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#1 mizzoueng

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Posted 25 August 2006 - 03:28 PM

my boss gave me this project to do and then left for the day. I have several condensate lines running through what he says are heaters, i also have a pumped condensate line going through this heater.

do i treat this heater like an exchanger to get my T and enthalpy on each side, or am i missing something?

#2 Timber

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Posted 25 August 2006 - 03:31 PM

Sounds like a question for a Thermo expert.

#3 DVINNY

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Posted 25 August 2006 - 03:54 PM

Multiply T by 1 minus the inverse tangent and you'll get an answer that is linearly related to enthalpy vs. an exponencially related answer that would be derived from the negative third root of the base T multiplied by the heat loss factor that can be obtained from the Penske charts.

Does that help?

#4 DVINNY

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Posted 25 August 2006 - 03:55 PM

BTW, I had Themo I, II, and III in college and got A's in all three. Can ya tell?

#5 mizzoueng

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Posted 25 August 2006 - 04:08 PM

:suicide:

#6 Sammy

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Posted 21 December 2008 - 06:21 AM

QUOTE (mizzoueng @ Aug 25 2006, 04:08 PM) <{POST_SNAPBACK}>
suicide.gif

yah, condensers can be treated as heat exchangers if you substitute the capacity rate as infinite on the side the condensation takes places( i.e with a capacity ratio, cmin/cmax=0)




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