Calculate Chilled Water Flow 2

[JRob71]

Hi,

I am studying for an exam through the Association of Energy Engineers and a practice question has me completely stumped. I am hopeful that someone could point me in the right direction and I will figure the rest out. To be honest, I do not even understand the question and I have gone down several rabbit holes trying to figure this out.

An absorption system with a COP of 0.8 is powered by hot water that enters at 200 F and exits at 180 F at a rate of 25 gpm. The chilled water operates on a 10 F temperature difference. Calculate the chilled water flow. You do not need to know how an absorption chiller works to solve this problem. Use COP = Q out/Q in.

The answer is 45 GPM.

Any ideas on how I can figure this out? Please do not solve this for me, (at least not yet) but I would be grateful if anyone could point me in the right direction. The part of the confuses me about the question is use the Q out of what and the Q in of what?

Please do not confuse this as posting coursework, as this is a self study for a certification and I am not looking to cheat, especially since I already have the answer. However, I would like to learn how to get the answer.

Thanks!

 

[IRstuff]

I deleted my post, mainly because I keep getting 40 gpm. In any case, you should start with the mass transport and the 20F drop vs. 10F rise

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[IRstuff]

This study guide says the answer is iii for problem 10:

https://studylib.net/doc/8651325/part-c--study-guide---association-of-energy-engineers

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[JRob71]

IRstuff: Thank you for your effort. I will see if I can adopt the thought process you suggested and see where it leads me. I will post back if and when I figure out how we get to 45gpm. Some of these questions have been wrong, so your 40 gpm might be correct. Thanks for the tip!!!

 

[JRob71]

I also see what you mean. Answer III is 40 gpm. Good catch and sorry I put the wrong answer. Now I just have to figure out how to get to 40 gpm and will use your suggested approach. I'm very guilty of getting so wrapped up in trying to figure out the hard part, I miss an easy detail. Thanks for pointing it out.

 

[JRob71]

OK, so I did some math and I came up with 40 GPM. Now is it coincidence or is the the way to go about solving this word problem.

We have 20 degrees going out and 10 degrees coming in. Therefore using q-out / q-in we have 20/10=2. The word problem says COP is .8 and I think the word problem is saying that the hot side is running at 25 GPM. Does it stand to reason that the solution is .8(COP) X 2 (q-in / q-out) X 25 GPM = 40 GPM?

I'm really not sure what I am looking at here but the math seems to work and come up with an answer of 40 GPM for the cold side. Is this solved correctly or is it incorrect and a coincidence that I am arriving at the known answer of 40 GPM?

Thanks again for the help!

 

[xnuke]

If you aren't sure what you're looking at, you don't understand the solution you came up with. That isn't good. Try following this, with subscripts c and h for cold and hot, respectively:
COP = Q[sub]c[/sub]/Q[sub]h[/sub]
COP = m[sub]c[/sub]c[sub]p[/sub]ΔT[sub]c[/sub]/(m[sub]h[/sub]c[sub]p[/sub]ΔT[sub]h[/sub])
COP = v[sub]c[/sub]ρ[sub]c[/sub]c[sub]p[/sub]ΔT[sub]c[/sub]/(v[sub]h[/sub]ρ[sub]h[/sub]c[sub]p[/sub]ΔT[sub]h[/sub])
Assuming ρ[sub]c[/sub] = ρ[sub]h[/sub] (which they really don't, but the answer they give requires this assumption because neither the cold water temps nor the densities are given) and canceling:
COP = v[sub]c[/sub]ΔT[sub]c[/sub]/(v[sub]h[/sub]ΔT[sub]h[/sub])
Therefore:
v[sub]c[/sub] = COPv[sub]h[/sub]ΔT[sub]h[/sub]/ΔT[sub]c[/sub]
which gives 40 gpm when the numbers are substituted.

xnuke
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[JRob71]

xnuke: this is terrific. I still have to work with this a little more to be absolutely sure I understand it and it looks pretty straight forward. Thank you so much!

 

[xnuke]

You're welcome. Sorry I couldn't put the dots over the Q, m, and v terms to show they are rates.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
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[JRob71]

xnuke: I am pleased to let you know that I have worked through your response and have learned a lot from your guidance. You assumed that I knew some things that I did not, but I went and figured it out, which is a good thing. I struggled with the density and constant pressure (P and CP) as it has been about 30 years since I have been exposed to these concepts and did not even know what they stood for. So I now understand your comment about canceling them out. If we assume the values are the same for both the hot and the cold side, then they just cancel out. I'm glad you pointed this out because if the values are provided, I will know I can just plug in the values. I am taking a class for at the end of May and trying to get a head start. Thanks again and I can finally move past this problem.

Stay Safe and thanks for sharing the knowledge!!!

 

[xnuke]

Glad to hear you're figuring it out. Just so you know, c[sub]p[/sub] is the specific heat capacity of the fluid at a constant pressure, not constant pressure. Good luck with your class.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[JRob71]

Thanks for the correction and sending the luck!!!

 

[IRstuff]

Mass transport balance is something you should already know; it's not any different than balancing chemical reaction equations.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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[JRob71]

Thanks IRstuff,

Thanks for the comparison to balancing chemical reaction equations and I will check them out.

I'm an industrial Energy Manager and have not done a thing with Chemistry since college days a few decades ago. My field in energy management requires a cursory knowledge of some engineering concepts relative to buildings such as building envelopes, HVAC, lighting, electric motors, power factor correction, Transformers, Generators, and anything else that consumes energy or water. My career also requires knowledge of financing, engineering and statistics.

Someday, there will be an Energy Manager forum, but till now, engineering and math forums are most closely matched to my background.

Thanks for your feedback and help!

Best...

 

[waross]

Heat in:
(200F - 180F) x 25 GPM x X(BTU per gallon) = 500X heat content.
Heat out, Flow out:
(500X / 10F) x .8 = 40 GPM
The BTUs per gallon cancel out so you may use a constant, without knowing the actual value.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[JRob71]

Waross: Nice simplification of the problem. I like it! With all these great answers, I will be an expert in no time!
Thanks...