# HVAC&R Practice Problem of the week



## Slay the P.E. (Mar 13, 2018)

Here's a good one to practice some psych chart navigation. The more adventurous TFS folks might want to give this one a shot too. Have fun:

An air conditioning unit handles 111,000 SCFM of 100% outdoor air at 80F, 70% r.h. A single coil performs cooling and dehumidification. The coil's apparatus dew point is 55F, and the condensate is noted to flow out of the unit at a rate of 5 gallons per minute.  Under these conditions, the coil bypass factor is most nearly:

(A) 0.10

(B) 0.15

(C) 0.20

(D) 0.25


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## mongolianbbq (Mar 13, 2018)

adventurous TFS person here. I got B. I might need to invest in a ruler (assuming finding the intersect for the humidity ratio exit between dew point and condition at the entrance was the correct way to solve this)


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## Slay the P.E. (Mar 13, 2018)

mongolianbbq said:


> adventurous TFS person here. I got B. I might need to invest in a ruler (assuming finding the intersect for the humidity ratio exit between dew point and condition at the entrance was the correct way to solve this)


This is, unfortunately, not correct. How did you obtain the humidity ratio at the coil exit?


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## mongolianbbq (Mar 13, 2018)

I converted the volumetric flow rate of the air and water to mass flow rates and used the equation m_w = m_a (w_1 - w_2) and solved for w_2. Am I on the right path?


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## MikeGlass1969 (Mar 13, 2018)

I think it's (C)...  

Convert 5gpm to lbm/hr..  Then find grains of moisture at the outlet of the coil from MR=cfm/1000 X .642 X delta grains.  Draw coil line from inlet air to ADP.  Read DB temp at the intersection of coil line and outlet grains.    Plug into the bypass  equation.


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## Vel2018 (Mar 13, 2018)

EDIT!

Sorry was 0.12!!! I took a better look at the graph. And that's where it should be. So A?

While cooling and dehumidifying, following the saturation line until you reach W2 = 0.0102 calculated from condensate mass. Tdbout is 58F that is where condensation stops.


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## Slay the P.E. (Mar 13, 2018)

mongolianbbq said:


> I converted the volumetric flow rate of the air and water to mass flow rates and used the equation m_w = m_a (w_1 - w_2) and solved for w_2. Am I on the right path?


That's right. What did you get for w_2?


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## Slay the P.E. (Mar 13, 2018)

MikeGlass1969 said:


> I think it's (C)...
> 
> Convert 5gpm to lbm/hr..  Then find grains of moisture at the outlet of the coil from MR=cfm/1000 X .642 X delta grains.  Draw coil line from inlet air to ADP.  Read DB temp at the intersection of coil line and outlet grains.    Plug into the bypass  equation.


Correct!


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## Slay the P.E. (Mar 13, 2018)

Vel2018 said:


> EDIT!
> 
> Sorry was 0.12!!! I took a better look at the graph. And that's where it should be. So A?
> 
> While cooling and dehumidifying, following the saturation line until you reach W2 = 0.0102 calculated from condensate mass. Tdbout is 58F that is where condensation stops.


Not quite right. I agree with your value of W2, but... the process path is a straight line from state 1 towards the ADP. The process ends where this line intersects the W2=0.0102 line.

Here's why the process is a straight line from state 1 to ADP.

A fraction of the air does get some intimate contact with the coil and thus ends up at the ADP by first cooling at constant W and then sliding down the 100% rh line all the way to ADP. However, there is fraction of the air that never touches the coil, and thus is at state 1. Therefore, at the exit of the coil we have the adiabatic mixing of some air at the ADP and some air at state 1. Adiabatic mixing in the psych chart is a straight line joining the two states being mixed.


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## mongolianbbq (Mar 13, 2018)

Slay the P.E. said:


> Not quite right. I agree with your value of W2, but... the process path is a straight line from state 1 towards the ADP. The process ends where this line intersects the W2=0.0102 line.
> 
> Here's why the process is a straight line from state 1 to ADP.
> 
> A fraction of the air does get some intimate contact with the coil and thus ends up at the ADP by first cooling at constant W and then sliding down the 100% rh line all the way to ADP. However, there is fraction of the air that never touches the coil, and thus is at state 1. Therefore, at the exit of the coil we have the adiabatic mixing of some air at the ADP and some air at state 1. Adiabatic mixing in the psych chart is a straight line joining the two states being mixed.


I also got .0102. I drew a straight line from state 1 (80 F, 70 rh) to ADP (55 F on the saturation curve) and ended up with a dry bulb temp around 58 (.12 bf)or 59 (.16 bf). Looks like I got around the same answer as vel. I’ll try using a different chart maybe?


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## Vel2018 (Mar 13, 2018)

Slay the P.E. said:


> Correct!


OMG lol we did all the same calculation, so the W2 is rounded off to 0.013? From 0.010258? This is the only way I see that line hitting close to 60F. 

I used 0.0102 for W2 since rules of rounding off, you round off odd numbers next to 5's. I did get 0.16 earlier doing same steps, close answer was B, and said was wrong. 

Even if you follow this by scale 0.01026 still answer should be closest to 0.168 which is still most nearly to 0.15 than 0.20.

The options are rounded off to hundreds, so I hope its not like 0.168 rounded off to 0.20 LOL


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## Slay the P.E. (Mar 13, 2018)

5 gpm of water = 2502.5 lbm/hour.

111,000 SCFM = (111,000 ft^3/min)*(60 min/hour)/(13.3 lbm/ft^3) = 500,752 lbm/hour of air.

m_cond = m_air(w1 - w2) thus:

w2 = w1 - m_cond/m_air = 0.0154 - 2502.5/500752 = 0.0104 lbm/lbm = 72.82 grains/lbm

From the chart, then the exit temperature is 60F




Then from the BF equation we get BF=0.2


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## Vel2018 (Mar 13, 2018)

Slay the P.E. said:


> 5 gpm of water = 2502.5 lbm/hour.
> 
> 111,000 SCFM = (111,000 ft^3/min)*(60 min/hour)/(13.3 lbm/ft^3) = 500,752 lbm/hour of air.
> 
> ...


Why did you use v 13.3ft^3/lbm to calculate mass of air from point 1? Shouldn't you use 13.95ft^3/lbm to get the initial amount of dry air?

If you calculate the initial mass of air from v=13.3ft^3/lbm how then can you dehumidify it? Also how then did you find that point with v=13.3 if you are in the process of calculating W2?

I think it should be v at point 1 in the calculation of mass of air.

v at point 1 should be 13.95ft^3/lbm.

So then, mass of air should be 477,419.4lbm not 500,752lbm and W2 should be 0.0102 not 0.0104.

The Temperature at point 2 should be 58.8. This corresponds to 0.152 BF.

Please let me know, I'm so confused now which one is which LOL


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## Slay the P.E. (Mar 13, 2018)

Vel2018 said:


> Why did you use v 13.3ft^3/lbm to calculate mass of air from point 1? Shouldn't you use 13.95ft^3/lbm to get the initial amount of dry air?
> 
> If you calculate the initial mass of air from v=13.3ft^3/lbm how then can you dehumidify it? Also how then did you find that point with v=13.3 if you are in the process of calculating W2?
> 
> ...


Because the flow rate is given in *SCFM*: Standard Cubic Feet per Minute. ASHRAE defines the standard condition as dry air at 20°C and 101.325 kPa (68°F and 14.7 psia). Under that condition the density of dry air is about 1.204 kg/m3 (0.075 lbm/ft3) and the specific volume is 0.83 m3/kg (13.3 ft3/lbm). If you were interested in the Actual CFM at the inlet, then you would calculate that as 111,000 * (13.95/13.3) = 116,425 ACFM.


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## Vel2018 (Mar 13, 2018)

Slay the P.E. said:


> Because the flow rate is given in *SCFM*: Standard Cubic Feet per Minute. ASHRAE defines the standard condition as dry air at 20°C and 101.325 kPa (68°F and 14.7 psia). Under that condition the density of dry air is about 1.204 kg/m3 (0.075 lbm/ft3) and the specific volume is 0.83 m3/kg (13.3 ft3/lbm). If you were interested in the Actual CFM at the inlet, then you would calculate that as 111,000 * (13.95/13.3) = 116,425 ACFM.


I see! NICE! Thanks for explaining that!


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## GR8 PLUMENG (Mar 14, 2018)

Good questions. I toiled for a while but realized condensate only happens when steam is changed to water.

So i used Q=m*hg= (5gal/min*8.4lbs/gal*60min/hr)*960Btu/lbm = 2419200btu/hr

then find Delt T. 

2419200btu/hr / (1.08*111,000)=20F

so the Tout of the coil is 80-20=60F

BF=(Tout-ADP)/(Tin-ADP)=(60-55)/(80-55)=.2

My approach seems to be different than most. 

Good Question.


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## Slay the P.E. (Mar 14, 2018)

GR8 PLUMENG said:


> So i used Q=m*hg= (5gal/min*8.4lbs/gal*60min/hr)*960Btu/lbm = 2419200btu/hr


Hmmm...

I have several comments about this approach. I hope you don't mind we take it one comment at a time. First of all, what is this 960 Btu/lbm?

Thanks,


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## GR8 PLUMENG (Mar 14, 2018)

Sorry,  970Btu/lbm. (heat of evap at 14.7psia) Actually, using 960, which is totally wrong gave me the right answer. 

Therefore my solution is

 Q=m*hg= (5gal/min*8.4lbs/gal*60min/hr)*970Btu/lbm = 2444400btu/hr

then find Delt T. 

2444400btu/hr / (1.08*111,000)=20.39F

so the Tout of the coil is 80-20=60F

BF=(Tout-ADP)/(Tin-ADP)=(60-55)/(80-55)=.2

My approach seems to be different than most.


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## Slay the P.E. (Mar 14, 2018)

GR8 PLUMENG said:


> Sorry,  970Btu/lbm. (heat of evap at 14.7psia) Actually, using 960, which is totally wrong gave me the right answer.


Ok. But, the water vapor in air is not at 14.7 psia. It is at a much lower pressure (the partial pressure of water vapor in atmospheric air is typically 0.15 to 0.5 psia.) and in fact the water vapor in air is superheated vapor. In this case it is superheated vapor at 80F (because it is at such a low pressure). Nevertheless, in HVAC approximations it is typical to neglect the superheat and just say it is saturated vapor at 80F. So if you want to calculate the *latent* load associated with the condensation, then the enthalpy change would be [email protected] = 1048.7 BTU/lbm.

So, you can calculate this latent load. But then you used the equation 1.08xCFMxDT, but this equation is for* sensible* loads. This equation is valid only to calculate enthalpy change for a constant humidity ratio process, which is NOT what we have here. The process here has both sensible and latent loads. In such processes the “1.08” equation is used with the sensible component of the total load. So it is not correct to plug in a latent load in there.

Does this help?


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## GR8 PLUMENG (Mar 15, 2018)

This absolutely helps. Thanks for the reply and the explanation. Really appreciate it.


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## Slay the P.E. (Mar 16, 2018)

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


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## Slay the P.E. (Mar 21, 2018)

So... nobody is going to take a stab at this problem?


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## MikeGlass1969 (Mar 21, 2018)

I did it...  I just didn't want to spoil it for anyone else.

(D)


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## Slay the P.E. (Mar 21, 2018)

MikeGlass1969 said:


> I did it...  I just didn't want to spoil it for anyone else.
> 
> (D)


Yes!


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## Vel2018 (Mar 21, 2018)

Slay the P.E. said:


> Yes!


I have to stop for now haha

Need to finish my schedule of practice test.


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## Audi Driver P.E. (Mar 21, 2018)

A good ruler is a must.  I highly recommend this one: https://www.amazon.com/Staedtler-12-Inches-Engineer-Triangular-987M1834BK/dp/B00094GVM8/ref=sr_1_17?ie=UTF8&amp;qid=1521674579&amp;sr=8-17&amp;keywords=engineering+ruler


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## Slay the P.E. (Mar 22, 2018)

Happy Thursday. Here's this week's problem:

*SPOILER ALERT: *Try to solve it before scrolling down and reading the discussion.

The manufacturer table shown below present gross compressor refrigeration capacity (qr) in thousands of Btu/hr (MBtu/hr), the required input power (kW), and the refrigerant mass flow rate (mr) for a hermetic scroll compressor with R-22 refrigerant. 

A refrigeration system with an evaporating temperature of 40°F and condensing temperature of 120°F uses this compressor. Additionally, there is a 20°F compressor suction superheat and a 15°F liquid subcooling at the condenser discharge. Assuming 100% efficiency for the motor and compressor, the heat rejected at the condenser (MBtu/hr) is most nearly:

(A) 39.1

(B) 42.3

(C) 50.1

(D) 75.2


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## MikeGlass1969 (Mar 22, 2018)

Refrigeration Capacity + Power Input = Condenser Heat Of Rejection.  39.1 MBH + 3.22kW x 3412 btu/hr/kW / 1000 btu/hr/MBH = 50.1 MBH

(C)


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## Slay the P.E. (Mar 22, 2018)

MikeGlass1969 said:


> Refrigeration Capacity + Power Input = Condenser Heat Of Rejection.  39.1 MBH + 3.22kW x 3412 btu/hr/kW / 1000 btu/hr/MBH = 50.1 MBH
> 
> (C)


Correct!


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## Slay the P.E. (Mar 29, 2018)

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
(D) 220


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## Slay the P.E. (Apr 4, 2018)

Last one before the big day on Friday. Good luck to all!!!

A pump is located 12 feet below the free surface of water in the condensate collection basin of a condenser, where the pressure is a vacuum of 1 inch of mercury. If the minor and friction losses in the suction line are 3.5 feet of water, the NPSHA (ft) for the pump is most nearly:

(A) 1.1

(B) 7.4

(C) 8.5

(D) 9.6


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## MikeGlass1969 (Apr 4, 2018)

(B)


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## Slay the P.E. (Apr 4, 2018)

MikeGlass1969 said:


> (B)


I disagree, Mike.

Show your work and let's figure it out...


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## MikeGlass1969 (Apr 4, 2018)

NPSHA = Hatm +Hz - Hvp - Hf

Hatm = Hvp so they cancel out.

12-3.5 = 8.5 feet  Answer (C)

What I originally did was not use absolute pressures and I also ignored vapor pressure.  Now I know better.


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## Slay the P.E. (Apr 4, 2018)

MikeGlass1969 said:


> NPSHA = Hatm +Hz - Hvp - Hf
> 
> Hatm = Hvp so they cancel out.
> 
> 12-3.5 = 8.5 feet  Answer (C)


Yup. Correct.


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## mongolianbbq (Apr 5, 2018)

MikeGlass1969 said:


> NPSHA = Hatm +Hz - Hvp - Hf
> 
> Hatm = Hvp so they cancel out.
> 
> ...


How does Hatm = Hvp? How do you know what either is if you dont know the temperatures? If the water is 70 degrees, Hvp is .84. For 1 inch of mercury I'm getting a head of around 1 or 2 feet depending on the temperature.

Also, if you say that the pressure is a vacuum of 1 in HG, wouldnt the absolute pressure be P_abs = P_atm - P_vac? I can get the right answer if I use Pvac, but if I convert to absolute, the numbers are a lot higher than any of the answer choices.


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## Vel2018 (Apr 5, 2018)

mongolianbbq said:


> How does Hatm = Hvp? How do you know what either is if you dont know the temperatures? If the water is 70 degrees, Hvp is .84. For 1 inch of mercury I'm getting a head of around 1 or 2 feet depending on the temperature.
> 
> Also, if you say that the pressure is a vacuum of 1 in HG, wouldnt the absolute pressure be P_abs = P_atm - P_vac? I can get the right answer if I use Pvac, but if I convert to absolute, the numbers are a lot higher than any of the answer choices.


Patm is negative 1inHg, you only use 14.7psi when its open to atmosphere, in this case the pressure in the condenser is vacuum, Pvp is negative 1inHg also becomes positive on the equation so it cancels out. Even if you convert both to absolute it will still cancel out since both are vacuum.


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## mongolianbbq (Apr 5, 2018)

Vel2018 said:


> Patm is negative 1inHg, Pvp is negative 1inHg becomes positive on the equation so it cancels out. Even if you convert both to absolute it will still cancel out since both are vacuum.


How is Pvp negative 1inHg? Is that because this is a condenser? The MERM says vaporization and condensation at constant temperature are equilibrium processes. If this was a different type of system the Pvp would be typically be different than the gage pressure right?


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## Vel2018 (Apr 5, 2018)

mongolianbbq said:


> How is Pvp negative 1inHg? Is that because this is a condenser? The MERM says vaporization and condensation at constant temperature are equilibrium processes. If this was a different type of system the Pvp would be typically be different than the gage pressure right?


Pvp was shown on the figure vacuum, patm was stated on the problem as vacuum as well.


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## MikeGlass1969 (Apr 5, 2018)

Maybe I should have used the term absolute Pressure head  instead of hatm...   They gave you a pressure and a condenser condensate is at saturation.  The important thing to realize here is when you are at saturated conditions the Pressure head and vapor pressure head negate each other.  

If the liquid was sub-cooled then we would have to find the vapor head of the subcooled liquid.


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## Slay the P.E. (Apr 5, 2018)

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.


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## Slay the P.E. (Apr 5, 2018)

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


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## medieval_pancake (Apr 7, 2018)

Slay the P.E. said:


> 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.
> 
> ...


@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.


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## Vel2018 (Apr 7, 2018)

medieval_pancake said:


> @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???


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## Slay the P.E. (Apr 7, 2018)

medieval_pancake said:


> @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.


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## Slay the P.E. (Apr 7, 2018)

Vel2018 said:


> 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.


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## Vel2018 (Apr 7, 2018)

Slay the P.E. said:


> 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


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## Slay the P.E. (Apr 7, 2018)

Vel2018 said:


> 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.


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## Vel2018 (Apr 7, 2018)

Slay the P.E. said:


> 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.


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## Slay the P.E. (Apr 7, 2018)

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?


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## Vel2018 (Apr 7, 2018)

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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## Slay the P.E. (Apr 7, 2018)

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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## Vel2018 (Apr 8, 2018)

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) ====&gt; 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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## medieval_pancake (Apr 9, 2018)

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


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## Jimbo Three (Apr 9, 2018)

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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## Slay the P.E. (Apr 9, 2018)

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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## Jimbo Three (Apr 9, 2018)

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.. 



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
> ...


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## Slay the P.E. (Apr 9, 2018)

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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## Vel2018 (Apr 9, 2018)

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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## Slay the P.E. (Apr 9, 2018)

@Vel2018

200 is not correct.

How did you find T_high using COP?


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## Vel2018 (Apr 9, 2018)

Slay the P.E. said:


> @Vel2018
> 
> 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.?


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## Slay the P.E. (Apr 9, 2018)

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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## Vel2018 (Apr 9, 2018)

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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## Slay the P.E. (Apr 9, 2018)

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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## Vel2018 (Apr 9, 2018)

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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## Vel2018 (Apr 9, 2018)

@Slay the P.E.

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.


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## Slay the P.E. (Apr 9, 2018)

Vel2018 said:


> @Slay the P.E.
> 
> Ok so is this some kind of trial and error? COP = Qin/Win say 4=h1-h4/h1-h2
> 
> ...


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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## Vel2018 (Apr 9, 2018)

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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## ManojD (May 16, 2018)

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


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## Slay the P.E. (May 16, 2018)

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&amp;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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## Slay the P.E. (Sep 28, 2018)

Happy Friday all. Brand new practice problem (also posted in the TFS thread):

The flow rate through the pipe system is 5,100 cubic feet per hour and the Darcy friction factor is known to be 0.02 through the entire system. Branch 1 is 3.5” ID, and Branch 2 is 5” ID. The total length of straight pipe in Branch 1 is 85 feet. The length of straight pipe in Branch 2 is also 85 ft. More information is given in the sketch. The difference in elevation (ft) between the water level of the two reservoirs is most nearly:

(A)   35
(B)   70
(C)   85
(D)   98


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## Slay the P.E. (Oct 6, 2018)

An indoor swimming pool room has a 6 ft × 4 ft double-pane, outside wall window with an effective R-value of 3 (°F ft2 )/(Btu/h). The R-value for the indoor air film is 0.2 (°F ft2 )/(Btu/h), and the convection coefficient for the side exposed to outdoor air is 12 (Btu/h)/(°F ft2 ). The indoor air is maintained at 75°F with a relative humidity of 85%. Under these conditions, the winter outdoor air temperature (°F) below which condensation on the inner face of the window is expected to form is most nearly:

(A) -4
(B)  0
(C)  4
(D) 12


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## MikeGlass1969 (Oct 8, 2018)

I got something less than (D)


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## MikeGlass1969 (Oct 8, 2018)

Slay the P.E. said:


> Happy Friday all. Brand new practice problem (also posted in the TFS thread):
> 
> The flow rate through the pipe system is 5,100 cubic feet per hour and the Darcy friction factor is known to be 0.02 through the entire system. Branch 1 is 3.5” ID, and Branch 2 is 5” ID. The total length of straight pipe in Branch 1 is 85 feet. The length of straight pipe in Branch 2 is also 85 ft. More information is given in the sketch. The difference in elevation (ft) between the water level of the two reservoirs is most nearly:
> 
> ...


Slay, I tried this...   However I converted the straight pipe, for each branch, into a K and summed all the K's for each branch.  Plugged into the Bernoulli equation.  And Came up with something in the area of (B).

But, in the TFS the answer was reported as (D).  In the solution posted in the other forum, added the head losses from friction in the energy equation.  Why?  Shouldn't they be subtracted?   

I found a mistake,  got 93ft which is close to (D)


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## mcc515 (Oct 9, 2018)

Slay the P.E. said:


> An indoor swimming pool room has a 6 ft × 4 ft double-pane, outside wall window with an effective R-value of 3 (°F ft2 )/(Btu/h). The R-value for the indoor air film is 0.2 (°F ft2 )/(Btu/h), and the convection coefficient for the side exposed to outdoor air is 12 (Btu/h)/(°F ft2 ). The indoor air is maintained at 75°F with a relative humidity of 85%. Under these conditions, the winter outdoor air temperature (°F) below which condensation on the inner face of the window is expected to form is most nearly:
> 
> (A) -4
> (B)  0
> ...


I'm coming up with -7F, so I'll go with (A).


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## Slay the P.E. (Oct 10, 2018)

MikeGlass1969 said:


> Slay, I tried this...   However I converted the straight pipe, for each branch, into a K and summed all the K's for each branch.  Plugged into the Bernoulli equation.  And Came up with something in the area of (B).
> 
> But, in the TFS the answer was reported as (D).  In the solution posted in the other forum, added the head losses from friction in the energy equation.  Why?  Shouldn't they be subtracted?
> 
> I found a mistake,  got 93ft which is close to (D)


Mike, the "extended" Bernoulli equation is actually a statement of conservation of energy. as such, it is essentially this (for steady state):

The rate at which energy *enters* a control volume = The rate at which energy *leaves* the control volume​
In this problem a control volume is defined with boundaries just below the water level in the reservoirs. It is at these locations where mass enters/leaves the control volume.

The left hand side of the energy balance will have the enthalpy (pressure head, which is zero), kinetic energy (zero), and potential energy (height) entering the control volume at the reservoir in the right side of the drawing. That is all we have on the left side of the equation. (If we had a pump, the pump head would be on this side, because it is energy entering the control volume)

The right hand side of the energy balance will have the enthalpy (pressure head, which is zero), kinetic energy (zero), and potential energy (height) leaving the control volume at the reservoir in the left side of the drawing. The right hand side of the equation will also contain the energy that is "lost" (stray heat is an enthalpy, which in hydraulics ends up being pressure loss) so the friction and minor losses are forms of energy leaving the control volume. These terms therefore, belong on the right side of the equation.

A much shorter way to say all this is that in the extended Bernoulli equation, the left side of the equation (subscript 1) is the inlet reservoir, and the right side (subscript 2) is the discharge reservoir. Pump head (if non-zero) goes in the left side of the equation. Turbine head (if non-zero) goes in the right side of the equation. Friction and minor losses go in the right side of the equation.


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## Slay the P.E. (Oct 10, 2018)

@mcc515

@MikeGlass1969

Thanks for working on the double pane window problem. Here is my solution. I get -4F which is different from what both of you are getting. Let me know if you see anything wrong


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## mcc515 (Oct 11, 2018)

@Slay the P.E. I used the same method, but with a Tdp of 70F instead of 70.2F to get -7. If I sub 70.2 in for Tsurface, I also end up with -3.82F

Rtotal = 0.2+3+(1/12) = 3.28

Rair= 0.2

Ts=70F

(Tin-Toa)/Rtotal = (Tin-Ts)/Rair

Toa = 75F - (3.28/.2)*(75F-70F)

Toa = -7F


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## Slay the P.E. (Oct 11, 2018)

mcc515 said:


> @Slay the P.E. I used the same method, but with a Tdp of 70F instead of 70.2F to get -7. If I sub 70.2 in for Tsurface, I also end up with -3.82F
> 
> Rtotal = 0.2+3+(1/12) = 3.28
> 
> ...


Thanks.

I’m going to have to re-write some of the given data in that problem so the final answer isn’t so sensitive to small variations of Tdp.

I’ll make sure that using 70F or 70.2F yields roughly the same answer.


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## MikeGlass1969 (Oct 11, 2018)

Slay the P.E. said:


> Thanks.
> 
> I’m going to have to re-write some of the given data in that problem so the final answer isn’t so sensitive to small variations of Tdp.
> 
> I’ll make sure that using 70F or 70.2F yields roughly the same answer.


I used 71F for the Tdp...  I got 9.3F.   But when I substitute in 70.2F I get -3.8F  Pretty large swing. 

All answers were possible in this problem


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## Slay the P.E. (Oct 11, 2018)

MikeGlass1969 said:


> I used 71F for the Tdp...  I got 9.3F.   But when I substitute in 70.2F I get -3.8F  Pretty large swing.
> 
> All answers were possible in this problem


Yep. Its badly crafted as is. The factor (3.28/.2) or (.1368/.-0083) is too big and makes the final answer too sensitive to Tdp. I need to change the relative magnitude of the given R-values to improve the problem.  

I think the problem itself is still pretty cool... just needs better numerical data. We shouldn't get a -7F to 9.3F swing by changing Tdp from 70F to 71F


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