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 DataCooling water flow rate - 4134 M3/hrCooling water inlet Temp - 44.0 °CCooling water exit Temp - 35.0 °CInlet air-wet bulb - 30.0 °CInlet air-dry bulb - 38.8 °CExit air-wet bulb - 40.7 °CExit air-dry bulb - 42.0 °C

Now follow step by step procedure for the calculation.

Step-1Calculate waterside actual heat load, which is as below

Qw = 4134 x 1000 x (44 – 35) / 1000000= 37.21 Gcal/Hr

Step-2Calculate 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-3Calculate 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-4Find 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-5Based 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-6Similarly find out the moist air enthalpy at exit condition, which is

Hex = 41.630 Kcal/Kg of wet air

Step-7Similarly, 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-8Calculate 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-9Now 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 °CLatent 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 PartI 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....This is in continuation of my previous post on this topic. In this part, I will explain the calculation of NTU for cooling towers, yes NTU which is very important & is similar to NTU in absorption towers.

It helps in indentifying the performance, capacity & effciiency of your cooling tower. In next part of this post I will explain How to use these calculations for mesurement of efficiencies, prediction for new conditions etc.

Now we will see the NTU calculation & efficiency of tower, use of NTU method for predictions etc.

Step-1First consider the cooling water exit temperature ‘twex’ in column A in excel sheet so i.e. 35°C in this case. All the data is given in Part-I...So Check it First.

Put h’ in column B which is the enthalpy of saturated air at twex and can be calculated by the equation

h’=9.446443x10-13x(twex^8)-1.433603766x10-10x(twex^7)+5.39506924*10-9(twex ^6)+3.02962638*10-7(twex^5)-0.0000272854755*(B7^4)+0.00096596975*(B7^3)-0.005340108*(B7^2)+0.458708485*B7+2.219286635

Put tawet in column C starting with actual wet bulb temperature of entering air, which is 30°C in this case.

Put w as absolute humidity at tawet in column D that is calculated from the same formula as shown in Part-I of this post.

Put hcal as humidity at tawet using the formula given above for h’ in column E.

Put ha as humidity at actual wet bulb temperature of entering air, which is 30°C in this case. Yes, that means initially in the first row of calculation sheet hcal & ha will be same. This is in column F.

Now put calculation of difference of h’ – ha in column G.

Step-2In first row G will be automatically zero.Now in second row consider the twex 2 = (Twin – Twex)/19 + twex 1

i.e. twex 2 = (44 – 35 ) / 19 + 35= 0.474 + 35 = 35.474°C

Copy this formula in column A for next 19 rows. This gives you incremental evaluation of tower step by step along the total tower height from 35° at exit at bottom to 44° at inlet at the top.

Copy h’ formula in column B for the same no of rows.

Step-3Now put any assumed figure for tawet in column C, w in column D, hcal in column E.

Now calculation for ha will change which will come from actual L/G ratio of tower calculated in Part-I.

Use the following formula for ha in second row onwards.

ha 2 = ha1 + L/G * (twex 2 - twex 1) + (w 2 – w 1) / 1000 * twex 1= ha1 + 1.715 * (35.474 – 35.00) + (w 2 – w 1) / 1000 * 35.0

Based on other figures it will vary.

Now since you have assumed tawet, hcal will be different from ha. Put this difference in next column G.

Now either change tawet manually to make the difference Zero in column G or use goal seek from excel. This will give you tawet, which is supposed to be the actual wet bulb temperature of air exiting from the tower at the top finally.

This will complete first part of NTU calculation after completing all the rows.

Step-4Now in next column i.e. H; put (h’ – ha) value which is Column B – Column F and copy it down till the last row.

Put reciprocal of column H in column I. This will give you 1/ (h’ – ha) value and copy it down till the last row.

Now in next column J, leave first row blank & start from second row where you should put average of first & second row in column I. This will give you average of 1 / (h’ – ha) for first & second value. Copy this formula also down till the end of rows.

Step-5Now in column K, put NTUL as calculated below (From second row as column J starts from second row).

NTUL = Column J x (twex 2 - twex 1)= Column J x (35.474 – 35.0)

Copy this formula in all rows.

In column L, put progressive summation of NTUL calculated in column K i.e. in each row of column L, use previous row of column L + same row of column K.

This value at the end of last row will give your towers total NTU for liquid side.

Step-6Repeat all calculations in next two columns for NTUG similar to Step-5 above and find out final value of gas transfer units. The only difference is to use the following formula to calculate NTUG in column M.

NTUG = Column J x (ha 2 - ha 1) ha is in column F.

Use progressive sum again in column N.

Now I will give you guidelines on using these calculations for prediction of performance, prediction of new conditions, calculation of existing system and how to improve it in the next part of this post.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?

9:30 PM 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.

1:37 PM 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.

2:10 PM 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

2:50 PM 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.

9:55 AM 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?

10:00 AM 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

1:34 PM 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.

regardssuresh

11:02 PM 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.

10:04 AM 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

12:19 AM Roshanlal said...

Hi ProfmasterThanks 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

6:57 PM 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 CI need to calculate the following1. Adiabatic Sat temp2. Water circulation rate3. 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

3:25 PM 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.

9:48 AM 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

7:19 AM profmaster said...

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

9:58 PM Anonymous said...

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

4:12 PM profmaster said...

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

8:14 PM 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

4:54 PM profmaster said...

Dear JBCooling 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.

1:26 PM .

Dear sirWe are Farabard Company located at Shiraz/south of Iran. Farabard is one of the designer and constructor of wet counter flow concrete cooling towers.We studied your performance calculation about cooling towers (NTU calculation). Your method is very good but there are some ambiguities for us. We invite you to come Iran and hope that you clear these ambiguities for us. Also we will be spent you all of your travel expenditures (include: air ticket, hotel, …).Your prompt action will be appreciated.Best RegardPlease contact us:info@farabard.ir+987116294813+987116266067

1:00 PM farabard said...

dear sirIm sepideh samghani.I study mechanics engineering in shiraz university.I work as a trainee in farabard company for 2 mounths.I have studied your performance calculation of cooling tower and I have some question about it. could you please help me to understand them.these are my quastion ...1-how did you calculate the enthalpy content of moist air inlet,based on %saturation?2-how did you calculate the amount of evaporated water?please submit your direct tel/mobile number or fax or E-mail.your quick response is highly appreciated.

Best Regardse_mail:info@farabard.irtel:+98-711-6294813fax:+98-711-6266067

2:10 PM profmaster said...

Regarding your Questions....1. First I used empirical equation which is already given in the post to find out the enthalpy of completely dry air and completely wet air at given dry bulb temp. Now by diff you have vapor enthalpy which can be added to dry air enthalpy after multiplying by %Saturation. So you have wet air enthalpy.

2. It is found by trial & error method based on enthalpy condition of air at inlet & exit.

I prefer only Email ocntact which you can do thru comments on this post.

11:52 AM TheServant said...

Hi profmaster,I am confused with some of the wording here.At the end of Step 1 you say make h'-ha in column G, and that that is "automatically zero" in step 2. I don't think it is, do you mean hcal-ha as you suggest in Step 3?

Also, I am unclear about what to put in tawet column (Step 3) where you say any assumed figure? Does it vary for the number of rows?

Finally, the number you use is 19, and I have forgotten, is that equivalent to column height you refer to with that number?

Thanks for your help.

7:06 AM profmaster said...

Oops Yes dear,You got it right. It is hcal-ha Not h'-ha in Column G.

No Tawet is any assumed figure at different intervals to make the balance by trial & error. So initially U assume it & then make difference in G column zero by varying it. That is your final tawet figure.

19 is a no based on my assumtion of 0.5° interval steps for 35 to 44°C. You can increase it for better results but this is generally sufficient. So if U use 0.1° interval U need 91 rows.

9:34 AM Anonymous said...

Hi Profmaster,I am refrigeration engineer and I tray to optimize an cooling tower as you suggested. But with this equation if 99.9% is ok it is not good as you know 100% is good.So can you send me a general(or an example will be better) excel sheet with what you propose here?My email address is poem_ro@yahoo.comThanks for your article.

5:15 PM profmaster said...

If you are an engineer then first thing you should understand that no engineering calculation is 100% accurate, they are not even 99.9% correct. Almost all corelations are 95-99% accurate & therefore design margins are always considered.

Regarding sheet I need to know your own efforts first in this direction.

12:26 PM Balaji said...

Dear Sir,

I was checking on cooling tower efficiency and found your useful 2 part calculations. However you have mentioned you will explain how to use these calculations to measure efficiency. But I am not able to find the post. Kindly help me with the link to that post. Thank you.

9:36 PM Anonymous said...

Dear Sir,

I got the task to design a cooling tower in detail, can these calculation performance help me plzzzzzz

7:12 PM antony said...

Sir please post the third part of the calculation soon sir.I have to finish my project work with this sir.

2:35 PM Darien said...

Sorry but I dont really understand Step 1 "Put hcal as humidity at tawet using the formula given for h' in column E". Is it the formula for hcal is the same for h' ?? but h' is enthalpy and hcal is humidity. Please advice. Thanks

1:48 AM profmaster said...

Dear DarienThe formula for h' is already given (Long formula mentioned in the post also). In that column h; is calculated at twex which means temperature of water at exit condition.

Now for hcal use tawet in place of twex in the same formula. tawet means wet air temperature.

I hope this clears. h is enthalpy & not humidity.

1:32 PM