Lab 2-Water Cooling Tower

TABLE OF CONTENTS:

No. Title Page

1.0 Abstract 3

2.0 Introduction 4

3.0 Objectives 5

4.0 Theory 5 - 14

5.0 Apparatus & Materials 15

6.0 Procedures 16

7.0 Results & Calculations 17 – 23

8.0 Discussion 24 – 25

9.0 Conclusion 25

10.0 Recommendations 26

11.0 References 27

12.0 Appendices 28 - 29

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1.0 ABSTRACT

This experiment was conducted to perform energy and mass balance on the cooling

tower system and to observe the effects of one of the process variables on the exit

temperature of water. For water cooling tower experiment, there are several parameters that

can be adjusted to observe its effects on the evaporation of water. The parameters are

temperature and flow rate of water, relative humidity and flow rate of air and cooling load. In

this experiment, we choose the cooling load as variable while water flow rate and flow rate as

constant parameters. The steady flow equations which is energy and mass balances were

employed in order to provide an insight on the amount of energy transferred between phases

under different conditions. The energy transfer calculated from the experiment for cooling

load of 0kW, 0.5kW, 1.0kW and 1.5kW are 0.0175kJ/s, 0.0143kJ/s, 0.107kJ/s and 0.325kJ/s

respectively.

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2.0 INTRODUCTION

In this experiment, SOLTEQ Water Cooling Tower (Model: HE 152) has been used. This

model has been designed to allow engineering students to get familiarized with all the

processes related to industrial force draught cooling tower. The construction, design and

operational characteristic of this model are as same as the modern cooling system, only it is

in much smaller scale. The basic unit can be used with another four types of packing column

(optional) which were designed to facilitate study of water and air conditions at different

situation. The difference is mainly on the surface contact area of the water with the air in

different column and also the objective of the experiment.

SOLTEQ Water Cooling Tower (Model: HE 152) is supplied with a packed column having

packing density of approximately 110 m2/m3. The unit mainly consists of a load tank with a

total of 1.5 kW electric heater, an air distribution chamber, a make-up tank and a test column.

Warm water is pumped from the load tanks to the top of the column before being uniformly

distributed over the top packing. The thin film of water is cooled, as it passes downward, due

to evaporation. The cooled water falls into the basin before going back into the load tank

where it is re-heated and re-circulated. The load tank water level is maintained by means of a

make-up tank. A blower is installed on the unit to deliver air into the air distribution chamber.

The air passes a wet and dry bulb thermometer before entering the column. As the air passes

up the column, its moisture content increases and the water is cooled. At the top of the

column, the air passes a mist eliminator before being discharged to the atmosphere.

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

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3.0 OBJECTIVES

The objectives of this experiment are:

1. To study the effects of heater power on the exit temperature of water.

2. To perform energy and mass balance on the cooling tower system.

4.0 THEORY

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. Convection and conduction 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 a liquid

form which the energy required is taken.

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 pressure 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”.

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For a cooling tower of infinite size and with adequate air flow, the water leaving will be at

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

effectiveness of the cooling tower, is the difference between the temperature of water leaving

the tower and the wet bulb temperature of the entering air.

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. Water flow rates

ii. Water temperatures

iii. Airflow rate

iv. Inlet air relative humidity

The effects of these factors will be studied in depth by varying it. In this way, students will

gain and overall view of the operation of the cooling tower.

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 0.01oC), the specific enthalpy of saturated water is

assumed to be zero, which is taken as datum. The specific enthalpy of saturated water (hf) at a

range of temperature above the datum conditions can be obtained from thermodynamic

tables.

The specific enthalpy of compressed liquid is given by

(1)

The correction for pressure is negligible for the operating condition of the cooling tower,

therefore we can see that h hf at a given temperature.

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Specific heat capacity (Cp) is defined as the rate of change of the enthalpy with respect to

temperature (often called the specific heat transfer at constant pressure). For the purpose of

the experiment using bench top cooling water, we may use the following relationship:

(2)

and

(3)

where Cp = 4.18 kJkg-1

Dalton’s and Gibbs Laws

It is commonly known that air consists of a mixture of “dry air” (O2 , N2 and other gases) and

water vapour. Dalton and Gibbs law describes the behaviour of such mixtures as:

a) The total pressure of the air is equal to the sum of then 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 same temperature.

The Absolute or Specific Humidity is defined as follows:

(4)

The Relative Humidity is defined as follows:

(5)

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The Percentage Saturation is defined as follows:

(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 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.

Psychometric Chart

The psychometric chart is very useful in 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 properties 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.

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/vapour mixture is given by:

(7)

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The mass flow rate of air and steam mixture through the orifice is given by

(8)

Where,

= Mass flow rate of air/vapour mixture

= Actual specific volume end

x = Orifice differential in mmH2O

Thus,

(9)

The mass flow rate of dry air,

(10)

A simplification can be made since in this application, the value of ω is unlikely to exceed

0.025. as such, neglecting would not yield significant error.

Application of Steady Flow Energy Equation

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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) Heat transfer at the load tank and possibly a small quantity to surroundings.

b) Work transfer at the pump.

c) Low humidity air enters at point A.

d) High humidity air leaves at point B.

e) Make-up enters at point E, the same amount as the moisture increase in the air stream.

From the steady flow equation,

(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)

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Note:

a) The mass flow rate of the air 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 can usually be neglected since its value is relatively small.

Under steady state conditions, by conservation of 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,

(13)

and

or

(14)

The ratio of steam to air (ω) is known for the initial and final state points on the psychometric

charts. Therefore,

and (15)

(16)

Therefore,

(17)

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Say, we redefine the cooling tower system 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.

Again from steady flow energy equation,

and

P=0

may have a small value due to heat transfer between the unit and its surroundings.

(18)

Rearranging,

(19)

Again, the term can be neglected.

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:

(20)

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where

Cpw = Specific heat capacity of water

mw = Mass flow rate of water per unit plan area of packing

T = Water temperature

K = Mass transfer coefficient

a = Area of contact between air and water per unit volume of packing

V = Volume occupied by packing per unit plan area

∆h = Difference in specific enthalpy, between the saturated boundary layer and the bulk

air

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,

(21)

Integration upper equation,

(22)

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The numerical solution to the integral expression upper equation using Chebyshev numerical

method gives,

(23)

Where

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(ha2-ha1) (24)

or,

(25)

Where,

L/G = Liquid to gas mass flow ratio

T1 = Cold water temperature

T2 = Hot water temperature

ha2 = Enthalpy of air-water vapour mixture at exhaust wet-bulb temperature

ha1 = Enthalpy of air-water vapour mixture at inlet wet-bulb temperature

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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 temperature. Although the area ABCD should remain constant, it

actually decreases about 2% for every 10 °F increase in hot water temperature above 100°F.

To account for this decrease, an “adjusted hot water temperature” is used in cooling tower

design.

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5.0 APPARATUS AND MATERIALS

SOLTEQ Bench Top Cooling Tower Unit (Model: HE152)

1. Orifice

2. Water Distributor

3. Packing Column

4. Flow Meter

5. Receiver Tank

6. Air Blower

7. Make-up Tank

8. Differential Pressure Transmitter

9. Load Tank

10. Control Panel

Stopwatch

Deionised water

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6.0 PROCEDURES

1. Valves V1 to V6 were checked and ensured to be closed and valve 7 to be partially

opened.

2. The load tank was filled with deionised water. Firstly, the make-up tank was removed and

deionised water was poured through the opening at the top of the load tank. The make-up

tank was replaced onto the load tank and the nuts were lightly tightened. Then, the tank

was filled with deionised water up to the zero mark n the scale.

3. Deionised water was added to the wet bulb sensor reservoir to the fullest.

4. All appropriate tubing was connected to the differential pressure sensor.

5. The appropriate cooling tower packing was installed for the experiment.

6. Temperature set point of temperature controller was set to . The 1.0kW water heater

was switched on and water was heated up until approximately .

7. The pump was switched on and the control valve V1 was slowly opened and the water

flow rate was set to 2.0LPM.

8. The damper was fully opened and the fan was switched on.

9. Blower switch was switched on after the water already went through the cooling tower.

10. The unit was run for 20 minutes to ensure float valve correctly adjusted the level in the

load tank. The makeup tank was refilled as required.

11. The damper and the flow rate were set to be constant.

12. The 1.0kW water heater was switched off to set the power as 0kW.

13. Record all the data required after 10 minutes to ensure the unit stabilized for first trial and

another 10 minutes for second trial.

14. To measure the differential pressure across the orifice, valves V4 and V5 were opened

while valves V3 and V6 were closed.

15. To measure the differential pressure across the column, valves V3 and V6 were opened

while valves V4 and V5 were closed.

16. The water heater then was set to 0.5kW, 1.0kW and 1.5kW.

17. After all the experiments were done, the heaters were switched off and the water was let

to circulate through the cooling tower system for 3-5 minutes until the water cooled

down.

18. The fan was switched off and the fan damper was closed fully.

19. The pump and power supply were switched off.

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20. The water was completely drained from the unit.

7.0 RESULTS AND CALCULATIONS

DESCRIPTION UNIT0 0.5 1.0 1.5

TRIAL 1 2 1 2 1 2 1 2Air Inlet Dry Bulb, T1 oC 29.1 28.9 28.7 28.7 29.2 29.2 29.4 29.5

Air Inlet Wet Bulb, T2

oC 26.8 26.6 26.4 26.2 26.3 26.2 26.1 26.3

Air Outlet Dry Bulb, T3

oC 28.2 26.2 26.9 27.5 29.3 30.6 33.0 34.3

Air Outlet Wet Bulb, T4

oC 28.0 27.0 26.7 26.9 27.9 29.3 31.2 32.9

Water Inlet Temperature, T5

oC 31.5 28.4 30.8 31.9 35.3 37.2 41.8 43.3

Water Outlet Temperature, T6

oC 29.4 27.0 28.1 28.6 30.8 31.9 34.4 35.5

Orifice Differential DP1

Pa 5 5 5 6 5 5 5 5

Water Flow Rate, FT1

LPM 2 2 2 2 2 2 2 2

Heater Power, Q1 W 0 0 424 422 794 795 1200 1209

Pressure Drop Across Packing, DP2

Pa 24 14 8 12 30 18 25 22

Packing Density m-1 200

Overall dimension:

Height: 1.25 m

Width: 0.91 m

Depth: 0.45 m

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By taking the data obtained when 0% air flow fully closed and 0% heater power:

First trial

Inlet air wet bulb T = 26.8 oC

Outlet water temperature =29.4 oC

Approach to wet bulb =2.6 oC

Specific volume of air at outlet (by plotting air outlet dry bulb and air outlet wet bulb on the Psychometric Chart) = 0.853 m3kg-1

Air mass flow rate =

(x= 5 Pa x 1 mmH2O/9.8 Pa) = 0.51 mmH2O

ṁ =

= 0.0106 kgs-1

Air volumetric flow rate =

= 0.0106 x 0.853

= 0.00904 m3s-1

Cross sectional area of empty tower A = 0.15 x 0.15

= 0.0225 m2

Air Velocity =

=

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= 0.4018 ms-1

DESCRIPTION UNIT HEATER POWER (kW)

0 0.5 1.0 1.5TRIAL 1 2 1 2 1 2 1 2

Approach to wet bulb

K 2.6 0.4 1.7 2.4 4.5 5.7 8.3 9.2

Specific volume of air at outlet

m3/kg 0.853 0.85 0.85 0.852 0.855 0.859 0.865 0.87

X mm H2O

0.51 0.51 0.51 0.612 0.51 0.51 0.51 0.51

ṁ kg/s 0.0106 0.0106 0.0106 0.0116 0.0106 0.01056 0.01052 0.0105m3/s 0.00904 0.00902 0.00902 0.00989 0.00947 0.00907 0.0091 0.00913

Cross sectional area

m2 0.0225

Air velocity m/s 0.4018 0.4089 0.4089 0.439 0.402 0.403 0.405 0.406

For heater power 0:

From psychometric chart:

H2 = 0.0233 kg water/kg air, H1 = 0.022 kg water/kg air

L2 = 2(kg/min)/1 min/60 s Þ L2 = 0.3 kg/sec

* Water mass balance:

L1 = 0.03 - (0.0106)*(0.0233 – 0.022)

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L1 = 0.0299kg water /sec

* Energy balance:

HY1 = (1.005 + 1.88*0.022)*(26.7) + 2501.4*0.022

HY1 = 82.969 KJ/kg air

HY2 = (1.005+0.0233*1.88)*(27.5) + (2501.4*0.0223)

HY2 = 84.623 kJ/kg air

Q = 0.0106*(84.623– 82.969) Þ Q = 0.0175 kJ/sec

For heater power 0.5 kW:


H2 = 0.0223 kg water/kg air, H1 = 0.022kg water/kg air



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L1 = 0.03 - (0.0106)*(0.0223 – 0.022)

L1 = 0.03 kg water /sec

* Energy balance:

HY1 = (1.005 + 1.88*0.022)*(26.3) + 2501.4*0.022


HY2 = (1.005+0.0223*1.88)*(26.8) + (2501.4*0.0223)


Q = 0.0111*(83.839– 82.55) Þ Q = 0.0143 kJ/sec






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L1 = 0.03 - (0.0106)*(0.025– 0.022)

L1 = 0.03kg water /sec

* Energy balance:

HY1 = (1.005 + 1.88*0.022)*(26.25) + 2501.4*0.022


HY2 =(1.005+0.025*1.88)*(28.6) + (2501.4*0.025)


Q = 0.0106*(92.622– 82.498) Þ Q = 0.107 kJ/sec






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L1 = 0.03 - (0.0106)*(0.0314 – 0.0217)

L1 = 0.03 kg water /sec

* Energy balance:

HY1 = (1.005 + 1.88*0.0217)*(26.2) + 2501.4*0.0217


HY2 = (1.005+0.0314*1.88)*(32.05) + (2501.4*0.0314)

HY2 = 112.646 kJ/kg air

Q = 0.0105*(112.646– 81.68) Þ Q = 0.325 kJ/sec

Description Heater Power (kW)

0 0.5 1.0 1.5

Energy Transfer,

Q (kJ/s)

0.0175 0.0143 0.107 0.325

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8.0 DISCUSSION

SOLTEQ Water Cooling Tower (Model: HE 152) is used to run this experiment. This

experiment is conducted to study the effects of variables changes on the temperature of the

product (water). There are three variables that will affect the water temperature which are

heater power, water flow rate, and the blower. For this experiment we only focused on the

heater power as the changing variable, meanwhile the water flow rate and the blower were set

as constant. For every change in heater power energy, the time interval for the process to

occur is 10 minutes. This is to ensure that the system is in steady state operation.

Heat is transferred from a body with a higher temperature to lower temperature. As

for this cooling tower experiment, we need to produce a product which has a lower

temperature (cooler) than the media being used for the transfer of heat in the system. This is

the difference between heat exchanger and cooling tower application. The concept is the

same, but the product that will be produced is different although both products will have a

reduction in temperature. For the heat exchanger; the temperature of the outlet cooled water

cannot be lower than the temperature of the cooling air. Whereas for the water cooling tower;

the temperature of the outlet water can be lower than the temperature of the cooling air. The

cooling of the hot water was in the form of forced convection by which ambient air was

blown over the hot water.

In the cooling tower, water is cooled by the process known as evaporation. In the

process, heat energy is being transferred between the water and air which having different

temperature. As the energy in the water molecules is transfer to the air flowing through the

water, the bond of the water molecules becomes weaker then it will slowly evaporate to the

air. It can be prove from the result of this experiment, whereby the water outlet temperature is

lower than the inlet. For example: Water outlet = 35.5 , Water inlet = 43.3 . The tower

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inner structure is filled with “fill” that is arranged in the form of labyrinth. The fill is used in

order to increase the time contact of the water and the air flowing in the system.

From the data obtained at the end of the experiment, changes in heater power will

affect the energy being transferred within the system and surrounding. Larger heater power

will cause higher temperature in water. As for this experiment, the temperature of the water is

assumed higher than the air flowing through the system (without considering the changes in

heater power). As the difference in the temperature between the water and air increase, the

heat energy being transferred rate is therefore higher than the lowered heater power. It can be

proved by the data obtained where for 0kW heater power, the heat energy transferred rate is

0.0175kJ/s, then followed by 0.5kW (0.0143kJ/s), 1.0kW (0.107kJ/s), 1.5kW (0.325kJ/s).

Therefore, it can be said that heater power is directly proportional to heat energy transfer rate.

9.0 CONCLUSION

This experiment is conducted to study the effects of heater power on the exit

temperature of water and to perform energy and mass balance on the cooling tower system.

From the data obtained, we have calculated that the rate of heat energy transferred for 0kW

heater power is 0.0175kJ/s, then followed by 0.5kW (0.0143kJ/s), 1.0kW (0.107kJ/s), 1.5kW

(0.325kJ/s). As the difference in temperature between the water and air increase, the rate of

heat energy being transferred is higher than the lowered heater power. Therefore, it can be

said that heater power is directly proportional to heat energy transfer rate. Hence, this

experiment has been a success since the objectives of the experiment were achieved and we

have gained knowledge regarding to the water cooling tower operation.

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10.0 RECOMMENDATIONS

There are several recommendations that can be deduced from this experiment to

improve the results of this experiment. The water that can be used in the unit only deionised

water or distilled water. When using tap water in the system, the impurities that exist in the

tap water may cause the depositing in the cover tower and thus lower the efficiency of the

cooling tower.

When installing the cooling tower column or make-up tank, make sure that the nuts

were not tighten too hard because it may cause crack. Also not too tight lightly which can

cause spilling of water out from the column.

Other than that, ensure that the pressure tubings for differential pressure measurement

are connected correctly which orifice pressure tapping point to V4, column’s lower pressure

tapping to V6, column’s higher pressure tapping point to V3 and V5 leave to atmosphere.

Also, make sure that there is no water in the pressure tubing for accurate differential pressure

measurement.

Besides that, after completed the experiment, ensure that that the water inside the

tower was cooled down first before draining it off because if water still hot when draining off

the water, the unit’s efficiency will decreased.

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11.0 REFERENCES

1. Richard M. Felder, Ronald W. Rousseau. (2005). Elementary Principles of

Chemical Process, 3rd Ed. John Wiley & Sons, Inc.

2. Frank P. Incropera, David P. Dewitt, Theodore L. Bergman, Adrienne S. Lavine.

2007. Fundamentals of Heat and Mass Transfer, 6th Ed. John Wiley & Sons (Asia)

Pte Ltd.

3. Cooling Technology Institute. (2012). Retrieved from

http://www.cti.org/whatis/coolingtowerdetail.shtml

4. .docstoc. (1/12/2010). Retrieved from

http://www.docstoc.com/docs/22066533/WATER-COOLING-TOWER

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http://www.docstoc.com/docs/22066533/WATER-COOLING-TOWER

http://www.cti.org/whatis/coolingtowerdetail.shtml

12.0 APPENDICES

1. Basic water cooling tower flow diagram

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2. The Psychometric Chart

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Lab 2-Water Cooling Tower

Documents

john wiley

022kg water kg air

air distribution chamber

lowered heater power

mass flow rate

wet bulb temperature

dry bulb temperature

water flow rate