NEET Class neet Answered
For a carnot engine operating in between 27°C and 327°C. i) Calculate heat rejected for each joule of work done. ii) Also calculate change in temperature of source to reduce the efficiency by 20%.
Asked by patra04011965 | 21 Apr, 2020, 01:39: PM
Expert Answer
Quantity of heat rejected in Carnot engine, Q2 = R T2 log(V4 / V3 ) ....................(1)
where R is universal gas constant , T2 is sink temperature , V3 is inital volume of isothermal compression and V4 is final volume .
Workdone W = Q1 - Q2 , where Q1 is the quantity of heat absorbed from heat source
Q1 = R T1 log(V2 / V1) ..................(2)
Where T1 is source temperature , V2 and V1 are final and initial volume in isothermal expansion.
In Carnot engine , Volume ratio ( V2/V1 ) of isothermal expansion process equals the volume ratio ( V4 / V3 ) of isthermal compression.
Hence, ratio of heat rejected for the workdone Q2 / [ Q1 - Q2 ] = T2 / [ T1 - T2 ] = 300/ (600-300) = 1
-------------------------
Efficiency = 1- ( T2 / T1 ) = 1 - ( 300 / 600 ) = 0.5 or 50 %
If we consider to reduce the efficiency by 20% as 30% efficency of changed system, then
1 - (300 / T1 ) = 0.3
By solving above equation, we get T1 = 429 K or 156 °C
Answered by Thiyagarajan K | 21 Apr, 2020, 08:48: PM
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