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Water Cooling Problem

Water Cooling Problem

Water Cooling Problem

(OP)
I am working on a problem where I need to figure out the GPM of a cooling water.  A customer wants to dilute 130°F  circulating water (125 GPM) with 50°F water in order to cool it to 75°F.  This is a continually flowing system.

The 75°F water enters the process, which heats the water to 130°F. He then wants to add the 50°F water directly to the system to bring it back down to 75°F before re-entering the process.

I've worked with heat exchangers before, but this is a little different.  I know this should be easy to figure, but I'm at a loss.  

Thanks.

 

RE: Water Cooling Problem

Simple energy balance (assuming you are sending a flow to drain equal to the in-flow of cooling water).

Flow_Hot x Cp x (130-75) = Flow_cold x Cp x (75-50)

RE: Water Cooling Problem

(OP)
Is Cp a constant? In some heat exchanger calculations I've used Cn.  Are these the same?

RE: Water Cooling Problem

Cp is the heat capacity of water, BTU/lb or BTU/gal.

Over the small range of temperatures of your problem, it may be considered a constant.

RE: Water Cooling Problem

(OP)
Many thanks.  That seemed to give me what I needed.

RE: Water Cooling Problem

I thought you said dilute.

If you dump 86 GPM of 130F water and then inject 86 GPM of 50F water, your end up with 125 GPM of 75F water. No exchanger required.

RE: Water Cooling Problem

(OP)
Danberry,

What formula are you using?

RE: Water Cooling Problem

Oops, my bad on earlier post.  That's what I get for doing things in my head rather than drawing the picture.

Correct formula is

Flow_Hot x Cp x (130-75) = Flow_cold x Cp x (130-50)

Resulting in flow_cold = 86 gpm to agree with Danberry.

RE: Water Cooling Problem

Mintjulep has it. Note that Cp is irrelevant.

An exchanger will require more cold water but avoids dumping the hot process water; an advantage if it is contaminated.

RE: Water Cooling Problem

(OP)
Thanks guys.  This really helps some guys out in the field making some modifications.

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