Heat Transfer Miscellaneous


Direction: An un-insulated 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/m3. For laminar flow Nusselt number is 3.4 for constant wall temperature conditions and, for turbulent flow, Nu = 0.023 Re0.8Pr0.33

  1. 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 Re0.8Pr0.33
    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 Re0.8Pr0.33
    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


  1. The Reynolds number for the flow is









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    Lc =
    4A
    = 4 ×
    1 × 0.5
    = 0.666
    D2 × 1.5

    Reynolds number,
    Re =
    ρLcV
    =
    1.2 × 0.66 × 10
    = 4.4 × 105 .
    μ18 × 10-6

    Correct Option: C

    Lc =
    4A
    = 4 ×
    1 × 0.5
    = 0.666
    D2 × 1.5

    Reynolds number,
    Re =
    ρLcV
    =
    1.2 × 0.66 × 10
    = 4.4 × 105 .
    μ18 × 10-6



  1. 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 δv and the thermal boundary layer thickness is δT. 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 (δv) > thermal boundary thickness (δt)

    Correct Option: A

    Prandtl number =
    Molecular diffusivity of momentum
    Molecular diffusivity of heat

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


  1. For the three-dimensional 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/m2K. The ambient temperature is 30°C. Heat is uniformly generated inside the object at the rate of 100 W/m3. Assuming the face PQRS to be at uniform temperature, its steady state temperature is









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    Given data;
    ho = W/m2K
    ho = 30 °C
    qG = 100 W/m3
    Volume, V = 2×1×2=4 m3

    Heat generated, Q = qG × V = 100 × 4 = 400 W
    Convection heat transfer from face PQRS,
    Q = ho A (Ts – To)
    400 = 10 × 2 × 2 (Ts – 30)
    Ts = 40°C

    Correct Option: D

    Given data;
    ho = W/m2K
    ho = 30 °C
    qG = 100 W/m3
    Volume, V = 2×1×2=4 m3

    Heat generated, Q = qG × V = 100 × 4 = 400 W
    Convection heat transfer from face PQRS,
    Q = ho A (Ts – To)
    400 = 10 × 2 × 2 (Ts – 30)
    Ts = 40°C



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

    Correct Option: A

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