t9 Cooling Tower

KNJ 2251 Mechanical and Manufacturing Engineering Laboratory 3Faculty of EngineeringUniversity Malaysia Sarawak

TABLE OF CONTENT

Content Page

Title 2

Theory / Introduction 3

Experiment 9.1

- Objective

- Procedure

- Result and Discussion

- Conclusion

16

16

16

17

19

Experiment 9.2

- Objective

- Procedure


- Conclusion

20

20

20

21

26

Experiment 9.3

- Objective

- Procedure


- Conclusion

26

26

26

27

29

Overall Summary 29

1


TITLE:

T9:

BENCH TOP COOLING

TOWER

2


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. Conduction and convection depend on the temperature

difference, the surface area, air velocity, etc. the effect 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

form which the required energy 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 vapor pressure at the liquid surface and the vapor

pressure in the surrounding air. The vapor 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 vapor pressure in the surrounding air. Such

evaporation process in an enclosed space shall continue until the two vapor 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

higher temperature. Under adiabatic conditions, this equilibrium temperature is the “wet bulb

temperature”.

For 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 towers. Thus, “Approach to Wet Bulb”, an

important parameter of cooling towers, is the difference between the temperature of water

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

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). Water flow rates

(ii). Water temperatures

(iii). Airflow rate

(iv). Inlet air relative humidity

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.

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 the

thermodynamic properties are presented below.

At the triple point (i.e. 0.00602 atm and 0.019 C), the 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

temperatures above the datum condition can be obtained from thermodynamic tables. The

specific enthalpy of compressed liquid is given by

h = hf + vf ( p – psat ) (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.

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

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

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

∆h = Cp∆T (2)

And

h = CpT (3)

Where Cp = 4.18kJ.kg-1

4


Dalton’s and Gibbs Law

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

and water vapor. Dalton and Gibbs law describes the behavior 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 same temperature.

The Absolute or Specific Humidity is defined as follows:

Specific Humidity ,ω=mass of water vapormass of dry air

(4)

The Relative Humidity is defined as follows:

Relative Humidity ,ϕ= ∂ pressure of water vapour∈t heairsaturation pressure of water vapour at t h e same temperature

(5)

The Percentage Saturation is defined as follows:

Percentage Saturation= mass of water va pour∈a given volume f airmass of same volumeof saturated water vapour at the same temperature

(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.

5


Psychometric Chart

The psychometric chart is very useful in determining the properties of air/water vapor

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.

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 1kg of dry air the stated total

pressure. However, for every 1kg of dry air, there is ω kg of water vapour, yielding the total

mass of 1 + ω kg. Therefore, the actual specific volume of the air/vapour mixture is given by:

va=vab

1+ω(7)

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

m=0 . 0137√ Xva

(8)

Where,

m=mass flow rate of air /vapour mixture

va=actual specific volume

X=orifice differential∈mmH2 O

6


Thus,

m=0.0137 √ X (1+ϖ )vab

(9)

The mass flow rate of dry air,

ma=1

1+ϖ×mass flow rate ofair /vapour mixture (0)

ma=1

1+ϖ×0.0137√ X (1+ϖ )

v ab

(0)

ma=0.0137√ Xv ab

(1+ϖ ) (0)

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

0.025. As such, neglecting wb would not yield significant error.

Application of Steady Flow Energy Equation

Consider System A for the cooling tower defined as in Figure 1. It can be seen that the 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 enters at point B

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

7


Figure 2: System A

From the steady flow equation,

Q – P = H exit - H entry

Q−P=(ma hda+m shs)B – (ma hda+ms hs)A– mE hE (11)

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:

Q – P = ma (hB – hA) – mE hE (12)

Under steady state conditions, by conservation of mass, the mass flow rate of the dry air and

of water (as liquid or vapor) must be the same at inlet and outlet to any system. Therefore,

(ma)A = (ma)B

8


And

(ms) A + mE=¿ (ms) B or

mE ¿ (ms) B – (ms) A (14)

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

charts. Therefore,

(ms) A = ma ϖ A and (15)

(ms ¿ B = ma ϖB (16)

Therefore,

mE = ma(ϖ B– ϖ A) (17)

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.

Figure 3: system B

Again from the steady flow energy equation,

Q−P= ˙H exit− ˙H entry and

P=0

9


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

Q=mahB+mw hD−(mah A ˙+mw hC ˙+mE hE) (18)

Rearranging

Q= ˙ma(¿hB−hA)+mw(hD−hC) ˙−mE hE ¿

Q= ˙ma(¿hB−hA)+mw Cp( tD−tC) ˙−mE hE ¿ (19)

The term mE hE 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 3, the energy balances of water and air stream in the tower are related to the

mass transfer by the following equation:

CPw mw dT=KadV (∆ h) (20)

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.

10


Figure 4: schematic representation of the air and water stream 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,

KaVmW

=CPw dT

∆ h (21)

Integrating equation 21

KaVmW

=CPwKaVmW

=CPw∫T 1

T 2

dThW−ha

(22)

The numerical solution to the integral expression equation 22 using Chebyshev numerical

method gives,

KaVmW

=CPw∫T 1

T 2

dThW−ha

=¿T 2−T 1

4 [ 1∆ h1

+ 1∆ h2

+ 1∆ h3

+ 1∆ h4 ]¿ (23)

Where;

KaVmW

=tower characteristic

∆ h1=value of hW−ha at T 2+0.1(T1−T2)

∆ h2=value of hW−ha at T 2+0.4(T 1−T 2)

∆ h3=value of hW−ha at T 1+0.4(T 1−T 2)

11


∆ h4=value of hW−ha at T 1+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(ha 2−ha 1) (24)

Or

LG

=(ha 2−ha 1)

(T 2−T 1 ) (25)

Where,

LG

=liquid to gas mass flow ratio

T 1=cold water temperature

T 2=hot water temperature

ha2=enthalpy of air-water vapor mixture at exhaust wet-bulb temperature

ha1=enthalpy of air-water vapor mixture at inlet wet-bulb temperature

12


Figure 5: graphical representation of tower characteristics

The following represent 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 4. Increasing heat load would have the following effects on the

diagram in figure 4:

1. Increase in the length of line CD and a CD line shift to the right

2. Increase in hot and cold water temperature

3. Increase in range and approach areas

The increase 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˚F increase in hot water temperature above 100˚F. to account

for this decrease, an “adjust hot water temperature” is used in cooling tower design.

Useful information

1. Orifice Calibration Formula

Mass flow rate of air and vapour mixture,

m=0.0137 √ x (1+ϖ )vab

The mass flow rate of dry air,

m=0.0137 √ vab

x (1+ϖ )

Where,

X=orifice differential in mm H 2 o

vab=specific volume of air at the outlet

ϖ=humidity ratio of the mixture

13


2. Pump work input=80W (0.08kW)

3. Column inner dimension= 150 mm×150mm×600mm

14


APPARATUS:

1 Orifice 6 Air blower2 Water

distributor7 Differential pressure

transmitter3 Packed column 8 Make-up tank4 Flow meter 9 Control panel5 Receiver tank 10 Load tank

15

1

2

3

4

5

10

8

9

6

7


SAFETY PRECAUTION/START-UP/ SHUT-DOWN PROCEDURES:

General start-up procedure:

1. The valves are checked properly to ensure that the valves V1 to V6 are closed and V7

is partially opened.

2. The load tank is filled 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. The make-up tank is replaced onto the load tank and lightly the nuts. The

tank is filled with distilled or deionised water.

3. The distilled water or deionised water is added to the wet bulb sensor reservoir to the

fullest.

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

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

6. Then, the temperature is set to the set point of temperature controller to 50˚C. The

1.0kW water heater is switched on and the water is heated up until approximately

40˚C.

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

flow rate is set to 2.0 LPM. A steady operation is obtained where the water is

distributed and flowing uniformly through the packing.

8. The fan damper is fully opened, and then the fan is switched on. The differential

pressure sensor is checked whether it is giving reading when the valve manifold is

switched to measure the orifice differential pressure.

9. For about 20 minute, the unit is left to run for the float valve to correctly adjust the

level in the load tank. The makeup tank is refilled as required.

10. The unit is ready or be use.

General shut-down procedure:

1. The heater is switched off and let the water to circulate through the cooling tower

system for 3-5 minutes until the water cooled down.

2. The fan is switched off and fully closes the fan damper.

3. The pump and power supply is switch off.

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.

16


T9.1: END STATE PROPERTIES OF AIR AND STEADY FLOW EQUATIONS

OBJECTIVES:

1. To determined the “end state” properties of air and water from tables or charts.

2. To determined Energy and mass balances using the steady flow equation on the

selected systems.

PROCEDURES:

1. The systems under the following conditions are set and stabilizing is allowed for

about 15 minutes.

Water flow rate : 2.0 LPM

Air flow rate : Maximum

Cooling load : 1.0 kW

2. The make-up tank is filled up with distilled water up, the initial water level is

recorded and then the stop watch are started.

3. The make-up water supplies are determined in an interval of 10 minutes.

4. In this 10 minutes interval, a few sets of the measurements (i.e. temperatures (TI-T7),

orifice differential pressure (DP1), water flow rate (FT1) and Heater Power (Q1)) are

recorded, and then the mean value for calculation and analysis are obtained.

5. The quantity of makeup water that has been supplied during the time interval are

determined by noting the height reduction in the make-up tank.

6. The observation at different conditions may be repeated, i.e. at different water flow

rates, or different air flow rates and with different load.

17


RESULT AND DISCUSSIONS:

Initial water level : __20___cm

Final water level :___11.9__cm

Time interval :___10___cm

Packing Density m-1 110

Air Inlet Dry Bulb, T1 ˚C 22.1

Air Inlet Wet Bulb, T2 ˚C 22.0

Air Outlet Dry Bulb, T3 ˚C 23.3

Air Outlet Wet Bulb, T4 ˚C 22.6

Water Inlet Temperatures, T5 ˚C 30.9

Water Outlet Temperatures, T6 ˚C 22.4

Water Make-Up Temperature, T7 ˚C 22.9

Orifice Differential, DP1 Pa 118.75

Water Flow Rate, FT1 LPM 1.6

Heater Power, Q1 Watt 939

Calculation:

1. The make-up rate

Water flow rate = 1.6 LPM

1.6 LPM = 0.0016 m3 per minute = 2.67 x 10-5 m3/sec

Make-up rate,

= FT1 x density of water

= 2.67 x 10-5 m3/sec x 1000 kg/m3

= 0.0267 kg/sec

18


2. The energy and mass balance

Orifice differential in mm

φ= pρg

φ= 118.751000 x 9.81

=0.012105 m=12.105 mm

From the psychometrics chart : Vab= 0.864

Mass flow rate of air and vapour mixture,

m=0.0137 √ x (1+ϖ )vab

m=0.0137 √ 12.105(1+0.018)0.864

mw=0.05174 kg /s

The mass flow rate of dry air

m=0.0137 √ xvab

(1+ϖ )

m=0.0137 √ 12.1050.864 (1+0.018)

ma=0.05082 kg/s

The energy balance equation:

Q−P=H exit−H entry∧¿

P=0

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

Q=mahB+mw hD−(mah A+ma hC+me hE)

19


¿ ma (hB−hA )+mw C P (t D−tC )−mE hE

(The term mE hE can be neglected)

˙Q=ma (hB−hA )+mw CP (tD−tC )

= 0.05082kg/s (70.9-66.1) kJ/kg dry air + (0.05174kg/s)( 0.4615kJ/kg.C°)(22.4-30.9°C)

= 0.041 kJ/s

DISCUSSION:

(Esther Frederick Uyo - 26237)

From the result obtained above, the water flow rate is remained constant at 1.6 LPM

throughout the experiment. The result obtained has slightly error due to the human error

when handling the cooling tower machine. From the data above, we use the psychometric

chart to find enthalpy at given temperature. Hence, we use the energy and mass balance

equation to find the energy done of the system per seconds.

(Mohd Nur Hilmi bin Sahlan – 27011)

Based on the result, we know that the water flow rate is constant at 1.6 LPM or 2.67 x 10 -5

m3/s. We also obtain the energy flow rate by using the energy and mass balance equation and

the enrgy flow rate is Q = 0.041 kJ/s. In this result also we get the water level drop from 20

cm to 11.9 cm in 10 minutes. But the result has some error due to human error in reading and

recording the temperature value.

CONCLUSION:

In this experiment, we are asked to find the “end state” properties of air and water from the

tables and psychometric chart. From the data above, we can determine the energy and mass

balance using the steady flow equation. Q=0.041kJs

.

20


T9.2: 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

PROCEDURES:

1. Systems were set under following conditions and stabilizing is allowed for about 15

minutes.

Water flow rate: 2.0 LPM

Air flow: Maximum

Cooling load: 1.0 kW

2. A few sets of measurement (i.e. temperature (T1-T6), orifice differential pressure

(DP1), water flow rate (FT1). Heater Power (Q1) and pressure drop across packing

(DP2)) is recorded after the system stabilizes, then the mean value for calculation

analysis obtained.

3. The test repeated 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.

4. The cross sectional area of the column is measured.

5. The test may repeated:

a. At another constant load

b. At another constant water flow rate

21



Description UnitAir Flow

100% 75% 50% 25%

Packing Density m-1 110 110 110 110

Air Inlet Dry Bulb, T1 oC 22.1 22.4 22.4 22.6

Air Inlet Wet Bulb, T2 oC 22.1 22.2 22.3 22.4

Air Oulet Dry Bulb, T3 oC 23.5 23.6 23.6 24.1

Air Outlet Wet Bulb, T4 oC 23.2 23.2 23.4 23.6

Water Inlet Temperature, T5 oC 30.8 30.2 30.3 30.5

Water Outlet Temperature, T6 oC 22.8 23.1 23.3 24.3

Orifice Differential, DPI

(when V4 and V5 open)Pa 118.75 89.00 59.38 29.69

Water Flow Rate, FT1 LPM 1.6 1.6 1.6 1.6

Heater Power, Q1 Watt 913 911 913 912

Pressure Drop Across Packing,

DP2 (when V3 and V6 open)Pa 43.75 31.25 25.00 12.50

Calculations:

1. Calculate the normal velocity of air and “approach to wet bulb”.

By taking the data obtained when 100% of air flow was employed:

Air inlet wet bulb = 22.1oC

Water outlet temperature = 22.8oC

Approach to wet bulb = 22.8oC – 22.1oC = 0.7 oC

Air Outlet Dry Bulb = 23.5oC

Specific volume of air at outlet by placing Air Outlet Dry Bulb and Attribute Psychometric

Chart = 0.867 m3/kg

22


(x = 118.75 Pa X 1 mmH2O / 10.13mmH2O = 11.72 Pa)

Mass flow rate, m=0 . 0137√ Xva

¿0 .0137√ 11 . 720.867

= 0.05037 kgs-1

Air Volume Flow Rate = m va

= (0.05037 kgs-1)(0.867 m3/kg)

= 0.04367 m3/s

Cross Sectional Area of empty tower A = 0.15 X 0.15 = 0.0225 m2

Air Velocity, VA

=0.043670.0225

=1.94ms

For 75% of air flow was employed, Va= 0.865 m3/kg

(x = 89 Pa X 1 mmH2O / 10.13mmH2O = 8.78 Pa)

Mass flow rate, m=0 .0137√ Xva

¿0 .0137√ 8.780.865

= 0.04365 kgs-1


= (0.04365 kgs-1)(0.865 m3/kg)

= 0.03776 m3/s


23


Air Velocity, VA

=0.037760.0225

=1.68ms


(x = 59.38 Pa X 1 mmH2O / 10.13mmH2O = 5.86 Pa)

Mass flow rate, m=0 . 0137√ Xva

¿0 .0137√ 5.860.864

= 0.03568 kgs-1


= (0.03568 kgs-1)(0.864 m3/kg)

= 0.03083 m3/s


Air Velocity, VA

=0.030830.0225

=1.37ms


(x = 29.69 Pa X 1 mmH2O / 10.13mmH2O = 2.93Pa)

Mass flow rate, m=0 .0137√ Xva

¿0 .0137√ 2.930.868

= 0.02517 kgs-1


= (0.02517 kgs-1)(0.868 m3/kg)

24


= 0.02185 m3/s


Air Velocity, VA

=0.021850.0225

=0.97ms

Air Flow 100% 75% 50% 25%

Nominal Velocity, m/s 1.94 1.68 1.37 0.97

Approach to Wet Bulb, K 273.7 273.9 274.0 274.9

Packing Pressure Drop , Pa 43.75 31.25 25.00 12.50

Graph 2.1

0.8 1 1.2 1.4 1.6 1.8 2273

273.2

273.4

273.6

273.8

274

274.2

274.4

274.6

274.8275

273.7

273.9274

274.9

Approach to wet bulb against norminal veloc-ity

Norminal velocity (m/s)

Appr

oach

to w

et b

ulb

(K)

25


Graph 2.2

0.8 1 1.2 1.4 1.6 1.8 205

101520253035404550

43.75

31.25

25

12.5

Packing pressure drop against norminal ve-locity

Y-Values

Norminal velocity (m/s)

Pack

ing

Pres

sure

Dro

p (P

a)

DISCUSSION:

(Esther Frederick Uyo – 26237)

From this experiment, we can find out the influence of air flow with the temperature of the air

and water and the pressure drop across packaging. The air flow is represented by the nominal

velocity. The pressure drop across the packing increases with the increasing in the nominal

velocity of the air. However, there is some error of value in the temperature obtained due to

the human error. This causing there are some errors result obtained in the wet bulb approach.


Based on the graph, we know that the nominal velocity or air flow velocity will affect the

temperature on wet bulb approach and the pressure drop through the packing. The higher the

air flow velocity, the temperature will drop and the pressure will increase. In this result also

we did have slightly error in recording the temperature that will affect the temperature in wet

bulb approach due to human error.

26


CONCLUSION:

In this experiment, we are to find the effect of air velocity to the wet bulb approach and the

pressure drop through the packing. When the air flow is high, the greater the pressure drops

through the packing. While the wet bulb approach decreases with the increasing in the

nominal velocity of air.

27


T9.3: INVESTIGATING OF THE RELATIONSHIP BETWEEN COOLING LOAD AND COOLING RANGE

OBJECTIVE:

To investigate the relationship between cooling load and cooling range

PROCEDURE:

1. The system is set under the following conditions and allowed stabilizing for about 15

minutes:

Water flow rate : 2.0 LPM

Air flow rate : Maximum

Cooling load : 0.0kW

2. After the system stabilized, a few sets of measurements [i.e. temperature (T1-T6),

orifice differential pressure (DPI), water flow rate (FT1) and heater power (Q1)] are

recorded, the mean value then obtained for calculation and analysis.

3. Without changes in the condition, the cooling load increased to 0.5kW. When the

system stabilized, all data are recorded.

4. The experiment is repeated at 1.0kW and 1.5kW.

5. The test may be repeated:

i. At other water flow rates

ii. At other air flow rate

28



Description Unit 0.0kW 0.5kW 1.0kW 1.5kWPacking Density m-1 110 110 110 110Air Inlet Dry Bulb, T1 oc 21.9 21.9 21.7 22.0Air Inlet Wet Bulb, T2 oc 21.8 21.7 21.8 21.8Air Outlet Dry Bulb, T3 oc 20.1 19.9 20.0 20.8Air Outlet Web Bulb, T4 oc 20.1 19.9 19.9 20.4Water Inlet Temperature, T5 oc 20.3 19.9 21.8 25.0Water Outlet Temperature, T6 oc 19.1 18.8 19.1 20.1Orifice Differential, DP1 Pa 119.5 119.5 119.5 119.5Water Flow Rate, FT1 LPM 1.6 1.6 1.6 1.6Heater Power, Q1 Watt 0 463 925 1390

Cooling Load (kW) Cooling range (oc)Average cooling temperature

(oc)0.0 20.3 – 19.1 1.20.5 19.9 – 18.8 1.11.0 21.8 – 19.1 2.71.5 25.0 – 20.1 4.9

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

1

2

3

4

5

6

1.2 1.1

2.7

4.9

cooling range against cooling load

cooling load (oc)

cool

ong

rang

e (o

c)

DISCUSSION:

29


(Esther Frederick Uyo – 26237)

We can clearly obtain the explanation by observing the graph above. The cooling range is

influence by the cooling load. The greater the cooling loads, the greater the cooling range of

the tower. There are some slightly error in the cooling range obtained, this is due to the

human error handling and obtain the data from the cooling tower machine.


From the graph obtained, we can see that cooling load will affect the cooling range. If we

increase the cooling load, the cooling range will also increase. However, the graph shows a

slightly drop in cooling range at the beginning and increase after that. This is once again due

to the error in recording the temperature.

CONCLUSION:

We can found out the relationship of cooling range and cooling load from this experiment.

The cooling range increases with the increasing in the cooling load.

OVERALL SUMMARY:

Bench top cooling tower is used to study the performance of a cooling tower. From this

experiment, we obtained the entire desired objective. We use the psychometric chart,

properties table, energy and mass balance equation to solve the entire question given in this

experiment. Through this experiment, we can verify the effect of the following factor on the

performance of the cooling tower:

11. Water flow rates

12. Water temperature

13. Air flow rate

14. Inlet air relative humidity

30

Date post:	28-Oct-2015
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t9 Cooling Tower

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