# Thermo reversible heat engines physics question

## Main Question or Discussion Point

I am having trouble with this question...It is from "applied thermodynamics for engineering technologists" by Eastop (Q5.2 if anyone has it).....

The Question:
Two reversible heat engines operate in series between a source at 527°C and a sink at 17°C. If the engines have equal efficiencies and the first rejects 400kJ to the second, calculate:
(i) the temp at which heat is supplied to the 2nd engine ALL DONE :-) =208.7°C
(ii) The heat taken from the source
(iii) The work done by each engine

I have done part (i) and I know it's right but I have no idea what equations to use for the next 2 parts because heat and work are dependant on each other so I dont understand how I can find one without the other....

Dave :-)

## Answers and Replies

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$$\eta_{rev}=1-\frac{T_L}{T_H}[/itex] [tex]\eta_{actual}=\frac{W}{Q_{in}}[/itex] [tex]Q_{in}=W+Q_{out}[/itex] (1st law) [tex]\eta_{rev}=1-\frac{T_L}{T_H}=.319=\eta_{actual}=\frac{W}{Q_{in}}[/itex] [tex]\frac{Q_{1_{in}}}{400}-.319=1[/itex] Using the above formulas, I ended up with the work for each engine at 127.6kJ. Total heat from the source 527.6kJ. However, I took the mid resovoir temp to be 272 deg C, the average of the high and low. Can you tell me how you got 208.7? [tex]\eta_{rev}=1-\frac{T_L}{T_H}[/itex] [tex]\eta_{actual}=\frac{W}{Q_{in}}[/itex] [tex]Q_{in}=W+Q_{out}[/itex] (1st law) [tex]\eta_{rev}=1-\frac{T_L}{T_H}=.319=\eta_{actual}=\frac{W}{Q_{in}}[/itex] [tex]Q_{in}=W+Q_{out}----->Q_{in}-W=400kJ --->1-.319=\frac{400}{Q_{in}}[/itex] I MESSED my first response up. Using the above formulas, I ended up with the work for each engine at 187.4kJ. Total heat from the source 587.4kJ. However, I took the mid resovoir temp to be 272 deg C, the average of the high and low. Can you tell me how you got 208.7? The answer in the book is 208.7 Deg C....but I got it using:- [tex]\eta = 1-(T2/T1)$$

1 - (T2/T3) = 1 - (T3/T1)

1 - (800/T3) = 1 - (T3/290) - Figures converted to Kelvin

232000 = T3$$^{}2$$

T3 = 481.66K = 208.66 Deg C

Ohh....derrr. My bad (I guess I've been paying more attention to the latex than the math). One more time.

[tex]\eta_{rev}=1-\frac{T_L}{T_H}=1-\frac{481.7K}{800K}=.398=\eta_{actual}=\frac{W}{Q_{in}}[/itex]

[tex]Q_{in}=W+Q_{out}----->Q_{in}-W=400kJ --->1-.398=\frac{400kJ}{Q_{in}}[/itex]

[tex]Q_{Total Input}=664.5kJ[/itex]

[tex]W_{1}=664.5kJ-400kJ=264.5kJ[/itex]

[tex]W_{2}=400kJ*.398=159.2kJ[/itex]

Last edited:
omg you're a champion!

I totally didnt get the -W = 1 - 0.398 by using the T3 from part one
Now it just all fits together...it's great

Thankyou so much :-)