76. A solar collector receiving solar radiation at the rate of 0.6 kW/m² transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is used to run a heat engine which rejects heat at 313 K. If the heat engine is to deliver 2.5 kW power, the minimum area of the solar collector required would be
    

1. 1. 83.33 m²
      

   
   
   

   2. 16.66 m²
      

   
   
   

   3. 39.68 m²
      

   
   
   

   4. 79.36 m²
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   10. Given; Receiving solar radiation at the rate of 0.6 kW/m²
   

   Internal energy of fluid after absorbing solar radiation = 0.6 ×	1	kW / m2 = 0.3 kW / m2
   	2

   
   

   ηengine = 1 -	315	= 0.1
   	350

   
   

   0.1 =	W
   	Q1

   
   

   ∴ Q1 =	2.5	= 25 k W
   	0.1

   
   Let A be minimum area of collector
   
   ∴ Q₁ = 0.3 × A or 25 kW = 0.3 kW/m²
   

   or A =	25	= 83.33 m2
   	0.3

   
   

   Correct Option: A

   10. Given; Receiving solar radiation at the rate of 0.6 kW/m²
   

   Internal energy of fluid after absorbing solar radiation = 0.6 ×	1	kW / m2 = 0.3 kW / m2
   	2

   
   

   ηengine = 1 -	315	= 0.1
   	350

   
   

   0.1 =	W
   	Q1

   
   

   ∴ Q1 =	2.5	= 25 k W
   	0.1

   
   Let A be minimum area of collector
   
   ∴ Q₁ = 0.3 × A or 25 kW = 0.3 kW/m²
   

   or A =	25	= 83.33 m2
   	0.3

   
   

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77. A cycle heat engine does 50 kJ of work per cycle. If the efficiency of the heat engine is 75%. The heat rejected per cycle is
    

1. 1. 16	2	kJ
      	3

      
      

   
   
   

   2. 33	1	kJ
      	3

      
      

   
   
   

   3. 37	1	kJ
      	2

      
      

   
   
   

   4. 66	1	kJ
      	2

   
   
   

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1. View Hint View Answer Discuss in Forum

   η = 0.75
   

   ∴ ηHE =	W	= 0.75
   	Q1

   
   
   or, Net heat input,
   

   Q1 =	50	=	200
   	0.75		3

   
   Also , W = Q₁ - Q₂
   

   or 50 =	200	- Q2
   	3

   
   

   ∴ Net heat required Q2 =	50	= 16.66 kJ
   	3

   
   

   Correct Option: A

   η = 0.75
   

   ∴ ηHE =	W	= 0.75
   	Q1

   
   
   or, Net heat input,
   

   Q1 =	50	=	200
   	0.75		3

   
   Also , W = Q₁ - Q₂
   

   or 50 =	200	- Q2
   	3

   
   

   ∴ Net heat required Q2 =	50	= 16.66 kJ
   	3

   
   

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78. For two cycles coupled in series, the topping cycle has an efficiency of 30% and the bottoming cycle has an efficiency of 20%. The overall combined cycle efficiency is
    

1. 1. 50%
      

   
   
   

   2. 44%
      

   
   
   

   3. 38%
      

   
   
   

   4. 55%

   
   
   

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1. View Hint View Answer Discuss in Forum

   η₀ = η₁ + η₂ - η₁η₂
   η₀ = 0.3 + 0.2 – (0.3) (0.2) = 0.44 = 44%

   Correct Option: B

   η₀ = η₁ + η₂ - η₁η₂
   η₀ = 0.3 + 0.2 – (0.3) (0.2) = 0.44 = 44%

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79. A solar energy based heat engine which receives 80 kJ of heat at 100°C and rejects 70 kJ of heat to the ambient at 30°C is to be designed. The thermal efficiency of the heat engine is
    

1. 1. 70%
      

   
   
   

   2. 18.8%
      

   
   
   

   3. 12.5%
      

   
   
   

   4. Indetermine

   
   
   

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1. View Hint View Answer Discuss in Forum

   ηth =	( 80 - 70 )	× 100 =	1	× 100 = 12.5%
   	80		8

   Correct Option: C

   ηth =	( 80 - 70 )	× 100 =	1	× 100 = 12.5%
   	80		8

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80. In the case of a refrigeration system undergoing an irreversible cycle, ∮	δQ	is ....(< 0/ = 0/ > 0)
    	T

1. 1. ∮	δQ	< 0
      	T

   
   
   

   2. ∮	δQ	> 0 clausius inequality
      	T

   
   
   

   3. δQ > 0

   
   
   

   4. δQ = 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∮	δQ	< 0 clausius inequality
   	T

   Correct Option: A

   ∮	δQ	< 0 clausius inequality
   	T

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