Reversible Composite Heat Engine 3 Reservoirs

[sanitykey]

Homework Statement

A reversible composite heat engine operates between three reservoirs at temperatures of 400K, 300K and 200K. One engine operates between the 400K and 300K reservoirs and a second engine operates between the 300K and 200K reservoirs and is synchronised with the first. In one cycle 1200 J of heat is extracted from the 400K reservoir and 100 J of heat is rejected to the 200K reservoir. Calculate the heat exchanged per cycle with the 300K reservoir and the net work done by the engine. Find the total entropy change.

Homework Equations

S_\mathrm{in}=\frac{Q_H}{T_H}=\frac{Q_C}{T_C}=S_\mathrm{out}

W=Q_H-Q_C

where Q_H>Q_C, T_H>T_C

The Attempt at a Solution

I've attatched my attempt at a solution as a scanned image onto this post which is mostly the same as what I've put below but with what i think makes an acceptable diagram (in fact i didn't use a ruler it's more of a quick sketch =P ). I've had a few attempts at this solution and although i think this is better than the others i made i still think it's wrong so if somebody could show me where I've went wrong i'd appreciate it very much.

Thanks in advance for any help =)

S_\mathrm{in}=\frac{1200 J}{400K}=\frac{Q_2}{300K}=S_\mathrm{out}

Q_2=900 J

W_1=1200 J - 900 J = 300 J


S_\mathrm{in2}=\frac{Q_3}{300K}=\frac{100J}{200K}=S_\mathrm{out2}

Q_3=150 J

W_2=150 J - 100 J = 50 J


Net work done by the system W_T=W_1+W_2=300 J + 50 J = 350 J

The total change in entropy is:

S_\mathrm{in} - S_\mathrm{out2} = 3 J/K - 0.5 J/K = 2.5 J/K

 

[sanitykey]

Can anyone help me? I mean no response could be either my approach is so horrendous that it's completely wrong and not worth commenting on or everything could be ok? XD

 

[sanitykey]

I think I'm wrong on the last bit, since the whole system is reversible the total entropy change should be zero i think and what i should of done is added all the changes in entropy to find the total change in entropy rather than taking the difference between the entropy input into the first part of the system and the entropy output in the second part of the system.

So i think it should look something like:

\Delta S = \Delta S_1 + \Delta S_2 = 0 + 0 = 0

where \Delta S is the total change in entropy, \Delta S_1 is the change in entropy between the 400K and 300K reservoirs and \Delta S_2 is the change in entropy between the 300K and 200K reservoirs.

As for the rest of it i hope it's right.

 

[sanitykey]

I got my marks back the other day and everything was ok ^_^

 

[Andrew Mason]

I think I'm wrong on the last bit, since the whole system is reversible the total entropy change should be zero i think and what i should of done is added all the changes in entropy to find the total change in entropy rather than taking the difference between the entropy input into the first part of the system and the entropy output in the second part of the system.

So i think it should look something like:

\Delta S = \Delta S_1 + \Delta S_2 = 0 + 0 = 0

where \Delta S is the total change in entropy, \Delta S_1 is the change in entropy between the 400K and 300K reservoirs and \Delta S_2 is the change in entropy between the 300K and 200K reservoirs.

As for the rest of it i hope it's right.

Your heat calculations are correct but the entropy calculations are not.

For the first engine, since it is a reversible (Carnot) cycle, the efficiency is given by:
\eta = W/Q_h = (Q_h - Q_c)/Q_h = (T_h - T_c)/T_h = (400-300)/400 = .25 where Qh is the heat flow out of the hot reservoir and Qc the heat flow into the cold.

Since Qh = 1200 J, Qc = 900 J.

Now, dS = dQ/T. But you have to give the heat a sign depending on whether it is flowing into (+) or out of (-) the reservoir.

So the entropy change for the hot reservoir (heat leaving) is: \Delta S_h = \int dQ_h/T_h = -1200/400 = -3 J/K and the entropy change for the cold reservoir is \Delta S_c = \int dQ_c/T_c = +900/300 = +3 J/K .

Therefore, the total change in entropy is the sum of the entropy changes for each reservoir:

\Delta S_{total} = \Delta S_h + \Delta S_c = \text{-3 + 3 = 0 J/K}

AM

 

[Andrew Mason]

Your heat calculations are correct but the entropy calculations are not.

For the first engine, since it is a reversible (Carnot) cycle, the efficiency is given by:
\eta = W/Q_h = (Q_h - Q_c)/Q_h = (T_h - T_c)/T_h = (400-300)/400 = .25 where Qh is the heat flow out of the hot reservoir and Qc the heat flow into the cold.

Since Qh = 1200 J, Qc = 900 J.

Now, dS = dQ/T. But you have to give the heat a sign depending on whether it is flowing into (+) or out of (-) the reservoir.

So the entropy change for the hot reservoir (heat leaving) is: \Delta S_h = \int dQ_h/T_h = -1200/400 = -3 J/K and the entropy change for the cold reservoir is \Delta S_c = \int dQ_c/T_c = +900/300 = +3 J/K .

Therefore, the total change in entropy is the sum of the entropy changes for each reservoir:

\Delta S_{total} = \Delta S_h + \Delta S_c = \text{-3 + 3 = 0 J/K}

AM

Repost with Latex working.