Heat Transfer Miscellaneous


Direction: Water (specifie heat, c = 4.18 kJ/kgK) enters a pipe at a rate 0.01 kg/s and a temperature of 20°C. The pipe, of diameter 50 mm and length 3 m, is subjected to a wall heat flux q"w in W/m2.

  1. If q"w = 5000 and the convection heat transfer coefficient at the pipe outlet is 1000 W/m2K, the temperature in °C at the inner surface of the pipe at the outlet is









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    q"w = 5000 w/m2
    h = 1000 w/m2k
    Heat transfer through pipe wall, Q = q"w × surface area
    = 5000 × π × d × l
    = 5000 × 3.14 × 0.05 × 3 = 2355 W
    Q = mCP (T0 –Ti), amount of heat gained by the water
    2355 = 0.01 × 4180 (T0 – 20)
    T0 = 76.33°C

    q"w = h( Tso - To)
    5000 = 1000 (Tso – 76.33)
    Tso = 81°C

    Correct Option: D

    q"w = 5000 w/m2
    h = 1000 w/m2k
    Heat transfer through pipe wall, Q = q"w × surface area
    = 5000 × π × d × l
    = 5000 × 3.14 × 0.05 × 3 = 2355 W
    Q = mCP (T0 –Ti), amount of heat gained by the water
    2355 = 0.01 × 4180 (T0 – 20)
    T0 = 76.33°C

    q"w = h( Tso - To)
    5000 = 1000 (Tso – 76.33)
    Tso = 81°C


  1. If q"w = 2500 x, where x is in m and in the direction of flow (x = 0 at the inlet), the bulk means temperature of the water leaving the pipe in °C is









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    At the inlet of pipe
    x = O
    q "wi = 2500 × 0 = 0
    At the exit of pipe,
    x = l

    q "wi = 2500 × 1 × 3 = 7500
    W
    m2

    Average heat transfer,
    q "avg
    q "wi + q "we
    =
    0 + 7500
    = 3750
    w
    22m2

    Net heat transfer
    Q = q "avg × surface area = 3750 × πdl
    = 3750 × 3.14 × 0.05 × 3 = 1766.25 W
    Also Q = mCp (T0 – Ti)
    1766.25 = 0.01× 4180 (T0 –20)
    By solving above equation, we get
    T0 = 62° C

    Correct Option: B

    At the inlet of pipe
    x = O
    q "wi = 2500 × 0 = 0
    At the exit of pipe,
    x = l

    q "wi = 2500 × 1 × 3 = 7500
    W
    m2

    Average heat transfer,
    q "avg
    q "wi + q "we
    =
    0 + 7500
    = 3750
    w
    22m2

    Net heat transfer
    Q = q "avg × surface area = 3750 × πdl
    = 3750 × 3.14 × 0.05 × 3 = 1766.25 W
    Also Q = mCp (T0 – Ti)
    1766.25 = 0.01× 4180 (T0 –20)
    By solving above equation, we get
    T0 = 62° C


  1. A slender rod of length L, diameter d (L >> d) and thermal conductivity k1 i s joined with another rod of identical dimensions, but of thermal conductivity k 2, to form a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is









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    As per question, heat transfer in radial direction and contact resistance are negligible. In series connection,
    (Rth) eq = R1 + R2

    2L
    =
    L
    +
    L
    AkeqAK1AK2

    2
    =
    1
    +
    1
    KeqK1K2

    ⇒ Req =
    2K1K2
    K1 + K2

    Correct Option: A


    As per question, heat transfer in radial direction and contact resistance are negligible. In series connection,
    (Rth) eq = R1 + R2

    2L
    =
    L
    +
    L
    AkeqAK1AK2

    2
    =
    1
    +
    1
    KeqK1K2

    ⇒ Req =
    2K1K2
    K1 + K2


  1. A hollow cylinder has length L, inner radius r1 outer radius r2, and thermal conductivity k. The thermal resistance of the cylinder for radial conduction is









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    In(r2/r1)
    2πkL

    Correct Option: A

    In(r2/r1)
    2πkL


  1. Consider a long cylindrical tube of inner and outer radii, ri and r0, respectively, length L and thermal conductivity, k. Its inner and outer surfaces are maintained at Ti, and T0, respectively (Ti > T0). Assuming one dimensional steady state heat conduction in the radial direction, the thermal resistance in the wall of the tube is









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    Ar = 2πrL
    From Fourier ’s Law

    qr = kAr
    dT
    dr

    qr = 2π krL
    dT
    dr

    Boundary conditions:
    T = Ti at r = ri
    T = T0 at r = r0
    q =
    2πkL(Ti - T0)
    In(r0/ri)


    Rth =
    In(r0/ri)
    2πkL

    Correct Option: C

    Ar = 2πrL
    From Fourier ’s Law

    qr = kAr
    dT
    dr

    qr = 2π krL
    dT
    dr

    Boundary conditions:
    T = Ti at r = ri
    T = T0 at r = r0
    q =
    2πkL(Ti - T0)
    In(r0/ri)


    Rth =
    In(r0/ri)
    2πkL