HVAC&R Practice Problem of the week

[Slay the P.E.]

A little clarification seems to be in order here. 

Equation 18.30 in MERM 13 (see attached photo) is based on the conditions at the fluid surface at the top of an open (to atmosphere) fluid source (e.g. tank, reservoir, etc). As this problem shows, that situation is by no means the only one that can be presented. We recommend using only equation 18.31 which is the actual definition of NPSHA: the actual total fluid energy at the pump inlet. Equation 18.31 is based on the conditions at the immediate entrance (suction, subscript s) to the pump. 

In our practice problem (the one with the condenser above the pump) we don't know anything about the pump suction but we need the sum of pressure head h_p,s and velocity head h_v,s at the pump suction. However, we can get (h_p,s + h_v,s) by applying the extended Bernoulli equation between the free surface inside the condenser and the pump inlet.

 

[Slay the P.E.]

Cross-posting to the TFS thread as HVAC&R folk should be comfortable with this kind of problem too:

 

[medieval_pancake]

Happy Friday! Here's a fresh practice problem.

The sketch shows a packaged unit dedicated to conditioning the air in a natatorium (an indoor swimming pool room). The unit uses a silica gel desiccant dehumidification wheel. The unit's heating coil provides the reactivation energy for the desiccant dehumidification process. The unit's cooling coil cools and dehumidifies the air prior to entering the desiccant wheel. The accompanying table provides the known information for the summer design condition. 

Assume all the moisture removed by the wheel from the air at state 2 is transferred to the air exhausted to the atmosphere. Under the conditions described above, the dew point temperature (°F) at state 6 is most nearly:

(A)   51

(B)   67

(C)   72

(D)   75

View attachment 10933

View attachment 10934

@Slay the P.E., I'm not sure what I did wrong. I came up with a dew point temp of 73degrees and I would have chosen answer (C) wrongly...See the picture I've uploaded of my work. I found that 22gr/lb of moisture is transferred from point 2-3, so that means at point 6, the humidity ratio would bee 122gr/lb (100+22) correct? @122grb the dew point temp = 73F.

 

[Vel2018]

@Slay the P.E., I'm not sure what I did wrong. I came up with a dew point temp of 73degrees and I would have chosen answer (C) wrongly...See the picture I've uploaded of my work. I found that 22gr/lb of moisture is transferred from point 2-3, so that means at point 6, the humidity ratio would bee 122gr/lb (100+22) correct? @122grb the dew point temp = 73F.

View attachment 11040

I just did it today. Never did try it before coz I really don't like much psychrometrics jajaja

So what I did here was I get first the temperature and enthalphy change from point 2 to 3. 

Then I equate both to points 5 to 6. So I did get T6 around 88Fdb so I drew a vertical line there. Then from point 5, I deducted the enthalpy change and make a point 1.8delta from enthalpy at point 5 and then drew a line. 

At that intersection, thats the state of air at point 6, then draw a line straight to saturation thats the dew point? I got about 75.6 F???

 

[Slay the P.E.]

@Slay the P.E., I'm not sure what I did wrong. I came up with a dew point temp of 73degrees and I would have chosen answer (C) wrongly...See the picture I've uploaded of my work. I found that 22gr/lb of moisture is transferred from point 2-3, so that means at point 6, the humidity ratio would bee 122gr/lb (100+22) correct? @122grb the dew point temp = 73F.

First of all, congratulations on creating the coolest screen name I've seen here.

One thing I see wrong is that you don't seem to be accounting for the fact that the mass flows of the streams crossing the wheel and exchanging moisture have different flow rates. There are 6,000 CFM at 2 but only 4,000 CFM at 5. So, though you are correct that 22 grains of moisture per pound at state 2 are moved to the air at 5, the moisture content of the air at 5 will NOT increase by 22 grains per pound. You have to do a formal water mass balance for a control volume encompassing the wheel.

Take a look at our solution, attached here.

 

[Slay the P.E.]

So what I did here was I get first the temperature and enthalphy change from point 2 to 3. 

Then I equate both to points 5 to 6.

hmmm...

Not quite accurate, though. The change in temperature and enthalpy would be the same only if the flow rates at 5 and 2 were the same -- but they're not.

 

[Vel2018]

hmmm...

Not quite accurate, though. The change in temperature and enthalpy would be the same only if the flow rates at 5 and 2 were the same -- but they're not.

oh of course not, I did account the flows, so say in point 1-2 I did get only enthalpy change of 1.2 , then same principle as Q1-2=Q5-6. I got enthalpy change of 1.8 on 5-6 and so on with the delta T. 

Thats how I plotted it. So that is wrong? but I arrived at 75.6.

To be clearer since its SCFM values right. Mass ratios would be the same so I just did the Flows for easier to type:

For Temperature: 6000(69-51) = 4000 (115-T6)

T6=88

For Enthalpy 6000(deltah=1.2) = 4000(deltah5-6)

deltaH5-6 = 1.8

 

[Slay the P.E.]

oh of course not, I did account the flows, so say in point 1-2 I did get only enthalpy change of 1.8, then same principle as Q1-2=Q5-6. I got enthalpy change of 1.8 on 5-6 and so on with the delta T. 

Thats how I plotted it. So that is wrong? but I arrived at 75.6.

I'm sorry. Its hard to follow what you did. What's this 1.8? No units.. confusing. Are you saying h2-h1=1.8 Btu/lbm? That doesn't seem right. Then you say you got "enthalpy change 1.8 on 5-6" so it seems to me you are saying h2-h1 = h6-h5 which is not correct. Again, I'm sorry you may be doing everything right. I'm just having a hard time following your description. Can you post a shot of your work?

It's important to clear this up because we don't want you to be arriving at the right answer using wrong techniques.

 

[Vel2018]

I'm sorry. Its hard to follow what you did. What's this 1.8? No units.. confusing. Are you saying h2-h1=1.8 Btu/lbm? That doesn't seem right. Then you say you got "enthalpy change 1.8 on 5-6" so it seems to me you are saying h2-h1 = h6-h5 which is not correct. Again, I'm sorry you may be doing everything right. I'm just having a hard time following your description. Can you post a shot of your work?

It's important to clear this up because we don't want you to be arriving at the right answer using wrong techniques.

To be clearer since its SCFM values right. Mass ratios would be the same so I just did the Flows for easier to type:

For Temperature: 6000(69-51) = 4000 (115-T6)

T6=88

For Enthalpy 6000(deltah=1.2) = 4000(deltah5-6)

deltaH5-6 = 1.8

SORRY for earlier lol was a typo...ofcourse they are not equal lol

So was this correct or not? Its principle is energy balance. I hope it is correct.

 

[Slay the P.E.]

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?

 

[Vel2018]

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? 

 

[Slay the P.E.]

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

 

[Vel2018]

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

 

[medieval_pancake]

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

 

[Jimbo Three]

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.. 

 

[Slay the P.E.]

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.

 

[Jimbo Three]

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.. 

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

 

[Slay the P.E.]

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.  

 

[Vel2018]

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.?

 

[Slay the P.E.]

@Vel2018

200 is not correct.

How did you find T_high using COP?