How Is Thermo Efficiency Calculated in a Monatomic Gas Cycle?

[AriAstronomer]

Thermo Efficiency question!

Homework Statement

Figure P22.53 represents n mol of an ideal monatomic
gas being taken through a cycle that consists of two
isothermal processes at temperatures 3Ti and Ti and two
constant-volume processes. For each cycle, determine, in terms of n, R, and Ti , (a) the net energy transferred
by heat to the gas and (b) the efficiency of an engine
operating in this cycle.

Homework Equations

The Attempt at a Solution

Now, I've solved part a), and get W = 2nRTln2 (difference between top and bottom portions), can't get part b). I know that efficiency = energy sought/ energy cost = W/Q_h, but they don't really specify in the question where the energy intake is, or energy cost. The answer is 0.273... I've found that for each cycle:
the sides both equal dU = dQ = 3nRT, the top Q = 3nRTln2, bottom Q = nRTln2. Seems like the 'energy cost' based on e = .273 and W = 2nRTln2 should be 3+3ln2, no idea how that comes about though... perhaps somehow the 'cost' is over the top and side??

Ari

 

[Andrew Mason]

The Attempt at a Solution

Now, I've solved part a), and get W = 2nRTln2 (difference between top and bottom portions), can't get part b). I know that efficiency = energy sought/ energy cost = W/Q_h, but they don't really specify in the question where the energy intake is, or energy cost. The answer is 0.273... I've found that for each cycle:
the sides both equal dU = dQ = 3nRT, the top Q = 3nRTln2, bottom Q = nRTln2. Seems like the 'energy cost' based on e = .273 and W = 2nRTln2 should be 3+3ln2, no idea how that comes about though... perhaps somehow the 'cost' is over the top and side??

Determine the direction of the heat flow in each part of the graph. To do that you must be careful about the sign. Heat flow into the gas is positive. Heat flow out is negative. Total heat flow in is Qh. Total heat flow out is Qc. What is W in terms of Qh and Qc? From that you should be able to determine efficiency.

AM

 

[AriAstronomer]

Alright. Makes sense. W = Q_h - Q_l, and Q_h is all the positive contributions to the work, while Q_l is all the negative ones. When I do my calculations, I get the first two legs coming out positive, and the second two legs coming out negative, thus Q_h = 3nRT + 3nRTln2. Thanks Andrew.

Ari

 

[Andrew Mason]

Alright. Makes sense. W = Q_h - Q_l, and Q_h is all the positive contributions to the work, while Q_l is all the negative ones. When I do my calculations, I get the first two legs coming out positive, and the second two legs coming out negative, thus Q_h = 3nRT + 3nRTln2.

Generally correct. Strictly speaking Qh, being heat flow, consists of all the positive contributions to the heat flow, not the work. (This heat flow enables greater work to be done in the expansion phase).

You can determine the direction of heat flow qualitatively from the First Law. A constant volume increase in pressure requires heat flow into the gas (dQ = dU > 0). An isothermal expansion requires heat flow in (dQ = dW > 0).

AM

 

[rwisz]

ana, it seems like you have made some progress in solving the problem. However, it is difficult for me to provide a response without seeing the full problem and your attempted solution. Can you provide more information or the full problem so I can assist you better? Additionally, it would be helpful if you could clarify what exactly you are struggling with in part b) of the problem.