Heat Transfer Miscellaneous Easy Questions and Answers | Page - 19

Heat Transfer Miscellaneous

A solid sphere of radius r₁ = 20 mm is placed concentrically inside a hollow sphere of radius r₂ = 30 mm as shown in the figure.

The view factor F₂₁ for radiation heat transfer is

1. 2
  3
1. 4
  9
1. 8
  27
1. 9
  4

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F₁₁ + F₁₂ = 1, here F₁₁ = 0
∴ F₁₂ = 1

Now from reciprocating law
A₁ F₁₂ = A₂ F₂₁
∴ 22π (20)² × 1 = 2π (30)² × F₂₁
⇒ F₂₁ = 10 ² = 4
30 9

Correct Option: B

F₁₁ + F₁₂ = 1, here F₁₁ = 0
∴ F₁₂ = 1

Now from reciprocating law
A₁ F₁₂ = A₂ F₂₁
∴ 22π (20)² × 1 = 2π (30)² × F₂₁
⇒ F₂₁ = 10 ² = 4
30 9

The heat loss from a fin is 6 W. The effectiveness and efficiency of the fin are 3 and 0.75 respectively. The heat loss (in W) from the fin, keeping the entire fin surface at base temperature is _____.

1. 10 W
1. 8 W
1. 18 W
1. 16 W

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η_fin = q_actual
q_{maximum possible}

q_max = 6 = 8 W
0.75

Correct Option: B

η_fin = q_actual
q_{maximum possible}

q_max = 6 = 8 W
0.75

Which one of the following configurations has the highest fin effectiveness?

1. Thin, closely spaced fins
1. Thin, widely spaced fins
1. Thick, widely spaced fins
1. Thick, closely spaced fins

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NA

Correct Option: A

NA

A fin has 5 mm diameter and 100 mm length. The thermal conductivity of fin material is 400 Wm^–1K^–1. One end of the fin is maintained at 130°C and its remaining surface is exposed to ambient air at 30°C. If the convective heat transfer coefficient is 40 Wm^–2K^–1, the heat loss (in W) from the fin is

1. 0.08
1. 5.0
1. 7.0
1. 7.8

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Fin diameter, d = 5 mm
Fin length, L = 100 mm
k = 400 W/mK
T₀ = 130°C
T_∞ = 30°C
h = 40 W/m²K
∴ Heat loss from fin Q = √kAhP (T₀ - T_∞) tanh ml
where P = πd = 3.14 × 5 × 10^{– 3} = 0.015
A = π d² = π × (5 × 10^{– 3})² = 1.96 × 10^{– 5} m²
4 4

∴ m = √hp / kA = √(40 × 0.0157) / (400 × 1.96 × 10^{– 5}) = 8.949
∴ ml = 8.949 × 0.100 = 0.895
∴ Q = √40 × 0.0157 × 400 × 1.96 × 10^{– 5}(T₀ - T_∞) tanh (0.895) = 5.0 W

Correct Option: B

Fin diameter, d = 5 mm
Fin length, L = 100 mm
k = 400 W/mK
T₀ = 130°C
T_∞ = 30°C
h = 40 W/m²K
∴ Heat loss from fin Q = √kAhP (T₀ - T_∞) tanh ml
where P = πd = 3.14 × 5 × 10^{– 3} = 0.015
A = π d² = π × (5 × 10^{– 3})² = 1.96 × 10^{– 5} m²
4 4

∴ m = √hp / kA = √(40 × 0.0157) / (400 × 1.96 × 10^{– 5}) = 8.949
∴ ml = 8.949 × 0.100 = 0.895
∴ Q = √40 × 0.0157 × 400 × 1.96 × 10^{– 5}(T₀ - T_∞) tanh (0.895) = 5.0 W

Two rods, one of length L and the other of length 2L are made of the same material and have the same diameter. The two ends of the longer rod are maintained at 100°C. One end of the shorter rod is maintained at 100°C while the other end is insulated. Both the rods are exposed to the same environment at 40°C. The temp at the insulated end of the shorter rod is measured to be 55°C. The temperature at the mid point of the longer rod would be.

1. 40°C
1. 50°C
1. 55°C
1. 100°C

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NA

Correct Option: C

NA