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.
- 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°CCorrect 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
- 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 m2
Average heat transfer,q "avg q "wi + q "we = 0 + 7500 = 3750 w 2 2 m2
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° 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 m2
Average heat transfer,q "avg q "wi + q "we = 0 + 7500 = 3750 w 2 2 m2
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
- 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 + R22L = L + L Akeq AK1 AK2 2 = 1 + 1 Keq K1 K2 ⇒ 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 + R22L = L + L Akeq AK1 AK2 2 = 1 + 1 Keq K1 K2 ⇒ Req = 2K1K2 K1 + K2
- 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
- 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 Lawqr = kAr dT dr qr = 2π krL dT dr
Boundary conditions:
T = Ti at r = ri
T = T0 at r = r0q = 2πkL(Ti - T0) In(r0/ri) 
Rth = In(r0/ri) 2πkL Correct Option: C
Ar = 2πrL
From Fourier ’s Lawqr = kAr dT dr qr = 2π krL dT dr
Boundary conditions:
T = Ti at r = ri
T = T0 at r = r0q = 2πkL(Ti - T0) In(r0/ri) Rth = In(r0/ri) 2πkL
