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Calculating Fan Power Involved For A Cooling Tower


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#1 Guest_Shen_*

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Posted 18 March 2009 - 01:23 PM

Hi there, I'm involved a project where I must determine the power requirements of a forced draft cooling tower that uses a large fan. Unfortunately I'm pretty new at cooling towers and I'm having several issues with this:

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Issue 1

I've read a lot of material on cooling towers at this point where I came across this:

Fan HorsePower = A*(cfm air flowrate)^B where A and B are fan "box" constant.

This equation came from this site:
http://www.me.gatech...eth/four.htm#A2

Which references this paper: http://www.me.gatech...s/liuthesis.pdf

Only problem is I don't understand how you get the A and B constants for fan box size?

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Issue 2

I read that:
1. for a given condition of drybulb temp, %sat and wet bulb temperature a lower limit for the water exit temperature can be determined.

I'm guessing that the lower limit of the exit temp is the wet bulb temperature? Should I be setting the exit temp to something this low?

2. Then given the water flowrate and energy removal rate (probably the cooling duty of my condenser/cooler) the inlet temperature of the water can be found.

I have a duty and I know the hot stream temperature but I'm not sure how this gets me a water inlet temperature.. is it just Duty = (water flowrate)(cpofwater)(cooling water temp - hot stream temp) where -1*(cooling water temp - hot stream temp) is my cooling tower inlet temp? But I'm not sure about the cooling water temp going into this scenario in the first place (from step 1 I'm guessing)


3. The 3rd thing they say is that minimum air flowrate can be estimated by assuming that the air flowing out of the tower is 100% saturated and has the same exit temperature as the water inlet temperature. You can then take an air flow 20-40% higher in order to get an approximate working airflow.

What are the equations for each of these steps?

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Issue 3

I also read somewhere that PA = ΔH / (cp * ΔT)

Where
PA = pumparound rate (my water flowrate I'm guessing)
ΔH = my condenser/cooler duty
cp = heat capacity of water
ΔT = cooling range

What is cooling range? Is it my cooling water outlet temperature - cooling water inlet temp?

What the heck is approach temperature?

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The conditions I have are that my cooling duty is a mere -1.3501794 kW for a stream that needs to be cooled from 40 degC to 25 degC. Wet bulb temp = 13.3 degC. Of course I will have other streams to solve eventually.

My goal is to figure out how much that duty translates to electricity/utility costs. I can neglect evaporative and windage losses so the main utility consumption comes from the fan power I think. Unfortunately I'm a little green at this whole concept, can anyone elaborate on the things I found above?

Thanks!

#2 djack77494

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Posted 23 March 2009 - 04:31 PM

shen,
It is asking too much to think that a cooling system can be designed by a responder based on the provided information. Let's look at some simplistic answers to your most important questions.

I'm working through your questions backwards. Your electricity costs are due largely to powering the fan(s) as well as the circulating pump(s). Don't neglect to add some extra water capacity for a sidestream filter. Everything needs a driving force, so you never reach equilibrium conditions. The closer your approach, the more you'll pay for it in equipment and operating costs. You can attempt to get close to your minimum water temperature = wet bulb temperature of 13.3C. Let's say you design for 15C for your cool water temperature. Your approach to wet bulb is 15 - 13.3 = 1.7C. Now say you circulate enough cooling water to accomplish all your cooling needs while heating the water to only 20C. Regarding your exchanger, your approach is 25 - 20 = 5C, which is a bit low but still in the realm of reasonable. Though I've never used this term, the cooling range is 20 - 15 = 10C. There is no simple equation for calculating air flow. Beginning with the air's inlet temperature AND humidity, you can lookup or calculate its enthalpy. Same for exiting air saturated at (say) 15C. By energy balance, the energy given up by your water = energy gained by air. Only unknown is the air flowrate, so calculate it. Energy removed is the water flowrate * Cp * (Tout - Tin). Finally, your Issue 1 concerning the equation for power consumption by the fan(s). The equation you provided is an empirical equation and constants "A" and "B" must be provided by the fan manufacturer.

#3 breizh

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Posted 27 March 2009 - 09:29 PM

Good day Shen ,
Let you do some "google" work .There is a document issued by SPX "Cooling towers-fundamentals "which may help you to answer your queries.
Pierre

#4 bhumin2511

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Posted 12 March 2012 - 11:27 AM

how to find out cooling tower Total power consumption ?

power consumptopn of pump and fan both..


i am waiting for accurate reply




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