Heat Transfer Miscellaneous
 Which one of the following configurations has the highest fin effectiveness?

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NA
Correct Option: A
NA
 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 _____.

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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
 A solid sphere of radius r_{1} = 20 mm is placed concentrically inside a hollow sphere of radius r_{2} = 30 mm as shown in the figure.
The view factor F_{21} for radiation heat transfer is

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F_{11} + F_{12} = 1, here F_{11} = 0
∴ F_{12} = 1
Now from reciprocating law
A_{1} F_{12} = A_{2} F_{21}
∴ 22π (20)^{2} × 1 = 2π (30)^{2} × F_{21}⇒ F_{21} = 10 ^{2} = 4 30 9 Correct Option: B
F_{11} + F_{12} = 1, here F_{11} = 0
∴ F_{12} = 1
Now from reciprocating law
A_{1} F_{12} = A_{2} F_{21}
∴ 22π (20)^{2} × 1 = 2π (30)^{2} × F_{21}⇒ F_{21} = 10 ^{2} = 4 30 9
 Three sets of parallel plates LM, NR and PQ are given in Figures 1, 2 and 3. The view factor Fij is defined as the fraction of radiation leaving plate I that is intercepted by plate J. Assume that the val ues of FL M and FN R ar e 0.8 and 0.4, respectively. The value of FPQ (round off to one decimal place) is _____.

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F_{LM} = F(I) + F(II) + F(III)
Where F is shape factor of surface area I, II and III of M with surface area of ‘L’
0.8 = F(I) + F(II) + F(III)
∵ (I) and (III) are symmetrical in shape so, shape factor of these surface will be same i.e. F(I) = F(III)
So, 0.8 = 2F(I) + F(II)... (1)
In this area (II) is same as fig. (I) area II.
i.e. F_{NR} = F(II) = 0.4... (2)
From equation (1),F(III) = 0.8  0.4 = 0.2 2
Here, F_{PQ} = F(II) + F(III)
= 0.4 + 0.2 = 0.6
Correct Option: B
F_{LM} = F(I) + F(II) + F(III)
Where F is shape factor of surface area I, II and III of M with surface area of ‘L’
0.8 = F(I) + F(II) + F(III)
∵ (I) and (III) are symmetrical in shape so, shape factor of these surface will be same i.e. F(I) = F(III)
So, 0.8 = 2F(I) + F(II)... (1)
In this area (II) is same as fig. (I) area II.
i.e. F_{NR} = F(II) = 0.4... (2)
From equation (1),F(III) = 0.8  0.4 = 0.2 2
Here, F_{PQ} = F(II) + F(III)
= 0.4 + 0.2 = 0.6
 For flow of fluid over a heated plate, the following fluid properties are known: viscosity = 0.001 Pa.s; specific heat at constant pressure = 1 kJ/kgK; thermal conductivity = 1 W/mK. The hydrodynamic boundary layer thickness at a specified location on the plate is 1 mm. The thermal boundary layer thickness at the same location is

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Given:
µ =0.001 Pa – s
C_{P} = 1 kJ/kg K
K_{f} = 1W/mK (Fluid thermal conductivity)
Hydrodynamic boundary layer thickness, δ = 1 mmP_{r} = μC_{P} = 0.001 × 1000 = 1 K_{f} 1
δ _{t} = δ (C_{r})^{1/3}
= 1 × (1)^{1/3} = 1 mmCorrect Option: C
Given:
µ =0.001 Pa – s
C_{P} = 1 kJ/kg K
K_{f} = 1W/mK (Fluid thermal conductivity)
Hydrodynamic boundary layer thickness, δ = 1 mmP_{r} = μC_{P} = 0.001 × 1000 = 1 K_{f} 1
δ _{t} = δ (C_{r})^{1/3}
= 1 × (1)^{1/3} = 1 mm