LAB REPORT

APPLIED THERMODYNAMICS

EME 2146

TRIMESTER 1 (2013/2014)

AT 2: Cooling Tower

Name Vicknendran Vickneswaram

ID 1091102189

Date of Experiment 23/07/2013

OBJECTIVE:

To familiarize us with the operation of a cooling tower

PROCEDURE:

1) The cooling tower is started by switching on the power, the pump, the fan and lastly

the heater.

2) The water-flow-rate (W1) is set to 50L/hr and the heater (H1) is set to 960W while the

fan-speed is set to constant.

3) The data that was recorded based on W1 and H1 combination setting are:

a) T1, the water temperature at the tower outlet

b) T3 , the water temperature at the heater outlet

c) T5 , the water temperature at the tower inlet

d) T7 , the water temperature at the tank

e) T2 , the wet-bulb temperature of air at the column top

f) T4 , the dry-bulb temperature of air at the column top

g) T6 , the wet-bulb temperature of air at the column bottom

h) T8 , the dry-bulb temperature of air at the column bottom

i) Amp and Volt readings of the heater

j) Air speed

4) Step 2 and 3 was repeated for several different combinations. There are:

a) H1W2 b) H1W3 c) H2W1 d) H2W2

e) H2W3 f) H3W1 g) H3W2 h) H3W3

H1 = 2A, H2 = 4A, H3 = 6A; W1 = 50L/hr, W2 = 100L/hr

5) After the experiment is finished, the cooling tower is turned off step by step by

turning off the heater, the fan, the pump and lastly the power.

Note that sequences of the steps given must be followed in order to avoid overheating.

RESULTS:

W1H1 W1H2 W1H3 W2H1 W2H2 W2H3

V (V) 234 233 233 233 233 233I (A) 2 4 6 2 4 6

T1 (C) 23.9 24 24.1 24.3 24.9 25.1T2 (C) 25.0 25.7 26.3 27.3 27.4 26.9T3 (C) 41.2 38.5 40.0 40.0 33.7 36.0

T4 (C) 25.7 26.5 27.1 28.6 28.3 27.9T5 (C) 40.9 37.9 39.4 40.1 31.8 35.5T6 (C) 26.8 26.9 27.0 27.1 27.1 27.2T7 (C) 25.2 24.8 24.5 24.2 24.1 24.1T8 (C) 26.3 26.3 26.5 26.6 26.6 26.7Velocit

y(m/s)

0.73 1.17 2.46 2.74 2.14 2.36

DISCUSSION:

1. The conservation of mass :

[Total mass entering the CV during Δt] − [Total mass leaving the CV during Δt]

= [Net change in mass within the CV during Δt]

The mass flow rate of vapour:

ṁv1 is the vapor-flow-rate at the outlet

ṁv2 is vapor-flow-rate at the inlet

ṁf is the evaporation losses

The mass flow rate of dry:

ṁa1 is the dry air-flow-rate at the inlet

ṁ a2 is the dry air-flow-rate at the outlet

ṁa is the net dry air-flow rate.

ṁf = (w2-w1) ṁa

w1=ṁv 1

ṁa 1w2=

ṁv 2

ṁa 2

; w1 = w2 = humidity ratio.

The mass of the water lost due to the evaporation can be considered negligible, which also

supported by the constant water level of the reserve tank as there is no water evaporation.

However, error might still be encountered while conducting the experiment, which is

contributed by the condensation of water before air, leaves the cooling tower.

ṁv1 + ṁf = ṁv2

ṁa1 = ṁ a2 = ṁa

2. Based on energy conservation theory, the energy that enters the system must be equal to

the energy that exits the system. According to the First Law of Thermodynamic, a

balanced energy equation is:

w1 (hg1 – hf2) = cp (T2 – T1) + w2 hfg2

w1 the relative humidity of the air-vapour mixture that enter the tower

w2 is the relative humidity of the air-vapour mixture that exit the tower

(T2 – T1) is the temperature different of the water entering and exiting the cooling tower

cp is the specific heat of water and h is the changes of enthalpies

The heat energy that lost during the cooling process is calculable with the equation

above.

w1 (hg1 – hf2) - w2 hfg2 = Qout

The heater enables electrical energy to be converted into heat energy;

P = VI= Qin

P is power supplied to the heater

I is current

V is voltage supplied to the heater

3.

a) Heat transferred between the water droplets and the air:

Q = ṁwater (h1−h5 ¿ =ṁwater CP, water(T 1−T 5)

b) Heat transferred from the heater:

H = ṁwater(h3−h7 ¿ =ṁwater CP, water(T 3−T 7)

3. a) Heat transferred between the water droplets and the air:

Q = ṁwater (h1-h5) = ṁwater CP,water (T1-T5)

i) ṁ1 = 50L/hr = 1.389 x 10-5 m3/s = 0.01389kg/s

CP,water = 4.184 kJ/kg.K

1) H1 = 2A

Q1 = (0.01389kg/s) (4.184kJ/kg.K) (296.9K – 313.9K)

= 0.988kW

2) H2 = 4A

Q2 = (0.0138m9kg/s) (4.184kJ/kg.K) (297.0K – 310.9K)

= 0.8078kW

3) H3 = 6A

Q3 = (0.01389kg/s) (4.184kJ/kg.K) (297.1K – 312.4K)

= 0.8892W

ii) ṁ2 = 100L/hr = 2.778 x 10-5 m3/s = 0.02778kg/s

CP,water = 4.184 kJ/kg.K

1) H1 = 2A

Q1 = (0.02778kg/s) (4.184kJ/kg.K) (297.3K – 313.1K)

= 1.8365kW

2) H2 = 4A

Q2 = (0.02778kg/s) (4.184kJ/kg.K) (297.9K – 304.8K)

= 0.8020kW

3) H3 = 1920W

Q3 = (0.02778kg/s) (4.184kJ/kg.K) (298.1K – 308.5K)

= 1.2088kW

b) Heat transferred from the heater:

H = ṁwater(h3−h7 ¿ =ṁwater CP, water(T 3−T 7)

i) ṁ1 = 50L/hr = 1.389 x 10-5 m3/s = 0.01389kg/s

CP,water = 4.184 kJ/kg.K

4) H1 = 2A

Q1 = (0.01389kg/s) (4.184kJ/kg.K) (314.2K – 298.2K)

= 0.9299kW

5) H2 = 4A

Q2 = (0.01389kg/s) (4.184kJ/kg.K) (311.5K – 297.8K)

= 0.7962kW

6) H3 = 6A

Q3 = (0.01389kg/s) (4.184kJ/kg.K) (313.0K – 297.5K)

= 0.9008kW

ii) ṁ2 = 100L/hr = 2.778 x 10-5 m3/s = 0.02778kg/s

CP,water = 4.184 kJ/kg.K

4) H1 = 2A

Q1 = (0.02778kg/s) (4.184kJ/kg.K) (313.0K – 297.2K)

= 1.8365kW

5) H2 = 4A

Q2 = (0.02778kg/s) (4.184kJ/kg.K) (306.7K – 297.1K)

= 1.1158W

6) H3 = 6A

Q3 = (0.02778kg/s) (4.184kJ/kg.K) (309.0K – 297.1K)

= 1.3832kW

c) i) In general, for ṁ1 = 0.01389kg/s

Set

Heat transfer between the droplets and the air (kW)

Heat transfer from the heater (kW)

1 0.9880 0.92992 0.8078 0.76923 0.8892 0.9008

ii) In general, for ṁ2 = 0.02778kg/s

Set

Heat transfer between the droplets and the air (kW)

Heat transfer from the heater (kW)

1 1.8365 1.83652 0.8020 1.11583 1.2088 1.3832

0.9299 0.7692 0.90080

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Graph heat transfer between the droplets and the air vs heater transfer from the heater

ṁ1

Heater transfer from the heater

Heat

tran

sfer

bet

wee

n th

e dr

ople

ts a

nd th

e ai

r

1.8365 1.1158 1.38320

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Graph heat transfer between the droplets and the air vs heater transfer from the heater

ṁ2

Heater transfer from the heater

Heat

tran

sfer

bet

wee

n th

e dr

ople

ts a

nd th

e ai

r

Conclusion:The operation of cooling tower is related with the mass conservation theory and the first law

of thermodynamics from this experiment.

Reference Y.A. Cengal and M.A. Boles, “Thermodynamics: An Engineering Approach”, 6 th

Edition, McGraw-Hill 2006

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

W1H1 W1H2 W1H3 W2H1 W2H2 W2H3

V (V) 234 233 233 233 233 233I (A) 2 3 4 2 3 4

T1 (C) 23.9 24.1 24.3 24.5 24.7 25.0T2 (C) 25.0 25.2 25.4 25.5 25.7 25.9T3 (C) 28.2 29.7 31.0 27.5 28.1 29.4T4 (C) 24.5 24.6 24.9 24.8 25.0 25.4T5 (C) 28.2 29.4 30.6 27.6 28.3 29.5T6 (C) 26.2 26.4 26.3 26.7 26.7 26.6T7 (C) 24.6 24.4 24.3 24.6 24.6 24.3T8 (C) 25.8 26.0 26.1 26.2 26.3 26.2Velocit

y(m/s)

3.02 2.54 2.45 3.71 3.00 2.81