166. A vehicle powered by a spark ignition engine follows air standard Otto cycle (γ = 1.4). The engine generates 70 kW while consuming 10.3 kg/hr of fuel. The calorific value of fuel is 44000 kJ/kg. The compression ratio is (correct to two decimal places).

1. 1. γ = 8

   
   
   

   2. γ = 8.61

   
   
   

   3. γ = 7.61

   
   
   

   4. γ = 9

   
   
   

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1. View Hint View Answer Discuss in Forum

   r = 14
   BP = 70 kW
   Ṁ_f = 10.3 kg/hr
   C.V. = 44000 kJ/kg
   

   η =	BP	=	70
   	Ṁf × C.V.		10.3/3600 × 44000

   
   η = 0.556
   

   ηotto = 1 -	1	= 0.556
   	(r)γ-1

   
   ⇒ γ = 7.61

   Correct Option: C

   r = 14
   BP = 70 kW
   Ṁ_f = 10.3 kg/hr
   C.V. = 44000 kJ/kg
   

   η =	BP	=	70
   	Ṁf × C.V.		10.3/3600 × 44000

   
   η = 0.556
   

   ηotto = 1 -	1	= 0.556
   	(r)γ-1

   
   ⇒ γ = 7.61

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167. A system undergoes a State change from 1 to 2.According to the second law of thermodynamics, for the process to be feasible, the entropy change, S₂ – S₁ of the system

1. 1. is positive or zero

   
   
   

   2. is negative or zero

   
   
   

   3. is zero

   
   
   

   4. can be positive, negative or zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   Entropy of irreversible process always increases.

   Correct Option: D

   Entropy of irreversible process always increases.

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168. For an ideal gas the expression

     T

     δs

     - T

     δs

     is always equal to

     δT

     P

     δT

     V

1. 1. Zero

   
   
   

   2. C_P/C_V

   
   
   

   3. R

   
   
   

   4. RT

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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169. Consider a refrigerator and a heat pump working on the reversed Carnot cycle between the same temperature limits. Which of the following is correct?

1. 1. COP of refrigerator = COP of heat pump

   
   
   

   2. COP of refrigerator = COP of heat pump + 1

   
   
   

   3. COP of refrigerator = COP of heat pump –1

   
   
   

   4. COP of refrigerator = Inverse of the COP of heat pump

   
   
   

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1. View Hint View Answer Discuss in Forum

   COP of refrigerator = COP of heat pump –1

   Correct Option: C

   COP of refrigerator = COP of heat pump –1

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170. The figure shows a heat engine (HE) working between two reservoirs. The amount of heat (Q₂) rejected by the heat engine is drawn by a heat pump (HP). The heat pump receives the entire work output (W) of the heat engine. If temperatures, T₁ > T₃ > T₂, then the relation between the efficiency (η) of the heat engine and t he coeffi ci ent and t he coeffi ci ent of performance (COP) of the heat pump is
     

1. 1. COP = 1 + η

   
   
   

   2. COP = η – 1 – 1

   
   
   

   3. COP = η – 1

   
   
   

   4. COP = η

   
   
   

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1. View Hint View Answer Discuss in Forum

   Efficiency of heat engine,
   

   η =	W	=	Q1 - Q2
   	Q1		Q1

   
   

   (COP)H.P =	Q3
   	W

   
   ∵ Q₃ = Q₂ + W = Q₂ + Q₁ - Q₂
   Q₃ = Q₁
   

   so, (COP)H.P =	Q1
   	W

   
   

   (COP)H.P =	1	= η-1
   	η

   Correct Option: C

   Efficiency of heat engine,
   

   η =	W	=	Q1 - Q2
   	Q1		Q1

   
   

   (COP)H.P =	Q3
   	W

   
   ∵ Q₃ = Q₂ + W = Q₂ + Q₁ - Q₂
   Q₃ = Q₁
   

   so, (COP)H.P =	Q1
   	W

   
   

   (COP)H.P =	1	= η-1
   	η

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