Uncategories Heat and Thermodynamics :: Problem 6

Heat and Thermodynamics :: Problem 6

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Problem 6: The low temperature reservoir of a carnot heat engine is at 7oC and has an efficiency of 40%. It is desired to increase the efficiency to 50%. By how much degrees the temperature of hot reservoir be increased?

Solution:

Step:1 Overview

How to change Efficiency of a Heat Engine:
The efficiency of a heat engine can be altered by changing either temperature of any or both reservoirs. Hot body can also be referred as high temperature reservoir or hot reservoir and cold body can also be referred as low temperature reservoir of cold reservoir.

Step:2 Calculation

T2=7oC=280K
E=40%=0.4
T1=?
E=1-T2/T1
0.4=1-280K/T1
280K/T1=1-0.4
280K/T1=0.6
280K=0.6T1
T1=280K/0.6
T1=466.67K
Now we have to find T1 at E=50%(0.5):
E=1-T2/T1
0.5=1-280K/T1
280K/T1=1-0.5
280K/T1=0.5
280K=0.5T1
T1=280K/0.5
T1=560K
dT=T1final-T1initial
dT=560K-266.67K
dT=93.33K(Ans)

 

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