I am invariably finding many hits on cooling tower capacity & performance calculation and related queries. Therefore, I have decided to include the detailed calculation procedure in order to enable many students & process engineers who are interested in improving cooling towers performance by following these simple steps.
If you have any query, kindly post them in the comments section. I’ll try my level best to answer those queries as soon as possible.
First you should collect all the data as given below. Be sure that the data collected for these temperatures is most accurate because of lower absolute level of generally ~40°C average temperatures, an error of 0.5°C due to manual data collection & judgment will cause more than 1.2% error in the result at one calculation. Repeating such errors may result in cumulative errors of more than 10% in totality giving you totally absurd results.
So the basic point is that collect the data on regular basis, keep a watch to have a feel of real values & then proceed.
If you have any query, kindly post them in the comments section. I’ll try my level best to answer those queries as soon as possible.
First you should collect all the data as given below. Be sure that the data collected for these temperatures is most accurate because of lower absolute level of generally ~40°C average temperatures, an error of 0.5°C due to manual data collection & judgment will cause more than 1.2% error in the result at one calculation. Repeating such errors may result in cumulative errors of more than 10% in totality giving you totally absurd results.
So the basic point is that collect the data on regular basis, keep a watch to have a feel of real values & then proceed.
Actual Data
Cooling water flow rate - 4134 M3/hr
Cooling water inlet Temp - 44.0 °C
Cooling water exit Temp - 35.0 °C
Inlet air-wet bulb - 30.0 °C
Inlet air-dry bulb - 38.8 °C
Exit air-wet bulb - 40.7 °C
Exit air-dry bulb - 42.0 °C
Now follow step by step procedure for the calculation.
Step-1
Calculate waterside actual heat load, which is as below
Qw = 4134 x 1000 x (44 – 35) / 1000000
= 37.21 Gcal/Hr
Step-2
Calculate absolute humidity at wet bulb of inlet air, which is at 30°C in this case. This is a function of wet bulb temperature only.
The equation for the same is
1.4478310678E-10*(Tw^7)-2.6920*10e-8*(Tw^6)+1.99053*10e-6*(Tw^5)-6.65614*10e-5* (Tw^4)+0.00131879344*(Tw^3)+0.00125483272*(Tw^2)+0.291649083*Tw+3.802441
Where Tw is wet bulb temperature in °C. So,
H1 = 27.29 Kg/ ‘000Kg of dry air
Step-3
Calculate absolute humidity at dry bulb of inlet air, which is at 38.8°C in this case. It will give you saturation level of humidity, say H2.
Step-4
Find out &% Saturation. Of course it can be done from Psychometric charts but then you wont be able to use powerful Excel Tool for simulation of your cooling tower that’s why these equations are generated.
You can also use any good Excel Add-IN for Psycho properties if available.
Here, it will be %Sat = H1/H2
Step-5
Based on % Saturation find out the enthalpy content of moist air at inlet condition. Again I did it using self-developed equations ~10 years back.
I found it to be Hin = 26.196 Kcal/Kg of wet air.
Step-6
Similarly find out the moist air enthalpy at exit condition, which is
Hex = 41.630 Kcal/Kg of wet air
Step-7
Similarly, find out the absolute humidity at wet bulb for exit condition, which is 50.74 Kg/ ‘000 kg of dry air in this case.
Step-8
Calculate airflow based on heat load and enthalpy difference, which shall be as below
A = 4134000 x (44-35)/(41.630 – 26.196)
= 2410652 Kg/hr
Now based on Absolute Humidity difference, calculate amount of water evaporated as below
W = 2756000 X (50.74 – 27.29)/1000
= 64654 Kg/hr
Step-9
Now heat required for evaporation of this water can be calculated based on average latent heat of water evaporation at the inlet & exit temperature.
Average water temperature = 39.5 °C
Latent heat = 575.33 Kcal/Kg
Hev = 64654 x 575.33
= 37.20 Gcal/Hr
This is matching with the heat load of waterside hence, calculation is correct due to accurate temperature measurements.
So L/G comes out to be = 1.715 in this case.
Second Part
I will cover the NTU calculation & efficiency of tower, use of NTU method for predictions etc. in the next part of this post. Wait till then....















15 comments:
Would it be possible to get an English version of these calculations for the entire series on cooling towers, part 1,2,3?
What do you mean by English version. This all has been provided in English. Please make it more clear with your user id so that I can answer your specific query.
sir,
you have shown total head load on water side equalised by total enthalpy gain by latent heat of evaporation. how do you account for change in sensible heat.
you have equated total enthalpy on water side by total enthalpy due to latent heat on vapour side , what about the sensible heat part by which temperature of vapour is increased which is also part of total change in enthalpy
You are right, I have missed out the step of adding specific heat enthalpy of dry air as well.
Thanks for thorough understanding of the subject, probably it will now be more helpful to you.
Hi profmaster,
Very good tutorial. I am new to cooling tower efficiency calculation, but I was confused in the last step. How do you manage to get L/G? Is there calculations or is this a number from the design of the tower?
This has come from calculation as you know water flow & air flow is calculated in step-8. So you have both L & G
L/G is liquid to Gas Ratio of the cooling tower. the L/G is found out for a particular duty condition of the cooling tower based on the type of heat exchanger provided ( fill media) based on the type and orientation of the cooling tower ( counter flow or cross flow).
For a specific configuration of fill media selected and depending upon its heat and mass transfer characteristics we come to asess the supply part and we check it with the demand part which is generally available as blue book (counter flow)curves or black book (cross flow). thus a demand curve is genreated for a particular L/G wrt a particular KAV/L (COUNTERFLOW) OR KAY/L(CROSSFLOW). thus L/G selection is a variable depending upon mass trasfer K depending upon fill configuration.
regards
suresh
You are right, but here objective is not the design of the CT but is to estimate the performance & suitability for other conditions & optimization of tower operation.
Once you identify the effect of each component (Listed out in other CT articles on this blog) you can plan to improve its performance based on priority & impact.
Once you identify the effect you can also know the fills efficiency & can take a decision when to replace them.
Simple thing - Idea is not to design the tower, Idea is to assess the existing tower & how to improve it, where from it should be started & what impact you will get in final CW temperature.
Goodday Profmaster,
I am young upcoming chemical engineering and I am trying to run your calculation for a cooling tower performance evaluation. My first problem is that I tried to re-calculate the abosolute humidity H1; Firstly i have never seen that correlation between absolute humidity and wet bulb temperature only.
Also one other thing... I am guessing that cooling water flow rate given is cooling water flow rate IN .... corect me if I am wrong...
I do not know if it is possible but I would gladly appreciate your help... thanks in advance
Hi Profmaster
Thanks for ur good help to asist me in calculating the CT perfomance.
But I am doubtful at ur 8 th step.To find water evaporated ,the flow rate used is 275600 kg /hr.
But in the early step it is obtained as 2410652 kg /hr.
Please clear me
Sir I am dealing with industrial gas cooling tower. A gas mixture at 30 C is to be cooled to 6 C with child water at 5 C ; gas leaving temperature is 6 C
I need to calculate the following
1. Adiabatic Sat temp
2. Water circulation rate
3. Height of cooling tower.
I can send you the detail of data on a scanned copy if you could please provide mail ID.
Please provide me suitable suggestion.
Best Regards
Dear you have not mentioned your process requirement properly.
Anyway you can send all your data & requirement (Why do you want to use this process) at my mail ID techkasamba@gmail.com
But I will be able to answer only in between my busy schedule.
Is this part of a text book? Can you provide this in Inch/Pounds (IP) for us non-SI unit folks.
Any help is much appreciated.
Thanks
Method is derived from a text book.
Currently I am unable to convert it in other units.
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