Heat Transfer Miscellaneous
 A long glass cylinder of inner diameter = 0.03 m and outer diameter = 0.05 m carries hot fluid inside. If the thermal conductivity of glass = 1.05 W/mK, the thermal resistance (K/W) per unit length of the cylinder is

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Given data:
d_{1} = 0.03 m∴ r_{1} d_{1} = 0.3 = 0.015 m 2 2
d_{2} = .05 m
∴ r_{2} = .025 mK = 1.05 W mK
Thermal resistanceR_{t} = 1 log_{e} r_{2} 2πKL r_{1} = 1 log_{e} .025 2 × 3.14 × 1.05 × 1 .015
= 0.077 W/mCorrect Option: B
Given data:
d_{1} = 0.03 m∴ r_{1} d_{1} = 0.3 = 0.015 m 2 2
d_{2} = .05 m
∴ r_{2} = .025 mK = 1.05 W mK
Thermal resistanceR_{t} = 1 log_{e} r_{2} 2πKL r_{1} = 1 log_{e} .025 2 × 3.14 × 1.05 × 1 .015
= 0.077 W/m
 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
 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}
 For flow of viscous fluid over a flat plate, if the fluid temperature is the same as the plate temperature, the thermal boundary layer is

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NA
Correct Option: D
NA
 An industrial gas (c = 1 kJ/kgK) enters a parallel flow heat exchange at 250°C with a flow rate of 2 kg/s to heat a water stream. The water stream (c = 4 kJ/kgK) enters the heat exchange at 50°C with a flow rate of 1 kg/s. The heat exchange has an effectiveness of 0.75. The gas stream exit temperature will be

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ε = C_{h}(T_{h1}  T_{h2}) C_{min}(T_{hi}  T_{ci}) 0.75 = 2 ×(250  T_{h2}) 2 ×(250  50)
T_{h2} = 100° CCorrect Option: B
ε = C_{h}(T_{h1}  T_{h2}) C_{min}(T_{hi}  T_{ci}) 0.75 = 2 ×(250  T_{h2}) 2 ×(250  50)
T_{h2} = 100° C