CHEMICAL ENGINEERING DESIGN OF A PARTIAL CONDENSER The design of condensers for the condensation of vapors from noncondensing gases is complicated by unusual conditions of heat transfer not encountered in total condenser design. In condensing part of the vapors in a gas stream, all of the properties of the gas stream vary greatly as the condensable vapor is removed. The heat-transfer coefficient of the gas film, the mass rate of gas flow, and the physical properties of the gas stream can change considerably as condensation proceeds. The condensation depends upon the diffusion of the vapor molecules through the gas mixture to the condensing surface. This involves two types of diffusion: 1. Molecular diffusion 2. Eddy diffusion Molecular diffusion involves the movement of individual molecules of condensable vapor from the main bulk stream onto the condensate film under the influence of concentration (partial pressure) gradient. Eddy diffusion is the movement of groups of molecules of the bulk stream by turbulent motion to the condensate surface. The main resistance to condensation now occurs between the bulk stream of vapor and the surface of the condensate. Therefore, mass-transfer coefficients must be considered as well as heat-transfer coefficients in the mechanism of condensation rate. The overall heat-transfer relationship for the heat transfer occurring in partial condensers must be written as:
dA=
dq U ∆t
Where ∆t
= (tc – tw)
tc = condensate-gas interface temperature, 0C
tw = tube wall temperature, 0C U = all resistances but the gas film q
A=∫ 0
dq U ∆t
(1)
In general, it is not possible to integrate this relationship formally using analytical expressions for both U and ∆t as functions of q. The method of Colburn and Hougen is generally accepted as the basis for obtaining rigorous design of cooler-condensers. The method is tedious since it involves successive trial and error substitutions. The rate of transfer of sensible heat from the gas stream on the shell side of the exchanger to the outside of the tubes is given by: d qs =h0 (t g−t c ) dA
(2)
The rate of transfer of latent heat from the gas stream on the shell side of the exchanger to the fin side of the tubes due to mass transfer of condensable material to the tube surface is given by: d qL =k m ( p g− p c ) dA
(3)
The total rate of heat transfer, given by the sum of (2) and (3) above, must be transferred through the tube and water film: q A
= h0 (t g −t c )
Where:
+
k m ( p g − pc )
= Ucomb.
(t c −t w )
= U∆t
(4)
h0 = heat-transfer coefficient in gas film, J/hr.m2.0C t g = temperature of main bulk of the gas vapor mixture, 0C t c = temperature of condensate at gas-condensate interface, 0C km
= mass-transfer coefficient, kg/(hr.m2) per unit pressure
ƛ = latent heat of condensation, J/kg
Pv = partial pressure of vapour in main bulk of the gas vapour mixture,
kg sq . m.
PC = Vapour pressure at tc, kg/ sq. m. tw = water temperature, oC Ucomb = Combined conductance other than the gas film
=
1 A0 A A + 0 r i+( 0 ) Am Ai Aihi
( h1 )+ r + r +r ( ) ( ) 0
cond
f
m
(5)
Where hcond = condensing film coefficient. The gas film heat transfer and mass transfer coefficients (h0 and k respectively) are obtained in the following manner. The fin side gas film heat transfer coefficient could be obtained from literature. The relationship for 19 fin per inch tubes in unbored shells is: 1/ 3 h0 De DeG 0.6 c p μ μ 0.14 =0.155( ) ( ) ( ) k μ k μw
(6)
The fin side gas film mass transfer coefficient is obtained using the heat transfer coefficient obtained from the above equation and the J factor relationships for heat and mass transfer. The heat transfer J factor is defined as:
jh =
h0 c p μ 2 /3 μw ( ) cp G k μ
0.14
( )
(7)
The mass transfer J factor is defined as:
jm =
μ ρ Dv ¿ ¿ k m M m P gf ¿ G MV
(8)
The mass transfer coefficient is obtained as:
km =
μw 2/ 3 ¿ μ c p ρ D v 2/3 ¿ ¿ k h0 M v ¿ c p M m Pgf
Where jh = heat-transfer ‘j’ factor (dimensionless) h0 = gas film heat-transfer coefficient, J/hr-m2-0C CP = Specific heat at constant pressure, J/kg-0C G = mass velocity, kg/hr-m2 µ = viscocity, kg/m-hr
(9)
k = thermal conductivity, J/hr-m-0C jm = mass transfer j factor (dimensionless) km = mass transfer coefficient, kg/hr-m2 per unit of pressure Mm = molecular weight of gas-vapour mixture (average) MV = molecular weight of vapour Pgf = log mean partial pressure of noncondensable gas across the film, kg/sq.m p (¿¿ g) at− ( p g ) at tc ( pg ) at tg Pgf = ( ln ( p g ) at tc ¿
Pg = partial pressure of the noncondensable gas in th main body, kg/sq.m ⍴ = vapour density, kg/cu.m DV = diffusion coefficient, sq.m/hr T 3/ 2 DV = 0.0166 (
1 3 A
1 3 2 B
P(V +V )
¿
√
1 1 + MA MB
Where: T = absolute temperature, 0K T = 0Rankine, change constant to 0.0069 P = Pressure, atmosphere
Va, Vb = molecular volumes Ma, Mb = molecular weights of gases and condensing vapours. The combined coefficient, Ucomb, defined for finned tubes is determined in the following manner. The condensing coefficient, hcond, can be determined by the methods given by in the second article of this series. The inside and outside fouling factors can be determined in the usual manner. The fin resistance of the tube can be obtained. The inside coefficient and metal resistance are determined in the usual manner. The condensing and water film coefficients vary from the inlet end to discharge end of the exchanger. These resistances are usually small in comparison to the gas film resistance and the variation can ordinarily be neglected. To obtain an exact design it is necessary to graphically solve for the heat transfer area. Partial condensers can have either a saturated vapour-gas feed or superheated vapour-gas feed. A different approach to the problem must be taken for the two cases.
CHEMICAL ENGINEERING DESIGN OF A FLASH EVAPORATOR The three principal elements involved in evaporator design are namely:
Heat transfer Vapour-liquid separation and Energy utilization
The vapour-liquid separators are variously called bodies, vapour heads, or flash chambers. The term flash chambers is also used to denote the minimum building block of an evaporator, comprising one heating element (the fluid with sensible heat) and one vapour head. This type is the choice here in this design project. Heat transfer is the most important single factor in evaporator design since the heating surface represents the largest part of evaporator cost. Equipment costs are usually correlated as function only of heating-surface area, materials of construction, and evaporator type. Other things being equal, the evaporator selected is the one having the highest transfer coefficient under operating conditions in terms of amount of energy per hour per degree temperature per cost of installation. Vapour liquid separation may be important for a number of reasons. Most important is usually prevention of entrainment because of value of product lost, pollution, contamination of the condensed vapour or fouling or corrosion of the surfaces on which the vapour is condensed. The thermodynamic efficiency of energy utilization in an evaporator is very low since the minimum energy requirement is only equal to the heat that will be liberated if the feed were reconstituted by mixing product and liquid solvent. Consequently, evaporator performance is rated on the basis of steam economy. Product quality considerations may require low hold up time and low temperature operation to avoid thermal degradation. The low hold up time eliminate some types of evaporator and some types are also eliminated because of poor heat transfer characteristic at low temperature. Product
quality may also dictate special materials of construction to avoid metallic contamination or a catalytic effect on decomposition of the product. Flash evaporators- as the feed to evaporation ratio is increased in a forward- feed evaporator having the feed heated by vapour blend from each effect, a point is reached where all the vapours is needed to preheat the feed and none is available to heat the succeeding effect. Then all ther heating surface is in the feed heaters and the evaporator themselves becomes merely flash chambers. This heating case is called a flash evaporator Calculations The calculation of the heat and material balance on a flash evaporator is relatively easy once it is understood that the temperature rise in each heater and temperature drop in each flash must all be substantially equal. This equality is almost exact if the condenser from each heater is flashed to the following heater. The steam sensory (E/S) may be approximated from ; E 1.1 ∆ T = S ∆T A+ R+ N
Where ∆ T
is the total temperature difference 0F, between feed to the flasher
A is the approach between vapour and temperature from the flasher and the liquid leaving the heater where the vapour condensed N is the number of stages R is the boiling point rise in the flash
Evaporator Accessories Condenser
The vapor from the effect of an evaporator is usually removed by a condenser. Surface condensers will be employed in this design, because the mixing of condensate and condenser cooling water is not desired. Surface condensers use more cooling water and are so much more expensive that they are never used where a direct contact condenser is suitable. The ratio of water consumption to vapour condensed can be determined from the following equation: H −(T 2−32) water = v vapor flow T 2−T 1
Where Hv = vapour enthalpy (J/kg) T2 = water temperature entering and leaving the condenser, 0C
Vent systems Non- condensable gases may be present in the evaporator vapor as a result of leakage air dissolved in the feed, or decomposition reactions in the feed. As the non-condensable increases, they tend to impede the heat transfer. In any event, non-condensable gases should be vented well before their concentration reaches 10% since gas concentration are difficult to measure the usual thing is to over-vent.
Evaporator Costs Capital costs Approximate selling prices of various type of evaporator are given by Zimmerman and Lavine (1958). These prices include all auxiliary equipment that a manufacturer would normally supply such as vapour piping barometric condenser, steam jet, condensate flash tanks, and in some cases, liquor piping and pumps. Installed costs The installed cost of a number of types of evaporator is given by Chilton (1949). The costs include foundation steel work, evaporator assembly, pumps, piping, insulation, painting, and a moderate of instrumentation. It is usually impossible to estimate the effect of a change in body material. In some cases, welded alloy bodies are cheaper than cast iron bodies. Operating costs Operating labour requirement depend mainly on the proximity of the evaporator to other process unit where occasional assistance and maintenance help can be obtained. Occasional maintenance labour will be required for the repacking of pumps and valve and repair of piping.
CHEMICAL ENGINEERING DESIGN OF A DISTILLATION COLUMN Calculation of Minimum Number of Plates:
The minimum number of stages Nmin is obtained from Fenske equation which is,
Where LK is the average relative volatility of the light key with respect to the heavy key, and xLK and xHK are the light and heavy key concentrations. The suffixes d and b denote the distillate (tops) (d) and the bottoms (b), Average geometric relative volatility LK = 1.51 Acetaldehyde = (7416.67/44) = 168.56 kmol Water = 0 kmol
Crotonaldehyde = (240.80/70) = 3.44 kmol Water = (220.79/18) = 12.27 kmol
At the distillate section acetaldehyde is the light component, xLK = 1.00, meanwhile water is the heavy component xHK = 0.00 For the bottoms, crotonaldehyde is lighter so, xLK = 0.2190, and xHK =0.7810 Therefore, Nmin = 20
Calculation of Minimum Reflux Ratio Rm: Using Underwood equations
1= the relative volatility of component I = 0.96 Rm = the minimum reflux ratio, X1 = concentration of component 1 = 1.00 2 = 1.51 X2 = 0.2190
By trial, = 0.85
Putting all values we get, Rm = 1.29
Actual Reflux Ratio, R: The rule of thumb is: R = (1.2 ------- 1.5) R min R = 1.3 R min R = 1.68
Theoretical no. of Plates: Gilliland related the number of equilibrium stages and the minimum reflux ratio and the no. of equilibrium stages with a plot that was transformed by Eduljee into the relation;
N N min
R Rmin 0.75 1 N 1 R 1
0.566
Recall R min = 1.29 , R = 1.68, and Nmin = 20
From which the theoretical number of stages to be, N = 40
Calculation of actual number of stages:
Actual number of stages = theoretical number of stages Efficiency
Overall Tray Efficiency Eo: Eduljee (1958) has expressed the O’Connell correlation in the form of an equation: Eo = 51 - 32.5 log(aαa) where a =the molar average liquid viscosity, by interpolation at the average temperature of the column,81oC = 0.1413mNs/m2,
αa = average relative volatility of the light key = 0.891 so, Eo = 80.25 % = 0.8025 So, No. of actual trays = 40/0.8025 = 50
Column diameter The principal factor which determines column diameter is the vapour flow-rate.
----------------------------------Eq2 where uv = maximum allowable vapour velocity, based on the gross (total) column cross-sectional area, m/s, Lt = plate spacing, m, 0.62 is chosen. V = density of vapour product,acetaldehyde = 0.7791g/cm3= 777.91 kg/m3 L = density of liquid product, (crotonaldehyde) + (H20) = = Total mass/(Total volume) Mass of crotonaldehyde formed = 240.80 kg/hr Mass of water = 220.79 kg/hr Density of water = 1000kg/m3 Density of crotonaldehyde = 0.846 g/cm3 846 kg/m3
Volume = Mass/(density) Volume of water = (220.29 kg/hr)/(1000kg/m3) = 0.2201 m3/hr Volume of crotonaldehyde = (240.80 kg/hr)/(846 kg/m3) = 0.2846 m3/hr Total volume = volume of crotonaldehyde + volume of water = 0.5047 m3/hr Total of mass = 220.29 + 240.80 = 461.09 kg/hr Density of liquid product, L =( Total mass)/(Total volume) = 913.59 kg/m3
By Eq2, Uv = maximum allowable vapour velocity {(-0.171*0.622)+ (0.27*0.62) – 0.047} *{(913.59-779.1)/779.1}1/2 0.0547 * 0.4155 =0.0227 m/hr.
Capacity Parameter: Assumed tray spacing = 18 inch (0.5 m) The flooding velocity can be estimated from the correlation given by Fair (1961):
K1 = a constant obtained from a chart of liquid-vapor flow factor against k = 0.1876 L= 913.59 kg/m3 and V = 0.7791g/cm3 = 777.91 kg/m3
By this the flooding velocity Uf = 0.0783 m/hr Assume 90% of flooding then actual vapor velocity Vn = 0.9 * 0.0783 = 0.07051 m/hr
Net column area used in separation is An = mv/Vn Volumetric flow rate of vapors = mv @
43oC vapor density of acetaldehyde (vapor) = 1.891atm = 191636.4 N/m2
Mass vapor flow rate = 7416.67 kg/hr
mv = (mass vapor flow rate /(191636.4) mv = 0.0387 m3/hr Now, net area An = mv/Vn = 0.5489 m2 Assume that downcommer occupies 15% of cross sectional Area (Ac) of column thus: Ac = An + Ad Where, Ad = downcommer area
Ac = An + 0.15(Ac) Ac = An / 0.85 Ac=0.6458 m2
So Diameter of Column Is Ac =(π/4)D2 D = (4Ac/π) D = 0.8223 meter (based upon bottom conditions)
Provisional Plate Design: Column Diameter Dc = 1.4513 m Column Cross-sectional Area (Ac) = 0.6458 m2 Down comer area Ad = 0.15Ac = 0.0969 m2 Net Area (An) = Ac - Ad =0.5489 m2 Active area Aa= Ac-2Ad = 0.4520 m2 Hole area Ah take 10% Aa = 0.1 × 0.4520 =0.0452 m2 Weir length Ad / Ac = 0.0969 / 0.6458 = 0.1500 0.15
From figure 11.31 Coulson & Richardson 6th volume 4th edition, which gives the relation between downcomer area and weir length, Lw / Dc = 0.80
Lw = 1.4513 *0.80 = 1.1616m Weir length should be 60 to 85% of column diameter which is satisfactory Take weir height, hw= 50 mm Hole diameter, dh = 5 mm Plate thickness = 5 mm
Weir liquid crest For a segmental downcomer the height of the liquid crest over the weir can be estimated using
Where Lw = weir length, m, how = weir crest, mm liquid, Lw = liquid flow-rate, kg/s. Maximum liquid rate “Lm” = (220.79 kg/hr + 240.80 kg/hr ) = 461.59 kg/hr = 0.128 kg/s Minimum Liquid Rate At 70% turn down ratio = 0.0896 kg/sec Recall weir length Lw = 1.1616m and L= 913.53 kg/m3
At Maximum rate (how) = 2.7868 mm Liquid At Minimum rate (how) = 2.2012 mm Liquid At minimum rate hw + how = 50 + 2.2012 = 52.20 mm Liquid
From fig 11.30 (weep- point correlation(Eduljee, 1959)), Coulson and Richardson Vol.6 K2 = 30.1
Now Check Weeping: in order to calculate minimum design vapor velocity.
Where Uh = minimum vapour velocity through the holes (based on the hole area), m/s, dh = hole diameter, mm= 5 mm K2 = a constant, dependent on the depth of clear liquid on the plate = 30.1 Also recall v =777.91 kg/m3. By that Uh = 0.4209 m/sec
DESIGN OF REFRIGERATION SYSTEM OF A COMPRESSOR. Compressors are typically of two types: reciprocating and rotary (screw or scroll). Scroll compressors are limited to lower capacity halocarbon systems. Rotary screw and scroll are increasingly popular due to lower maintenance costs. Screw compressors dominate the refrigeration market. This is mainly due to their high reliability, usually capable of operating over 50,000 hours between overhauls, and the selection of capacities of commercially available equipment. Commercially available motor driven capacities range from 20 kilowatts (25 horsepower) to over 1250 kilowatts (1675 horsepower).
The recommended compressor for refrigeration service is a screw type compressor that comes as a package unit. The screw compressor package units consist of screw compressor, motor, coupling, oil separator, local logic controller, oil pump and filter.
Scroll compressors are also positive displacement compressors. The refrigerant is compressed when one spiral orbits around a second stationary spiral, creating smaller and smaller pockets and higher pressures. By the time the refrigerant is discharged, it is fully pressurized.
A compressor is considered to be single stage when the entire compression is accomplished with a single cylinder or a group of cylinders in parallel. Many applications involve conditions beyond the practical capability of a single compression stage. Too great a compression ratio (absolute discharge pressure/absolute intake pressure) may cause excessive discharge temperature or other design problems. Two stage machines are used for high pressures and are characterized by lower discharge temperature (140 to 1600C) compared For practical purposes most plant air reciprocating air compressors over 100 horsepower are built as multi-stage units in which two or more steps of compression are grouped in series. The air is normally cooled between the stages to reduce the temperature and volume entering the following stage. (National Productivity Council, 1993). Reciprocating air compressors are available either as air-cooled or water-cooled in lubricated and non-lubricated configurations, may be packaged, and provide a wide range of pressure and capacity selections.
Performance Assessment of Refrigeration of a compressor The cooling effect produced is quantified as tons of refrigeration (TR). TR of refrigeration = 3024 kCalmol/hr heat rejected = 12657.86 kJmol/hr
The refrigeration TR is assessed as TR = Q x Cp x (T0– Ti) /12657.86 Where Q is molar flow rate of inlet component in kmol/hr Ti is inlet temperature in °C T0 is outlet temperature in °C. For the condenser to be designed, Cp for Acetaldehyde = 89.05J/mol.K Q = 184.27 kmol/hr (component entering the compressor) Ti = 350C, To = 430C Substituting values, TR = 184.27 x 89.05 x (43-35)/12657.86 TR = 10.37 kJmol/hr The above TR is also called as chiller tonnage.
Compressor Efficiency Several different measures of compressor efficiency are commonly used for the design : volumetric efficiency, adiabatic efficiency, isothermal efficiency and mechanical efficiency. Adiabatic and isothermal efficiencies are computed as the isothermal or adiabatic power divided by the actual power consumption. The figure obtained indicates the overall efficiency of a compressor and drive motor.
Compressor Power The power of a compressor is given by; P=
Work / Kmol x Inlet flowrate Efficiency
Work/Kmol = Zi Ti R
n P2 ⌊ n−1 P 1
( )
n−1 n
−1 ⌋
where Z = compressibility factor (1 for an ideal gas), R = universal gas constant, 8.314 J/K. mol T1 = inlet temperature, K P2 = outlet pressure Where T2 = 43 0C = 316 K, T1 = 35 0C = 308 K From the relationship,
T 2 P2 = T 1 P1
=
316 308
= 1.025
The value of n will depend on the design and operation of the machine. The design is more effective in a polytropic process, hence n is approximately 1.64 Substituting values, Work/Kmol = 1 x 308 x 8.314 x
1.64 ⌊ ( 1.025 ) 1.64−1
1.64−1 1.64
−1 ⌋
Work/Kmol = 0.206 kJ/Kmol
Inlet flowrate into the condenser = 184.27 kmol/hr In kmol/ sec = 184.27/3600 = 0.051 kmol /sec Volumetric flow rate of the compressor = 0.051 x 22.4 x
308 273
= 1.29 m3/s
From fig 3.6 (graph of compressor efficiency, Ep against volumetric flow rate) in Richardson and Coulson vol.6, the corresponding flow rate is 65% From Heuristics, Compression ratio is about the same in each stage of a multistage unit, ratio = (Pn/P1) x 1/n, with n stages. Efficiencies of reciprocating compressors: 65% at compression ratio of 1.5, 75% at 2.0, and 8085% at 3-6. Efficiency of large centrifugal compressors at suction is 76-78%. The compression ratio is 1.5 and the compressor is not very large.
0.206 x 267.73 65
Power =
= 84.85 kJ /s = 84.85 kW
Energy required per hour = 84.85 x 3600 = 305,459.34 kJ = 305.46 MJ
The specific power consumption kW/TR is a useful indicator of the performance of refrigeration system. By measuring refrigeration duty performed in TR and the kiloWatt inputs, kW/TR is used as a reference energy performance indicator.
Therefore, Specific power consumption =
power consumption TR
84.85 = 10.37
= 8.18
Effectively, the overall energy consumption would be towards: Compressor kW Chilled water pump kW Condenser water pump kW Cooling tower fan kW, for induced / forced draft towers The specific power consumption for certain TR output would therefore have to include: Compressor kW/TR Chilled water pump kW/TR Condenser water pump kW/TR Cooling tower fan kW/TR The overall kW/TR is the sum of the above.
Integrated Part Load Value (IPLV)
Although the kW/ TR can serve as an initial reference, it should not be taken as an absolute value since this value is derived from 100% of the equipment's capacity level and is based on design conditions that are considered the most critical. These conditions occur may be, for example, during only 1% of the total time the equipment is in operation throughout the year. Consequently, it is essential to have data that reflects how the equipment operates with partial loads or in conditions that demand less than 100% of its capacity. To overcome this, an average of kW/TR with partial loads ie Integrated Part Load Value (IPLV) have to be formulated. The IPLV is the most appropriate reference, although not considered the best, because it only captures four points within the operational cycle: 100%, 75%, 50% and 25%. Furthermore, it assigns the same weight to each value, and most equipment usually operates at between 50 % and 75% of its capacity. This is why it is so important to prepare specific analysis for each case that addresses the four points already mentioned, as well as developing a profile of the heat exchanger's operations during the year.
Leak quantification For rotary compressors, there is an easy way to estimate the amount of leakage in the system. This method involves starting the compressor when there are no demands on the system (when all the air -operated, end-use equipment is turned off). A number of measurements are taken to determine the average time it takes to load and unload the compressor. The compressor will load and unload because the air leaks will cause the compressor to cycle on and off as the pressure drops from air escaping through the leaks. Total leakage (percentage) can be calculated as follows:
Leakage (%) =
T x 100 T+t
Where T = on-load time (minutes)
t = off-load time (minutes) Leakage will be expressed in terms of the percentage of compressor capacity lost. The percentage lost to leakage should be less than 10 percent in a well- maintained system. Poorly maintained systems can have losses as high as 20 to 30 percent of air capacity and power. For accuracy, take ON & OFF times for 8 – 10 cycles continuously. For a leakage of 5% and a load time of 1.5 mins
Leakage (5 %) =
1.5 x 100 1.5+t
t = 28.5 mins For the compressor capacity of 1.29 m3/s = 77.4 m3/min Leakage capacity =
1.5 x 77.4 1.5+28.5
= 3.87 m3/min
