## January 22, 2008

### Cooling Tower: Performance calculation - I

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.

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....

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Anonymous said...

Would it be possible to get an English version of these calculations for the entire series on cooling towers, part 1,2,3?

profmaster said...

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.

Suresh said...

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.

Suresh said...

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

profmaster said...

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.

TheServant said...

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?

profmaster said...

This has come from calculation as you know water flow & air flow is calculated in step-8. So you have both L & G

Suresh said...

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

profmaster said...

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.

Kiev said...

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

Roshanlal said...

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.

Anonymous said...

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
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.
Best Regards

profmaster said...

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.

Anonymous said...

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

profmaster said...

Method is derived from a text book.
Currently I am unable to convert it in other units.

Anonymous said...

I'm a foreigner. i' very interested this formular.
Thanks. If I have any question, Can I ask You?

profmaster said...

you can always ask any question & I will try to help you out of my busy schedule.

JB said...

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

profmaster said...

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.

Anonymous said...

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.

John said...

Hello profmaster,

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"

profmaster said...

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.

Suresh said...

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

profmaster said...

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.

By the way how will you define the proposed method?

Suresh said...

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.

profmaster said...

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.

Whitfields said...

interesting post

Anonymous said...

@ 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.

Anonymous said...

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

profmaster said...

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.

i'm_loving_it said...

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. :)

arun said...

i think your calculation works for my cooling tower products

Jike said...

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

profmaster said...

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.

Anonymous said...

Profmaster, I was wondering if these calculations could be used on both a forced draft and a natural draft cooling tower.

profmaster said...

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.

Mist Eliminator said...

Thanks. This looks easy but its too lengthy.

Plumbing Supplies said...

Mechanical engineering students like my sister are really into that stuff of computing for cooling tower heat in and outs.

Cooling Tower Chemicals said...

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).

Industrial chiller said...

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.

Sunil Arya said...

I want required the Calculation for Water Evaporation Loss in Cooling Tower...Kidly tell me -what Parameter are rwquired to calculate water evaporation Loss.

profmaster said...

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.

Aleksandar Stefanov said...

Dear profmaster

I am a chemical engineer and now I am working on a project for improvement of an existing old cooling tower.
I have study your equation and double check with my data.
My question is – could I use the equation 8 from your statement for airflow based on heat load on water side and enthalpy difference in order to evaluated the required airflow for certain CWT?

profmaster said...

Yes You can but I could not understand your doubt....when it is already described in the post.

Anonymous said...

How to derive Total Static Load of 200TR Cooling Tower... 200TR Tower size 3m x 3m x 3.3m Ht with 10HP Motor & 2390mm Fan dia

Aleksandar Stefanov said...

Dear profmaster,

I have the follwing data for one cooling tower - see below.
After doing the same estimation as described from you here, I get a L/G = 0.54, while by ration between measured air flow and water flow I have L/G 1.40. What should be the reason for this difference, incorrect measurments of the temperatures or I should repeat all in summre conditions? Normaly the cooling tower is designed for DBT 35-38 degree, WBT 30-33 degree and HWT 35, CWT 25-26, but has never more then 3-5degree range (dT)

1. HWT = 20.7◦C
2. CWT = 15.7◦C
3. WBT inlet = 13.39◦C
4. WBT outlet = 15.00◦C
5. DBT inlet = 17.9◦C
6. DBT outlet = 19.0◦C
7. Water flow = 700m3/h
8. Airflow measured with anemometer = 417 894m3/h – for 5 cells totally
9. Water flow 700m3/h
10. L/G = 1.40
11. Characteristic of filling material per cell – 8.0meter x 8.0meter x 1.2meter height , specific surface154m2/m3

Aleksandar Stefanov said...

Hi Profmaster,

Do you have an idea whether I could do a trial with the cooling tower filling materail (the range of water distribition ) by measuring of dp of it. Good water distribution over the filling materail will means higher dp, then bad water distribution. I could plote different L/G data versus this indication.

Unknown said...

I presume that each cell is having a water flow of 700m3/hr. with the L/G of 1.4 you roughly deliver about 500000 kg/hr of air = 132 m3/sec of air per cell.
with the measurment of 417894 m3/hr for totally 5 cells each cell would get 417894 /5 = 83579 m3/hr per cell = 23.21 m3/ sec per cell. the design air flow is 132 m3/s and tested is 23.21 m3/hr which really does not sound logical. since i presume that for a 8.0 m length and width the fan installed is 6.0 m hence an area of 28.27 m2 hence to get a flow of 23.21 the velocity measured has to be 0.86 m/sec which does not sound correct. please confirm so that we could try to work out some details.
i feel that the air flow per cell has to be around 417894 which you have said as total for 5 cells.
regards
suresh sarma

Anonymous said...

Pretty! This was an incredibly wonderful post.
Many thanks for providing these details.
My page ; elecommunications tower

profmaster said...

Dear Stefanov, Sorry for delay in the reply. Suresh has raised valid points in this measurement.

On the first look, it seems that you are talking about only one measurement which is not logical because a large error is observed if any sensitive data like Temp or Velocity measurement varies. So you need to measure it several times before coming to any conclusion and check repeatable nature of data points.

Anyway whatever be the L/G, if the data given is actual data you are operating pretty efficiently at ATE of <2 C, So I suggest dont change any parameter, just use them for your understanding, on the other hand I doubt this data as it is practically next to impossible and is very close to perfect operation which is rare.

You should check it on daily basis at a fixed time schedule for at least 2-6 months and plot it on a graph that is the best method to KNOW it.

Regarding your second point I don't know what do you mean by dp of fills (& how r u going to measure it?? Let me know also its interesting to learn) but yes you can measure height of water level on water deck.

Anonymous said...

in step 2
H1 = 27.29 Kg/ ‘000Kg of dry air

is it 1000 kg of dry air?

profmaster said...

Yes it is 1000 kg of dry air.

Aleksandar Stefanov said...

Dear Suresh,

After double check of the data at the plant I am sending you the right info
Flow rate of water per cell L is 140m3/h , air flow per cell is 207000m3/h . From here I have around 0.67L/G.
If I starting from the other way – calculation of airflow based on heat load and enthalpy difference
I have about 0.57 L/G. So the difference was in my measuring.

Aleksandar Stefanov said...

Dear Profmaster ,

You are right , I did the measurement of the flow in a profile – at every 10 cm from the fan diameter and was big difference at the and.
Looking the data like now it really looks like it is perfect operation. But is it true that I must see the same data also in other conditions - DBT 35-37 and then to see again the ATE?

About the second point – we have about 45 Pa – total pressure and 40 Pa static just below the fan , about 25 Pa – total pressure and 20 Pa static above the filling media (distribution material) and 15 Pa – total pressure and 10 Pa static below the filling media (distribution material). This let say is sensible for a good air – water transfer.
But if we assume that the L/G is very low, which means not uniformly wet filling media (distribution material), then all the air will pass without any resistance and what we have above as a pressure similar will have below. I just hear about this method of study, do not really check it

Techkasamba Agarwal said...

Good to see your feedback post.
Now make sure that you are operating at ATE of <2 C by measuring daily data for at least a fortnight or more.

Because it seems that you are operating at very high air flow rate which can be reduced for saving lot of energy.

Yes, as I always say if you have lot of applications of CW in process cooling than don't save energy in Cooling Tower if your cooling water temperature goes up by >0.2 C or so because in that case you will be loosing much more on process side.

Good Work

Suresh Sarma said...

Dear Stefanov,
For achieving this close approach a Low L/G ratio is required since the demand side KaV/L also is pretty high at near about 3.1.at the operating L/G. I guess you are operating at good efficiency with not much losses. of course the what cold water you desire is based on your process requirement.
The best performance is always on most uniform distribution and thus uniform exit temperature profile above the drift eliminator. so keep it the same and do not disturb the distribution.
regards
suresh sarma

Aleksandar Stefanov said...

Dear Suresh ,
Dear Profmaster

Haw I could find info for the CTI curves (Blue Book). Are the curves available or must be ordered and paid.
I must give an answer to the following question - in respective HWT(max), DBT(max),WBT(max) and L/G (existing) what could be the possible minimum CWT achievable with the existing cooling tower and L/G and what could be the possible minimum CWT in case of lowering the L/G .
I think I could proceed in the following way: I have the L/G, WBT and Kav/L. Using them, from the CTI curves will find the range and approach of operation, from which will find the CWT. After that will try with lowering of L/G (higher air flow or lower water flow) or higher Kav/L (adding of more fill media) and will see the new approach and range which could be obtained.

Cooling Towers in India said...

Hello,
I found the post helpful for my study purposes as well as in the practical work of cooling solutions technology. I'm doing internship right now and looking for practical information. If you can give me some calculations related to FRP cooling towers? Thanks in advance.

Anonymous said...

Sir I am want to know that how we increase our mechanical induced draft cooling tower efficiency.
What are the factors in which efficiency depend.

Anonymous said...

If. You have any pdf or any study material related to cooling tower efficiency please send me on my mail id
Prashanttiwari72@yahoo.com

Anonymous said...

Hi,

Could you tell me how I can find the outlet air temperature. I am designing a cooling tower and need to find this value.

Cooling Water inlet T = 40
Cooling Water outlet T = 32
Inlet Air T (dry) = 34
Inlet Air T (wet) = 31.5

Is there a way to use the psychrometric chart to find the outlet Air Temperature??

Thank youu!

vishwas said...

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.

Cooling Tower Efficiency Calculations said...

Hi Profmaster,

Very useful information... Please have a look on our website also. Hope it will be useful like yours..

Keep posting such useful things

Anonymous said...

We manufacture and export experimental water cooling tower instruments for school, college and teaching laboratory since 1954.We are based in Ambala.

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