16. A solid cylinder (surface 2) is located at the centre of a hollow sphere (surface 1). The diameter of the sphere is 1 m, while the cylinder has a diameter and length of 0.5 m each.'The radiation configuration factor F₁₁ is

1. 1. 0.375

   
   
   

   2. 0.625

   
   
   

   3. 0.75

   
   
   

   4. 1

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   Diameter of sphere = 1 m
   Diameter of aylinder = 0.5 m
   F₂₁ + F₂₂ = 1 (∵ F₂₂ = 0)
   F₂₂ = 1
   we know that
   A₁ F₁₂ = A₂ F₂₁
   

   F12 =	A2
   	A1

   
   

   ∴ F11 = 1 - F12 = 1	- A2
   	A1

   
   

   ∴ F11 = 1 -	A2
   	A1

   
   

   F11 = 1 -	[(π × 0.5 × 0.5) + ( 2 × π/4 × 0.5 × 0.5)]	= 0.625
   	4π × 0.5 × 0.5

   Correct Option: B

   
   Diameter of sphere = 1 m
   Diameter of aylinder = 0.5 m
   F₂₁ + F₂₂ = 1 (∵ F₂₂ = 0)
   F₂₂ = 1
   we know that
   A₁ F₁₂ = A₂ F₂₁
   

   F12 =	A2
   	A1

   
   

   ∴ F11 = 1 - F12 = 1	- A2
   	A1

   
   

   ∴ F11 = 1 -	A2
   	A1

   
   

   F11 = 1 -	[(π × 0.5 × 0.5) + ( 2 × π/4 × 0.5 × 0.5)]	= 0.625
   	4π × 0.5 × 0.5

---

17. A hollow enclosure is formed between two inf initely long concentric cylinders of radii 1 m and 2 m, respectively. Radiative heat exchange takes place between the inner surface of the larger cylinder (surface-2) and the outer surface of the smaller cylinder (surface-1). The radiating surface are diffuse and the medium in the enclosure is non- participating. The fraction of the thermal radiation leaving the larger surface and striking itself is
    

1. 1. 0.25

   
   
   

   2. 0.5

   
   
   

   3. 0.75

   
   
   

   4. 1

   
   
   

---

1. View Hint View Answer Discuss in Forum

   F₁₁ + F₁₂ = 1
   F₂₁ + F₂₂ = 1
   F₁₁ = 0 due to concave surface
   F₁₂ = 1 Now
   A₁ F₁₂ = A₂ F₂₁
   ⇒ F₂₁ = F₁₂ A₁/A₂
   ⇒ F₁₂ = 1
   

   ∴ F21 =	A1	2πr1l	r1	1
   	A2	2πr2l	r2	2

   
   F₂₁ = 0.5
   F₂₂ = 1 – F₂₁ = 0.5

   Correct Option: B

   F₁₁ + F₁₂ = 1
   F₂₁ + F₂₂ = 1
   F₁₁ = 0 due to concave surface
   F₁₂ = 1 Now
   A₁ F₁₂ = A₂ F₂₁
   ⇒ F₂₁ = F₁₂ A₁/A₂
   ⇒ F₁₂ = 1
   

   ∴ F21 =	A1	2πr1l	r1	1
   	A2	2πr2l	r2	2

   
   F₂₁ = 0.5
   F₂₂ = 1 – F₂₁ = 0.5

---

---

18. Consider two infinitely long thin concentric tubes of circular cross section as shown in figure. If D₁ and D₂ are the diameters of the inner and outer tubes respectively, then the view factor F₂₂ is given by
    

1. 1. 	D2		- 1
      	D1

   
   
   

   2. Zero

   
   
   

   3. 	D1
      	D2

   
   
   

   4. 1 -		D1
      		D2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   F₂₂ = 1 - F₂₁ =
   

   = 1 -	A1	= 1 -	D1
   	A2		D2

   Correct Option: D

   F₂₂ = 1 - F₂₁ =
   

   = 1 -	A1	= 1 -	D1
   	A2		D2

---

19. For the same inlet and exit temperatures of the hot and cold fluids, the log mean temperature difference (LMTD) is

1. 1. Greater for parallel flow heat exchanger than the counter flow heat exchanger

   
   
   

   2. Greater for counter flow heat exchanger than the parallel and cross flow heat exchanger

   
   
   

   3. Same for both parallel and counter flow heat exchangers

   
   
   

   4. Depending on the properties of fluid

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---

---

20. In a condenser of a power plant, the steam condenses at a temperature of 60°C. The cooling water enters at 30°C and leaves at 45°C. The Logarithmic Mean Temperature Difference (LMTD) of the condenser is

1. 1. 16.2°C

   
   
   

   2. 21.6°C

   
   
   

   3. 30°C

   
   
   

   4. 37.5*C

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Flow configuration in condenser as shown below.
   
   ΔT₁ = 30°C (Q₁)
   ΔT₂ = 15°C (Q₂)
   

   LMTD =	ΔT1 - ΔT2				=	30 - 15
   	In		ΔT1			In		30
   			ΔT2					15

   
   = 21.6°C

   Correct Option: B

   Flow configuration in condenser as shown below.
   
   ΔT₁ = 30°C (Q₁)
   ΔT₂ = 15°C (Q₂)
   

   LMTD =	ΔT1 - ΔT2				=	30 - 15
   	In		ΔT1			In		30
   			ΔT2					15

   
   = 21.6°C

---