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ricebowl07
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The figure shows the cycle for a heat engine that uses a gas having gamma =1.25. The initial temperature is T1=300K, and this engine operates at 20 cycles per second.
a.) What is the power output of the engine?
b.) What is the engine's thermal efficiency?
The link to this image is here:
http://session.masteringphysics.com/problemAsset/1074111/4/knight_Figure_19_54.jpg
This is how I attempted to solve this problem:
Given: T1 = T2 = 300K
P2 = P3
P1 = 1 atm = 1.013 * 10^(5) Pa
gamma = 1.25
V1 = V3 = 600 cm^(3)
V2 = 200 cm^(3)
R = 8.31 J/mol*K or 0.08206 atm *L/mol*K
Relevant equations:
PV = nRT
η = Wout/QH
Power = Wnet/time
Q_12 = W_12
Q_23 = nc_pΔT
Q_31 = nc_vΔTFinding number of moles
n = PV/RT
n = [1.013 * 10^(5) Pa(0.0006 m^(3))]/[(8.31J/(mol*K))(300K)]= 0.024 mol
Finding P2 and P3
P1V1 = P2V2
P2 = [P1V1]/V2 = [1atm(0.6L)]/0.2L
P2 = 3atm
P2 = P3
Finding T3
P3V3 = nRT3
T3 = P3V3/nR = 900K
Finding W_12
W_12 = nRT(ln(vf/vi)) = 0.024mol(8.31J/mol*K) (300K)(ln(0.0006m^(3)/0.0002m^(3))) = 65.73 J
Finding W_23
W_23 = P(vf-vi) = 3atm[(0.0006 - 0.0002)m^(3)] * 101325J = 121.59 J
W_31 = 0 because it is an isochoric process
Wnet = 187.32J
Power
P = Wnet/time = 187.32J(20(s^(-1))) = 3746.4 W = 3.746 kW
I got it wrong. Why was gamma given? I thought it was only used for adiabatic processes. I also don't know which c_p and c_v to use since the type of gas is not given.
a.) What is the power output of the engine?
b.) What is the engine's thermal efficiency?
The link to this image is here:
http://session.masteringphysics.com/problemAsset/1074111/4/knight_Figure_19_54.jpg
This is how I attempted to solve this problem:
Given: T1 = T2 = 300K
P2 = P3
P1 = 1 atm = 1.013 * 10^(5) Pa
gamma = 1.25
V1 = V3 = 600 cm^(3)
V2 = 200 cm^(3)
R = 8.31 J/mol*K or 0.08206 atm *L/mol*K
Relevant equations:
PV = nRT
η = Wout/QH
Power = Wnet/time
Q_12 = W_12
Q_23 = nc_pΔT
Q_31 = nc_vΔTFinding number of moles
n = PV/RT
n = [1.013 * 10^(5) Pa(0.0006 m^(3))]/[(8.31J/(mol*K))(300K)]= 0.024 mol
Finding P2 and P3
P1V1 = P2V2
P2 = [P1V1]/V2 = [1atm(0.6L)]/0.2L
P2 = 3atm
P2 = P3
Finding T3
P3V3 = nRT3
T3 = P3V3/nR = 900K
Finding W_12
W_12 = nRT(ln(vf/vi)) = 0.024mol(8.31J/mol*K) (300K)(ln(0.0006m^(3)/0.0002m^(3))) = 65.73 J
Finding W_23
W_23 = P(vf-vi) = 3atm[(0.0006 - 0.0002)m^(3)] * 101325J = 121.59 J
W_31 = 0 because it is an isochoric process
Wnet = 187.32J
Power
P = Wnet/time = 187.32J(20(s^(-1))) = 3746.4 W = 3.746 kW
I got it wrong. Why was gamma given? I thought it was only used for adiabatic processes. I also don't know which c_p and c_v to use since the type of gas is not given.
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