Answers

2015-04-21T19:40:08+05:30
In case of efficiency 40 %--------η=40/100=T₁-T₂/T₁

2/5=T₁-300/T₁
T₁=500k
T₁=227degree celsius
if efficiency increased by10 %  then
50/100=T₁¹-T₂/T₁¹
1/2=T₁¹-300/T₁¹
T₁¹=600k
    =327 degree celsius
T₁¹-T₁=327-227
         =100 degree celsius
0
2015-04-22T14:31:59+05:30

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Carnot Engine in Thermodynamics:

efficiency η = 40%
T₁ = 27 degC = 300 °K  = heat sink temperature
T₂ = ? = temperature of the heat source

  40 % = η = 1 - T₁ / T₂ = 1 - 300 / T₂
     T₂ = 300 / 0.60  = 500 °K =  227 °C

 If the efficiency is increased by 10% to 50 %

   Then    1 -  T₁ / T₂ = η = 0.50
               T₂  = 2  T₁  = 600 °K  =  327 °C

Thus the temperature of the source is to be increased by 100 °C  to increase the efficiency by 10 %  for an ideal Carnot machine.

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