A gas refrigeration system using air as the working fluid has a pressure ratio of 5. Air enters the compressor at 0°C. The high-pressure air is cooled to 35°C by rejecting heat to the surroundings. The refrigerant leaves the turbine at -80°C and then it absorbs heat from the refrigerated space before entering the regenerator. The mass flow rate of air is 0.4 kg/s. Assuming isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using constant specific heats at room temperature, determine...
a.) The effectiveness of the regenerator
b.) The refrigeration load
c.) The COP of the cycle
d.) The refrigeration load and the COP if this system operated on the simple gas refrigeration cycle. Use the same compressor inlet temperature as given, the same turbine inlet temperature as calculated, and the same compressor and turbine efficiencies.
26 comments:
How can the effluent of the turbine be at 80(deg C) when it's the lowest point on the TS diagram and absorbs heat from both the space and the regenerator to return to 0(C)?
Go_Huskies 7:41 PM
Doh ! That is a typo. The turbine effluent is at MINUS 80 degC.
so how do we use the air tables at -80C the tables don't go that low, are there bigger tables somewher or are we supposed to use something else
Brendan 7:12 PM
I should have included in the hints: Cp,air = 1.005 kJ/kg-K. You can consider this to be a constant value. This makes life easier since deltaH = Cp deltaT.
The Shomate Eqn from NIST is only accurate down to 25 degC. We would need to use a different heat capacity equation in order to create an IG property table for ait that went down to -80 degC.
how do i find the effiency of the regenerator? I'm lost on this one.
i have the table filled out i'm just not sure what to do to find this.
I'm completely lost here. We've got three temperatures and mdot. I can look up H and Snaught for one of the temperatures only. I'm also still not entirely sure about the relationship between Hnaught and H, and Snaught and S. I see that we have deltaH=CpdeltaT, but I'm not sure how to use that. Finally, although P2,3,4 are equal and P1,5,6 are equal, I don't know how to find either. Any more general hints would be great. Though I may come to office hours if I can
First of all,
is Eregen=(H4-H3)/(H6-H3)?
Secondly,
how do we go about solving for T6?
Lastly,
I just wanted to make sure that Cp=1.005kJ/kg (from problem 4).
Thanks.
Does T2 = 472.224K ; T4 = 288.78K? I found these using isentropic efficiencies and constant heat capacity.
Brendan 1:06 PM
Regenerator effectiveness is the ratio of het transferred in the actual HEX to the Q in an infinitely large HEX.
In an infinitely large HEX, stream 3 would be cooled all the way down to T6 (using stream numbers from the diagram on page 274 of the TCD Notebook).
Since Mdot is the same in all streams in this cycle, we can us the 1st Law to show that effectiveness = (H3-H4)/(H3-H6). But deltaH = Cp * deltaT, so this becomes...
effectiveness = (T3-T4)/(T3-T6)
Anon 5:06 PM
The keys here are that you know gamma = 1.4 and Cp = 1.005 kJ/kg-K and the compressor and turbine are isentropic.
Don't use the IG Property tables for this problem because the heat capacities (and gamma) are constant.
You do not need to know the individual pressures because only the pressure ratio show up in Gibbs 2nd Eqn or in the Polytropic relationship (you can use either to determine the T at the compressor and turbine outlets).
See my comment to Brendan above about regenerator effectiveness.
Remember that deltaH = Cp * deltaT when Cp is constant. Then apply the 1st Law and the definition of isentropic efficiency and you should be in good shape.
Yoda 2:52 AM
1- Yes, Eregen=(H4-H3)/(H6-H3).
2- Apply the 1st Law to the regenerator.
3- Yes, Cp=1.005kJ/kg-K
Good luck !
Yoda 7:39 AM
I hope you slept since your last comment !
Your T's are very close. 0 degC = 273.15 K.
New Answers...
It is hard to explain why, but using the approach I have described here, you should get the following answers:
Eregen ~ 0.35
Qc ~ 24 kW
(c) COPref ~ 0.6
(d) COPref ~ 0.6
The regenerator doesn't effect the COP much, but it lets you get down to 193 K instead of just 212 K.
I found T4=281.3K, and T2=472, but am stuck on T6. Applying the first law yields: Q=mdot(H6-H1). This has two unknowns, H6(essentially T6 after manipulation) and Q.
Washburn 10:55 AM
Apply the MIMO 1st Law to the regenerator to determine H6. You know T1, T3 and T4 !
Good work !
I can't get COP = .6 for part d; I keep getting a value of .7. The refrigeration load I'm getting is 30kW.
The diagram I'm using is that of page 271 in Thermo CD. The temperatures I'm using are:
T1 = 198.3K
T2 = 273.15K (given/from a,b,c)
T3 = 472.2K
T4 = 288.28K (given/from a,b,c)
The pressure ratio is still 5, right? With this and gama=1.4, I found T1s and T3s. Then I used isentropic efficiencies to determine T1 and T3. Are the values I found wrong? Thanks.
yoda and graham,
I am not sure if what went rong for you. Your T1 and T4 are a little bit off. T2 and T3 are correct.
Could someone give us a few points as to where to start for this problem. We have no idea even how to start.
graham,
How did you get T4?
efficiency=(Tx-Ty)/(Tx-Tys)=0.85
x and y can be your inlet and exit (1,2,3,or 4).
subtitle s represets the isentropical value.
Graham,
I see what you mean. have you tried using T3, T3s and 0.85 to get T4? The T4 calculated this way may be a little bit different from the T4 as calculated in part a.
graham,
not sure if T4 we are talking about is the same thing or not. in part a, T4 is the inlet temp entering turbine and T5 is the exit temp leaving turbine. In which you get T4=281.3K
but in part d, the inlet temp for the turbine is defined as T3 (308K) and exit temp defined T4. here T4s=T3(1/r)^(k-1)/k
graham,
yes, in part d, if u set T3 as the inlet temp, T4 is 211. but efficiency is 0.6. Qref is mCp(T1-T4) (24.7KW), and Wnet=mCp(T2-T1)-mCp(T3-T4)(41.3KW).
graham, no, I am not using T4 as the the inlet temp in both cases. all these numbers depend on how we draw the diagram and name the stream number.
I'm not getting the right efficiency for the heat exchanger. I haven't seen anyone post what they got as their T6 value but I got 246.31K because I figured (T3-T4)=(T1-T6) using
T1 = 273.15
T4 = 281.31
T3 = 308.15
These seem to be the same temperatures everyone else is getting for T1-T3, but I don't get the same efficiency. (I get .434)
Kuester 11:51 AM
You did the problem correctly.
There is a quirk to this problem.
There are two ways to use isentropic efficency. The way you did it is just fine. It may be better than the way I did. But do not worry about. Send me an email if you want to learn some more about it.
Hello Chirag,
I am sorry, but it is not my policy to post complete solutions.
Dr. B
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