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Q. A heat engine working on carnot cycle absorbs heat from three thermal reservoirs at 1000k, 800k, and 600k. The engine does 10 kw of net work and rejects 400kj/min of heat to a sink at 300k. If heat supplied by the reservoir at 1000k is 60% of the heat supplied by the reservoir at 600k, make calculations for the quantity of heat absorbed by each reservoir.

Category asked Dec 23, 2016
edited Jan 11, 2017 | 863 views
Always "Provide your research efforts to solve this homework problem."
Have already solve for the answer but I needed clarification....

And also I want people to ask a lot of questions.....for me to solve but I see less.....am sorry for putting that question there
You can edit your question and add sections which need clarification.
It is always good to be curious and ask questions. You can even post answers for known questions with tag self reference,so that it would be useful to other users who go through your post.

Thanks
rightly said by @ppk, questions should have research efforts.Hope you provide them for your questions in future :) , have a look for answer

Heat from reservoirs is absorbed at temperatures:

T1=1000 K

T2=800 K

T3=600 K

Let heat from those reservoirs be Q1,Q2 and Q3 respectively.

Now, Workdone by Engine :

W = 10 kw

Heat Rejected : Q4=400 kj/min

At Temp : T4=300 K

Net Heat -> Q = Q1+Q2+Q3

But we dont have Q1, Q2 or Q3.

Q = W + Q4 = 10kw + 400 kj/min

1 kw=60 000 J / min -> 10 kw = 600 KJ / min

Q = 600 + 400 = 1000 KJ/min

Also from question :

Heat from reservoir at 1000K = 60% of heat from reservoir at 600K

that is : Q1=0.6Q3

Now,

Q = Q1 +Q2 + Q3

1000 = 0.6Q+Q+Q3

Q2=1000 - 1.6Q3

Apply Clausius theorem to find values of Q1 , Q2 and Q3

Q1/T1 + Q2/T2 + Q3/T3 = Q4/T4

0.6Q3/1000 + (1000 - 1.6Q3)/80 + Q3/600 = 400/300

Q3 = 312.5 kJ / min

Q1 = 0.6 * 312.5 = 187.5 kJ/min

we have , Q2=Q - (Q1+Q3)

Q2 = 1000 - (312.5 + 187.5 )

Q2 = 500 kJ / min

Thus we have heat absorbed from each reservoir as Q1,Q2 and Q3 .

Thanks

answered Jan 7, 2017