HVAC&R Practice Problem of the week

[Slay the P.E.]

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

 

[Slay the P.E.]

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

 

[MikeGlass1969]

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

(D)

 

[Slay the P.E.]

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

(D)

Yes!

 

[Vel2018]

Yes!

I have to stop for now haha

Need to finish my schedule of practice test. 

 

[Audi Driver P.E.]

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&qid=1521674579&sr=8-17&keywords=engineering+ruler

 

[Slay the P.E.]

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

 

[MikeGlass1969]

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)

 

[Slay the P.E.]

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!

 

[Slay the P.E.]

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

 

[Slay the P.E.]

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

 

[MikeGlass1969]

(B)

 

[Slay the P.E.]

(B)

I disagree, Mike.

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

 

[MikeGlass1969]

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.

 

[Slay the P.E.]

NPSHA = Hatm +Hz - Hvp - Hf

Hatm = Hvp so they cancel out.

12-3.5 = 8.5 feet  Answer (C)

Yup. Correct.

 

[mongolianbbq]

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.

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.

 

[Vel2018]

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.

 

[mongolianbbq]

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?

 

[Vel2018]

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. 

 

[MikeGlass1969]

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.