Heat Transfer Miscellaneous
Direction: Water (specifie heat, c = 4.18 kJ/kgK) enters a pipe at a rate 0.01 kg/s and a temperature of 20°C. The pipe, of diameter 50 mm and length 3 m, is subjected to a wall heat flux q"w in W/m^{2}.
 If q"w = 2500 x, where x is in m and in the direction of flow (x = 0 at the inlet), the bulk means temperature of the water leaving the pipe in °C is

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At the inlet of pipe
x = O
q "_{wi} = 2500 × 0 = 0
At the exit of pipe,
x = lq "_{wi} = 2500 × 1 × 3 = 7500 W m^{2}
Average heat transfer,q "_{avg} q "_{wi} + q "_{we} = 0 + 7500 = 3750 w 2 2 m^{2}
Net heat transfer
Q = q "_{avg} × surface area = 3750 × πdl
= 3750 × 3.14 × 0.05 × 3 = 1766.25 W
Also Q = mC_{p} (T_{0} – T_{i})
1766.25 = 0.01× 4180 (T_{0} –20)
By solving above equation, we get
T_{0} = 62° CCorrect Option: B
At the inlet of pipe
x = O
q "_{wi} = 2500 × 0 = 0
At the exit of pipe,
x = lq "_{wi} = 2500 × 1 × 3 = 7500 W m^{2}
Average heat transfer,q "_{avg} q "_{wi} + q "_{we} = 0 + 7500 = 3750 w 2 2 m^{2}
Net heat transfer
Q = q "_{avg} × surface area = 3750 × πdl
= 3750 × 3.14 × 0.05 × 3 = 1766.25 W
Also Q = mC_{p} (T_{0} – T_{i})
1766.25 = 0.01× 4180 (T_{0} –20)
By solving above equation, we get
T_{0} = 62° C
 A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in a water environment at 300 K. The convective heat transfer coefficient is 1000 W/m^{2}K. Thermal conductivity of steel is 40 W/m K. The time constant for the cooling process is 16 s. The time required (in s) to reach the final temperature is

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Biot Number = hL_{c} = 0.458 K For sphere L_{c} = Volume = d surface area 6 ∴ Bi = hd = 1000 × 0.01 6K 6 × 40
= 0.0416 < 0.1
Hence lumped heat analysis is used
T  T_{∞} = exp hA_{s}t = e t T_{i}  T_{∞} _{ρ V Cp} ^{ṫ}
Thermal time constant,t^{*} = ρ V C_{p} = 16 sec hA_{s} ∴ 350  300 = e _{t} 1000  300 _{16}
⇒ t = 42.2249 secCorrect Option: A
Biot Number = hL_{c} = 0.458 K For sphere L_{c} = Volume = d surface area 6 ∴ Bi = hd = 1000 × 0.01 6K 6 × 40
= 0.0416 < 0.1
Hence lumped heat analysis is used
T  T_{∞} = exp hA_{s}t = e t T_{i}  T_{∞} _{ρ V Cp} ^{ṫ}
Thermal time constant,t^{*} = ρ V C_{p} = 16 sec hA_{s} ∴ 350  300 = e _{t} 1000  300 _{16}
⇒ t = 42.2249 sec
 Consider the radiation heat exchange inside an annulus between two very long concentric cylinders. The radius of the outer cylinder is R_{0} and that of the inner cylinder is R_{i}. The radiation view factor of the outer cylinder onto itself is

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F_{1– 1} = 0
F_{1– 2} = 1
Reciprocacy theorem
A_{2} F_{2– 1} = A_{1} F_{1– 2}⇒ F_{2– 1} = A_{1} F_{1– 2} A_{2} = 2πR_{i}L = R_{i} 2πR_{o}L R_{o}
F_{2– 1} + F_{2– 2} = 1∴ F_{2– 2} = 1  F_{2– 1} = 1  R_{i} R_{o}
Correct Option: D
F_{1– 1} = 0
F_{1– 2} = 1
Reciprocacy theorem
A_{2} F_{2– 1} = A_{1} F_{1– 2}⇒ F_{2– 1} = A_{1} F_{1– 2} A_{2} = 2πR_{i}L = R_{i} 2πR_{o}L R_{o}
F_{2– 1} + F_{2– 2} = 1∴ F_{2– 2} = 1  F_{2– 1} = 1  R_{i} R_{o}
 Sphere 1 with a diameter of 0.1 m is completely enclosed by another sphere 2 of diameter 0.4 m. The view factor F_{12} is

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F_{11} = 0
F_{12} = 1Correct Option: D
F_{11} = 0
F_{12} = 1
 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