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scavok
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Homework Statement
A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes:
Process 1-2: Constant volume, [tex]V_1=0.028m^3[/tex], [tex]U_2-U_1=26.4kJ[/tex]
Process 2-3: Expansion with pV=constant, [tex]U_3=U_2[/tex]
Process 3-1: Constant pressure, [tex]p_2=1.4 bar, W_{31}=-10.5kJ[/tex]
There are no significant changes in kinetic or potential energy.
a) Calculate the net work for the cycle, in kJ
b) Calculate the heat transfer for process 2-3, in kJ
c) Calculate the heat transfer for the process 3-1, in kJ
d) Is this a power cycle or a refrigeration cycle?
Homework Equations
[tex]W=p(V_2-V_1)[/tex]
[tex]\Delta U=Q-W[/tex]
Cycle Energy Balance: [tex]Q_{cycle}=W_{cycle}[/tex]
Power Cycles: [tex]W_{cycle}=Q_{in}-Q_{out}[/tex]
Refrigeration Cycles: [tex]W=Q_{out}-Q_{in}[/tex]
The Attempt at a Solution
I think my problem is more in the method of solving these problems and not so much in the details, but I missed the lecture and there are no examples in my book or that I can find on the internet.
My first idea was to use the equation [tex]W_{31}=p_2(V_1-V_2)[/tex] and find V2. With that I could find the constant in the pV=constant equation, and use it to find p1 in process 2-3:
[tex]W_{31}=p(V_1-V_2)[/tex]
[tex]-10.5kJ=1.4bar(0.028m^3-V_2)[/tex]
After some algebra and unit conversions..
[tex]V_2=0.103m^3[/tex]
pV=constant
[tex]p_2V_2=1.4bar\ast0.103m^3=0.144bar\ast m^3[/tex]
[tex]p_1V_1=0.144bar\ast m^3[/tex]
[tex]p_1=5.15bar[/tex]
Then I plugged p1 into the work equation to get the work done in process 2-3:
[tex]W_{23}=p_1(V_2-V_1)=0.530bar\ast m^3-0.144bar\ast m^3=0.386bar\ast m^3=38.6kJ[/tex]
a)
[tex]W_{net}=W_{12}+W_{31}=38.6kJ-10.5kJ=28.1kJ[/tex]
b)
[tex]\Delta U=Q_{23}-W{23}[/tex]
[tex]26.4kJ=Q_{23}-38.6kJ[/tex]
[tex]Q_{23}=65kJ[/tex]
c)
[tex]Q_{cycle}=W_{cycle}[/tex]
[tex]W_{net}=Q_{23}+Q_{12}[/tex]
[tex]Q_{12}=W_{net}-Q_{23}=-36.9kJ[/tex]
d) No idea. If it's either of those then the equation I used in part c) was probably wrong.
I'm pretty sure none of this is right. Any advice or help would be appreciated.
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