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/m^{2}.
 If q"w = 5000 and the convection heat transfer coefficient at the pipe outlet is 1000 W/m^{2}K, the temperature in °C at the inner surface of the pipe at the outlet is

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q"_{w} = 5000 w/m^{2}
h = 1000 w/m^{2}k
Heat transfer through pipe wall, Q = q"_{w} × surface area
= 5000 × π × d × l
= 5000 × 3.14 × 0.05 × 3 = 2355 W
Q = mC_{P} (T_{0} –T_{i}), amount of heat gained by the water
2355 = 0.01 × 4180 (T_{0} – 20)
T_{0} = 76.33°C
q"_{w} = h( T_{so}  T_{o})
5000 = 1000 (T_{so} – 76.33)
T_{so} = 81°CCorrect Option: D
q"_{w} = 5000 w/m^{2}
h = 1000 w/m^{2}k
Heat transfer through pipe wall, Q = q"_{w} × surface area
= 5000 × π × d × l
= 5000 × 3.14 × 0.05 × 3 = 2355 W
Q = mC_{P} (T_{0} –T_{i}), amount of heat gained by the water
2355 = 0.01 × 4180 (T_{0} – 20)
T_{0} = 76.33°C
q"_{w} = h( T_{so}  T_{o})
5000 = 1000 (T_{so} – 76.33)
T_{so} = 81°C
 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 = lq "_{wi} = 2500 × 1 × 3 = 7500 W m^{2}
Average heat transfer,q "_{avg} q "_{wi} + q "_{we} = 0 + 7500 = 3750 w 2 2 m^{2}
Net heat transfer
Q = q "_{avg} × surface area = 3750 × πdl
= 3750 × 3.14 × 0.05 × 3 = 1766.25 W
Also Q = mC_{p} (T_{0} – T_{i})
1766.25 = 0.01× 4180 (T_{0} –20)
By solving above equation, we get
T_{0} = 62° CCorrect Option: B
At the inlet of pipe
x = O
q "_{wi} = 2500 × 0 = 0
At the exit of pipe,
x = lq "_{wi} = 2500 × 1 × 3 = 7500 W m^{2}
Average heat transfer,q "_{avg} q "_{wi} + q "_{we} = 0 + 7500 = 3750 w 2 2 m^{2}
Net heat transfer
Q = q "_{avg} × surface area = 3750 × πdl
= 3750 × 3.14 × 0.05 × 3 = 1766.25 W
Also Q = mC_{p} (T_{0} – T_{i})
1766.25 = 0.01× 4180 (T_{0} –20)
By solving above equation, we get
T_{0} = 62° C
 A slender rod of length L, diameter d (L >> d) and thermal conductivity k_{1} 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,
(R_{th}) _{eq} = R_{1} + R_{2}2L = L + L Ak_{eq} AK_{1} AK_{2} 2 = 1 + 1 K_{eq} K_{1} K_{2} ⇒ R_{eq} = 2K_{1}K_{2} K_{1} + K_{2} Correct Option: A
As per question, heat transfer in radial direction and contact resistance are negligible. In series connection,
(R_{th}) _{eq} = R_{1} + R_{2}2L = L + L Ak_{eq} AK_{1} AK_{2} 2 = 1 + 1 K_{eq} K_{1} K_{2} ⇒ R_{eq} = 2K_{1}K_{2} K_{1} + K_{2}
 A hollow cylinder has length L, inner radius r_{1} outer radius r_{2}, and thermal conductivity k. The thermal resistance of the cylinder for radial conduction is

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In(r_{2}/r_{1}) 2πk_{L} Correct Option: A
In(r_{2}/r_{1}) 2πk_{L}
 Consider a long cylindrical tube of inner and outer radii, r_{i} and r_{0}, respectively, length L and thermal conductivity, k. Its inner and outer surfaces are maintained at T_{i}, and T_{0}, respectively (T_{i} > T_{0}). 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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A_{r} = 2πrL
From Fourier ’s Lawq_{r} = kA_{r} dT dr q_{r} = 2π krL dT dr
Boundary conditions:
T = T_{i} at r = r_{i}
T = T_{0} at r = r_{0}q = 2πkL(T_{i}  T_{0}) In(r_{0}/r_{i}) R_{th} = In(r_{0}/r_{i}) 2πkL Correct Option: C
A_{r} = 2πrL
From Fourier ’s Lawq_{r} = kA_{r} dT dr q_{r} = 2π krL dT dr
Boundary conditions:
T = T_{i} at r = r_{i}
T = T_{0} at r = r_{0}q = 2πkL(T_{i}  T_{0}) In(r_{0}/r_{i}) R_{th} = In(r_{0}/r_{i}) 2πkL