Engine Efficiency: Find ΔW & ΔQh

[physics123]

Homework Statement

An ideal gas with Cv = 5 2R, and γ = 1.4 starts at a volume of 1.5m3 , a pressure of 2.0×105Pa ,and a temperature of 300K. It undergoes an isobaric expansion until the volume is V , then undergoes an adiabatic expansion until the volume is 6.0m3 , and finally undergoes an isothermal contraction until it reaches the original state.

Homework Equations

What is the efficiency of this heat engine?

The Attempt at a Solution

I am aware that the answer is 0.19, and that efficiency is ∆W/∆Qh but I am unsure of how to find ∆W and ∆Qh in this scenario.

 

[DrClaude]

Make a PV diagram and calculate W and Q for each leg of the cycle.

 

[physics123]

There is usually a formula for W, but what is the formula for Q?

 

[physics123]

Make a PV diagram and calculate W and Q for each leg of the cycle.

I did the calculations for the W and Q for each PV leg and got:
W= 146000 for isobaric, W = 346524 for adiabatic, and W = -415888 for isothermic.
Q = 511000 Q= 0 Q = W = -415888
Adding these up I get 94636/95112 = 0.99. I know the answer is not that. Any ideas of what I could be doing wrong?

 

[Ekullabran]

Make a PV diagram and calculate W and Q for each leg of the cycle.

What is the next step after I calculate the W and Q for each leg of the cycle?

 

[Chestermiller]

I did the calculations for the W and Q for each PV leg and got:
W= 146000 for isobaric, W = 346524 for adiabatic, and W = -415888 for isothermic.
Q = 511000 Q= 0 Q = W = -415888
Adding these up I get 94636/95112 = 0.99. I know the answer is not that. Any ideas of what I could be doing wrong?

Please show the details of your work.

 

[physics123]

Please show the details of your work.

Isobaric: W = P∆V = (2x10^5)(2.23-1.5) = 146000J Q = (C/R)(Vf-Vi)p = 3.5(2.23-1.5)2x10^5 = 511000J
Adiabatic: W = (K (Vf^1-gamma - Vi^1-gamma) )/(1-gamma) where K = PV^1-gamma = (2x10^5)2.23^1.4 = 614690
W = 614690 ( 0.4884-0.72556) / -0.4 = 364524J Q = 0
Isothermic: W = PiVi ln(vf/vi) = (6)(5x10^4) ln(1.5/6) = -415888 Q = -W

Now to add the Ws, and the Qs. W = 146000 + 364524 + -415888 = 94636 Q = 511000 + 415888 = 926888
e = W/Q = 94636/926888 = 0.10
which is incorrect. Any ideas?

 

[TSny]

Isobaric: W = P∆V = (2x10^5)(2.23-1.5) = 146000J Q = (C/R)(Vf-Vi)p = 3.5(2.23-1.5)2x10^5 = 511000J
Adiabatic: W = (K (Vf^1-gamma - Vi^1-gamma) )/(1-gamma) where K = PV^1-gamma = (2x10^5)2.23^1.4 = 614690
W = 614690 ( 0.4884-0.72556) / -0.4 = 364524J Q = 0
Isothermic: W = PiVi ln(vf/vi) = (6)(5x10^4) ln(1.5/6) = -415888 Q = -W

OK. For the isothermic you have that Q = -W. Should that be Q = +W?

Now to add the Ws, and the Qs. W = 146000 + 364524 + -415888 = 94636 Q = 511000 + 415888 = 926888
e = W/Q = 94636/926888 = 0.10
which is incorrect. Any ideas?

The definition of efficiency is W_cycle/Q_in where Q_in is the amount of heat that went into the engine during the cycle rather than the total Q for the entire cycle. So, identify the part or parts of the cylce where heat went into the engine.

 

[physics123]

OK. For the isothermic you have that Q = -W. Should that be Q = +W?


The definition of efficiency is W_cycle/Q_in where Q_in is the amount of heat that went into the engine during the cycle rather than the total Q for the entire cycle. So, identify the part or parts of the cylce where heat went into the engine.

i found the answer, thanks everyone

 

[Chestermiller]

i found the answer, thanks everyone

Actually, to do this, all you have to do is. to get the three heats:

Q1 =511000

Q2 = 0

Q3 = -415888

So, the work is W = Q1 + Q2 +Q3 = 95112

So the efficiency is 95112/511000