Calculating Net Work, Heat Transfers in a Thermodynamic Cycle

[scavok]

Homework Statement

A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes:
Process 1-2: Constant volume, V_1=0.028m^3, U_2-U_1=26.4kJ
Process 2-3: Expansion with pV=constant, U_3=U_2
Process 3-1: Constant pressure, p_2=1.4 bar, W_{31}=-10.5kJ

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

W=p(V_2-V_1)
\Delta U=Q-W
Cycle Energy Balance: Q_{cycle}=W_{cycle}
Power Cycles: W_{cycle}=Q_{in}-Q_{out}
Refrigeration Cycles: W=Q_{out}-Q_{in}

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 W_{31}=p_2(V_1-V_2) 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:

W_{31}=p(V_1-V_2)
-10.5kJ=1.4bar(0.028m^3-V_2)
After some algebra and unit conversions..
V_2=0.103m^3

pV=constant
p_2V_2=1.4bar\ast0.103m^3=0.144bar\ast m^3
p_1V_1=0.144bar\ast m^3
p_1=5.15bar

Then I plugged p₁ into the work equation to get the work done in process 2-3:

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

a)
W_{net}=W_{12}+W_{31}=38.6kJ-10.5kJ=28.1kJ

b)
\Delta U=Q_{23}-W{23}
26.4kJ=Q_{23}-38.6kJ
Q_{23}=65kJ

c)
Q_{cycle}=W_{cycle}
W_{net}=Q_{23}+Q_{12}
Q_{12}=W_{net}-Q_{23}=-36.9kJ

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.

 

[Andrew Mason]

Homework Statement

A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes:
Process 1-2: Constant volume, V_1=0.028m^3, U_2-U_1=26.4kJ
Process 2-3: Expansion with pV=constant, U_3=U_2
Process 3-1: Constant pressure, p_2=1.4 bar, W_{31}=-10.5kJ

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?

First, draw a PV diagram of the cycle. You will need to know the Cv for this gas.
You have to use the energy changes to find the pressure: eg P_2V_2 - P_1V_1 = nR(T_2-T_1) and U_2-U_1 = nC_v(T_2-T_1). Work that out to find P2. That will give you all the points on the PV graph.

In a) you have to find the sum of the areas under each section (note: the area is positive as V increases and negative as V decreases). The only difficult part is 2-3 since the graph will be a curve. You have to do a bit of calculus to find that area.

In b) you have to find the work done by the gas and use the first law. There is no change in internal energy in going from 2 to 3 so \Delta Q = W. Again, to find W from 2-3 you have to do a bit of calculus.

In c) again use the first law to find the heat Q. Note: \Delta U \ne 0. \Delta Q = (U_1-U_3) + P_1(V_1-V_3)

In d) is work done by the gas or is work done on the gas, over the whole cycle?

AM

 

[najla]

please i have the same problem and i can't solve it ;( i need help ;(
A gas undergoes a thermodynamic cycle consisting of three processes beginning at an
initial state where P1=1 bar, V1=1.5 m³, and U1=512 kJ. The processes are as follows:
Process 1-2: Compression with PV=constant to P2=2 bars, U2=690 kJ
Process 2-3: W23=0, Q23= –150 kJ
Process 3-1: W31= +50 kJ
Ignoring the kinetic and potential energies, determine the heat transfers Q12, Q31, and the total Q
in the cycle in kJ.

 

[Andrew Mason]

I would suggest that 1. you start a new thread and follow the format that it will provide you, and 2. show us your work and tell us your ideas for solution.

AM