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Thread: Thermodynamics of Isochoric Cycle

  1. #1 Thermodynamics of Isochoric Cycle 
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    May 2009
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    So I am studying for a test and am having some problems with this thermodynamics problem and hopefully somebody can clear up some of my issues.

    Here goes:

    1 kg of Air passes through the following changes in a plant (assume ideal gas).

    State 1-2...Adiabatic Comp from 10^5 Pa and 298K with Volume ratio (V1/V2) of 10.
    State 2-3...Isochoric thermal energy addition to T=1500K
    State 3-4...Adiabatic Expansion to original Volume
    State 4-1...Isochoric thermal rejection until original P and T are reached.

    I'm having problems calculating the thermal energy input and output...thus the efficiency of the cycle.

    Here is what I have for my values:

    T1: 298K P1: 10^5 Pa V1: 0.85526m^3 using ideal gas law

    T2: 118.71 P2: 2.5E6 Pa V2: 0.085526 m^3 T & P found using Adiabatic relations

    T3: 1500 P3: 5.04E6 V3: .085526 T given, P from ideal gas

    T4: 3765 P4: 2.011E3 V4: .85526 T&P from adiabatic relations

    Now adiabatic relations are the polytropic relations: T2=T1(V1/V2)^(1-y) and P2=P1(V1/V2)^(y) where y=Cv/Cp

    Cv=.718 kJ/kg-K
    Cp=1.005kJ/kg-K

    Now my data 1-3 I believe is correct, but T4 seems too high. Using this value, my Qin becomes much less than Qout leaving my efficiency somehwere around 300%. If anybody can steer me away from my misunderstanding, I would greatly appreciate it.

    Oh, my Q I am using is just mCpdT where m is 1 kg and air MW=28.9 kg/kmol...if you like, R (gas constant) =.287 kJ/kg-K

    Thanks in advance.


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