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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HVAC&R Practice Problem of the week
- Thread starter Slay the P.E.
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Help Support Professional Engineer & PE Exam Forum:
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.?
Slay the P.E.
Well-known member
COPref=TL/TH-TL Tlow is in 30psi thats in 0deg line with amonia ph chart. So I got Thigh 140deg.
Slay the P.E.
Well-known member
Maybe not, I'll try again later, I'm pretty beat up now. I think need to rest. xD
@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.
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.
Well-known member
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.
Slay the P.E.
Well-known member
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.
Slay the P.E.
Well-known member
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
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.
Well-known member
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
(A) -4
(B) 0
(C) 4
(D) 12
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MikeGlass1969
Well-known member
I got something less than (D)
MikeGlass1969
Well-known member
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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Slay the P.E.
Well-known member
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.
Slay the P.E.
Well-known member
@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
@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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@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
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
Slay the P.E.
Well-known member
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.
MikeGlass1969
Well-known member
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
