Heat Transfer Miscellaneous
 The ratios of the laminar hydrodynamic boundary layer thickness to thermal boundary layer thickness of flows of two fluids P and Q on a flat plate are 1/2 and 2 respectively. The Reynolds number based on the plate length for both the flows is 10^{4}. The Prandtl and Nusselt numbers for P are 1/8 and 35 respectively. The Prandlt and Nusselt numbers for Q are respectively

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δ_{t} = 1 × P_{r}^{1/3} δ 1.026 For fluid Q: 1/2 = δ_{t} = 1 × P_{r}^{1/3} δ 1.026
⇒ P_{r} = 8
For fluid P: Laminar flow over flat plate Nu = 0.664 Re_{L}^{1/2} P_{r}^{1/3} = 35
Similarly for fluid Q: Nu = 0.664 Re_{L}^{1/2} P_{r}^{1/3}
= 0.664 (10^{4})^{1/2} 8^{1/3} ≃ 140Correct Option: A
δ_{t} = 1 × P_{r}^{1/3} δ 1.026 For fluid Q: 1/2 = δ_{t} = 1 × P_{r}^{1/3} δ 1.026
⇒ P_{r} = 8
For fluid P: Laminar flow over flat plate Nu = 0.664 Re_{L}^{1/2} P_{r}^{1/3} = 35
Similarly for fluid Q: Nu = 0.664 Re_{L}^{1/2} P_{r}^{1/3}
= 0.664 (10^{4})^{1/2} 8^{1/3} ≃ 140
 The total emissive power of a surface is 500 W/ m^{2} at temperature T_{1} and 1200 W/m^{2} at temperature T_{2}, where the temperature are in Kelvin. Assuming the emissivity of the surface to be constant, the ratio of the temperature T_{1} /T_{2} is

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Emissive power ∝ T^{4}
⇒ 500 = T_{1} ^{4} 1200 T_{2} ∴ T_{1} = 0.8034 T_{2} Correct Option: C
Emissive power ∝ T^{4}
⇒ 500 = T_{1} ^{4} 1200 T_{2} ∴ T_{1} = 0.8034 T_{2}
 An infinitely long function 0.5 m × 0.4 m cross section is shown in the figure below. Consider all surfaces of the furnace to be black. The top and bottom walls are maintained at temperature T_{1} = T_{3} = 927°C while the side walls are attemperature T_{2} = T_{4} = 527°C. The view factor, F_{1  2} is 0.26.The net radiation heat loss or gain on side 1 is ______ W/m.
StefanBoltzmann constant = 5.67 × 10^{8} W/ m^{2}K^{4}.

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T_{1} = 927°C = 1200 K,
T_{2} = 527°C = 800 K,
F_{12} = F_{14} = 0.26
F_{11} + F_{12} + F_{13} + F_{14} = 1
F_{13} = 0.48
Q = Q_{12} + Q_{13} + Q_{14}
Q_{13} = 0 since the temperatures are same
Q = Q_{12} + Q_{14} = 2 × σ_{b} × A × F_{12} ×(T_{1}^{4}  T_{2}^{4})
Q = 2 × 5.67 × 10^{8} × (0.5 × 1) × 0.26 × (1200^{4} – 800^{4})
= 24530.688Correct Option: A
T_{1} = 927°C = 1200 K,
T_{2} = 527°C = 800 K,
F_{12} = F_{14} = 0.26
F_{11} + F_{12} + F_{13} + F_{14} = 1
F_{13} = 0.48
Q = Q_{12} + Q_{13} + Q_{14}
Q_{13} = 0 since the temperatures are same
Q = Q_{12} + Q_{14} = 2 × σ_{b} × A × F_{12} ×(T_{1}^{4}  T_{2}^{4})
Q = 2 × 5.67 × 10^{8} × (0.5 × 1) × 0.26 × (1200^{4} – 800^{4})
= 24530.688
 Two large parallel plates having a gap of 10 mm in between them are maintained at temperatures T_{1}, = 1000 K and T_{2} = 400 K. Given emissivity values, ε_{1} = 0.5, ε_{2} = 0.25 and StefanBoltzmann constant σ = 5.67 × 10^{8} W/ m^{2}K^{4}, the heat transfer between the plates (in kW/m^{2}) is ______

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Q + 2 = σ(T_{1}^{4}  T_{2}^{4}) 1 + 1  1 ε_{1} ε_{2} = 5.76 ×10^{8}(1000^{4}  400^{4}) 1 + 1  1 0.5 0.25
= 11049.696 W/m^{2}
= 11.049 kW/m^{2}Correct Option: A
Q + 2 = σ(T_{1}^{4}  T_{2}^{4}) 1 + 1  1 ε_{1} ε_{2} = 5.76 ×10^{8}(1000^{4}  400^{4}) 1 + 1  1 0.5 0.25
= 11049.696 W/m^{2}
= 11.049 kW/m^{2}
 Two black surfaces, AB and BC, of length 5m and 6 m, respectively, are oriented as shown. Both surfaces extend infinitely into the third dimension. Given that view factor F_{12} = 0.5. T_{1} = 800 K, T_{2} = 600 K, Tsurrounding = 300 K and Stefan Boltzmann constant, a = 5.67 × 10^{8} W/ (m^{2}K^{4}), the heat transfer rate from Surface 2 to the surrounding environment is _____kW.

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F_{12} = 0.5
T_{1} = 800 K, T_{2} = 600 K
Tsurrounding = 300 K
Let (3) denote the surrounding environment (Assume unit width perpendicular to plane of Fig.)
From summation rule:
F_{21} + F_{22} + F_{23} = 1
F_{21} + F_{23} = 1
∴ F_{22} = 0
or F_{23} = 1 – F_{21}
A_{1} F_{12} = A_{2} F_{21} = 6 × 1 × 0.5
= (5 × 1) × F_{21}F_{21} = 6 × 0.5 = 0.6 5
∴ F_{23} = 1 – F_{21} = 0.4
Since surrounding environment being large. No surface has a surface resistance
Heat transfer rate from (2) and (3)= E_{b2}  E_{b3} 1 A_{2}F_{23} = σ(T_{2}^{4}  T_{3}^{4}) 1 A_{2}F_{23} = 10^{8}(600^{4}  300^{4}) 1 5 × 1 × 0.4
= 13.778 kW/metre
Assume that heat exchange is per unit width perpendicular to plane of figure.Correct Option: C
F_{12} = 0.5
T_{1} = 800 K, T_{2} = 600 K
Tsurrounding = 300 K
Let (3) denote the surrounding environment (Assume unit width perpendicular to plane of Fig.)
From summation rule:
F_{21} + F_{22} + F_{23} = 1
F_{21} + F_{23} = 1
∴ F_{22} = 0
or F_{23} = 1 – F_{21}
A_{1} F_{12} = A_{2} F_{21} = 6 × 1 × 0.5
= (5 × 1) × F_{21}F_{21} = 6 × 0.5 = 0.6 5
∴ F_{23} = 1 – F_{21} = 0.4
Since surrounding environment being large. No surface has a surface resistance
Heat transfer rate from (2) and (3)= E_{b2}  E_{b3} 1 A_{2}F_{23} = σ(T_{2}^{4}  T_{3}^{4}) 1 A_{2}F_{23} = 10^{8}(600^{4}  300^{4}) 1 5 × 1 × 0.4
= 13.778 kW/metre
Assume that heat exchange is per unit width perpendicular to plane of figure.