111. Heat transfer coefficients for free convection in gases, forced convection in gases and vapours, and for boiling water lie, respectively, in the range of

1. 1. 5-15; 20-200 and 3000-50000 W/m²K

   
   
   

   2. 20-50,200-500 and 50000-100000 W/m²K

   
   
   

   3. 50-100, 500-1000 and 100000-100000 W/m²K

   
   
   

   4. 20-100, 200-1000 and a constant 1000000 Wm²K

   
   
   

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1. View Hint View Answer Discuss in Forum

   5-15; 20-200 and 3000-50000 W/m²K

   Correct Option: A

   5-15; 20-200 and 3000-50000 W/m²K

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112. Three slabs are joined together as shown in the figure. There is no thermal contact resistance at the interfaces. The cent er slab experience a nonuniform internal heat generation with an average value equal to 10000 Wm^-3, while the left and right slabs have no internal heat generation. All slabs have thickness equal to 1 m and thermal conductivity of each slab is equal to 5 Wm^-1 K^-1. The two extreme faces are exposed to fluid with heat transfer coefficient 100 Wm^-2K ^-1 and bulk temperature 30°C as shown. The heat transfer in the slabs is assumed to be one dimensional and steady, and all properties are constant. If the left extreme face temperature T₁ is measured to be 100°C, the right extreme faced temperature T₂ is___________°C.
     

1. 1. 60°C

   
   
   

   2. 50°C

   
   
   

   3. 51°C

   
   
   

   4. 52°C

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   Correct Option: A

   
   

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113. Consider a laminar boundary layer over a heated flat plate. The free stream velocity is U_∞. At some distance x from the leading edge the velocity boundary layer thickness is δ_v and the thermal boundary layer thickness is δ_T. If the Prandtl number is greater than 1, then

1. 1. δ_v > δ_T

   
   
   

   2. δ_T > δ_v

   
   
   

   3. δ_v~ δ_T~(U_∞x)^–1/2

   
   
   

   4. δ_v ~ δ_T~x^–1/2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Prandtl number =	Molecular diffusivity of momentum
   	Molecular diffusivity of heat

   
   From question, since prandtl number > 1
   ∴ Velocity boundry thickness (δ_v) > thermal boundary thickness (δ_t)

   Correct Option: A

   Prandtl number =	Molecular diffusivity of momentum
   	Molecular diffusivity of heat

   
   From question, since prandtl number > 1
   ∴ Velocity boundry thickness (δ_v) > thermal boundary thickness (δ_t)

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114. The Reynolds number for the flow is

1. 1. 444

   
   
   

   2. 890

   
   
   

   3. 4.44 × 10⁵

   
   
   

   4. 5.33 ×10⁵

   
   
   

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1. View Hint View Answer Discuss in Forum

   Lc =	4A	= 4 ×	1 × 0.5	= 0.666
   	D		2 × 1.5

   
   Reynolds number,
   

   Re =	ρLcV	=	1.2 × 0.66 × 10	= 4.4 × 105 .
   	μ		18 × 10-6

   Correct Option: C

   Lc =	4A	= 4 ×	1 × 0.5	= 0.666
   	D		2 × 1.5

   
   Reynolds number,
   

   Re =	ρLcV	=	1.2 × 0.66 × 10	= 4.4 × 105 .
   	μ		18 × 10-6

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115. The heat transfer per meter length of the duct, in watts, is

1. 1. 3.8

   
   
   

   2. 5.3

   
   
   

   3. 89

   
   
   

   4. 769

   
   
   

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1. View Hint View Answer Discuss in Forum

   As Re > 4000, the flow is turbulent
   Nu = 0.023 Re^0.8Pr^0.33
   hL/k = 0.023 × (4.44 × 10⁵)^0.8 × (0.73)^0.33
   h × 0.0667/0.25 = 683.173
   h = 25.60 W/m²K
   Surface area of duct
   A = 2 × al + 2 × bl
   where, l = length of duct
   ∴ A = 3l
   Heat transfer rate, Q = h A(T₀ – T)
   Q = 25.60 × 3l × (30 – 20)
   Q/l = 769W

   Correct Option: D

   As Re > 4000, the flow is turbulent
   Nu = 0.023 Re^0.8Pr^0.33
   hL/k = 0.023 × (4.44 × 10⁵)^0.8 × (0.73)^0.33
   h × 0.0667/0.25 = 683.173
   h = 25.60 W/m²K
   Surface area of duct
   A = 2 × al + 2 × bl
   where, l = length of duct
   ∴ A = 3l
   Heat transfer rate, Q = h A(T₀ – T)
   Q = 25.60 × 3l × (30 – 20)
   Q/l = 769W

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