Its a cooling tower presentation

1. Faculty of Engineering & Technology
Department of Chemical & Metallurgical Engineering
Cooling Towers
Surname, initials & student No
1. Mashao PA 221026681
2. Masilela SP 221003525
3. Phaswana T 220177791
4. Toyi M 221020195
5.Tshinavhe F 221003274
Name of subject : Chemical Engineering Laboratory 2
Level of study : Third year S5
Lab Technician/s : S Muthubi, P Mwenge, G Mathebula, N
Hlatshwayo, L keele
1
DATE: 27/04/2023

2. 2
Presentation Outline
1. Introduction
1.1 Background
1.2 Main and Specific Objectives
1.3 Scope
2. Methodology
2.1 Chemicals, Materials and Equipment
2.2 Experimental Procedures
3. Results and Discussion
4. Conclusion
5. Recommendations
6.References

3. 3
1. Introduction
1.1 Background
Cooling towers are specialized heat exchangers that remove waste heat from industrial processes,
power plants and other facilities. They work by using water as a coolant, which is then circulated
through the system and exposed to air. The airflow removes heat from the water, causing it to
evaporate, and resulting cool water is then recirculated back into the system. They represent a
relatively cheap and a dependable means of removing low-grade heat. Cooling towers are designed
to be highly efficient, with a low energy consumptions and a minimal environmental impact. They
operate in a hot water that flows downward overfill surfaces which helps to increase the contact
time between the water and the air that is fed at the bottom of tower, this helps to increase heat
transfer between the two. The make up water is needed to replace water lost through
evaporation.(Marley,2019)

4. 4
1.2 Main objectives
The purpose of the experiment is to investigate how a cooling tower operates under different
operational conditions.
1.3 Specific objectives
• Determine the effect of heating load on the performance of the cooling tower.
• Determine the effect of water flow on the performance of the cooling tower.
1.4 Scope
To achieve this objectives the experiment will involve measuring the temperatures of both experiments
to calculate the energy content of air and water at different heating loads and water flowrates. The
experiment will be conducted over a range of heating loads including 1.5kw,1kw,0.5kw and 0kw at
different flowrates. The results of this experiment will provide insight into how the cooling tower
preforms under varying operational conditions. The experiment is intended to provide a better results of
how their performance can be optimized for different applications.
1. Introduction

5. 5
2. Methodology
2.1 Chemicals, Materials and Equipment

6. 2.2 Experimental Procedure

7. 7
3. Results and Discussion
Table for constant flowrate
Run 2 3 4
Temperature ( Degrees Celsius)
Air inlet Wet Bulb Temperature(t3) 13.9 14 14.2
Air inlet Dry Bulb Temperature(t4) 21.4 21.6 217
Air outlet Wet Bulb Temperature(t6) 26.3 25.8 24.5
Air outlet Dry Bulb Temperature(t5) 21.5 20.5 20.3
Water inlet Temperature (t8) 36.2 33.8 31.4
Water outlet Temperature (t7) 20.6 20.8 20
Make up water Temperature (t1) 20 20.2 20.4
Sampling time (min) 5 5 5
Water flow rate (L/h) 120 120 120
Heading Load (Kw) 1 0.5 0

8. 8
3. Results and Discussion
Range
1st run
𝑅𝑎𝑛𝑔𝑒 = 𝑇8 − 𝑇7
𝑅𝑎𝑛𝑔𝑒 = 36.2 − 20.6
Range= 15.6°𝐶
Approach
1st run
𝐴𝑝𝑝𝑟𝑜𝑎𝑐ℎ = 𝑇7 − 𝑇3
=20.6-13.9
=6.7°𝐶
Cooling Efficiency
1st run
cooling efficiency =
T8 − T7
T8 − T3
Cooling efficiency =
36.2 − 20.6
36.2 − 13.9
= 0.7 = 70%
2 3
Range 13°C 11.4°C
Approach 6.8°C 5.8°C
Cooling
efficiency
66% 66%
Table of calculated results

9. 9
SUBSTANCES C1 C2 C3 C4
WATER 5.45
9
0.30542 647.13 0.081
𝜌 =
𝐶1
𝐶2(1+(1−
𝑇
𝐶3
)𝐶4
∗ mw
1st run
𝑇8 = 36.2 + 273 = 309.2𝐾
5.459
0.30542(1+(1−
309.2
647.13
)0.081)
∗ 18.015
p =
55.069𝑘𝑚𝑜𝑙
𝑚3
∗
18.015𝑘𝑔
𝑘𝑚𝑜𝑙
𝑝 =
992.077𝑘𝑔
𝑚3
𝐹𝑙𝑜𝑤𝑟𝑎𝑡𝑒 =
120𝑙
ℎ
𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 =
120𝐿
ℎ
∗
1ℎ
3600𝑠
×
1𝑚3
1000𝑙
= 3.33× 10−5 𝑚3
𝑠
mhot = 𝜌 × 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒
1st run
𝑚ℎ𝑜𝑡 = 992.077 × 3.33 × 10−5
=
0.033kg
s
3. Results and Discussion
No of runs 2 3
density 992.209kg/m^3 992.852kg/m^3
Mass hot water 0.033𝑘𝑔/𝑠 0.033𝑘𝑔/𝑠

10. Point 3
Air inlet
Mass of air
At T5 and T6 ,using a psychometric chart
specific volume was assumed to be 𝑣 =
0.837𝑘𝑔/𝑠
pressure=16mmH20
𝑚 = 0.0165 ∗
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑣𝑜𝑙𝑢𝑚𝑒
m3 = 0.0165
16
0.837
= 0.0721𝑘𝑔/𝑠
Mass balance
𝑚1ℎ𝑜𝑡 =
0.033kg
s
𝑚3𝑎𝑖𝑟 =
0.0716kg
s
𝑤3 =
0.0098𝑘𝑔
𝑘𝑔
dry air … … … … . . from chart
𝑤4 =
0.0203𝑘𝑔
𝑘𝑔
dry air … … . . from chart
Inlet=Outlet
𝒎𝟑𝒘𝟑 + 𝒎𝟏 = 𝒎𝟑𝒘𝟒 + 𝒎𝟐
𝑚2 = 𝑚1 + 𝑚3(𝑤3 − 𝑤4)
𝑚2 = 0.033 + 0.0721 0.0098 − 0.0203 =
0.0322𝑘𝑔
𝑠
𝑀5 = 𝑀1 − 𝑀2 = 0.033 − 0.032 =
0.001𝐾𝑔
𝑆
3. Results and Discussion

11. 11
ENERGY BALANCE
SUBST
ACES
C1 C2 C3 C4 C5
WATE
R
2.76
∗ 105
−2.091
∗ 103
8.125 −1.4116
∗ 10−2
9.37
∗ 10−6
𝑇𝑎𝑣𝑔 =
𝑇8 + 𝑇7
2
=
36.2 + 20.6
2
= 28.2 + 273 = 301.4𝐾
𝐶𝑃
= 2.76 × 105
+ −2.091 × 103
∗ 301.4 + 8.125 ∗ (301.4)2
+ −1.4116 ∗ 10−2
× (301.4)3
+ 9.37
× 10−6
∗ (301.4)4
3. Results and Discussion
𝐶𝑃 = 𝐶1 + 𝐶2 × 𝑇 + 𝐶3 × 𝑇2
+ 𝐶4 × 𝑇3
+ 𝐶5 × 𝑇4
1ST run
h3=
46.5𝑘𝐽
𝑘𝑔
and h4=
73𝑘𝐽
𝑘𝑔
dry air from a chart
C𝑝 =
74694.43𝐽
𝑘𝑚𝑜𝑙. 𝑘
∗
1
18.015
kmol
kg
=
4146.235J
kg. k
Table of heat capacity

12. 12
Energy balance around the cooler
𝐻3 =
46.5𝑘𝑗
𝑘𝑔
dry air
𝐻4 =
73𝑘𝑗
𝑘𝑔
dry air
Energy of air=m3(h3-h4)
𝐸𝑛𝑒𝑟𝑔𝑦𝑎𝑖𝑟 = 0.0721 46.5 − 73 = −
1.9107𝑘𝑗
𝑠
𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 𝑚1𝑐𝑝 𝑇8 − 𝑇7 = 4146.235 ∗ 0.033 36.2 − 20.6 =
2134.482𝑗
𝑠
= 𝐸𝑛𝑒𝑟𝑔𝑦𝑤𝑎𝑡𝑒𝑟 =
2134.482𝑗
𝑠
∗
1𝑘𝑗
1000𝑗
=
2.1345𝑘𝑗
𝑠
𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 > 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑎𝑖𝑟
3. Results and Discussion

13. 13
3. Results and Discussion
Run 1 2 3 4
Temperature ( Degrees Celsius)
Air inlet Wet Bulb Temperature(t3) 14.0 15.9 14.1 14.1
Air inlet Dry Bulb Temperature(t4) 21.9 22.3 22.5 22.9
Air outlet Wet Bulb Temperature(t6) 21.1 25.5 27.1 27.6
Air outlet Dry Bulb Temperature(t5) 19.9 21.6 21.6 21.3
Water inlet Temperature (t8) 45.8 40.2 36.8 34.4
Water outlet Temperature (t7) 15.8 20.2 20.9 20.9
Make up water Temperature (t1) 20.2 20.3 20.5 20.6
Sampling time (min) 3 3 3 3
Water flow rate (L/h) 40 80 120 160
Heading Load (Kw) 1.5 1.5 1.5 1.5
Table for changes in a flowrate and constant heating load

14. 14
3. Results and Discussion
Range
1st run
𝑇8 − 𝑇7 = 45.8 − 15.8 = 30℃
Approach
1st run
𝑇7 − 𝑇3 = 15.8 − 14 = 1.8℃
Cooling Efficiency
1st run
cooling efficiency =
𝑇8−𝑇7
𝑇8−𝑇3
=
45.8−15.8
45.8−14
=0.952 =
95.2%
No of
runs
2 3 4
Range 20°C 15.9°C 13.5°C
Approach 4.3°C 6.8°C 6.8°C
Cooling
efficiency
82.30% 70.04% 66.5%

15. POINT 1
Density:
SUBST
ANCES
C1 C2 C3 C4
WATER 5.459 0.30542 647.13 0.081
𝜌 =
𝐶1
𝐶2
1+ 1−
𝑇
𝐶3
𝐶4
For 1st run
T8=45.8+273= 318.8K
𝜌 =
5.459
0.30542(1+(1−
318.8
647.13
)0.081)
× 18.015
= 989.474kg/m^3
Flowrate
1st run:
40𝑙
ℎ
×
1ℎ
3600𝑠
×
1𝑚^3
1000𝑙
=1.11×
10−5𝑚3
𝑠
Mass flow
mhot = 𝜌 × 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒
= 989.474 ∗ 1.11 × 10−5
=
0.01098𝑘𝑔
𝑠
No of runs 2 3 4
density 992.590kg
/m^3
991.917kg
/m^3
992.558kg
/m^3
Mass hot
water
0.01098K
G/S
0.033𝑘𝑔
/𝑠
0.033𝑘𝑔
/𝑠
FLOWAR
ATE
1.11*10^-
5*m^3/S
2.22*10^-
5m^3/S
4.44*10^-
5m^3/S

16. 16
3. Results and Discussion
Mass balance
𝑚1ℎ𝑜𝑡 =
0.011kg
s
𝑚3𝑎𝑖𝑟 =
0.073kg
s
𝑤3 =
0.010𝑘𝑔
𝑘𝑔
dry air … … … … . . from chart
𝑤4 =
0.017𝑘𝑔
𝑘𝑔
dry air … … . . from chart
Inlet=Outlet
𝒎𝟑𝒘𝟑 + 𝒎𝟏 = 𝒎𝟑𝒘𝟒 + 𝒎𝟐
𝑚2 = 𝑚1 + 𝑚3(𝑤3 − 𝑤4)
𝑚2 = 0.011 + 0.073 0.010 − 0.017
=
0.01𝑘𝑔
𝑠
𝑀5 = 𝑀1 − 𝑀2 = 0.011 − 0.010 =
0.001𝐾𝑔
𝑆
Point 3
Mass of air
At T5 and T6 ,using a psychometric chart
specific volume was assumed to be 𝑣 = 0.8270.
𝑘𝑔
𝑠
pressure=16mmH20
𝑚 = 0.0165 ∗
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑣𝑜𝑙𝑢𝑚𝑒
m3 = 0.0165
16
0.827
= 0.073𝑘𝑔/𝑠

17. 17
3. Results and Discussion
𝐶𝑝 = 𝐶1 + 𝐶2 ∗ 𝑇 + 𝐶3 ∗ 𝑇2
+ 𝐶4 ∗ 𝑇3
+ 𝐶5 ∗ 𝑇4
∗ 𝑚𝑟𝑤𝑎𝑡𝑒𝑟
𝑇𝑎𝑣𝑒 =
𝑇8 + 𝑇7
2
=
45.8+15.8
2
= 30.8°𝐶
𝑇 = 30.8 + 273 = 303.8𝐾
𝐶𝑝 =
2.76 ∗ 105
−2.091 ∗ 103
∗ 303.8 + 8.125 ∗ 100
∗ 303.82
+ −1.4116 ∗ 10−2
∗ 303.83
+ 9.37 ∗ 10−6
∗ 303.34
18
c𝑝 =
4144.5135𝑗
𝑘𝑚𝑜𝑙. 𝑘
SUBSTACES C1 C2 C3 C4 C5
WATER 2.76 ∗ 105 −2.091 ∗ 103
8.125 −1.4116 ∗ 10−2
9.37 ∗ 10−6
Energy balance

18. 18
Energy balance around the cooler
𝐻3 =
47.5𝑘𝑗
𝑘𝑔
dry air
𝐻4 =
63.4𝑘𝑗
𝑘𝑔
dry air
Energy of air=m3(h3-h4)
𝑬𝒏𝒆𝒓𝒈𝒚𝒂𝒊𝒓 = 0.073 47.5 − 63.4 = −
1.161𝑘𝑗
𝑠
𝑬𝒏𝒆𝒓𝒈𝒚 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓 = 𝑚1𝑐𝑝 𝑇8 − 𝑇7 = 4144.514 ∗ 0.011 45.8 − 15.8
=
1367.690𝑗
𝑠
= 𝐸𝑛𝑒𝑟𝑔𝑦𝑤𝑎𝑡𝑒𝑟 =
1367.690𝑗
𝑠
∗
1𝑘𝑗
1000𝑗
=
1.368𝑘𝑗
𝑠
𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 > 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑎𝑖𝑟
3. Results and Discussion

19. 19
Discussion
heat from water is removed by air in cooling tower. When warm water combines with cold air, latent heat is
removed. Experiment 1 ,flowrate was kept constant at 120L/h at decreasing heating load of 1-0kw, the
energy of air was -1.9107kj/s and water energy was 2.134kj/s. Experiment 2 the heating load was kept
constant at 1.5kw with increasing flowrate of 40-160l/h ,energy of air was found to be at -1.161kj/s while
the energy of water was find to be 1.368kj/s.. negative energy of air in a cooling tower is a normal
occurrence and is a desired outcome. This indicates that the air has less energy after absorbing heat from
water, which is a purpose of a cooling tower system. The temperatures in both inlet and outlet dry and wet
increases with flowrates in experiment 2 than experiment 1. The comparison of the energy of water to the
energy of air indicates the effectiveness of the cooling towers. If the energy of water is greater than the
energy of air, it suggests that the cooling towers are effective in transferring heat from the water to the air,
and the water is being effectively cooled.

20. 20
4. Conclusion
Both experiments were successfully conducted because the objective of the experiment were
achieved, the comparison of the energy of water to the energy of air indicates the effectiveness
of the cooling towers. If the energy of water is greater than the energy of air, it suggests that the
cooling towers are effective in transferring heat from the water to the air, and the water is being
effectively cooled. According to the findings of the cooling tower experiment, it can be said that
the air has a negative energy, which indicates that it is releasing energy into its surroundings.
This is to be expected since the cooling tower uses air to reduce the temperature of the hot
water. It can be seen that the energy of the water in the cooling tower in the experiment is more
than in the air experiment. This is also to be expected because water can absorb more heat than
air because it has a higher heat capacity.
However, since water has positive energy, it is possible that cooling towers are successfully
removing heat from the water and dispersing it into the atmosphere. A cooling tower's ability to
effectively transfer heat from the water to the air is demonstrated by the fact that the energy of
water is higher than the energy of air.

21. 21
5. Recommendations
Even if the experiment was successful , several errors may occur but can be fixed, firstly all the heating
load must operate, problem encountered was that J2 caused E.LC.B switch to trip and this might as well
affect results. Water quality such as pH, conductivity and total dissolved solids is required to be checked
to ensure the water is suitable for cooling. Measurements should be put on the make up water tank to
allow us to read the amount of water lost from a system. The make up water should not be all used up in
a tank. After completed experiment, ensure that water in a tower was cooled down before draining it, if it
was drained off while still hot the units efficiency may be affected. Computer automated thermometer
should be introduced to record temperatures at a desired time.

22. 22
References
• Retrieved from http://chem.engr.utc.edu/webres/435F/3T-CT/3T-CT.html Marley. (2019).What
Is A Cooling Towers. Retrieved fromhttps://spxcooling.com/coolingtowers
• Cooling Towers: Understanding Key Components of Cooling Towers and How They Work."
Energy.gov. U.S. Department of Energy, Marley. Web. 29 Apr. 2023.
https://www.energy.gov/eere/amo/cooling-towers

23. 23