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Hello out there,
I'm working on a project in which i am expected to find the quanity of water evaporated by the cooling towers used for cooling where i work.
We have a number of pumps that pump cooled water from the towers (3 in number) to different process areas, with a capacity totaling about 1000m3/h, a wet bulb temp of 82.4F, inlet and outlet temps of 107.6 and 89.6F respectively. Two of the CT are said to have a mass flow rate of 250m3/h,and the third 350m3/h (Pray,i'm confused about this mass flow rate)
I'm at my wits end. Any one to help???
Thanks
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Cooling Towers
Started by aderemijohn, Jun 05 2003 02:20 PM
8 replies to this topic
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#2
mbeychok
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Posted 05 June 2003 - 02:40 PM
Cubic meters per hour (m3/hr) is a volumetric flow rate ... it is not a mass flow rate.
Two towers each at 250 m3/hr plus a third tower at 350 m3/hr amounts to a total of 850 m3/hr. Thus your total capacity of 1000 m3/hr is either in error or you have not given us the correct information.
Two towers each at 250 m3/hr plus a third tower at 350 m3/hr amounts to a total of 850 m3/hr. Thus your total capacity of 1000 m3/hr is either in error or you have not given us the correct information.
#3
Guest_Arthur L. Mayclin_*
Guest_Arthur L. Mayclin_*
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Posted 24 June 2003 - 03:34 PM
You only provide flow rates to process areas that most likely return to the cooling tower. The flow rate you actually need is the make-up water flow rate to the cooling tower basin. You loose water in two ways from the tower: Evaporation and carry-over. Carry over is the result of small water droplets being carried with the vapor.
If you monitor the makeup flow into the cooling tower, and any blowdown flow (you often continuously blow down the basin to maintain a set conductivity (or cycles of concentration), the difference will be the amount you are losing due to evaporation and carry-over. If you don't know these flows then you have to do a heat balance and calculate the amount of vaporization required to disipate that heat. That is of course, more complicated.
Don't worry that you only have volumetric flow rates. You can calculate the mass flow rate from the provided water temperatures. The small error from not knowing the exact chemical composition of the water, or its presure, will be negligible.
Good luck...........
If you monitor the makeup flow into the cooling tower, and any blowdown flow (you often continuously blow down the basin to maintain a set conductivity (or cycles of concentration), the difference will be the amount you are losing due to evaporation and carry-over. If you don't know these flows then you have to do a heat balance and calculate the amount of vaporization required to disipate that heat. That is of course, more complicated.
Don't worry that you only have volumetric flow rates. You can calculate the mass flow rate from the provided water temperatures. The small error from not knowing the exact chemical composition of the water, or its presure, will be negligible.
Good luck...........
#4
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Posted 10 July 2003 - 10:25 PM
Evaporation Rate = Make up water rate Rate - Blowdown Rate
#5
Guest_Milton Beychok_*
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Posted 11 July 2003 - 09:47 AM
The response by Nazreenb is not quite correct. For an evaporative cooling tower system:
Evaporation = makeup - blowdown - windage losses
Evaporation = makeup - blowdown - windage losses
#6
Guest_durga prasad_*
Guest_durga prasad_*
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Posted 24 July 2003 - 09:58 AM
QUOTE (Milton Beychok @ Jul 11 2003, 09:47 AM)
The response by Nazreenb is not quite correct. For an evaporative cooling tower system:
Evaporation = makeup - blowdown - windage losses
Evaporation = makeup - blowdown - windage losses
[FONT=Times][SIZE=1][COLOR=blue]
Can u lease tell me how to calculate the windage losses
#7
mbeychok
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Posted 24 July 2003 - 11:08 AM
Durga Prasad:
For older induced draft cooling towers: W = 0.1 to 0.3 % of C
For older natural draft cooling towers: W = 0.3 to 1 % of C
For modern cooling towers: W = may be as low as 0.01 % of C
where: W = windage loss and C = circulating water rate
The best way to determine the windage loss of your cooling tower is to ask the manufacturer of your cooling tower. In any event, the windage loss is quite small.
For older induced draft cooling towers: W = 0.1 to 0.3 % of C
For older natural draft cooling towers: W = 0.3 to 1 % of C
For modern cooling towers: W = may be as low as 0.01 % of C
where: W = windage loss and C = circulating water rate
The best way to determine the windage loss of your cooling tower is to ask the manufacturer of your cooling tower. In any event, the windage loss is quite small.
#8
Guest_moha_irzeeqat@hotmail.com_*
Guest_moha_irzeeqat@hotmail.com_*
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Posted 24 July 2003 - 11:44 AM
Hello out there,
I'm working on a project in which i am expected to find the quanity of water evaporated by the cooling towers.
with a capacity totaling about 3m3/s, the air inlet at 35C, 65% relative huomidity. inlet and outlet water temps of 65 and 30C respectively.
I'm at my wits end. Any one to help???
Thanks
I'm working on a project in which i am expected to find the quanity of water evaporated by the cooling towers.
with a capacity totaling about 3m3/s, the air inlet at 35C, 65% relative huomidity. inlet and outlet water temps of 65 and 30C respectively.
I'm at my wits end. Any one to help???
Thanks
#9
mbeychok
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Posted 24 July 2003 - 12:10 PM
Moha Irzeeqat:
From a simplified heat balance around the cooling tower:
(E) (Hv) = ( C )(delta T) (cp)
where (in USA customary units):
E = evaporation rate in gallons/minute
Hv = latent heat of vaporization of water in BTU/pound = 1,000
C = cooling water circulation rate in gallons/minute
delta T = inlet water temperature, °F - outlet water temperature, °F
cp = specific heat of water in BTU/pound/°F = 1.00
Note that E and C do not require being converted to mass flow rate, because the same conversion would appear on both sides of the equation.
I will leave the conversion of the above heat balance equation into metric units to you.
From a simplified heat balance around the cooling tower:
(E) (Hv) = ( C )(delta T) (cp)
where (in USA customary units):
E = evaporation rate in gallons/minute
Hv = latent heat of vaporization of water in BTU/pound = 1,000
C = cooling water circulation rate in gallons/minute
delta T = inlet water temperature, °F - outlet water temperature, °F
cp = specific heat of water in BTU/pound/°F = 1.00
Note that E and C do not require being converted to mass flow rate, because the same conversion would appear on both sides of the equation.
I will leave the conversion of the above heat balance equation into metric units to you.
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