Heat Transfer Miscellaneous
Direction: An uninsulated air conditioning duct of rectangular cross section 1 m × 0.5 m, carrying air at 20°C with a velocity of 10 m/s, is exposed to an ambient of 30°C. Neglect the effect of duct construction material. For air in the range of 20 – 30°C, data is as follows: thermal conductivity = 0.025 W/mK; viscosity = 18 μPa.s; Prandtl number = 0.73; density =1.2 kg/m^{3}. For laminar flow Nusselt number is 3.4 for constant wall temperature conditions and, for turbulent flow, Nu = 0.023 Re^{0.8}Pr^{0.33}
 The heat transfer per meter length of the duct, in watts, is

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As Re > 4000, the flow is turbulent
Nu = 0.023 Re^{0.8}Pr^{0.33}
hL/k = 0.023 × (4.44 × 10^{5})^{0.8} × (0.73)^{0.33}
h × 0.0667/0.25 = 683.173
h = 25.60 W/m^{2}K
Surface area of duct
A = 2 × al + 2 × bl
where, l = length of duct
∴ A = 3l
Heat transfer rate, Q = h A(T_{0} – T)
Q = 25.60 × 3l × (30 – 20)
Q/l = 769WCorrect Option: D
As Re > 4000, the flow is turbulent
Nu = 0.023 Re^{0.8}Pr^{0.33}
hL/k = 0.023 × (4.44 × 10^{5})^{0.8} × (0.73)^{0.33}
h × 0.0667/0.25 = 683.173
h = 25.60 W/m^{2}K
Surface area of duct
A = 2 × al + 2 × bl
where, l = length of duct
∴ A = 3l
Heat transfer rate, Q = h A(T_{0} – T)
Q = 25.60 × 3l × (30 – 20)
Q/l = 769W
 The Reynolds number for the flow is

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L_{c} = 4A = 4 × 1 × 0.5 = 0.666 D 2 × 1.5
Reynolds number,R_{e} = ρL_{c}V = 1.2 × 0.66 × 10 = 4.4 × 10^{5} . μ 18 × 10^{6} Correct Option: C
L_{c} = 4A = 4 × 1 × 0.5 = 0.666 D 2 × 1.5
Reynolds number,R_{e} = ρL_{c}V = 1.2 × 0.66 × 10 = 4.4 × 10^{5} . μ 18 × 10^{6}
 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

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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})
 For the threedimensional object shown in the figure below, five faces are insulated. The sixth face (PQRS), which is not insulated, interacts thermally with the ambient, with a convective heat transfer coefficient of 10W/m^{2}K. The ambient temperature is 30°C. Heat is uniformly generated inside the object at the rate of 100 W/m^{3}. Assuming the face PQRS to be at uniform temperature, its steady state temperature is

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Given data;
h_{o} = W/m^{2}K
h_{o} = 30 °C
q_{G} = 100 W/m^{3}
Volume, V = 2×1×2=4 m^{3}
Heat generated, Q = q_{G} × V = 100 × 4 = 400 W
Convection heat transfer from face PQRS,
Q = h_{o} A (T_{s} – T_{o})
400 = 10 × 2 × 2 (T_{s} – 30)
T_{s} = 40°CCorrect Option: D
Given data;
h_{o} = W/m^{2}K
h_{o} = 30 °C
q_{G} = 100 W/m^{3}
Volume, V = 2×1×2=4 m^{3}
Heat generated, Q = q_{G} × V = 100 × 4 = 400 W
Convection heat transfer from face PQRS,
Q = h_{o} A (T_{s} – T_{o})
400 = 10 × 2 × 2 (T_{s} – 30)
T_{s} = 40°C
 Heat transfer coefficients for free convection in gases, forced convection in gases and vapours, and for boiling water lie, respectively, in the range of

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515; 20200 and 300050000 W/m^{2}K
Correct Option: A
515; 20200 and 300050000 W/m^{2}K