26. Hot oil is cooled from 80 to 50°C in an oil cooler which uses air as the coolant. The air temperature rises from 30 to 40°C. The designer uses a LMTD value of 26°C. The type of heat exchanger is
    

1. 1. parallel flow
      

   
   
   

   2. double pipe
      

   
   
   

   3. counter flow
      

   
   
   

   4. cross flow

   
   
   

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1. View Hint View Answer Discuss in Forum

   θ =	θ1 - θ2
   	ln	θ1
   		θ2

   
   θ₁ = 50° , θ₂ = 10°
   θ₁ = 40° , θ₂ = 20°
   As given value is between these two, so cross flow is the best option.
   

   Correct Option: D

   θ =	θ1 - θ2
   	ln	θ1
   		θ2

   
   θ₁ = 50° , θ₂ = 10°
   θ₁ = 40° , θ₂ = 20°
   As given value is between these two, so cross flow is the best option.
   

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27. In a heat exchanger, it is observed that ∆T₁ = ∆T₂ where ∆T₁ is the temperature difference between the two single phase fluid streams at one end and ∆T₂ is the temperature difference at the other end. This heat exchanger is
    

1. 1. a condenser
      

   
   
   

   2. an evaporator
      

   
   
   

   3. a counter flow heat exchanger
      

   
   
   

   4. a parallel flow heat exchanger
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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28. In certain heat exchanger, both the fluids have identical mass flow rate specific heat product. The hot fluid enters at 76°C and leaves at 47°C, and the cold fluid entering at 28°C leave at 55°C, the effectiveness of the HE is
    

1. 1. 0.16
      

   
   
   

   2. 0.60
      

   
   
   

   3. 0.72
      

   
   
   

   4. 1.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   ε =	Ch(Th1 - Th2)
   	Cmin(Th1 - Tc1)

   
   Here , C_h = C_c'
   

   ε =	Th1 - Th2	=	76 - 47	= 0.604
   	Th1 - Tc1		76 - 28

   
   

   Correct Option: B

   ε =	Ch(Th1 - Th2)
   	Cmin(Th1 - Tc1)

   
   Here , C_h = C_c'
   

   ε =	Th1 - Th2	=	76 - 47	= 0.604
   	Th1 - Tc1		76 - 28

   
   

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29. Two rods, one of length L and the other of length 2L are made of the same material and have the same diameter. The two ends of the longer rod are maintained at 100°C. One end of the shorter rod is maintained at 100°C while the other end is insulated. Both the rods are exposed to the same environment at 40°C. The temp at the insulated end of the shorter rod is measured to be 55°C. The temperature at the mid point of the longer rod would be.
    

1. 1. 40°C
      

   
   
   

   2. 50°C
      

   
   
   

   3. 55°C
      

   
   
   

   4. 100°C

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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30. A fin has 5 mm diameter and 100 mm length. The thermal conductivity of fin material is 400 Wm^–1K^–1. One end of the fin is maintained at 130°C and its remaining surface is exposed to ambient air at 30°C. If the convective heat transfer coefficient is 40 Wm^–2K^–1, the heat loss (in W) from the fin is
    

1. 1. 0.08
      

   
   
   

   2. 5.0
      

   
   
   

   3. 7.0
      

   
   
   

   4. 7.8

   
   
   

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1. View Hint View Answer Discuss in Forum

   Fin diameter, d = 5 mm
   Fin length, L = 100 mm
   k = 400 W/mK
   T₀ = 130°C
   T_∞ = 30°C
   h = 40 W/m²K
   ∴ Heat loss from fin Q = √kAhP (T₀ - T_∞) tanh ml
   where P = πd = 3.14 × 5 × 10^{– 3} = 0.015
   

   A =	π	d2 =	π	× (5 × 10– 3)2 = 1.96 × 10– 5 m 2
   	4		4

   
   ∴ m = √hp / kA = √(40 × 0.0157) / (400 × 1.96 × 10^{– 5}) = 8.949
   ∴ ml = 8.949 × 0.100 = 0.895
   ∴ Q = √40 × 0.0157 × 400 × 1.96 × 10^{– 5}(T₀ - T_∞) tanh (0.895) = 5.0 W

   Correct Option: B

   Fin diameter, d = 5 mm
   Fin length, L = 100 mm
   k = 400 W/mK
   T₀ = 130°C
   T_∞ = 30°C
   h = 40 W/m²K
   ∴ Heat loss from fin Q = √kAhP (T₀ - T_∞) tanh ml
   where P = πd = 3.14 × 5 × 10^{– 3} = 0.015
   

   A =	π	d2 =	π	× (5 × 10– 3)2 = 1.96 × 10– 5 m 2
   	4		4

   
   ∴ m = √hp / kA = √(40 × 0.0157) / (400 × 1.96 × 10^{– 5}) = 8.949
   ∴ ml = 8.949 × 0.100 = 0.895
   ∴ Q = √40 × 0.0157 × 400 × 1.96 × 10^{– 5}(T₀ - T_∞) tanh (0.895) = 5.0 W

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