With the help of the indicator diagram. explain working of Carnot engine
temperature of source and sink are 227 C and 27 C respectively, calculate
141
efficiency of Carnot engine.
Answers
hange during heat addition?
3 ANSWERS
Naveen Kumar verma, Engineer at Research
Answered May 20, 2018
Here ,
Source temperature is (327+273=600k)
Sink temperature is (27+273=300k)
Work delivered by engine is 300 kJ .
We know that
Carnot efficiency = ( Higher temp — Lower tem)/Higher temprature
Carnot efficiency = (600-300)/600
Carnot efficiency=50%
Now we have to calculate heat Q1 provided by the source and heat rejected by the engine to the sink Q2.
Again ; carnot efficiency = (W/Q1)
Therefore ; (W/Q1) =(50/100)
Where , W= 300 KJ
Now , (300/Q1)=(.5) or( Q1= 600KJ)
We know that from first law of thermodynamics
dQ= dW ; now (W= Q1– Q2) or (300=600– Q2)
Therefore ; (Q2= 300KJ)
NOW change in entropy during heat addition is :
(dS=dQ/T) where dS is change in entropy
Now; heat rejected from the system at constant temprature therefore ( dQ is negative)
Here in this question source is system and engine is surrounding .when heat is rejected by the system then we consider it is negative and when heat is added to the system .it is positive,if heat added to the sink we consider postive value of heat .
[dS=Q2/T2]
[ dS=( 300KJ)/(300K)]
[dS= 1KJ/K ] ( ANSWER).
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RELATED QUESTIONS (MORE ANSWERS BELOW)
hange during heat addition?
3 ANSWERS
Naveen Kumar verma, Engineer at Research
Answered May 20, 2018
Here ,
Source temperature is (327+273=600k)
Sink temperature is (27+273=300k)
Work delivered by engine is 300 kJ .
We know that
Carnot efficiency = ( Higher temp — Lower tem)/Higher temprature
Carnot efficiency = (600-300)/600
Carnot efficiency=50%
Now we have to calculate heat Q1 provided by the source and heat rejected by the engine to the sink Q2.
Again ; carnot efficiency = (W/Q1)
Therefore ; (W/Q1) =(50/100)
Where , W= 300 KJ
Now , (300/Q1)=(.5) or( Q1= 600KJ)
We know that from first law of thermodynamics
dQ= dW ; now (W= Q1– Q2) or (300=600– Q2)
Therefore ; (Q2= 300KJ)
NOW change in entropy during heat addition is :
(dS=dQ/T) where dS is change in entropy
Now; heat rejected from the system at constant temprature therefore ( dQ is negative)
Here in this question source is system and engine is surrounding .when heat is rejected by the system then we consider it is negative and when heat is added to the system .it is positive,if heat added to the sink we consider postive value of heat .
[dS=Q2/T2]
[ dS=( 300KJ)/(300K)]