L6 Cooling Tower.pdf

SOLTEQ®

EQUIPMENT FOR ENGINEERING EDUCATION

EXPERIMENTAL MANUAL

WATER COOLING TOWER MODEL: HE 152

SOLUTION ENGINEERING SDN. BHD. NO.3, JALAN TPK 2/4, TAMAN PERINDUSTRIAN KINRARA, 47100 PUCHONG, SELANGOR DARUL EHSAN, MALAYSIA. TEL: 603-80758000 FAX: 603-80755784 E-MAIL: [email protected] WEBSITE: www.solution.com.my 227-0110-HE

Table of Contents Page List of Figures............................................................................................................................... i 1.0 . INTRODUCTION ……………………………………………………………….……………………1 2.0. GENERAL DESCRIPTIONS 2.1 Components of the HE152 Basic Cooling Tower Unit …………………..….............. 2 2.2 The Process Involved in the Operation ………………………………………………... 5 2.3 Overall Dimensions……………………………………………………………………….. 5 2.4 General Requirements……………………………………………………………………. 5 3.0 SUMMARY OF THEORY 3.1 Basic Principle ………………………………………………………………............... 6 3.2 Evaporation from a Wet Surface …………………………….................................. 6 3.3 Cooling Tower Performance………………………………………………………….. 6 3.4 Thermodynamic Property……………………………………………………………… 7 3.4.1 Dalton’s and Gibbs Laws…………………………………………………….. 7 3.4.2 Psychometric Chart…………………………………………………………… 8 3.5 Orifice Calibration…………………………………………………………………………. 9 3.6 Application of Steady Flow Energy Equation………………………………………….. 10 3.7 Characteristics Column Study………………………………………………………….. 13 3.8 Useful Information………………………………………………………………………… 16 4.0 EXPERIMENTAL PROCEDURE 4.1 General Operating Procedures………………………………………………………….. 17 4.1.1 General Start-up Procedures………………………………………………… 17 4.1.2 General Shut-Down Procedure……………………………………………… 18 4.2 Experiment 1: General Observation of the Forced Draught Cooling Tower…………. 18 4.3 Experiment 2: End State Properties of Air and Steady Flow Equations….………….. 19 4.4 Experiment 3: Investigation of the Effect of Cooling Load on Wet Bulb Approach … 20 4.5 Experiment 4: Investigation of the Effect of Air Velocity on Wet Bulb Approach and Pressure Drop through the Packing………………………………. 21 4.6 Experiment 5: Investigation of the Relationship between Cooling Load and Cooling Range …………………………………………………………….22 4.7 Experiment 6: Investigation of the Effect of Packing Density on the Performance of the Cooling Tower………………………………… 23 4.8 Experiment 7: Determination of Characteristic Equation of the Packing Characteristic Column…………………………………………………… 24 5.0 REFERENCES………………………………………………………………………………………. 25

APPENDIX A Experimental Data Sheets APPENDIX B Sample Results and Calculation APPENDIX C Components Properties and Diagram APPENDIX D Process Flow Diagram

List of Figures Page Figure 1

Parts Identification and Equipment Set-up of Bench Top Cooling Tower

4

Figure 2

System A

10

Figure 3

System B

12

Figure 4

Schematic Representations of the Air and Water Streams entering and leaving a Block of Packing

13

Figure 5

Graphical Representation of Tower Characteristics

15

i

SOLTEQ® BENCH TOP COOLING TOWER UNIT (MODEL: HE152)

1.0

INTRODUCTION The SOLTEQ® Basic Cooling Tower Unit (Model: HE152) has been designed to demonstrate students the construction, design and operational characteristics of a modern cooling system. The unit resembles a full size forced draught cooling tower and it is actually an "open system" through which two streams of fluid (in this case air and water) pass and in which there is a mass transfer from one stream to the other. The unit is selfcontained supplied with a heating load and a circulating pump. Once energy and mass balances are done, students will then be able to determine the effects on the performance of the cooling tower by the following parameters: a) Temperature and flow rate of water b) Relative Humidity and flow rate of air c) Cooling load Additionally, a Packing Characteristics Column (optional) is available for SOLTEQ® Basic Cooling Tower Unit (Model: HE152). This column is designed to facilitate study of water and air conditions at three additional stations (I, II and III) within the column. This enables driving force diagrams to be constructed and the determination of the Characteristic Equation for the Tower.

1


2.0

GENERAL DESCRIPTIONS 2.1

Components of the HE152 Basic Cooling Tower Unit The unit comes complete with the following main components: i) Load Tank The load tank is made of stainless steel having a capacity of approximately 9 liters. The tank is fitted with two cartridge heaters, 0.5 kW and 1.0 kW each, to provide a total of 1.5 kW cooling load. A make-up tank is fixed on top of the load tank. A float type valve at the bottom of the make-up tank is to control the amount of water flowing into the load tank. A centrifugal type pump is supplied for circulating the water from the load tank through a flowmeter to the top of the column, into the basin and back to the load tank. A temperature sensor and temperature controller is fitted to load tank to prevent overheating. A level switch is fitted to the load tank so that when a low level condition occurs, the heater and the pump will be switched off. ii) Air Distribution Chamber The stainless steel air distribution chamber comes with a water collecting basin and a one-side inlet centrifugal fan. The fan has a capacity of approximately 251 CFM of air flow. The air flowrate is adjustable by means of an intake damper. iii) Column and Packing Four packed columns (A, B, C and D) are available. Column A is a standard column that comes together with this unit whereas column B, C and D are the optional. The columns are made of clear acrylic with a square cross-sectional area of 225 cm2 and a height of 60 cm. Columns A, B and C comes with eight decks of inclined packing while column D is an empty column to allow user to design own packing. A top column that fitted on top of the column comes standard with a sharp edged orifice, a droplet arrester and a water distribution system. The Packing Characteristics Column (Column E) is also available as an accessory for this Basic Cooling Tower Unit. Packed column A: 110 m2/m3 Packed column B: 77 m2/m3 Packed column C: 200 m2/m3 Packed column D: Empty Packed column E: Packing Characteristics Column

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iv) Measurements Temperature sensors are provided to measure the inlet and outlet water temperatures as well as the make-up tank water temperature. In addition, temperature sensors have been installed to measure the dry bulb and wet bulb temperatures of inlet and outlet of the air. The followings show the list of codes assigned to each temperature sensors. T1 T2 T3 T4 T5 T6 T7 T8

Dry Bulb Temperature of the Inlet Air Wet Bulb Temperature of the Inlet Air Dry Bulb Temperature of the Outlet Air Wet Bulb Temperature of the Outlet Air Inlet Water Temperature Outlet Water Temperature Make up Tank Temperature Hot Water Tank Temperature

A differential pressure transmitter is provided for the measurement of pressure drop across the packed column. On the other hand, the differential pressure transmitter and the orifice are also used to determine the air flowrate. A flowmeter is provided for the measurement of water flowrate. The flowmeter is ranged at 0.4 to 4 LPM.

3


1 2

3 4

7 8

5

9

6

10

Figure 1: Parts Identification and Equipment Set-up of Bench Top Cooling Tower 1. 2. 3. 4. 5.

Orifice Water Distributor Packed Column Flowmeter Receiver tank

6. 7. 8. 9. 10.

4

Air Blower Differential Pressure Transmitter Make-up Tank Control Panel Load tank


2.2

The Process Involved in the Operation i) Water Circuit Water temperature in the load tank will be increased before the water is pumped through a control valve and flow meter to the column cap. Before entering the column cap, the inlet temperature of the water is measured and then the water is uniformly distributed over the top packing deck. This creates a large thin film of water, which is exposed to the air stream. The water gets cooled down, while passing downward through the packing, due to the evaporation process. The cooled water falls into the basin below the lowest deck and return to the load tank where it is re-heated before re-circulation. The outlet temperature is measured at a point just before the water flows back into the load tank. Evaporation causes the water level in the load tank to fall. The amount of water lost by evaporation will be automatically compensated by equal amount from the make-up tank. At steady state, this compensation rate equals the rate of evaporation plus any small airborne droplets discharged with the air. ii) Air Circuit A one-side inlet centrifugal fan draws the air from the atmosphere into the distribution chamber. The air flow rate is varied by means of an intake damper. The air passes a dry bulb temperature sensor and wet bulb temperature sensors before it enters the bottom of the packed column. While the air stream passes through the packing, its moisture content increases and the water temperature drops. The air passes another duct detector measuring its exit temperature and relative humidity, then through a droplet arrester and an orifice, and finally leaves the top of the column into the atmosphere.

2.3

Overall Dimensions Height Width Depth

2.4

: : :

1.25 m 0.91 m 0.45 m

General Requirements Electricity Water Supply

: 230VAC/1-phase/50Hz : Laboratory Water Supply

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3.0

SUMMARY OF THEORY 3.1

Basic Principle First consider an air stream passing over the surface of a warm water droplet or film. If we assume that the water is hotter than the air, then the water temperature will be cooled down by radiation, conduction and convection, and evaporation. The radiation effect is normally very small and may be neglected. Conduction and convection depend on the temperature difference, the surface area, air velocity, etc. The effect of evaporation is the most significant where cooling takes place as water molecules diffuse from the surface into the surrounding air. During the evaporation process, the water molecules are replaced by others in the liquid from which the required energy is taken.

3.2

Evaporation from a Wet Surface When considering evaporation from a wet surface into the surrounding air, the rate is determined by the difference between the vapour pressure at the liquid surface and the vapour pressure in the surrounding air. The vapour pressure at the liquid surface is basically the saturation pressure corresponding with the surface temperature, whereas the total pressure of the air and its absolute humidity determines the vapour pressure in the surrounding air. Such evaporation process in an enclosed space shall continue until the two vapour pressures are equal. In other words, until the air is saturated and its temperature equals the surface. However, if unsaturated air is constantly supplied, the wet surface will reach an equilibrium temperature at which the cooling effect due to the evaporation equals the heat transfer to the liquid by conduction and convection from the air, which under these conditions; will be at a higher temperature. Under adiabatic conditions, this equilibrium temperature is the "wet bulb temperature". For a cooling tower of infinite size and with an adequate air flow, the water leaving will be at the wet bulb temperature of the incoming air. Therefore, the difference between the temperature of the water leaving a cooling tower and the local wet bulb temperature is an indication of the effectiveness of the cooling tower. Thus, "Approach to Wet Bulb", an important parameter of cooling towers, is the difference between the temperature of the water leaving the tower and the wet bulb temperature of the entering air.

3.3

Cooling Tower Performance A study on the performance of a cooling tower can be done with the help of a bench top unit. Students shall be able to verify the effect of these factors on the cooling tower performance: (i) (ii) (iii) (iv)

Water flow rates Water temperatures Airflow rate Inlet Air Relative Humidity

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The effect of these factors will be studied in depth by varying it. In this way, students will gain an overall view of the operation of cooling tower. 3.4

Thermodynamic Property In order to understand the working principle and performance of a cooling tower, a basic knowledge of thermodynamic is essential to all students. A brief review on some of the thermodynamic properties is presented below. At the triple point (i.e. 0.00602 atm and 0.01°C), the specific enthalpy of saturated water is assumed to be zero, which is taken as datum. The specific enthalpy of saturated water (h f ) at a range of temperatures above the datum condition can be obtained from thermodynamic tables. The specific enthalpy of compressed liquid is given by h = h f + v f (p − p sat )

(1)

The correction for pressure is negligible for the operating condition of the cooling tower; therefore we can see that h ≈ h f at a given temperature. Specific heat capacity (C p ) is defined as the rate of change of enthalpy with respect to temperature (often called the specific heat at constant pressure). For the purpose of experiment using bench top cooling tower, we may use the following relationship: (2)

∆h = C p ∆T

and (3)

h = C pT

Where C p = 4.18 kJ.kg-1 3.4.1

Dalton’s and Gibbs Laws It is commonly known that air consists of a mixture of "dry air" (O 2 , N 2 and other gases) and water vapour. Dalton and Gibbs law describes the behaviour of such a mixture as: a) The total pressure of the air is equal to the sum of the pressures at which the "dry air" and the water vapour each and alone would exert if they were to occupy the volume of the mixture at the temperature of the mixture. b) The dry air and the water vapour respectively obey their normal property relationships at their partial pressures. c) The enthalpy of the mixture may be found by adding together the enthalpies at which the dry air and water vapour each would have as the sole occupant of the space occupied by the mixture and at the 7


same temperature. The Absolute or Specific Humidity is defined as follows: Specific Humidity , ω =

Mass of Water Vapour Mass of Dry Air

(4)

The Relative Humidity is defined as follows: Re lative Humidity , φ =

Partial Pr essure of Water Vapour in the Air Saturation Pr essure of Water Vapour at the same temperature

(5)

The Percentage Saturation is defined as follows: Percentage Saturation =

Mass of Water Vapour in a given Volume of Air Mass of same vol of Sat Water Vapour at the same Temp

(6)

At high humidity conditions, it can be shown that there is not much difference between the "Relative Humidity" and the "Percentage Saturation" and thus we shall regard as the same. To measure the moisture content of the atmosphere, this bench top cooling tower unit is supplied with electronic dry bulb and wet bulb temperature sensors. The temperature readings shall be used in conjunction with a psychometric chart. 3.4.2

Psychometric Chart The psychometric chart is very useful in determining the properties of air/water vapour mixture. Among the properties that can be defined with psychometric chart are Dry Bulb Temperature, Wet Bulb Temperature, Relative Humidity, Humidity Ratio, Specific Volume, and Specific Enthalpy. Knowing two of these properties, any other property can be easily identified from the chart provided the air pressure is approximately atmospheric. In the Bench Top Cooling Tower application, the air inlet and outlet sensor show the dry bulb temperature and wet bulb temperature. Therefore, the specific enthalpy, specific volume, humidity ratio and relative humidity can be readily read from the psychometric chart. The psychometric chart provided with this manual is only applicable for atmospheric pressure operating condition (1.013 bar). However, the error resulting from variation of local atmospheric pressure normally is negligible up to altitudes 500m above sea level.

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3.5

Orifice Calibration As mentioned above, the psychometric chart can be used to determine the value of the specific volume. However, the values given in the chart are for 1 kg of dry air at the stated total pressure. However, for every 1 kg of dry air, there is w kg of water vapour, yielding the total mass of 1 + w kg. Therefore, the actual specific volume of the air/vapor mixture is given by: va =

v ab

(7)

1 +ϖ

The mass flow rate of air and steam mixture through the orifice is given by m = 0.0137

x va

(8)

Where, m = Mass flow rate of air/vapor mixture v a = Actual specific volume and x = Orifice differential in mmH 2 0. Thus, x (1 + ϖ ) m = 0.0137 v ab

(9)

The mass flow rate of dry air, m a =

1 × Mass flow rate of air / vapor mixture 1 +ϖ

m a =

1 x (1 + ϖ ) × 0.0137 1 +ϖ v ab

m a = 0.0137

x v ab (1 + ϖ )

(10)

A simplification can be made since in this application, the value of ϖ is unlikely to exceed 0.025. As such, neglecting w b would not yield significant error.

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3.6

Application of Steady Flow Energy Equation Consider System A for the cooling tower defined as in Figure 2. It can be seen that for this system, indicated by the dotted line, a) b) c) d) e)

Heat transfer at the load tank and possibly a small quantity to surroundings Work transfer at the pump Low humidity air enters at point A High humidity air leaves at point B Make-up enters at point E, the same amount as the moisture increase in the air stream

B

m a

E

m E m a A

Work, P

Heat, Q

Figure 2: System A From the steady flow equation,

Q − P = H exit − H entry Q − P = (m a hda + m s hs ) B − (m a hda + m s hs ) A − m E hE

(11)

Note: The pump power, P is a work input. Therefore it is negative. If the enthalpy of the air includes the enthalpy of the steam associated with it, and this quantity is in terms of per unit mass of dry air, the equation may then be written as: (12) Q − P = m a (hB − hA ) − m E hE

10


Note: a) The mass flow rate of dry air ( m a ) through a cooling tower is a constant, whereas the mass flow rate of moist air increases as the result of evaporation process. b) The term m E hE can usually be neglected since its value is relatively small. Under steady state conditions, by conservation of mass, the mass flow rate of dry air and of water (as liquid or vapour) must be the same at inlet and outlet to any system. Therefore,

(m a ) A = (m a )B

(13)

and

(m s ) A + m E = (m s )B

or

m E = (m s )B − (m s ) A

(14)

The ratio of steam to air ( ϖ ) is known for the initial and final state points on the psychrometric charts. Therefore,

(m s )A = m aϖ A

and

(15)

(m s )B = m aϖ B

(16)

Therefore, m E = m a (ϖ B − ϖ A )

(17)

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Say, we re-define the cooling tower system to be as in Figure 3 where the process heat and pump work does not cross the boundary of the system. In this case warm water enters the system at point C and cool water leaves at point D.

B

m a

C

m w E

m E m a A D

Figure 3: System B Again from the steady flow energy equation, Q − P = H exit − H entry and P =0

Q may have a small value due to heat transfer between the unit and its surroundings. Q = m a hB + m w hD − (m a h A + m w hC + m E hE )

(18)

Rearranging,

Q = m a (hB − h A ) + m w (hD − hC ) − m E hE = m a (hB − h A ) + m w C p (t D − t C ) − m E hE Again, the term m E hE can be neglected.

12

(19)


3.7

Characteristics Column Study In order to study the packing characteristics, we define a finite element of the tower (dz) as shown in Figure 4, the energy balances of the water and air streams in the tower are related to the mass transfer by the following equation: C pW m W dT = K a dV (∆h )

(20)

where C pW = Specific heat capacity of water

= Mass flow rate of water per unit plan area of packing = Water Temperature = Mass Transfer Coefficient = Area of contact between air and water per unit volume of packing = Volume occupied by packing per unit plan area ∆h = Difference in specific enthalpy between the saturated boundary layer and the bulk air

m W T K a V

WATER INLET

T2 H2 mw

t2 h2 AIR ma OUTLET dz

z

WATER OUTLET

T1 H1 mw

t1 AIR h1 INLET ma

Figure 4: Schematic Representation of the Air and Water Streams entering and leaving a Block of Packing In this equation, we assume that the boundary layer temperature is equal to the water temperature T and the small change in the mass of water is neglected. Thus, from Equation 20, K a dV C pW dT = ∆h m W

(21)

13


Integrating Equation 21, Ka V = C pW m W

T2

∫h

T1

dT w − ha

(22)

The numerical solution to the integral expression Equation 22 using Chebyshev numerical method gives, Ka V = C pW m W

T2

∫

T1

T −T dT = 2 1 hw − ha 4

 1 1 1 1  + + +  ∆h1 ∆h 2 ∆h 3 ∆h 4

  

(23)

Where Ka V = Tower Characteristic m W

∆h1 = value of h w − h a at T2 + 0.1(T1 − T2 ) ∆h 2 = value of h w − h a at T2 + 0.4(T1 − T2 ) ∆h 3 = value of h w − h a at T1 − 0.4(T1 − T2 )

∆h 4 = value of h w − h a at T1 − 0.1(T1 − T2 )

Thermodynamics state that the heat removed from the water must be equal to the heat absorbed by the surrounding air. Therefore, the following equation is derived: L(T2 − T1 ) = G (h a 2 − h a1 )

(24)

or, L (h a 2 − h a1 ) = (T2 − T1 ) G

(25)

Where, L G T1 T2 ha 2 h a1

= Liquid to gas mass flow ratio = Cold water temperature = Hot water temperature = Enthalpy of air-water vapour mixture at exhaust wet-bulb temperature = Enthalpy of air-water vapour mixture at inlet wet-bulb temperature

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hw2 (Hot water Temp) Enthalpy Driving Force (hw-ha)

Enthalpy Water Operating Line C

ha2 (Air out) hw1 (Cold water Temp)

B

Saturation Curve ha1 (Air in)

D

Air Operating Line A

L/G

Approach

Range

Twb (In)

T1

Twb (Out)

T2

Temperature

Figure 5: Graphical Representation of Tower Characteristics The following represents a key to Figure 5: BA = Initial enthalpy driving force AD = Air operating line with slope L/G Referring to Equation 22, the tower characteristics could be found by finding the area between ABCD in Figure 5. Increasing heat load would have the following effects on the diagram in Figure 5: 1. Increase in the length of line CD, and a CD line shift to the right 2. Increase in hot and cold water temperatures 3. Increase in range and approach areas The increased heat load causes the hot water temperature to increase considerably faster than does the cold water temperature. Although the area ABCD should remain constant, it actually decreases about 2% for every 10 0F increase in hot water temperature above 100 0F. To account for this decrease, an "adjusted hot water temperature" is used in cooling tower design.

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3.8

Useful Information 1.

Orifice Calibration Formula: Mass flow rate of air and vapor mixture, x(1 + ϖ ) m = 0.0137 v ab The mass flow rate of dry air, x m a = 0.0137 v ab (1 + ϖ ) Where, x = orifice differential in mmH 2 0, v aB = specific volume of air at the outlet ϖ

= humidity ratio of the mixture

2.

Pump Work Input = 80W (0.08kW)

3.

Column Inner Dimension = 150 mm x 150 mm x 600 mm

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4.0

EXPERIMENTAL PROCEDURES 4.1

General Operating Procedures 4.1.1

General Start-up Procedures 1. Check to ensure that valves V1 to V6 are closed and valve V7 is partially opened. 2. Fill the load tank with distilled or deionised water. It is done by first removing the make-up tank and then pouring the water through the opening at the top of the load tank. Replace the make-up tank onto the load tank and lightly tighten the nuts. Fill the tank with distilled or deionised water up to the zero mark on the scale. 3. Add distilled/deionised water to the wet bulb sensor reservoir to the fullest. 4. Connect all appropriate tubing to the differential pressure sensor. 5. Install the appropriate cooling tower packing for the experiment. 6. Then, set the temperature set point of temperature controller to 50°C. Switch on the 1.0 kW water heater and heat up the water until approximately 40°C. 7. Switch on the pump and slowly open the control valve V1 and set the water flowrate to 2.0 LPM. Obtain a steady operation where the water is distributed and flowing uniformly through the packing. 8. Fully open the fan damper, and then switch on the fan. Check that the differential pressure sensor is giving reading when the valve manifold is switched to measure the orifice differential pressure. 9. Let the unit run for about 20 minutes, for the float valve to correctly adjust the level in the load tank. Refill the makeup tank as required. 10. Now, the unit is ready for use. Note: i. It is strongly recommended that ONLY distilled or deionised water be used in this unit. The impurities existing in tap water may cause the depositing in cover tower. ii. Check that the pressure tubings for differential pressure measurement are connected correctly. (Orifice pressure tapping point to V4; Column’s lower pressure tapping point to V6; Column’s higher pressure tapping point to V3; V5 leave to atmosphere) iii. To measure the differential pressure across the orifice, open valve V4 and V5; close valve V3 and V6. iv. To measure the differential pressure across the column, open valve V3 and V6; close valve V4 and V5. v. Always make sure that no water is in the pressure tubings for accurate differential pressure measurement.

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4.1.2

General Shut-Down Procedure 1. Switch off heaters and let the water to circulate through the cooling tower system for 3-5 minutes until the water cooled down. 2. Switch of the fan and fully close the fan damper. 3. Switch off the pump and power supply. 4. Retain the water in reservoir tank for the following experiment. 5. Completely drain off the water from the unit if it is not in used.

4.2

Experiment 1: General Observation of the Forced Draught Cooling Tower Objective: To observe the processes within a forced draught cooling tower 1. Perform the general start-up procedures and observe the forced draught cooling tower proves. 2. As the warm water enters the top of the tower, it is fed into channels from which it flows via water distribution system onto the packing. The channels are designed to distribute the water uniformly over the packing with minimum splashing. 3. The packing surfaces are easily wetted and the water spreads over the surfaces to expose a large area to the air stream. 4. The cooled water falls from the lowest packing into the basin and then is pumped to the simulated load in the load tank. 5. During the process, some water is lost due to the evaporation. Thus, "makeup" water must be supplied to keep the amount of water in the cooling system constant. The make-up is observed flowing past the float-controlled valve in the load tank. 6. A “droplet arrester”, or “mist eliminator” is fitted at the tower outlet to minimize loss of water due to escape of droplets of water (resulting from splashing, etc.) which is entrained in the air stream. This loss does not contribute to the cooling, but must be made good by "make-up". The droplet arrester causes droplets to coalesce, forming drops that are too large to be entrained and these falls back into the packing. 7. The fan drives the air upward through the wet packing. At air outlet, the air leaving the cooling tower is almost saturated, i.e. Relative Humidity is ~100%. The Relative Humidity at the air outlet is much higher than the Relative Humidity at the air inlet. The increase in the moisture content of the air is due to the evaporation of water into steam and the "latent heat" for this account for most of the cooling effect. 8. When the cooling load is switched off and the unit is allowed to stabilize, it is found that the water leaves the basin at temperature close to the wet bulb temperature of the air entering. Wet bulb temperature is lower than the dry bulb temperature and this varies according to the local atmospheric conditions (i.e. pressure and relative humidity). 9. With no load, the water would be cooled to the incoming wet bulb temperature. However, the condition cannot be achieved since the work done by the pump transfers about 40W to the water.

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4.3

Experiment 2: End State Properties of Air and Steady Flow Equations Objective: To determine the “end state” properties of air and water from tables or charts To determine Energy and mass balances using the steady flow equation on the selected systems Procedure: 1. Prepare and start the cooling tower with according to Section 4.1.1. 2. Set the system under the following conditions and allow stabilizing for about 15 minutes. Water flow rate : 2.0 LPM Air Flow : Maximum Cooling load : 1.0 kW Column installed : A 3. Fill up the make-up tank with distilled water up to zero mark at the level scale, and then start the stop watch. 4. Determine the make-up water supply in an interval of 10 minutes. 5. In this 10 minutes interval, record a few sets of the measurements (i.e. temperatures (T1–T7), orifice differential pressure (DP1), water flowrate (FT1) and Heater Power (Q1)), then obtain the mean value for calculation and analysis. 6. Determine the quantity of make up water that has been supplied during the time interval by noting the height reduction in the make-up tank. 7. The observation may be repeated at different conditions, i.e. at different water flow rates, or different air flow rates and with different load. Assignment: 1. Calculate the make-up rate. 2. Calculate the energy and mass balances by using the steady flow equation.

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4.4

Experiment 3: Investigation of the Effect of Cooling Load on Wet Bulb Approach Objective: To investigate the effect of cooling load on “Wet Bulb Approach” Procedure: 1. Prepare and start the cooling tower with according to Section 4.1.1. 2. Set the system under the following conditions and allow stabilizing for about 15 minutes. Water flow rate : 2.0 LPM Air Flow : Maximum Cooling load : 0 kW Column installed : A 3. After the system stabilizes, record a few sets of measurements (i.e. air inlet dry bulb and wet bulb temperature (T1 and T2), water outlet temperature (T6), orifice differential pressure (DP1), water flowrate (FT1) and Heater Power (Q1)), then obtain the mean value for calculation and analysis. 4. Without changes in the conditions, increase the cooling load to 0.5 kW. When the system stabilized, record all data. 5. Similarly, repeat the experiment at 1.0kW and 1.5kW. 6. Finally, measure the cross sectional area of the column. 7. The four tests may be repeated at another constant airflow. 8. The observation may also be repeated at different conditions, i.e. at different water flow rates, or different air flow rates and with different load. Assignment: 1. Calculate the “wet bulb Approach” and total cooling load. 2. Plot a graph to show that the relationship between cooling load and “approach to wet bulb” temperature.

20


4.5

Experiment 4: Investigation of the Effect of Air Velocity on Wet bulb Approach and Pressure Drop through the Packing Objective: To investigate the effect of air velocity on: (a) Wet Bulb Approach (b) The pressure drop through the packing Procedure: 1. Prepare and start the cooling tower with according to Section 4.1.1. 2. Set the system under the following conditions and allow stabilizing for about 15 minutes. Water flow rate : 2.0 LPM Air flow rate : Maximum Cooling load : 1.0 kW Column installed : A 3. After the system stabilizes, record a few sets of measurements (i.e. temperature (T1-T6), orifice differential pressure (DP1), water flowrate (FT1). heater power (Q1) and pressure drop across packing (DP2)), then obtain the mean value for calculation and analysis. 4. Repeat the test with 3 different sets of orifice pressure drop values (75%, 50% and 25% of the maximum value) without changing the water flow rate and cooling loads. 5. Finally, measure the cross sectional area of the column. 6. The test may be repeated: i. At another constant load ii. At another constant water flow rate Assignment: 1. Calculate the nominal velocity of air and find the “approach to wet bulb”. 2. Plot a graph to show that the relationship between “approach to wet bulb” and packing pressure drop versus nominal air velocity.

21


4.6

Experiment 5: Investigation of the Relationship between Cooling Load and Cooling Range Objective: To investigate the relationship between cooling load and cooling range Procedure: 1. Prepare and start the cooling tower with according to Section 4.1.1. 2. Set the system under the following conditions and allow stabilizing for about 15 minutes: Water flow rate : 2.0 LPM Air flow rate : Maximum Cooling load : 0.0 kW Column installed : A 3. After the system stabilized, record a few sets of measurements (i.e. temperature (T1-T6), orifice differential pressure (DP1), water flowrate (FT1) and heater power (Q1)), then obtain the mean value for calculation and analysis 9. Without changes in the conditions, increase the cooling load to 0.5 kW. When the system stabilized, record all data. 4. Similarly, repeat the experiment at 1.0kW and 1.5kW. 5. The tests may be repeated: i.At other water flow rates ii.At other air flow rate Assignment: 1. Plot a graph to show that the relationship between cooling loads and cooling range.

22


4.7

Experiment 6: Investigation of the Effect of Packing Density on the Performance of the Cooling Tower Objective: To investigate the effect of packing density on the performance of the cooling tower Procedure: 1. Prepare and start the cooling tower with according to Section 4.1.1. 2. Set the system under the following conditions and allow stabilizing for about 15 minutes: Water flow rate : 2.0 LPM Orifice differential : Maximum Cooling load : 1.0kW Column installed : A 3. After the system stabilizes, record a few sets of measurements (i.e. temperature (T1-T6), orifice differential pressure (DP1), water flowrate (FT1). heater power (Q1) and pressure drop across packing (DP2)), then obtain the mean value for calculation and analysis. 4. Without changing condition, change the column packing to column B. When stability is achieved, repeat the observation. 5. Repeat step 4 with column B and C. 6. The tests may be repeated: i. At other water flow rates ii. At other air flow rates Assignment: 1. Calculate the wet bulb temperature approach. 2. Plot a graph to show that the relationship between “approach to wet bulb” and packing density.

23


4.8

Experiment 7: Determination of Characteristic Equation of the Packing Characteristic Column Objectives: To determine the Characteristic Equation of the cooling tower using Packing Characteristic Column Procedures: 1. Install the Packing Characteristic Column (Column E) properly. 2. Prepare and start the cooling tower with according to Section 4.1.1. 3. Set the system under the following conditions and allow stabilizing for about 15 minutes: Water flow rate : 1.5 LPM Orifice differential : Maximum Cooling load : 1.0kW Column installed : E 4. After the system stabilizes, record a few sets of measurements (i.e. temperature (T1-T6 and T8-T15), orifice differential pressure (DP1), water flowrate (FT1) and heater power (Q1)), then obtain the mean value for calculation and analysis. 5. Without changing the air flow rate, and cooling load, change the water flow rate to 2.0 LPM. When stability is achieved, repeat the observation. 6. Similarly, repeat the observations at 2.5 LPM. Assignment: 1. Plot the Column driving Force KaV m 2. Calculate L/G, and w m w m a

24


5.0

REFERENCES

Perry, R.H., Green, D.W. and Maloney, J.O., “Perry’s Chemical Engineering Handbook”, 6th Edition, McGraw Hill, 1984.

25

Appendix A Experimental Data Sheets

Experiment 2: End State Properties of Air and Steady Flow Equations Results: Column Installed Initial water level Final water level Time Interval

: _______ : cm : cm : minutes

Packing Density

m-1

Air Inlet Dry Bulb, T1

˚C

Air Inlet Wet Bulb, T2

˚C

Air Outlet Dry Bulb, T3

˚C

Air Outlet Wet Bulb, T4

˚C

Water Inlet Temperature, T5

˚C

Water Outlet Temperature, T6

˚C

Water Make-Up Temperature, T7

˚C

Orifice Differential, DP1

Pa

Water Flow Rate, FT1

LPM

Heater Power, Q1

Watt

Experiment 3: Investigation of the Effect of Cooling Load on Wet Bulb Approach Results: Column Installed

:

_________

Description

Unit

Packing Density

m-1


˚C


˚C


˚C


Pa


LPM

Heater Power, Q1

Watt

0.0kW

0.5kW

1.0kW

1.5kW

Experiment 4: Investigation of the Effect of Air Velocity on Wet bulb Approach and Pressure Drop through the Packing Results: Column Installed

:

_____

Description

Unit

Packing Density

m-1


˚C


˚C


˚C


˚C


˚C


˚C


Pa


LPM

Heater Power, Q1

Watt

Pressure Drop Across Packing, DP2

Pa

Air Flow 100 %

75 %

50 %

25 %

Experiment 5: Investigation of the Relationship between Cooling Load and Cooling Range Results: Column Installed

:

____

Description

Unit

Packing Density

m-1


˚C


˚C


˚C


˚C


˚C


˚C


Pa


LPM

Heater Power, Q1

Watt

0.0kW

0.5kW

1.0kW

1.5kW

Experiment 6: Investigation of the Effect of Packing Density on the Performance of the Cooling Tower Results:

Description

Unit

Packing Density

m-1


˚C


˚C


˚C


˚C


˚C


˚C


Pa


LPM

Heater Power, Q1

Watt


Pa

Column A

B

C

Experiment 7: Determination of Characteristic Equation of the Packing Characteristic Column Result: Description

Unit

Top Air Outlet Dry Bulb, T3

˚C


˚C


˚C

Station III Air Dry Bulb, T8

˚C


˚C

Water Temperature, T14

˚C

Station II Air Dry Bulb, T10

˚C


˚C


˚C

Station I Air Dry Bulb, T12

˚C


˚C


˚C

Bottom Air Inlet Dry Bulb, T1

˚C


˚C


˚C


Pa


LPM

Heater Power, Q1

Watt

Water Flowrate (LPM) 1.5

2.0

2.5

Appendix B Sample Results and Calculation

Column A

Experiment 2: End State Properties of Air and Steady Flow Equations Results: Column Installed Initial water level Final water level Time Interval

: : : :

A . 0 cm 4.0 cm 10 minutes

Description

Unit

Value

Packing Density

m-1

110


˚C

31.6


˚C

27.1


˚C

30.2


˚C

29.5


˚C

37.4


˚C

30.8

Water Make-Up Temperature, T7

˚C

30.2


Pa

88


LPM

2.0

Heater Power, Q1

Watt

883

Calculation: By plotting the air inlet dry bulb and air inlet wet bulb (point A) and air outlet dry bulb and air outlet wet bulb (point B) on the Psychrometric chart, the enthalpy and humidity of air mixture is obtained and the result is showed as below: hA wA hB wB vaB

= 88.0 kJ/kg = 0.023 kg water/kg dry air = 96.0kJ/kg = 0.0268 kg water/kg dry air = 0.950 m3/(kg dry air)

From the orifice calibration,

x v aB (1 + wb )

m a = 0.0137 x

= 88Pa x 1mm H2O / 10.13Pa = 8.69 mm H2O 8.69 0.950(1 + 0.0268)

m a = 0.0137

m a = 0.0409kgs −1

Water Mass Loss,

mE =

πD 2 (h1 − h2 )ρ 4

Where, D = 74mm (I.D of Make up tank)

ρ

= 1000kg/m3 (Water Density)

and h1, h2 are level water units in cm

mE =

π .74 2 .(40 − 0 ).1000

4.10 9 = 0.172 kg Make-up rate, m m = E kg.s −1 ∆t

0.172 kg.s −1 600 = 2.867 x 10-4 kg.s-1

=

Specific enthalpy of make-up, hE = 126.586kJ/kg (at T7) Applying the Steady Flow Equation to the system indicated by the chain the (System A)

 − P = ∆H  − ∆K E Q Now, solving the left side, Q − P = 883W − (− 65W ) = 948W (Work by pump is estimate –65W) On the right side of the equation, ∆K E ≈ 0 (change of air velocity is negligible)

∆H = ∆H EXIT − ∆H ENTRY

= m a (hB − h A ) − m E hE = [0.0409 (96.0 – 88) – 2.867 x 10-4 x 126.586] kW = 0.2910 kW The discrepancy between the right hand side and left hand side of the equation may be attributed to errors and heat lost from the system. Mass Balance m E = m S B − m S A m E = m a (wB − w A ) m E = 2.867 x10 −4 kg.s −1 m a (wB − w A ) = 0.0409(0.0268 − 0.0230 ) = 1.554 x 10-4 kg.s-1 The values for both sides of the equation differ. This discrepancy may be attributed to leak, and errors.

Experiment 3: Investigation of the Effect of Cooling Load on Wet Bulb Approach Results: Column Installed

:

A

.

Description

Unit

0.0kW

0.5kW

1.0kW

1.5kW

Packing Density

m-1

-

110

110

110


˚C

-

31.5

31.7

31.8


˚C

-

27.1

27.2

27.3


˚C

-

29

30.6

32.1


Pa

-

89

89

89


LPM

-

2.0

2.0

2.0

Heater Power, Q1

Watt

-

471

899

1359

Calculation: The pump transfer approximately 65W to the water, and this should be added to the load imposed in the load tank. For example, when cooling load of 0.5kW was applied, Total cooling load = Applied load + Pump input = (471 + 65) W = 0.536kW “Approach to wet Bulb” = Water Outlet Temperature - Air Inlet Wet Bulb = 29.0 – 27.1 = 1.9 K

Specific volume at outlet = 0.950 m3/kg ( x = 89 Pa x 1mm H2O / 10.13 Pa = 8.786mm H2O ) m a = 0.0137

x vB

8.786 0.950 = 0.0417 kg/s = 0.0137

Cross sectional area of column (A) = 0.15 x 0.15 m2 = 0.0225m2

a m A 0.0417 = kgs −1 m − 2 0.0225 = 1.853kg s-1 m-2

Air mass flow per unit area =

W m A 1kg min ( m w = 2.0 LPM × = 0.03333 kgs-1) × L 60 s Water flow rate per unit area =

 W 0.03333 m = kgs −1 m − 2 A 0.0225 = 1.48 kgs-1m-2

Results: Description

1

2

3

110

110

110

Air Flow per Unit Area, kgs-1m-2

1.853

1.853

1.853

Total Cooling Load, kW

0.536

0.964

1.424

Approach to Wet Bulb, K

1.9

3.6

4.8

Packing Density, m-1

Experiment 4: Investigation of the Effect of Air Velocity on Wet bulb Approach and Pressure Drop through the Packing Results: Column Installed

:

A

.

Description

Unit

Air Flow 100 %

75 %

50 %

25 %

Packing Density

m-1

110

110

110

110


˚C

29.5

29.9

30.2

31.0


˚C

26.3

26.5

26.8

27.2


˚C

29.4

29.8

30.4

31.8


˚C

28.4

28.7

29.5

30.7


˚C

37.8

36.4

37.1

38.8


˚C

30.6

30.1

31.3

32.7


Pa

89

79

47

25


LPM

2.0

2.0

2.0

2.0

Heater Power, Q1

Watt

873

875

868

873

Pa

19

16

10

5

Pressure Drop across Packing, DP2

Calculation: By taking the data obtained when 100% of air flow was employed, Inlet wet bulb temperature = 26.3oC Outlet water temperature = 30.6 oC “Approach to wet bulb” = 30.6 – 26.3 K = 4.3K Specific volume of air at outlet (by plotting Air Outlet Dry bulb and Air Outlet Wet bulb on the Psychometric Chart) = 0.890m 3kg-1 Air mass flow rate = 0.0137

x vB

(x = 89 Pa x 1mm H2O / 10.13 Pa = 8.786 mm H2O)

8.786 0.890 = 0.04049 kgs-1 = 0.0137

Air volume flow rate v v

 vB =m = 0.04049 x 0.893 m3s-1 = 0.0362m3s-1

Cross sectional area of empty tower A Air velocity

v A

0.0362 0.0225 =1.609 ms-1

=

= 0.15 x 0.15 = 0.0225 m2

Results:

Nominal Velocity of Air, ms-1 Approach to Wet Bulb, K Pressure, mm H2O

1

2

3

4

1.7027

1.609

1.2387

0.9064

4.3

3.6

4.5

5.5

0.3949

0.8885

1.4808

1.7769

Experiment 5: Investigation of the Relationship between Cooling Load and Cooling Range Results: Column Installed

:

A

Description

. Unit

0.0kW

0.5kW

1.0kW

1.5kW

Packing Density

m-1

-

110

110

110


˚C

-

30.5

30.8

30.9


˚C

-

27.2

27.3

27.4


˚C

-

28.6

29.3

30.9


˚C

-

28.3

28.7

29.9


˚C

-

32.9

35.4

39.4


˚C

-

29

30

31.4


Pa

-

91

90

90


LPM

-

2.0

2.0

2.0

Heater Power, Q1

Watt

-

455

864

1321

Experiment 6: Investigation of the Effect of Packing Density on the Performance of the Cooling Tower Results:

Description

Unit

Column A

B

C

Packing Density

m-1

110

77

200


˚C

30.1

30.2

31.0


˚C

27.4

27.3

27.5


˚C

29.7

30.1

30.4


˚C

29.4

29.6

29.7


˚C

37.7

38.6

37.4


˚C

31.6

32.3

31.3


Pa

91

97

65


LPM

1.9

1.9

1.9

Heater Power, Q1

Watt

864

895

800

Pa

19

15

90


Results: 1

2

3

Packing Density, m-1

77

110

200

Approach to Wet Bulb, K

5

4.2

3.8

Experiment 7: Determination of Characteristic Equation of the Packing Characteristic Column Results: Description

Unit

Water Flowrate (LPM) 1.5

2.0

2.5

Top Air Outlet Dry Bulb, T3

˚C

29.9

30.3

30.3


˚C

29.1

29.8

29.7


˚C

39.3

38.5

36.2

Air Dry Bulb, T8

˚C

31.6

32.1

31.7


˚C

29.6

30.0

29.8


˚C

35.5

35.6

34.2

Air Dry Bulb, T10

˚C

29.7

30.5

32.5


˚C

28.5

29.2

29.0


˚C

33.3

33.7

33.0

Air Dry Bulb, T12

˚C

28.8

29.1

28.9


˚C

30.0

30.0

30.2


˚C

30.2

31.3

30.7


˚C

31.6

31.7

31.8


˚C

27.1

27.1

27.1


˚C

31.8

32.2

31.5


Pa

50

48

49


LPM

1.5

2.0

2.5

Heater Power, Q1

Watt

864

861

870

Station III

Station II

Station I

Bottom

Calculation: By taking the data obtained when 2.50 LPM of water was used, 1kg min m w = 2.50 LPM × × L 60s = 0.04167 kg s-1 vaB wb

= =

m a = 0.0137

0.8980 m3/(kg dry air) 0.0266 kg water/kg dry air

x v aB (1 + wb )

x = 49 Pa x 1mmH2O / 10.13Pa = 4.84 mmH2O

m a = 0.0314 kgs −1 For Air Operating Line, T1 = Water Outlet temperature, T6 (31.5°C) T2 = Water Inlet temperature, T5 (36.2°C)

∆h1

= value of h w − h a at T2 + 0.1(T1 − T2 )

∆h 2

= value of h w − h a at T2 + 0.4(T1 − T2 )

∆h 3 ∆h 4

= value of h w − h a at T1 − 0.4(T1 − T2 ) = value of h w − h a at T1 − 0.1(T1 − T2 )

T2 + 0.1(T1 − T2 ) = 35.73˚C T2 + 0.4(T1 − T2 )

= 34.32˚C

T1 − 0.4(T1 − T2 )

= 33.38˚C

T1 − 0.1(T1 − T2 )

= 31.97˚C

Therefore, from the graph, ∆h1

= 15.92 kJ/kg

∆h 2

= 14.23 kJ/kg

∆h 3

= 12.93 kJ/kg

∆h 4

= 10.09 kJ/kg

L (h a 2 − h a1 ) = (T2 − T1 ) G

Referring to the Driving Force Diagram Plotted 110.84 − 93.96 36.2 − 31.5 = 3.59

L/G =

KaV T2 − T1  1 1 1 1  = + + + m w 4  ∆h1 ∆h2 ∆h3 ∆h4 36.2 − 31.5 (0.3095) 4 = 0.3637 (Tower Characteristic)

=

′ m w = 1.33 ′ m a

  

Appendix C Components Properties and Diagrams

Appendix D Process Flow Diagram

WATER COOLING TOWER (MODEL: HE152)

Air Out T3

T4 Orifice

T5

Water Distributor V4 V5

V6

Atm

High

Low

DP 1 V3

Test Section (Packed Column)

FT1 Flowmeter

V1 Make-up Tank T1

T2 T7 V2

T6

Air In

V7

T8

Hot Water Tank

Heater TC Air Blower Receiver Tank

Temp Controller

V8 Pump Drain

L6 Cooling Tower.pdf

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