An air-standard cycle is executed in a closed system and is composed of the following four processes:
1-2 Isentropic compression from 100 kPa and 27°C to 1 MPa
2-3 P constant heat addition in amount of 2800 kJ/kg
3-4 Heat rejection at constant specific volume 100 kPa
4-1 Heat rejection at constant pressure back to the initial state
Assuming constant specific heats at room temperature...
a.) Show the cycle on PV and TS diagrams
b.) Calculate the maximum temperature in the cycle
c.) Determine the thermal efficiency.
20 comments:
In finding the Maximum Temperature, I used the relative pressures for the temperature increase from 1 to 2, but have no idea how to do step 2 to 3, though I think it should involve Cp and the heat input.. And what IS the ideal heat capacity of air at room temperature?
im having the same problem as above but I also have the problem of finding the efficiency, form thermo cd, I read that it is 1-1/pressure ratio^ gamma-1/gamma. I used 1.4 as gamma and the pressures 1MPA and 100kpa converting units of course and plugged in but i get like 48 percent ish
Washburn 4:02 PM
You can use the Shomate Eqn to determine Cp and then Cv = Cp - R. But, here they are: Cp = 1.005 kJ/kg-K and Cv = 0.718 kJ/kg-K and γ = 1.4.
Your approach to step 1-2 is fine.
Step 2-3, apply the 1st Law and use Cp to get T3.
Chris 9:57 PM
Although this is an Air-Standard Cycle, it is not a Brayton Cycle and the formula you selected does not apply.
Try determining efficiency as Nth = 1 Qc/Qh.
I calculated Qc = Cp(T4-T1). However, my efficiency is higher than expected. My T4 is about 2025 K...is that the right value? (I used relative pressures to find it.)
Man, i sure am lost on this problem! ideal gas is not the ideal thing for me right now...
so to find the max temp (t3) im using deltaH=Cp*deltaT - delta H is given as the heat input, cp is provided by dr.B, and my t2 calculated by myself to be 571.02 C
im only about 30 K off at the end of all that. what am i doing wrong?
thanks!
Carina 7:25 PM
I am not sure what happened, but here is what I think...
The highest T in the cycle is T3. Using the Cp I posted in an earlier comment, I got more than 3000 K for T3.
T4 is MUCH lower than T3. You can use the IG EOS at constant Vhat to determine T4 quickly from T3.
The TOTAL heat rejected in this cycle is Q34 + Q41. This is what counts for determining efficiency of the cycle.
AdamAnt 9:52 PM
Your T2 is close. I got closer to 580 K using a polytropic relationship since we are considering the Cp, Cv and gamma to be constant.
I think your method is correct. I am not sure why your answer is off by so much. Did you use the IG Air Tables ? That would explain the difference.
Graham, I was having the same problem. I think that you need to switch from using Cp for Q34 to using Cv, since it is at constant specific volume.
If I work backwards from the answer, I need Qc to be something like 2240K, but no matter what equation I try to use to get T4, I cannot seem to get that value. Using PV~ = RT yields a value that is WAY off. I really don't understand why my numbers are so far off. Advice, anybody?
Graham 3:13 Pm
You need to apply the 1st Law for closed systems to determine Q34 and Q41. Use deltaH = Cp deltaT and deltaU = Cv deltaT.
Qc = -(Q34+Q41) since heat is removed in BOTH steps.
Claire 11:19 PM
I think you are correct, but the reason is that this is a CLOSED system...think piston and cylinder.
1st Law: Q - Wb = deltaU
Carina 12:12 AM
With constant Cp, Cv and gamma, I suggest you use a polytropic relationship to get T2. Use T2 and the 1st law for step 2-3 to get T3 and then use the IG EOS for states 3 and 4 to determine T4.
Graham 9:39 PM
Your T4 is correct.
You got the rest of the problem correct as well, right ?
Good work.
this may be rather basic. but how do we determine T2?
T2 is determined by using relative pressures.
I found T2 to be about 300*C. On the ideal gas table for air, that gives me S2 = 0.6669Kj/Kg.
Then for T1 = 27*C, the table says S1 = 0.006168Kj/Kg.
But the problem statement says step 1-2 is suppose to be isentropic. How does this work?
I am having some problems with my efficicency... Qc = -(Q41 + Q34)...
Q41 = 1.005(300.15 - 62.52K) = 238.77kJ/kg
Q34 = 0.718 (62.57 - 3084.08K) = -2169.44kJ/kg
this gives me Qc = 1930.67kJ/kg..
n = 1 - Qc/Qh
Qh = 2800kJ
and thus an efficiency of 31.05... does anyone know where i am going wrong? thanks
T2 is determined by using relative pressures.
I found T2 to be about 300*C. On the ideal gas table for air, that gives me S2 = 0.6669Kj/Kg.
Then for T1 = 27*C, the table says S1 = 0.006168Kj/Kg.
But the problem statement says step 1-2 is suppose to be isentropic. How does this work?
For your questions above:
Note that the pressures here are not the pressures in the table given by Pr. So you cannot use the table directly.
Anonymous said...
I am having some problems with my efficicency... Qc = -(Q41 + Q34)...
Q41 = 1.005(300.15 - 62.52K) = 238.77kJ/kg
Q34 = 0.718 (62.57 - 3084.08K) = -2169.44kJ/kg
this gives me Qc = 1930.67kJ/kg..
n = 1 - Qc/Qh
Qh = 2800kJ
and thus an efficiency of 31.05... does anyone know where i am going wrong? thanks
For your questions above:
The T1, T2, T3 and T4 you got are not correct. Better check that. The equation you used for Q are right.
T2 is calculated from T2=T1(P2/P1)^(k-1)/k
From Qin to get T3, and ideal gas equation to get T4
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