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# HVAC&R Practice Problem of the week

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1 minute ago, Slay the P.E. said:

Ok. I see now. The enthalpy part is correct. That's how you obtain h6.

The way you are finding T6 is not correct, though. Where does that equation come from?

Temperature part. I don't need to convert the flow to mass don't I? Since its SCFM it would give the same effect. and cP of air I assumed same or if there's any difference it would be not a game changer.

mcdeltaT 1-2 = mCdeltaT5-6?

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3 minutes ago, Vel2018 said:

Temperature part. I don't need to convert the flow to mass don't I? Since its SCFM it would give the same effect. and cP of air I assumed same or if there's any difference it would be not a game changer.

mcdeltaT 1-2 = mCdeltaT5-6?

You’re right about the SCFM. No need to change to mass.

Moisture content can be a game changer with respect to cp of air. Don’t always assume that in Psychrometrics problems.

The proper thing to do is an energy balance to get h6 and a water mass balance to get w6

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1 minute ago, Slay the P.E. said:

You’re right about the SCFM. No need to change to mass.

Moisture content can be a game changer with respect to cp of air. Don’t always assume that in Psychrometrics problems.

The proper thing to do is an energy balance to get h6 and a water mass balance to get w6

I see. But I suppose, if in case point 2-3 did extract more moisture and hotter, it would reflect on both enthalpies and temperature on the other side of the flow since there is no loss. Unless there's a leak and moisture loss in the process then I would only do water balance.

So I redid the process to see the difference, using water balance so ended getting 75.4 Dew point using water balance.

Then I re-plotted using mcdeltaT,  it landed exactly at 75 F see photo in the link. I guess using mcdeltaT deviated by 0.4deg.

Can't upload anymore photos so here's the link I uploaded on imgur https://imgur.com/pJbZOEk

I tried an experiment I made point 3 at 75Fdb and W3=13grains.

This time it lost an enthalpy of 0.8 from point 2-3 unlike the first one where it "gained" so the enthalpy change now in this experiment on the other point (point5-6) is "gained" just energy transfer there, deltaH on point point 5-6 should be 1.2. Point 6 should have 165 grains with Tdp of 81.7F.

Now using mcdeltaT same approach 6000(75-51=4000(115-T6) ====> T6= 79Fdb this vertical line intersect with the point6 enthalpy line which is 44.7 and crossed paths exactly at 165gr. And it appears it condenses.

Just sayin, I think as long as it is assumed that there is no losses, leaks, change of phase, etc. It should arrive at about the same. Otherwise it will violate the law of energy balance. But anyway, I will mess around with this topic later. For now I will follow your advice, to just do enthalpy and water mass balance. xD

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@Slay the P.E., thanks for clearing that up! I understand the problem now!

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I've attempted the cruise ship ammonia refrigeration system problem posted on Mar. 29 and am stumped.. It would seem that more information is needed to determine the refrigeration cycle state points and then therefore be able to determine the compressor discharge pressure... Either evaporator water flow rate, condenser water delta T, or refrigerant mass flow rate? Can you privide a hint on this one without giving the whole solution away? I suspect that I am overthinking it somehow... Thanks..

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51 minutes ago, Jimbo Three said:

I've attempted the cruise ship ammonia refrigeration system problem posted on Mar. 29 and am stumped.. It would seem that more information is needed to determine the refrigeration cycle state points and then therefore be able to determine the compressor discharge pressure... Either evaporator water flow rate, condenser water delta T, or refrigerant mass flow rate? Can you privide a hint on this one without giving the whole solution away? I suspect that I am overthinking it somehow... Thanks..

Hi Jimbo.

You're not overthinking it. I think it might be just a tad too gnarly for the PE exam because it involves a trial-and-error approach -- something I don't think I've ever seen in any official NCEES test prep material. Remember you are given COP (which relates several enthalpies) and the quality at the condenser discharge.

I still encourage you to give it a shot, as the solution requires a good grasp of the vapor compression cycle.

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Thanks.. I do Naval/Marine HVAC design engineering, and this is the sort of problem I would encounter at work... my initial intuition was to set up an excel spreadsheet to brute force it.. I might circle back to it after Friday once I'm done with this GD exam and have more than six minutes a pop..

On 3/29/2018 at 11:28 AM, Slay the P.E. said:

Happy Thursday. This one is good for the more daring TFS folks too...

A refrigeration system for comfort air conditioning in a cruise ship operating with ammonia is schematically shown. The condenser is cooled with 450 gpm of seawater. A network of remotely located fan coil units (FCUs) uses chilled water-glycol provided by this system. You may assume that 1) any pressure loss within the ammonia system is negligible and 2) that the ammonia is discharged from the condenser as a saturated liquid. If the coefficient of performance, COP=4.0, the compressor discharge pressure (psia) is most nearly:
(A) 100
(B) 180
(C) 200

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14 minutes ago, Jimbo Three said:

Thanks.. I do Naval/Marine HVAC design engineering, and this is the sort of problem I would encounter at work... my initial intuition was to set up an excel spreadsheet to brute force it.. I might circle back to it after Friday once I'm done with this GD exam and have more than six minutes a pop..

If you didn't have answer choices you could plug-in and try, you would have to take that approach. Also, I think I should've put in the problem statement that the compressor may be assumed to operate isentropically.

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16 minutes ago, Slay the P.E. said:

If you didn't have answer choices you could plug-in and try, you would have to take that approach. Also, I think I should've put in the problem statement that the compressor may be assumed to operate isentropically.

Just saw this, I think it did pass little bit 200 but still very close to 200 so C? First find Thigh using COP then plot isentropic compression.

Did I do it right @Slay the P.E.?

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Posted (edited)

200 is not correct.

How did you find T_high using COP?

Edited by Slay the P.E.

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Posted (edited)
9 minutes ago, Slay the P.E. said:

200 is not correct.

How did you find T_high using COP?

Oh I bombed it. Sorry, was very tired now jajaja

So I redid it and I see that it should go up to Th = 140 + 20 deg superheat = 160deg with isentropic and then it lands to 100psi! So it should be A!

Right @Slay the P.E.?

Edited by Vel2018

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9 minutes ago, Vel2018 said:

Oh I bombed it. Sorry, was very tired now jajaja

So I redid it and I see that it should go up to Th = 140 + 20 deg superheat = 160deg with isentropic and then it lands to 100psi! So it should be A!

Right @Slay the P.E.?

100 psi is also incorrect. What's the 140? Where are you getting that?

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Just now, Slay the P.E. said:

100 psi is also incorrect. What's the 140? Where are you getting that?

COPref=TL/TH-TL Tlow is in 30psi thats in 0deg line with amonia ph chart. So I got Thigh 140deg.

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12 minutes ago, Vel2018 said:

COPref=TL/TH-TL Tlow is in 30psi thats in 0deg line with amonia ph chart. So I got Thigh 140deg.

Are you sure you can use that definition of COP here?

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9 minutes ago, Slay the P.E. said:

Are you sure you can use that definition of COP here?

Maybe not, I'll try again later, I'm pretty beat up now. I think need to rest. xD

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Posted (edited)

Ok so is this some kind of trial and error? COP = Qin/Win say 4=h1-h4/h1-h2

4 = 620btu/lbm - h4(I tried enthalpy at 180psi saturated liquid which is 140BTU/lbm)/620BTU/lbm - h2

I got h2 = 740 and then I plot isentropic line from P=30psi with 20deg superheat, at the point it crossed path with 740BTU it is align with 180psi.

I tried all other pressures in the choices and this is the only one that worked isentropically and aligned from the assumed value of h4 saturated liquid.

Edited by Vel2018

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8 minutes ago, Vel2018 said:

Ok so is this some kind of trial and error? COP = Qin/Win say 4=h1-h4/h1-h2

4 = 620btu/lbm - h4(I tried enthalpy at 180psi saturated liquid which is 140BTU/lbm)/620BTU/lbm - h2

I got h2 = 740 and then I plot isentropic line from P=30psi with 20deg superheat, at the point it crossed path with 740BTU it is align with 180psi.

I tried all other pressures in the choices and this is the only one that worked isentropically and aligned from the assumed value of h4 saturated liquid.

Yes! This is correct. It requires trial-and-error. Nice work.

This time you used the appropriate definition of COP. The one with absolute temperatures is only for Carnot cycles.

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9 minutes ago, Slay the P.E. said:

Yes! This is correct. It requires trial-and-error. Nice work.

This time you used the appropriate definition of COP. The one with absolute temperatures is only for Carnot cycles.

Got it! Thanks!

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I liked these problems. Do you have any book for practice questions?

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9 hours ago, ManojD said:

I liked these problems. Do you have any book for practice questions?

Hi ManojD. We have an e-book titled "Psychrometrics and Basic HVAC System Calculations Practice Problems for the HVAC&R Exam"

You can download free sample pages here: https://www.slaythepe.com/hvacr-psychrometrics.html and then if you like it, you can buy it from our store page.

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