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
 Onedimensional steady state heat conduction takes place through a sol i d whose cross sectional area varies linearly in the direction of heat transfer. Assume t her e is no heat generation in the solid and the thermal conductivity of the material is constant and independent of temperature. The temperature distribution in the solid is

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We know that onedimensional steady state heat conduction equation with no heat generation is given by,
d  kA dT = 0 dx dx ⇒ kA dT = C_{1} dx
∵ A = Cx + B (linear variation)
Logarithmic temperature distributionCorrect Option: C
We know that onedimensional steady state heat conduction equation with no heat generation is given by,
d  kA dT = 0 dx dx ⇒ kA dT = C_{1} dx
∵ A = Cx + B (linear variation)
Logarithmic temperature distribution
 Steady onedimensional heat conduction takes place across the faces 1 and 3 of a composite slab consisting of slabs A and B in perfect contact as shown in the figure, where k_{A}, k_{B} denote the respective thermal conductivities. Using the data as given in the figure, the interface temperature T_{2} (in °C) is _________

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Q = kA.(T_{1}  T_{2}) A LA = kB(T_{2}  T_{3}) LB 20(130  T_{2}) = 100(T_{2}  30) 0.1 0.3
∴ T_{2} = 67.5°CCorrect Option: A
Q = kA.(T_{1}  T_{2}) A LA = kB(T_{2}  T_{3}) LB 20(130  T_{2}) = 100(T_{2}  30) 0.1 0.3
∴ T_{2} = 67.5°C
 A plastic sleeve of outer radius r_{0} = 1 mm covers a wire (radius r = 0.5 mm) carrying electric current, Thermal conductivity of the plastic is 0.12 W/mK. The heat transfer coefficient on the outer surface of the sleeve exposed to air is 25 W/m^{2}K. Due to the addition of the plastic cover, the heat transfer from the wire to the ambient will

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r_{0} = 1 mm, k = 0.15 W/mk
h = 25 W/m^{2} Kr_{c} = k for cylindrical shape h_{0} = 0.15 × 1000 = 0.15 × 40 = 6mm 25
∴ r_{c} > r_{0}
⇒ The heat transfer from the wire to the ambient will increaseCorrect Option: A
r_{0} = 1 mm, k = 0.15 W/mk
h = 25 W/m^{2} Kr_{c} = k for cylindrical shape h_{0} = 0.15 × 1000 = 0.15 × 40 = 6mm 25
∴ r_{c} > r_{0}
⇒ The heat transfer from the wire to the ambient will increase
 A brick wall (k = 0.9 W/mK) of thickness 0.18 m separates the warm air in a room from the cold ambient air. On a particular Winter day, the outside air temperature is  5°C and the room needs to be maintained at 27°C. The heat transfer coefficient associated with outside air is 20 W/m^{2}K. Neglecting the convective resistance of the air inside the room, the heat loss, in (W/m^{2}), is

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Total thermal resistance1 = 1 R_{th} 1 + 1 h k 1 = 1 = 4 w/m^{2}K R_{th} 1 + 0.18 20 0.9 Q = ∆T [27  (5)] × 4 = 128 W/m^{2} R_{th} Correct Option: C
Total thermal resistance1 = 1 R_{th} 1 + 1 h k 1 = 1 = 4 w/m^{2}K R_{th} 1 + 0.18 20 0.9 Q = ∆T [27  (5)] × 4 = 128 W/m^{2} R_{th}