Heat Transfer Miscellaneous
 A metal ball of diameter 60 mm is initially at 220°C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m^{2}K. The specific heat, thermal conductivity and density of the metal ball are 400 J/kgK, 400 W/mK and 9000 kg/m^{3}, respectively. The ball temperature (in °C) after 90 seconds will be approximately

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C_{P} = 400 J /kgK
K = 400 W/mK
ρ = 9000 kg/m^{3}
Time (τ) = 90 sechA τ ρVC_{P} T_{i}  T_{∞} = e T  T_{∞} Put A = 3 = 100 v R hA τ = 200 × 3 × 1 × 1 × 90 = 0.5 ρVC_{P} 0.03 9000 400 220  20 = e^{0.5} T  20
By solving, we get T = 141.3°CCorrect Option: A
C_{P} = 400 J /kgK
K = 400 W/mK
ρ = 9000 kg/m^{3}
Time (τ) = 90 sechA τ ρVC_{P} T_{i}  T_{∞} = e T  T_{∞} Put A = 3 = 100 v R hA τ = 200 × 3 × 1 × 1 × 90 = 0.5 ρVC_{P} 0.03 9000 400 220  20 = e^{0.5} T  20
By solving, we get T = 141.3°C
 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}
 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
 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
 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