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....















42 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.
I'm a foreigner. i' very interested this formular.
Thanks. If I have any question, Can I ask You?
you can always ask any question & I will try to help you out of my busy schedule.
If a cooling tower is overloaded (for example, to a 30 TR unit, if 35 TR load is given) still the cooling tower works and only the approach increases (drift between wet bulb & cold water temp. increases). How to know the optimum cooling capacity of the tower
Dear JB
Cooling tower will not say no to you any time. I have seen cooling towers operating at as high as 20 C approach unknowingly.
They will NEVER say NO to you.
The only issue is that your cooling water return temp will keep on increasing & will settle at equilibrium based on the heat load from process side vs heat load removed by cooling tower in revised conditions.
So first you fix your objective & then try to find out the no from process side. Later on focus on your cooling tower.
Hi Profmaster,
I have difficulty calculating the 2nd step of the calculation into excel file.
I copy your formula, and change all Tw into inlet air bulb temperature (30 C). But somehow the final Qw result differ a lot from yours.
Please help. Thanks in advance.
Hello profmaster,
Thanks for this page, it's very useful.
I am writing a report on how to carry out a performance test for a cooling tower. Unfortunately, I don't think I'll be able to source your site, so I was hoping you could provide the textboox you adapted this from?
Thanks!
"profmaster said...
Method is derived from a text book.
Currently I am unable to convert it in other units.
9:58 PM"
Thanks John
You can source my site - I do not have any issue in putting it as a reference.
As far as I can remember this - 15 years old reference - It was probably authored by Baker.
Hope it will help.
this method is a bit difficult because in large mechanical draugt or natural draught cooling towers say for a 250mw power plant or say 12m.x 12m or even smaller cells the uniformity of the exit air temperature and hence the exit air enthalpy is very poor in fact it would be very difficult to measure at all the plan locations above the fill. if measured at the exit to the fan then there also we observe variations acrss the dia and effect of ambient temp is also felt. hence the calculations become very much a rough estimate. hence we need to have data on the fill media with which we can correlate all the readings and can come to a better conclusion.
suresh
Suresh you are theoretically right on first part that measurements have to be reasonably good enough to have good accuracy of the prediction But that is true for any troubleshooting or improvement exercise. This doesn't mean method is not good.
Second you have to use your judgement in measuring the data, its repeatability & accuracy. Its not so difficult if you do it carefully.
I have used the same for 16000 M3/Hr & 24000 M3/hr CW flow rate which is a reasonably large size (May not be good enough for bigger power plants).
Fill media data is not going to give you the correct picture wheras this method can even predict fills efficiency also when you count on all factors one by one.
In fact fill media data is not going to tell you what is the air flow rate, what is the liquid distribution variation, enthalpy, temperature etc. all need to be measured whether you use this method or not.
The real fact is that performance of 3 big towers have been improved by this method with following clarity.
1. Current status of ATE.
2. Breakup of ATE factors including fills & their contribution in overall ATE.
3. Each factor was attacked one by one.
4. Results after improvement were same as predicted.
5. Final result - ATE brought down to 3-4 C.
Now its your choice.
By the way how will you define the proposed method?
sir,
ideally a measurement of cwt,hwt, waterflow and inlet wbt can be made after callibrating all equipment. then air flow and exit temperatures can be measured. heavy variation in the exit temperatures could mean huge air and water flow variations. once the same is properly taken care of then exit air temperatures at top of fill and top of flume could be taken. all these readings in conjunction with the mass transfer coefficient of the fill (if available ) could help narrow down on the authenticity of the data. then work out an l/g and kav/l which is very closely satisfying all the above data and thus the heat balance including evaopration loss. on any approach line for a fixed wet bulb and range a number of l/g could give the same cold water, however with above data we could come close to the actual operating l/g. once this is done we need to project the same on the blue book curves and check at what l/g the derived kav/l satisfies the approach requirement.
now we also simaltaneously need to check whether the electrical and mechanical ratings would have sufficient margin to operate on the reduced l/g since the power requirement would increase in cube of the air flow rate increase. generally motor ratings have roughly 15% margins. the other possibility would be to increase mass transfer by operating higher efficiency fills, also improved nozzles and low resistance drift eliminators might help.
Suresh You are correct again. The difference is that the curve you are talking about and the operating line varies on daily basis due to change in ambient conditions.
Also the data on fills efficiency (Or mass transfer coefficient) can not help you at all because, the efficiency varies over a period of time due to displacements and due to channeling. Example, most of the towers installed in one time period or by one company will have same fills but will never have same ATE because of many other factors.
Basic heat balance will give you the accuracy of your measurements at first place. Also if you have some error in the measurements you will know it becasue there are two equations in this - one is heat transfer & the other is mass transfer.
Anyway the objective of this exercise was to provide a mathematical model for the evaluation of any CT with breakup of all contributing factors to poor ATE without using any curve. If you feel comfortable by the curve method you can use it, no issue.
Yes, further check on air flow measurement is available from motor & its power consumption data.
The margins available in the drives are to be checked as part of your complete recommendation in any case.
Its not so easy. I did the rigorous evaluation of each parameter and variation analysis on the data. It took me to solve the problem about 6 months of continuous monitoring for one tower.
interesting post
Quite informative and helpful post.
@ Suresh : I am doing various Cooling Tower projects in Delhi. Can I seek your consultation for testing and analysis of Counter flow cooling towers? My email address is atled@in.com.
Sir. I cant seem to make sense of the solution you provided. For an example the values for H1 and H3(wet bulb exit air) you calculated using the equation you provided seem close to a 100% humidity level(psychrometric chart) and thats just for the entering air,if humidity levels are that high then little(very) evaporation will occur due to this high moisture content of the entering air. I expexcted atleast +/- 10% diviation of the value from your equation an the chart. Again i tried using the formula you provided but my values arent even close to the value you have. I'm interested in hearing your point of view from my questions.
Kind regards BLU
First of all, there must be some confusion in entering the correct formula. Even if it is correct it is just to show the calculation method you can develop your own equations & use them.
Next yes many times during rainy season humidity level will be as close as 100% (practically 80%) is more frequent and at that time only sensible transfer is available and hence the CW temp is maximum due to very high wet bulb.
hi sir, i'm a mechanical engineering student. i'm doing a final year project about the experiment of the cooling tower. i have few questions about the calculation of cooling tower. my questions are as the following:
1.) what is the equation of make-up rate?
2.) how to calculate the energy and mass balance?
3.)what is the equation of the wet bulb temperature approach? and how to calculate it?
hope you can give me a good answer. you could answer me by reply an email (wai_chan1626@hotmail.com)
your help will be appreciated. :)
i think your calculation works for my cooling tower products
Hi Sir,
Can you help me calculate the savings i will derive if i introduce a much cooler air into the cooling tower.
Existing Data:
Inlet Air Dry Bulb = 36 degC
Inlet Air Wet Bulb = 28.3 degC
Exit Air Dry Bulb = 41.2 degC
Exit Air Wet Bulb = 38.9 degC
Cooling Water Flow rate = 1,890 CMH
Cooling Water Inlet Temp = 32 degC
Cooling Water Outlet Temp = 42.5 degC
Fan Air Capacity = 986,000 CMH
New Air Properties to be introduced:
Air Flow = 74,000 CMH
Inlet Air Wet Bulb = 18.3 degC
Inlet Air Dry Bulb = 25 degC
Hi Jike
I will try to help you out on this calculation. First I will predict the Return CW Temp & then savings will come mainly from process side not from cooling tower side.
So initially I will assume that process heat load is same for identifying Return & Supply temp.
But I need some time for this as I am busy in my assignments.
Profmaster, I was wondering if these calculations could be used on both a forced draft and a natural draft cooling tower.
Fundamentally the equations are same, the only difference is fan power in case of forced draft.
Since the velocities are different height of natural draft automatically becomes more.
Thanks. This looks easy but its too lengthy.
Mechanical engineering students like my sister are really into that stuff of computing for cooling tower heat in and outs.
The purpose of a cooling tower or condenser system is to reuse water and reduce water usage. Determining the conductivity set point of the water systemis important to maximumyour water reuse. Conductivity is the electrical resistance measurement of the dissolved solids, suspended solids, and dissolved gases of the water and is expressed in mico ohms (uomhs).
Cooling water systems are essential to production in many industrial plants, and they can require a considerable amount of water to run, to reduce chloride concentrations in the intake cooling water try to implement water treatment programs. Thanks a lot.
I want required the Calculation for Water Evaporation Loss in Cooling Tower...Kidly tell me -what Parameter are rwquired to calculate water evaporation Loss.
Dear Sunil
If you read it carefully you will find that step-8 is giving the amount of water evaporation rate.
So focus on each step.
Post a Comment