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a heat engine is operating on Carnot cycle and has thermal efficiency of 75% .the waste heat from.this engine nia rejected to nearby lake at 15°C at rate of 14KW.find power output of engine and the temperature of the source
Asked by manvirsingh2242 | 20 Jun, 2022, 12:32: PM
answered-by-expert Expert Answer
Thermodynamic efficiency η of carnot cycle
 
begin mathsize 14px style eta space equals space 1 space minus space T subscript L over T subscript H space equals space 1 space minus space Q subscript L over Q subscript H end style
where QL is heat energy rejected at constant sink temperature TL and
QH is heat energy absorbed at constant source temperature TH
 
If heat energy rejected at the rate of 14 kW , then we have
 
begin mathsize 14px style eta space equals space 1 space minus space 14 over Q subscript H space equals space 0.75 end style
from above expression, we get QH = 56 kW
 
Net Workdone = QH - QL = ( 56 - 14) kW = 42 kW
 
If heat energy rejected at temoerature 15o C = 288 K , then we have
 
begin mathsize 14px style eta space equals space 1 space minus space 288 over T subscript H space equals space 0.75 end style
from above expression, we get TH = ( 288 × 4 ) = 1152 K
 
Temperature of source = 1152 K
Answered by Thiyagarajan K | 20 Jun, 2022, 12:57: PM
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