Calculating Net Work, Heat Transfers in a Thermodynamic Cycle

In summary, a gas undergoes a thermodynamic cycle consisting of three processes with initial conditions of P1=1 bar, V1=1.5 m³, and U1=512 kJ. Process 1-2 involves compression with PV=constant to P2=2 bars and U2=690 kJ. Process 2-3 involves no work and a heat transfer of -150 kJ. Process 3-1 involves a work of 50 kJ. The heat transfers Q12, Q31, and the total Q in the cycle are yet to be determined.
  • #1
scavok
26
0

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.
 
Last edited:
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  • #2
scavok said:

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?
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 [itex]P_2V_2 - P_1V_1 = nR(T_2-T_1)[/itex] and [itex]U_2-U_1 = nC_v(T_2-T_1)[/itex]. 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 [itex]\Delta Q = W[/itex]. 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: [itex]\Delta U \ne 0[/itex]. [itex]\Delta Q = (U_1-U_3) + P_1(V_1-V_3)[/itex]

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

AM
 
  • #3
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.
 
  • #4
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
 
  • #5


I would first recommend checking your units and conversions to make sure they are correct. In your attempt at a solution, you have a mix of bar and kJ units, which may lead to incorrect results.

Next, I would suggest using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the net heat transfer into the system minus the net work done by the system. This can be written as:

ΔU = Q - W

Using this equation, we can solve for the unknown values in each process.

a) To calculate the net work for the cycle, we can use the equation:

W = p(V2-V1)

In process 1-2, the volume is constant, so V2 = V1. Therefore, the work done in process 1-2 is:

W12 = p1(V2-V1) = 0

In process 2-3, we are given that pV = constant, so we can use the relationship pV = nRT to find the final volume V3. Since the pressure and number of moles of gas are constant, the final volume is directly proportional to the temperature. We can write this as:

V3 = (T3/T2)V2

Since we are given that U3 = U2, we know that the change in internal energy (ΔU) is also 0. Therefore, the heat transfer in process 2-3 can be calculated as:

Q23 = W23 = p3(V3-V2) = p3(T3/T2)V2 - p2V2

Substituting in the values given in the problem, we get:

Q23 = (1.4 bar)(0.144 bar m^3) - (1.4 bar)(0.028 m^3) = 0.144 bar m^3

In process 3-1, we are given that the pressure is constant, so we can use the equation W = pΔV to calculate the work done. We can also use the relationship Q = mCΔT to calculate the heat transfer, where m is the mass of the gas and C is the specific heat capacity. Since there are no significant changes in kinetic or potential energy, we can assume that the change in internal energy (ΔU) is 0. Therefore, we can write:

W31 =
 

1. What is net work in a thermodynamic cycle?

Net work in a thermodynamic cycle is the total amount of energy transferred as work between the system and its surroundings. It is calculated by subtracting the work done by the surroundings on the system from the work done by the system on the surroundings.

2. How do you calculate net work in a thermodynamic cycle?

To calculate net work in a thermodynamic cycle, you first need to determine the work done by the system and the work done by the surroundings. The net work is then calculated by subtracting the work done by the surroundings from the work done by the system.

3. What is heat transfer in a thermodynamic cycle?

Heat transfer in a thermodynamic cycle is the transfer of thermal energy between the system and its surroundings. This transfer can occur in the form of heat entering or leaving the system.

4. How is heat transfer calculated in a thermodynamic cycle?

Heat transfer in a thermodynamic cycle can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

5. What is the relationship between net work and heat transfer in a thermodynamic cycle?

In a thermodynamic cycle, the net work done is equal to the difference between the heat transfer into the system and the heat transfer out of the system. This is based on the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred between systems.

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