Jump to content
Engineer Boards
​ ​
Slay the P.E.

HVAC&R Practice Problem of the week

Recommended Posts

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

Edited by Slay the P.E.

Share this post


Link to post
Share on other sites

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)

Share this post


Link to post
Share on other sites
51 minutes ago, 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?

Share this post


Link to post
Share on other sites

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?

Share this post


Link to post
Share on other sites

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.

 

  • Like 1

Share this post


Link to post
Share on other sites

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. 

Edited by Vel2018

Share this post


Link to post
Share on other sites
2 hours ago, 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?

Share this post


Link to post
Share on other sites
1 hour ago, 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!

Share this post


Link to post
Share on other sites
46 minutes ago, 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.

Share this post


Link to post
Share on other sites
5 minutes ago, 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?

Share this post


Link to post
Share on other sites
26 minutes ago, 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

Edited by Vel2018

Share this post


Link to post
Share on other sites

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

Screen Shot 2018-03-13 at 7.55.36 PM.png

 

 

Then from the BF equation we get BF=0.2

 

Edited by Slay the P.E.

Share this post


Link to post
Share on other sites
1 hour ago, 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.

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

Screen Shot 2018-03-13 at 7.55.36 PM.png

 

 

Then from the BF equation we get BF=0.2

 

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

20180313_214521.jpg

Edited by Vel2018
  • Like 1

Share this post


Link to post
Share on other sites
Just now, 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?

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

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.

 

Share this post


Link to post
Share on other sites
38 minutes ago, 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! 

 

Share this post


Link to post
Share on other sites

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.

Share this post


Link to post
Share on other sites
2 hours ago, 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,

Share this post


Link to post
Share on other sites

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.

 

 

Share this post


Link to post
Share on other sites
16 hours ago, 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 hfg@80F = 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?

Edited by Slay the P.E.

Share this post


Link to post
Share on other sites

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

natatorium (indoor pool) dehumidifier schematic.png

natatorium (indoor pool) dehumidifier performance data.PNG

Share this post


Link to post
Share on other sites

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

Share this post


Link to post
Share on other sites

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

(D)

  • Like 1

Share this post


Link to post
Share on other sites
44 minutes ago, MikeGlass1969 said:

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

(D)

Yes!

Share this post


Link to post
Share on other sites
2 hours ago, Slay the P.E. said:

Yes!

I have to stop for now haha

Need to finish my schedule of practice test. 

  • Like 1

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

×