## January 28, 2008

### Cooling Tower: Performance calculation - II

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-1
First 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-2
In 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-3
Now 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-4
Now 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-5
Now 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-6
Repeat 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.

## 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....

## January 03, 2008

### Improve efficiency of power transmission equipment

Happy New Year'2008 to All!

Power transmission equipment e.g. gears or couplings are the most neglected areas where nobody (including good energy auditors also) focuses for energy saving opportunities.

However, a certain approach or simple steps can improve their efficiency by at least 2% which may result in total annual cost saving in power bill by more than 6%.

Comparative lubricant studies have shown that power inefficiencies result more often when mineral-based lubricants are employed. The drawbacks of using mineral lubes are fluid churning, friction wear, higher operating temperatures, increased need to re-lubricate components, or even outright component replacement. Over time, these issues not only increase the amount of energy being released, but drive down the efficiency rates of production, potentially diminishing profitability.

Here is the comparison of efficiency for a Worm Gear with mineral lube & synthetic lube oil.

Choosing a synthetic over a mineral-based oil maximizes energy efficiency as the initial investment proves to be far less than the cost of energy down the line. When synthetics are tested in controlled experiments, they have shown a 2% to 8% improvement in energy conservation over mineral oils. This is due to the fact that synthetics provide a proprietary chemical makeup that protects itself from formula depletion thus extending the effective life of the lubricant.

Due to the performance capabilities of synthetics, companies operating in extreme hot or cold environments also achieve increased energy efficiency. A recent trial of 12 pump jacks in Canada, at bitter temperatures down to -51°C/-60°F, gained a 14.4% reduction in energy consumption from the product’s low viscosity and Pour Point. Other benefits include, extended gear performance by the elimination of friction, reduced start-up torques, lower amperage draw and improved lubricant pumpability. Furthermore, the formulation minimizes downtime, prevents pre-mature wear and requires change-outs after five years, supporting decreased lubricant use

So it means that the efficiency depends on ambient condition also.

Here is one simple comparison of Synthetic Vs Mineral.

Cost of Lube can be paid back in a years time.

So check out in your plant that all rotary machines are using right kind of lubricant for better energy efficiency.