the super saturation is achieved. In evaporative crystallizers conditions are approximately isothermal, and super saturation is achieved as a result of the removal of solvent. In cooling crystallizers, super saturation results from lowering of the temperature of the solution, and this can be effected either by means of exchange of sensible heat or by evaporative cooling; in the latter case, there is a small loss of solvent. Evaporative crystallization must of course be used where the solubility shows little variation with temperature. The simplest and cheapest type of crystallizer consists of an open tank, which can be used either as an evaporative or as a cooling crystallizer. In small scale of batch processing, such crystallizers are quire convenient because of their low first cost, simplicity of operation and flexibility. They are too wasteful of labor and give too uneven product to be attractive for large scale continuous processing 4. Experimental Set-up The experimental setup has a 5 liter capacity jacketed vessel made of SS 304, 1.6 mm thick, provided with a FHP variable speed std. make stirrer (Gear Motor, 150 RPM max., for better control of stirring and revolving the stirrer at low RPM during crystallization). 5. Experimental Procedure a) Take about 3-4 liter of water in the unit. Switch ON the power supply and pump and start circulating the water from the tank in to the jacket at fixed value using the rotameter. The exit water from the jacket shall be recycled back to the water supply tank. b) Set the temperature of the crystallizer content using the front panel of the Temperature Indicator-Controller (TIC) to the desired value and slowly start adding the solid to be crystallized and allow it to dissolve completely. c) Now set the temperature value of the crystallizer on the TIC to the value corresponding to the value at which the crystallization is to be carried out. d) Reduce the rate of circulation of hot water in the jacket and maintain it in the range of about 1.5-2.0 LPM and start draining the water coming out of the jacket. e) Add the cooling water available in the laboratory at the same rate to the water supply tank to make up the loss of water in the water supply tank and start recycling the exit water from the jacket back to the water supply tank. f) If ice addition is required to bring down the temperature of the Crystallizer obtained by the cooling water alone, slowly add finely crushed ice to the water supply tank so that the temperature of the crystallizer do not go below the set value. g) After 10-15 minutes stop circulation of the cold water and drain the water from the crystallizer using drain valve.

126


h) Filter and then weigh the water and crystals collected. Collect the crystals from the crystallizer. Dry them and weigh. 6. Observations and Calculations T1 = hot water inlet temperature / cold water inlet temperature T2= hot water outlet temperature / cold water outlet temperature T3=sample liquid inner temperature T4=sample liquid outlet temperature (if required) Initial weight of water taken (w1) Initial weight of Na2so4 added (m1) Temperature at which dissolution takes place(t3) Final weight of Na2so4 collected (m2) Final Cooling Temperature

= = = = =

% Yield of crystals = (m2 /Solubility at exp. Temp.) * 100 Yield of crystal =

Wt. of Na2so4 obtained as crystals Wt. of Na2so4 initially taken in feed

7. Result & Discussions i) Yield of crystals = ii) Plot the Solubility Curve of Na2SO4 in Water with Temperature 8. Conclusion

9. Precautions

127

gm gm C gm C


EXPERIMENT NO. 16(a)

STEAM DISTILLATION SETUP 1. Aim To study the characteristics of simple steam distillation using turpentine oil as a feed stock. 2. Objective To determine the vaporization efficiency, thermal efficiency and percentage recovery of simple steam distillation. 3. Theory Steam Distillation is a term used to describe a distillation process with open steam. Steam distillation is a method for distilling compounds which are heat-sensitive. It is specifically used where it is desired to separate substances at a temperature lower than their normal boiling point. The temperature of the steam is easier to control than the surface of a heating element, and allows a high rate of heat transfer without heating at a very high temperature. The vapor mixture is cooled and condensed, usually yielding a layer of oil and a layer of water. Steam distillation is commonly used in the following situations: 1. To separate relatively small amounts of volatile impurity from a large amount of material 2. To separate appreciable quantities of higher-boiling materials 3. To recover high-boiling materials from small amounts of impurity that has a higher boiling point. 4. Where the material to be distilled is thermally unstable or reacts with other components associated with it at the boiling temperature 5. Where the material cannot be distilled by indirect heating even under low pressure because of the high boiling temperature 6. Where direct-fired heaters cannot be used because of fire hazards In a steam distillation process, the liquid is distilled by feeding open steam to the distillation still. The steam carries with it vapors of volatile liquid and is then condensed to separate the liquid from water. The essential requirement for carrying out steam distillation is: 1. Substance does not react with steam. 2. Substance is in-soluble in water (immiscible). From the Hausbrand vapor pressure diagram for turpentine water system at 101.3 kN/m 2 pressure, obtain the distillation temperature, TD oC ( 95°C). From the Hausbrand diagram, curves for turpentine and water at atmospheric pressure cross at 95°C. Distillation: Turpentine oil at 101.3 kN/m2 pressure, (1.03kg/cm²) .The intersection of the two curves gives the distillation temperature. P (101.3 - PB) A

Vapour Pressure

Figure 1: Plot of (101.3 - PB) Vs

PA Vs T and T. 128

T

Distillation Temp.


Figure 2: Hausbrand vapor-pressure diagram for turpentine oil-water system

4. Experimental Setup 4.1 Description The apparatus consists of simple batch distillation using steam as a source of heat. The feed stock used is turpentine oil. Distillate is collected in a separating funnel for the formation of organic layer and an aqueous layer. Experimental Setup is shown in Fig 3.

Cooling Water In Cooling Water Out Condenser Turpentine oil In Live Steam

Distillate (Turpentine Oil + Water)

Residue Out Water In

Jacket Steam

PG

Heater

Fig 3: Experimental Setup

4.2.

Requirements 129


Steam supply, electricity, turpentine oil, demineralized water and separating funnel 5. Experimental Procedure 1. Charge the distillation still with 2kg of turpentine oil. 2. Adjust the jacket steam pressure to 170 kN/m2 (Pg,)/1.73 kg/cm² and start the cooling water supply to the condenser. Record the still temp. Collect the steam condensed in the jacket. 3. When the temperature in the still reaches 2°C below the distillation temp. (Td), the jacket steam is stopped and the flow of live steam is started through the steam sparger. 4. The live steam pressure is adjusted around 150 kN/m2 (Ps) /1.52 kg/cm². 5. Weigh the steam condensed in the jacket (WS), kg. 6. Continue the distillation process for sufficient time so that about 50-70% turpentine charged is distilled (around 1 hr). 7. Stop the steam supply and collect the distillate in the 2L separating funnel. Allow the formation of organic layer and an aqueous layer. Separate the two phases and weigh them, (WAD, WBD), kg. 8. Collect the residue, separate the two layers and weigh WAB, WBB (kg). 9. Stop the water supply to the condenser. 6. Observations and Calculation Mol. Wt. of turpentine oil MA =136.22 Normal boiling point of turpentine oil = 160oC Sp. heat of turpentine oil at 20°C = CpA = 1.8 kJ/kg °C Sp. heat of turpentine oil at 50°C = CpA =1.926 kJ/kg °C Sp. heat of turpentine oil at 100°C = CpA = 2.093 kJ/kg °C Latent heat of vaporization of turpentine oil, A = 74 kcal/kg = 309.84 kJ/kg Mol. wt. of Water = MB = 18 System Pressure, P = 101.3 KN/m2 /1.03kg/cm² Wt. of turpentine (feed) = 2 kg. Distillation temp., TD = 95 °C Wt. of Steam Condensed in jacket = WS, kg Steam pressure in the jacket, Pg = 170 KN/m2 Pressure of live steam, PS = 150 KN/m2 Steam pressure in the jacket = 172.21 kg/m2 = Pg Live pressure in the still = 151.95 kg/m2 = Ps Mass of steam condensed during pre-heating (feed change from 23oC to 95oC) WS = 410 g Distillation time = Initial temperature of feed charge, TR = °C Observed distillation temp., = °C Distillate readings (Turpentine oil + water): Wt. of turpentine oil in distillate WAD = Wt. of water in distillate WBD= Residue collected (Turpentine oil + water): Wt. of turpentine oil in residue WAB = Wt. of water in residue WBB = Ambient temperature Tr = 23°C Distillation time = 15 min approximately Observed distillation temperature = Converting Weights to moles: 130


No. of moles of turpentine oil in feed WAF NAF = ---------- = MA No. of moles of turpentine oil in distillate WAD NAD = ---------- = MA No. of moles of turpentine oil in Residue WAB NAF = ----------- = MA No. of moles of water in distillate WBD NBD = ------------ = MB No. of moles of water in residue WBB NBB = ---------- = MB Applying Material balance for turpentine oil, NAF = NAD + NAB Losses from the still turpentine oil = Steam Distillation Temp, TD = PAO= Partial Pressure of turpentine oil during distillation = P [NAD/ (NAD + NBD)] Where P =total pressure = 101.3 KN/m2  Vaporization efficiency = PAº  = --------- = PA WBD (P -  PA) MB And ---------- = ----------------x ---------WAD  MA where PA is the pressure of turpentine oil and water system at 95 °C from Hausbrand diagram. WBD / WAD in distillate = % Recovery = Calculation of Thermal efficiency: T Total heat output T = --------------------------------- x 100 Total heat input Heat Output: QO = WAF (TD – TR) CpA + A WAD = WS + A WAD TR = ambient temp TD = distillation temp. =92oC Heat Input: Qi Qi = Heat given by condensing steam in the jacket + Heat given by the condensing live steam during distillation. Qi = WSJ + (WBD + WBB) [S + Cp (TS – TD)] Jacket steam pressure = PJ (KN/m2) 131


J PS TS Cp PJ J PS S Cp  =

=latent heat of vaporization of steam corresponding to pressure PJ, kJ/kg = Pressure of live steam, KN/m2 = saturation temp. of steam at PS, °C corresponding to PS, kJ/kg = Specific heat of steam, kJ/kg°C. = 172 KN/m2 = 2216 kJ/kg = 151.95 KN/m2 = 2230 kJ/kg = Sp. heat of steam = 2.0 kJ/kgoK QO -------x 100 = Qi

NOMENCLATURE WAF = Wt. of turpentine in feed (kg) TD = Distillation temp. °C TR = Reference temp. (Ambient temp) °C Cp = Specific heat of turpentine oil (kJ/kgoC) A = Latent heat of evaporation of turpentine (kJ/kg) WAD = Wt. of turpentine in distillate, (kg) WS = Wt. of Steam Condensed in the jacket during pre-heating or charge (kg)  = Latent heat of steam corresponding to jacket steam pressure P (kN/m2) WBD = Wt. of water in distillate (kg) WBB = Wt. of water in residue, kg S = Latent heat of steam at live steam pressure PS (kN/m2) TS = saturation temperature of steam at PS Cp = Specific heat of steam (kJ/kgoC) 7. Results and Discussion Vaporization efficiency= Recovery= Thermal efficiency= 8. Conclusion References 1. McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”, 7th ed.McGraw Hill, NY, 2005. 2. Coulson, J.M., Richardson, J.F., “Coulson & Richardson’s Chemical Engineering Vol. - 1”, 5th ed., Asian Books ltd., ND, 1996. 3. Treybal, R.E., “Mass Transfer Operations”, 3rd ed., McGraw Hill, NY, 1981.

132


EXPERIMENT NO. 16(b)

VAPOR LIQUID EQUILIBRIUM SETUP 1. Aim & Objective Study of vapor-liquid equilibrium of Toluene – CCl4 mixture 2. Apparatus VLE setup consisting of heater, condenser, temperature indicator and refractometer, test tubes. 3. Chemicals Carbon Tetrachloride, Toluene 4. Theory The design of distillation and other contacting equipment requires reliable VLE data. Although relatively few ideal solutions are known whose equilibrium relations can be calculated from vapor pressure - temperature data of the pure components, by far the larger numbers of systems of industrial importance are non-ideal; and attempt to predict the equilibrium compositions of such mixtures from theoretical considerations alone have not proved successful. It has been the practice to determine such data experimentally under various conditions. Vapor liquid diagram shows relationship between the composition of the vapor and that of liquid in equilibrium with the vapor for a binary mixture at constant pressure or constant temperature. If liquid and vapor behave ideally, such curves are calculated as follows: From Raoult' law: p1 = P1 X1 (1) p2 = P2 X2 (2) Where p1 and p2 are partial pressures of components 1 and 2 in the mixture, P = total pressure P1 and P2 are vapor pressure of pure components at the same temperature as mixture. From Dalton's law of partial pressures: P1= P y 1 (3) P2= P y 2 (4) Where, y1 and y2 are the mole fractions of components in vapor. From these equations we have: P  p2 x1  (5) P1  P2 P x2 (6) y1  1 P Theoretical VLE curve can thus be calculated by choosing various boiling points of themixture and calculate x and y as shown above. Carbon tetrachloride - Toluene system closely follows ideal behavior. 5. Experimental Procedure 1. The vapor - liquid equilibrium still containing about 15O ml to 200 ml of known mixture (toluene – CCl4) is heated gently for at least 45 to 60 minutes (to ensure the mixture has attained equilibrium in above time) 2. The liquid and vapor are then sampled at noted temperature (T) and analyzed for molar composition of more volatile component in vapor (y) and liquid (x).

133


3. The heating is stopped and the feed liquid is replaced by another feed mixture (increasing gradually the amount of high boiling liquid), the liquid and vapor are once more allowed to come to equilibrium (indicated by constant temperature) and collecting the samples of liquid from the still and the condensate. The still liquid being hot, should be collected in an ice cooled test tube to avoid any change in composition due to vaporization. The above procedure should be repeated for at least six feed compositions. 4. The samples are viewed under a high precision refractometer, and their composition determined from the pre-determined calibration chart for CCl4 - toluene mixture, which converts refractive index to mole fractions. This way T-x-y and x-y diagrams can be plotted. 5. The calibration curve (nD vs. mole fraction of more volatile component, x) should be generated at a specific temperature (say 250C) by taking different mixtures of CCl4 toluene of known molar composition and recording the refractive index of the mixture. The samples collected should also be evaluated at the same temperature. 6. Observation & Calculation Correlation for CCl4 – Toluene system w.r.t refractive index is given by the equation y = -27.757*x + 41.529 ------------- (7) For VLE curve data S.No Liquid Mole Vapor Mole Temperature refractive fraction refractive fraction (°C) index (x) index (y)

7. Results & Discussion Plot the following graphs 1. y vs. x 2. Temperature vs. x 3. Temperature vs. y 8. Conclusion 9. Precautions 1. The still liquid being hot, should be collected in an ice cooled test tube to avoid any change in composition due to vaporization. 2. The above procedure should be repeated for at least six feed compositions. References: (a) McCabe, W.L., Smith, J.C., Harriott, P., “Unit Operations of Chemical Engineering”, 7th ed., McGraw Hill, NY, 2005. (b) W.L Badger & J.T. Banchero., “Introduction to Chemical Engineering”, 26th reprint, McGraw Hill, NY, 2011. 134


135


EXPERIMENT NO. 17

TWO PHASE FLOW IN A HORIZONTAL PIPE 1. Aim To Study Two phase flow in Horizontal Pipe. 2. Objective To study the two-phase flow in a horizontal pipe and to determine the pressure drop for the two phase flow. 3. Theory Concurrent gas-liquid flow is very common in chemical process equipment like reactors, reboilers, partial condensers, flash vaporizers gas pipe lines etc. As such it is always important to have a clear understanding of flow pattern and pressure drop in systems involving a twophase mixture. 3.1 Flow Pattern There are various flow regimes or “patterns” that are common to two phase flow systems, each having different characteristics and associated pressure drops. Two phase flow cannot simply be described as laminar, transitional and turbulent. The two-phase patterns vary not only with flow rates and fluid properties, but also depend upon pipe diameter and inclination. In horizontal pipe, the two phase flows regimes are (in order of lowest gas velocities to highest) bubble flow, plug flow, stratified flow, wavy flow, slug flow, annular flow and spray flow. In vertical pipe there are bubble, slug, churn, ripple, annular flow and mist flow. Most remediation professionals would recognize the appearance of these four patterns from conducting pilot tests and from observations gathered during full-scale system operation. However, most practitioners are unaware of the significance of that the type of flow pattern has on efficient system effectiveness. (Note: The term” Pattern” refers to visual observations and the flow “regime” applies to flow behavior that can be described by more quantitative expressions. Additionally, general estimations of flow pattern can be made using superficial gas and liquid velocities (Perry, 1999), Horizontal flow patterns can be roughly determined from gas and liquid superficial velocities, while vertical patterns can be estimated based on gas velocity only (Note: Superficial velocity is defined as the velocity of a phase as if it was flowing alone within the pipe. In horizontal pipe, flow patterns for fully developed flow have been reported in numerous studies. Transitions between flow patterns are gradual, and subjective owing to the visual interpretation of individual investigators. In some cases, statistical analysis of pressure fluctuations has been used to distinguish flow patterns. Bubble flow is prevalent at high ratios of liquid to gas flow rates. The gas is dispersed as bubbles which move at velocity similar to the liquid and tend to concentrate near the top of the pipe at lower liquid velocities. Plug flow describes a pattern in which alternate plugs of gas and liquid move along the upper part of the pipe. In stratified flow, the liquid flows along the bottom of the pipe and the gas flows over a smooth liquid/gas interface. Similar to stratified flow, wavy flow occurs at greater gas velocities and has waves moving in the flow direction. When wave crests are sufficiently high to bridge the pipe, they form frothy slugs which move at much greater than the average liquid velocity.

136


Slug flow can cause severe and sometimes dangerous vibrations in equipment because of impact of the high-velocity slugs against bends or other fittings. Slugs may also flood gas/liquid separation equipment. In annular flow, liquid flows as a thin film along the pipe wall and gas flows in the core. Some liquid is entrained as droplets in the gas core. At very high gas velocities, nearly all the liquid is entrained as small droplets. This pattern is called spray, dispersed, or mist flow

When a liquid and gas flows through a horizontal pipe, the following types of flow may occur. 1. Both the phases may be in viscous region. 2. Both the phases may be in turbulent region. 3. One of the phases may be in turbulent region and the other in viscous region. The type of flow of each phase is determined by the Reynolds number calculated on the basis of superficial velocity (superficial velocity = total flow rate of single phase/ total cross sectional area of flow) The pressure drop in a two-phase flow is the sum of the pressure drops due to friction and pressure drop due to acceleration. In a two phase flow, where gas is one of the phases, the pressure drop due to acceleration is important because the gas normally flows faster than the liquid phases and thus accelerates the liquid phase resulting in transfer of energy. The estimation the pressure drop is not possible analytically. The most common method to 137


estimate the pressure drop for two-phase flow is the one proposed by Lockhart and Martinelli (Page 6-26, fluid and particle dynamics, Perry’s chemical engineering). In this, each phase is considered separately and the combined effect is determined. Lockhart and Martinelli (ibid.) correlated pressure drop data from pipes 25 mm (1 in) in diameter or less within about 50 percent. In general, the predictions are high for stratified, wavy, and slug flows and low for annular flow. The correlation can be applied to pipe diameters up to about 0.1 m (4 in) with about the same accuracy. The pressure drop due to two-phase flow is the pressure drop of either of the two phases if it alone were present, multiplied by a factor. (P/Z)tp = YL (P/Z)L (P/Z)tp = Yg (P/Z)g where YL and Yg are the functions of dimension less parameter X defined as: X = [(P/Z)L/(P/Z)g]0.5 The two-phase flow analysis can also be expressed in terms of friction factors. By definition of pressure drop, P/Z = (2fV2/(gcD)) Friction factor, f = (P/Z) .g. D/2*V2) where, v is the superficial fluid velocity, D is pipe I.D,  is the fluid density and P/Z is the pressure drop per unit length Thus for gas only: fg = (P/Z)g .g. D/(2*Vg2g) for liquid only fl = (P/Z)l .g. D/(2*Vl2l) for two phase flow flpl = (P/Z)tp .g. D/(2*Vl2l) (based on superficial velocity of liquid) 2 flpg = (P/Z)lp .g. D/(2*Vg g) (based on superficial velocity of gas) Defining g flpg/fg = l flpl/fl = X2 = [(P/Z)L / (P/Z)g] = [g / l]2 Two phase pressure drop is then expressed as: (P/Z)tp = g2 (P/Z)g (P/Z)tp = g2 (P/Z)l Friction factors for single component flow (gas or liquid) can be easily estimated from: f = 16/Re (for Re < 2000) f = 0.079 (Re)-0.25 (for Re > 3000) Recently empirical co-relations have been derived from a large data bank for void fraction () and g and are given below:

=__________1_______________ (1+0.0904 X 0.548)2.82 g = (1+ X 2ln) n/2 and, n is obtained from the table given below:-

138


N 4.12 3.61 3.56 2.68 3.27

Liquid Flow Turbulent Viscous Turbulent Viscous Viscous

Gas Flow Turbulent Turbulent Viscous Viscous Viscous

If X < 1 If X > 1

Pure phase Reynolds numbers are defined as: Reg = DVg g / g Rel = DVl l / l Vl Superficial velocity of liquid = volumetric flow rate of liquid/cross-section area of = empty pipe. Vg = Superficial velocity of Gas = volumetric flow rate of gas/cross- section area of empty pipe. D is the I.D. of pipe = 25.4mm= 0.0254m P is the pressure drop across length = Z = 1 meter approximate. 4. Experimental Procedure 8. 9. Start the water pump and allow water to flow at a constant flow rate 10. Start the compressor and fix the air pressure in the air tank at 1.5 kg/cm² and allow air to flow at some constant flow rate. 11. Wait for the steady state and observe the flow pattern. Record it. Note the volumetric flow rates of water and air (rotameter readings). 12. Record the average reading of the manometer (cm of CCl4). Also record the temperature of water and air by thermometer. 13. Increase the gas flow rate gradually keeping water flow rate constant at the previous level and repeat steps 4 and 5. 14. Repeat step 6 for at least 6 gas flow rates. 15. Repeat steps 2 to 7 for at least 5 liquid flow rates 16. Stop the compressor. 17. Stop the water pump. 5. Observations System : Distance between manometer taps Manometer fluid = Water temperature = Air Temperature = Air Pressure = Density of manometer fluid = Density of water = Density of air = Viscosity of water (from data book) Viscosity of air (from data book) 139

=

m l a

Air-water 1000mm = 1Z, m mercury T ambient T ambient kg/cm² = 1660 kg/m3 = 1000 kg/m3 = 1.205 kg/m3 l = 0.00089 N-s/m2 g = 0.0000192 N-s/m2


Run No

Air rotameter reading, Water rotameter reading, Ql, Manometer reading R, Qg, LPM LPM cm of CCl4

6. Calculations I.D. of pipe

Cross sectional area of pipe

=

A

=

0.0254 D, m  ------- D², = 4

5.0670 x 10⁻⁴ m²

Volumetric flow rate of air Volumetric flow rate of Water Air superficial velocity Vg Water superficial velocity Vl Air Reynolds number Reg Water Reynolds number Rel Experimental two phase pressure drop: Two phase pressure drop Ptp Two phase pressure drop per unit length Two phase flow friction factor

= = = = = =

Qg LPM Ql , LPM ((Qg/60) *10-03)/A, m/s ((Ql/60) *10-03)/A, m/s DVg g / g DVl l / l

= = =

(R/100 [m-w)* g], N-s (P/Z) tp, N-s/m²/m ftpg or ftpl

ftpl

=

(P/Z)tp .g. D/(2*Vl2 l) (based on superficial velocity of liquid)

ftpg

=

(P/Z)tp .g. D/(2*Vg2 g) (based on superficial velocity of Gas)

7. Results & Discussion 8. Conclusion

140


EXPERIMENT NO. 18

MASS TRANSFER WITH CHEMICAL REACTION 1. Aim 1. To study the dissolution of benzoic acid in aqueous NaOH solution. 2. To compare the observe enhancement factor for mass transfer with those predicted by the film and boundary layer models. 2. Theory Solid-liquid mass transfer plays an important role in some industrial operations. The dissolution may occur with or without chemical reaction. In case dissolution is accompanied by solid-liquid reaction, it is desirable to know the enhancement in the rate of mass transfer due to chemical reaction. In the present experiment we aim at finding the enhancement in the rate of dissolution due to simultaneous reaction and compare it with the enhancement predicted on the basis of the film and boundary layer models. The system is: dissolution of benzoic acid in aqueous NaOH solution. The dissolution of a solid in a solution accompanied with instantaneous chemical reaction can be expressed as:

   B B  Product Where, A is the solid and B is the liquid phase reactant, assuming the reaction to be instantaneous so that A and B don’t coexist. The mechanism of solid dissolution involves dissolution of A in liquid followed by its reaction with species B diffusing from the bull liquid phase at a reaction plane. If the film model is applied to this situation, the enhancement factor,  defined as the ratio of the solid-liquid mass transfer coefficient with reaction, kr to the mass transfer coefficient without reaction, k, given by:  film 

kr k



B0  DA DB   DB DA  B A

--------- (1)

And for boundary layer model it is:

h

D  k  r   A  k  DB 

1/ 3

D   B  DA

  

2/3



Bo   B A

---------- (2)

3. Experimental Set-up The Set-up consists of a cylinder of benzoic acid mounted on a SS rod and driven by a D.C. motor. The operational range of rotation is between 10 to 30 rpm. The cylinder is immersed in an aqueous solution of sodium hydroxide of known concentration in a 500 ml vessel filled to 2/3rd its capacity. The position of the benzoic acid cylinder is so adjusted that the liquid level rises above the top surface by about 3 cm. The dimensions of the benzoic acid cylinder may be fixed at diameter: 2 to 3 cm, length: 5 to 7 cm. The cylinder can be prepared by pouring molten benzoic acid in the mould of desired dimensions with 4 to 5 mm SS rod located in the center of the mould in a vertical position. User can also use the cylinder 141


provided with the set-up. 4. Experimental Procedure 1. Record the dimensions of the benzoic acid cylinder (length and OD of SS rod) and then fix I in the vertical position with the D.C Motor. (Do not immerse it in solution as yet) 2. Fill the vessel with aqueous NaOH solution of known concentration of 2/3rd of its volume. Record the volume of aqueous NaOH solution added (V). 3. Start the water bath and fix the dissolution temperature (ambient to 50 0C), wait till the aqueous solution attains the desired temperature. Record the temperature (T). 4. Now fix the benzoic acid cylinder inside the vessel containing aqueous NaOH solution and start the motor at a fixed rotational speed. (N, rpm) 5. Run the experiment for 10 minutes. 6. Stop the motor and remove the benzoic acid cylinder. 7. Mix thoroughly the contents of vessel and analyze it for un-reacted NaOH concentration by titration against standard HCL solution. 8. Measure the benzoic acid cylinder dimensions again. 9. Repeat steps 1 to 8 for different concentrations of aqueous NaOH solutions. 10. Repeat steps 1 to 8 using de-ionized water. This run may be carried for about 45 to 60 minutes duration. During this period small samples (5 ml) should be withdraw at regular intervals of 10 minutes and analyzed for dissolved benzoic acid by titration against 0.02 kmol/m3 HCL solution. 5. Observations and Calculations Reaction: C6H5COOH + NaOH

 C6H5COONa + H2O

(For each mole of benzoic acid dissolved, 1 mole of NaOH is consumed) Thus rate of dissolution can be determined by measuring the fall in NaOH concentration. 1. Recommended concentration range for aqueous NaOH solution: 0.5 kmol/m3 to 1.5 kmol/m3. 2. Use the following standard solutions for titration: a. 0.02 kmol/m3 NaOH solution for the estimation of dissolved benzoic acid. b. 0.4 to 1 kmol/m3 standard HCL solution for the estimation of un-reacted NaOH. 3. For each titration in case of estimation of NaOH use 10 ml of solution and titrate against standard HCL solution using suitable indicator. 4. For each titration in case of estimation of dissolved benzoic acid use 5 ml of solution and titrate against 0.02 kmol/m3 standard NaOH solution. Let titer value for sample before the experiment Let titer value for sample after the experiment Let titer value for sample Time of dissolution Dissolution temp Hence, [NaOH]i [NaOH]f [C6H5OOH] The rate of dissolution of Benzoic acid:

142

= T1 (m3). = T2 (m3). = T3 (m3). = t (sec) = T oC = (T1/10) [HCL] = (T2/10) [HCL] = (T3/5) [NaOH]


R = ({[NaOH]i - [NaOH]f}) X Vol. of NaOH sol. in tank (V m3). Let the average dimensions (avg. of start and of expt.) of benzoic acid cylinder be Dav, Lav. Let S.S. rod diameter = dr.    2  Average surface area, As = Dav Lav  2   D 2 av  d r   4   Specific rate of dissolution = R / As (kmol/m2-s)



 exp





R A k

Where, A is the solubility of benzoic acid in water (0.0276 kmol/m3) k is obtained from physical dissolution run (i.e. with plain water) After integrating the material balance equation for a species, we have:

k

  A  As t   In 1   b   V   A 

Where, [Ab] is the bulk liquid cone of benzoic acid at time, t, in kmol/m3 and [A] is the solubility of benzoic acid in water, kmol/m3. After plotting t vs. ln [1 - Ab / A], we record the slope of the line given by: Slope = kAs / V; knowing As and V, we get k. The theoretical values for enhancement factor,  for the film model and boundary layer model is obtained from Eq.1 and Eq.2 respectively. The diffusivity of benzoic acid, DA, may be obtained from the literature = 1.04 x 10-9 m2/ at 30 oC. Value of DB may be taken as = 4.1 x 10-9 m2/ at 30 oC. Compare the theoretical values for  with the experimental values and observe that boundary layer model is more close to the experimental value.

6. Results and Discussion

7. Conclusion

143


EXPERIMENT NO. 19

ADSORPTION IN PACKED BED 1. Aim To understand the phenomena of adsorption in packed beds. 2. Objective To determine adsorption isotherm of acetic acid on activated charcoal. To determine the adsorption constant (k) and the maximum adsorbed substance amount of acetic acid per gram of charcoal (Amax) of Langmuir isotherm. 3. Theory Adsorption is a process that occurs when a gas or liquid solute accumulates on the surface of a solid or a liquid (adsorbent), forming a molecular or atomic film (the adsorbate). It is different from absorption, in which a substance diffuses into a liquid or solid to form a solution. The term absorption encompasses both processes, while desorption is the reverse process. Adsorption is operative in most natural physical, biological, and chemical systems, and is widely used in industrial applications such as activated charcoal, synthetic resins and water purification. Similar to surface tension, adsorption is a consequence of surface energy. In a bulk material, all the bonding requirements (be they ionic, covalent or metallic) of the constituent atoms of the material are filled. But atoms on the (clean) surface experience a bond deficiency, because they are not wholly surrounded by other atoms. Thus it is energetically favorable for them to bond with whatever happens to be available. The exact nature of the bonding depends on the details of the species involved, but the adsorbed material is generally classified as exhibiting physisorption or chemisorptions. Physisorption or physical adsorption is a type of adsorption in which the adsorbate adheres to the surface only through Van der Waals (weak intermolecular) interactions, which are also responsible for the non-ideal behavior of real gases. Chemisorption is a type of adsorption whereby a molecule adheres to a surface through the formation of a chemical bond, as opposed to the Van der Waals forces which cause physisorption. Adsorption is usually described through isotherms, that is, functions which connect the amount of adsorbate on the adsorbent, with its pressure (if gas) or concentration (if liquid). One can find in literature several models describing process of adsorption, namely Freundlich isotherm, Langmuir isotherm, BET isotherm, etc. We will deal with Langmuir isotherm in more details: Langmuir isotherm In 1916, Irving Langmuir published an isotherm for gases adsorbed on solids, which retained his name. It is an empirical isotherm derived from a proposed kinetic mechanism. It is based on four hypotheses: i. ii. iii.

The surface of the adsorbent is uniform, that is, all the adsorption sites are equal. Adsorbed molecules do not interact. All adsorption occurs through the same mechanism.

144


iv.

At the maximum adsorption, only a monolayer is formed: molecules of adsorption do not deposit on other, already adsorbed, molecules of adsorbate, only on the free surface of the adsorbent.

For liquids (adsorbate) adsorbed on solids (adsorbent), the Langmuir isotherm (Fig. 1) can be expressed by

------------------------ (1) Where m is the substance amount of adsorbate adsorbed per gram of adsorbent (or kg) of the unit of m is mol.g-1.resp. mol. kg-1

Amax is the maximal substance amount per gram (or kg) of the adsorbent. The unit of Amax is mol.g-1, resp. mol.kg -1 k is the adsorption constant (mol-1.dm3), c is the concentration of adsorbate in liquid (mol.dm-3). In practice, activated carbon is used as an adsorbent for the adsorption of mainly organic compounds along with some larger molecular weight inorganic compounds such as iodine and mercury. Activated Carbon - Activated carbon can be manufactured from carbonaceous material, including coal (bituminous, sub-bituminous, and lignite), peat, wood, or nutshells (i.e., coconut).The manufacturing process consists of two phases, carbonization and activation. The carbonization process includes drying and then heating to separate by-products, including tars and other hydrocarbons, from the raw material, as well as to drive off any gases generated. The carbonization process is completed by heating the material at 400–600°C in an oxygen-deficient atmosphere that cannot support combustion. 4. Experimental Setup The experimental set up consists of a three Glass Columns having different diameter and height, filled with activated carbon. A common feed inlet (manifold) is provided at the bottom of each column. Feed is supplied from feed tank by means of pump and rotameter. Flow rate can be varied by rotameter. A bye-pass arrangement is also provided to maintain level in feed tank (optional). Adsorption process can be studied by varying bed height and feed flow rate in each of the three columns turn-wise. The whole set-up is mounted on a powder coated sturdy MS frame. 145


Glass Column Borosilicate - ID 28mm height 500 mm, ID 38mm height 500mm and ID 52mm and height 750mm, Rotameter Make Eureka 2-20LPH. Pump make Promivac Model PMP-15, Max head 2.5 mtrs, RPM 2800, single phase 220 Volts @ 50 Hz, sump tank capacity 20 ltrs apprx. Chemicals Required (Not in supply scope): Acetic acid, NaOH, phenolphthalein indicator, activated charcoal granular form (supplied with set-up). 5.

Experimental Procedure First prepare the aqueous solution of acetic acid and water. Switch ON the set up and ensure pump discharge. Set the desired flow rate in Rotameter. Open the feed valve provided at the back of column 1 (a common manifold has been provided for feed routing to all the three columns). e) Maintain constant flow rate in Rota meter (in case of fluctuation adjust with valve) after feed starts to come out from top of column, collect the sample in flask. f) Add 4-5 drops of phenolphthalein indicator in added in each flask and titrate by standard NaOH solution. 0 g) Once the endpoint has been reached, read the burette. The volume of the base X i (ml) a) b) c) d)

that was required to reach the endpoint. h) Repeat procedure for different flow rates and bed heights. i) After completing taking readings of Column 1, start feed supply to column NO 2 and close feed valve of column NO 1. j) Repeat procedure from 4-5 as mentioned above for different flow rates and bed heights. k) After completing taking readings of Column 2, start feed supply to column NO 3 and close feed valve of column NO 2. l) After completing experiment switch off pump and disconnect set-up from mains. Drain the SET-UP. 6.

Observations and Calculations c i0 - Concentration of acetic acid before the adsorption reaction ci- Concentration of acetic acid after the adsorption reaction mi- Amount of acetic acid adsorbed per gram of charcoal X i0 -The volume of the titrant (NaOH)

Table 1 Flask No.

X i0 (ml)

Xi c i0 3 (mol/dm ) (ml)

ci (mol/dm3)

1 2 3 4 5 146

mi 1/ci l/mi (mmol/g) (dm3 /mol) (g/mmol)


6 Determination of the concentration of acetic acid before (ci0) and after (ci) Adsorption: X i0 cT 0 ------------------------- (2) ci  , V Where X i0 is the volume of the titrant (NaOH), cT is the concentration of the titrant, V is the volume of the analytic (acetic acid according to Tab. 2), i=1-6 is the number of flask. Calculate the concentration of acetic acid after adsorption (ci), using the Eq. 2 and data form Tab. 3 after adsorption. Determination of the substance amount of acetic acid adsorbed per gram of charcoal (mol.g1) in individual flask: (c 0  ci )V A --------------------------- (3) mi  i , g Where ci0 , ci are the concentrations of acetic acid before and after adsorption, respectively. VA is the volume of the liquid phase in the mixture charcoal – acetic Acid, g is the mass of the adsorbent – charcoal (in grams), i=1-6 is the number of flask. Eq. 3 supposes that VA is the same for i=1-6, and also the mass of the charcoal (g). Write down the obtained values of mi to the Tab. 3. Determination of k and Amax: The Eq. 1 one can rearrange into a form: 1 1 1 1   , m Amax k c Amax

Thus,

------------------------ (4)

1 1  f   should be a straight line. m c

1 1  f   , where c is the concentration of acetic acid m c after adsorption. Fit the experimental points with a linear function. The slope represents the 1 1 value of , and the intercept corresponds to Amax Amax k Calculate Amax and k from the slope and the intercept.

Use MS Excel to create the dependence

7.

Precautions i. Do not run the pump at low voltage i.e. less than 180 Volts. ii. Never fully close the Delivery line fully. iii. Always keep apparatus free from dust. iv. To prevent clogging of moving parts, Run Pump at least once in a fortnight. v. Always use clean water (use distill water if available in lab). vi. If apparatus will not in use for more than one month, drain the apparatus completely. vii. If pump gets jam, open the back cover of pump and rotate the shaft manually.

147


EXPERIMENT NO. 20

HUMIDIFICATION IN WETTED WALL COLUMN 1. Aim Humidification & Dehumidification study in wetted wall column. 2. Objective 1. To take, and learn to interpret psychometric (wet and dry bulb) data, and to understand more generally what is involved in transferring mass across gas-liquid interfaces. 2. To investigate the effects of air flow rates on the mass transfer coefficient, KG (kg mole of water transferred /m2.h.atm). 3. Theory A Wetted Wall Column is essentially a vertical tube with means of admitting liquid at the top and causing it to flow downwards along the inside wall of the tube, under influence of the gravity. Gas is admitted to the inside bottom of the tube, where it flows through the tower in contact with liquid (i.e. water). The mass transfer coefficient KG can be measured in the equipment in which the area of contact between the two phases is known and boundary layer separation. The measured flow of air at a measured humidity is brought in to contact with a film of water at certain temperature and vapor pressure. Moisture is absorbed in the air from the water film and the resultant humidity of the air and temperature & vapor pressure of the water at the entry and exit is measured. The rate of diffusion through the gas film, NA, is given by: NA = KG A (ΔP)m kg mole/h Where: KG = Gas film coefficient, kg mole of water transferred/m2 –h-atm. Partial pressure difference. A = wetted surface of column (ΔP)m = log mean partial pressure driving force across the ends of the column. 4. Experimental Setup 4.1. Schematic Diagram of Experimental Setup The schematic diagram of the experimental setup is shown below:

148


5. Experimental Procedure b) Feed water to the column from top at a rate at which complete wetting with minimum ripple formation takes place. c) Commence and operation with minimum air flow (H =3 cm) and after 10-15 minute the wet bulb and dry bulb temperature of the inlet air and outlet air at this flow rate is noted. d) Read the corresponding humidity (from psychometric chart) and vapor pressure of water corresponding to the entry and exit temperatures. e) Repeat step 2 and 3 for 3-4 air flow rates. f) Calculate experimental value of KG 6. Observations 1. Column diameter, 2. Column effective length, 3. Column pressure, 4. Area for mass transfer

D L P A

6.1. Observation Table Air temperature, °C

= 2.54 cm = 1.04 m = 760 mmHg/ 1 atm =  Dl = 3.14  2.54  104 = 829.9 cm2

Column Bottom

Column Top

A. HB

A.HT

Dry bulb Wet bulb Air humidity(Kg water/kg dry air)

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Water temperature, °C

7. Calculations: Air temperature, °C

Column Bottom

Column Top

Dry bulb Wet bulb Air humidity(Kg water/kg dry air) Water temperature, °C

Molar fraction of water in air at bottom, y1 = (A. H.B / 18) / (A. H.B /18 + 1/29) y1 = Molar fraction of water in air at top, y2 = (A. H.T / 18) / (A. H.T/18 + 1/29) y2 = Vapour pressure of water at inlet temperature, (P2) T = 0.0508 atm (Bhatt & Vora-343 pp.) Vapour pressure of water at outlet temperature, (P1) B = (ΔP)m = [( P1sat-P1)-( P2sat-P2)]/[ln(P1sat-P1/P2sat-P2) yi= (y1+y2) / 2 yi= (0.02+0.026) / 2= 0.023 KG (ΔP)m *A / V = [1 / (1-yi)]. In [{(yi-y1)/y2-yi)} {(1-y2) / (1-y1)}] KG(ΔP)m *A/V= X H = Head of water P = H (water - air) g= V = molar flow rate of gas = 0.65. [(2 P)0.5/ ] = Where;

[Co / (1-)4 =0.65]

m = Vo  /4  (do)2   = V = molar flow rate of gas V=mass flow rate of air / mol. Wt. of air

=

150


KG = (X * V) / (ΔP)m * A = Where, A. H.B= air humidity at bottom A. H.T= air humidity at top V = molar flow rate of gas V0 = mass flow rate of gas Molar fraction of water in air at bottom=y1, and at top =y2 8. Result & discussions Hence, the performance of the wetted wall column has been analysed and mass transfer coefficient is calculated. Mass Transfer Coefficient of the wetted wall column at flow rate----- is = 9. Conclusion 10. Precautions 1. Do not run the pump with the delivery valve closed for a long time to avoid damage of the pump. 2. Make sure during the experiment water tank should not be emptied totally. 3. Uniform wetting of the inner surface of the column ensures more valid data. 4. The column will require several minutes to come to steady state, especially with low water flow rate. Be sure to verify the actual attainment of this condition.

References: 1. Bird R. Stewart W. and Lightfoot E., “Transport Phenomena”, Wiley (2004) 2. Treybal, R. E., Mass Transfer Operations, McGraw Hill Company, New York, 1980 3. McCabe, W. L., Smith J. C., and Harriott, P., Unit Operations of Chemical Engineering, 6th ed., McGraw Hill Company, New York, 2001

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Lab manual for various experiments in chemical engineering

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