Heat Transfer Miscellaneous
 In the laminar flow of air (Pr = 0.7) over a heated plate, if δ and δ_{T} denote, respectively, the hydrodynamic and thermal boundary layer thicknesses, then

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When P_{r} < 1 ⇒ δ_{T} > δ
P_{r} > 1 ⇒ δ_{T} < δ
P_{r} = 1 ⇒ δ_{T} = δCorrect Option: C
When P_{r} < 1 ⇒ δ_{T} > δ
P_{r} > 1 ⇒ δ_{T} < δ
P_{r} = 1 ⇒ δ_{T} = δ
 If a foam insulation is added to a 4 cm outer diameter pipe as shown in the figure, the critical radius of insulation (in cm) is ________.

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K_{foam} = 0.1 W mK h_{0} = 2 W m²K Critical radius, r_{c} = K_{foam} for a pipe h_{0} = 0.1 = 0.05 m = 5 cm 2 Correct Option: C
K_{foam} = 0.1 W mK h_{0} = 2 W m²K Critical radius, r_{c} = K_{foam} for a pipe h_{0} = 0.1 = 0.05 m = 5 cm 2
Direction: Consider steady onedimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m³. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.
 The maximum temperature within the plate in °C is

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T^{max} = 165º [putting x = 5 × 10–3 is equation (1)]
Correct Option: B
T^{max} = 165º [putting x = 5 × 10–3 is equation (1)]
 With an increase in the thickness of insulation around a circular pipe, heat loss to surroundings due to

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With an increases in the thickness of insulation around a circular pipe, heat loss to surroundings due to convection increases and due to conduction decreases.
Correct Option: A
With an increases in the thickness of insulation around a circular pipe, heat loss to surroundings due to convection increases and due to conduction decreases.
 A fluid (Prandt/number, Pr = 1) at 500 K flows over a flat plate of 1.5 m length, maintained at 300 K. The velocity of the fluid is 10 m/s. Assuming kinematic viscosity. μ = 30 × 10^{–6} m²/s, the thermal boundary layer thickness (in mm) at 0.5 m from the leading edge is ______.

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Re_{x=0.5} = U_{x} = 10 × 0.5 = 166666.67 30 × 10^{6}
SinceRe_{x=0.5} ≤ 5 × 10^{5}
Then flow will be laminar.∴ δ = 5x = 5 × 0.5 = 6.1237 × 10^{3} √Re_{x} √166666.67 δ_{t} = (pr)^{1/3} ⇒ δt = (1)^{1/3} × δ = 1 × 6.1237 × 10^{3} m = 6.1237 mm K Correct Option: A
Re_{x=0.5} = U_{x} = 10 × 0.5 = 166666.67 30 × 10^{6}
SinceRe_{x=0.5} ≤ 5 × 10^{5}
Then flow will be laminar.∴ δ = 5x = 5 × 0.5 = 6.1237 × 10^{3} √Re_{x} √166666.67 δ_{t} = (pr)^{1/3} ⇒ δt = (1)^{1/3} × δ = 1 × 6.1237 × 10^{3} m = 6.1237 mm K