## Heat Transfer Miscellaneous

#### Heat-Transfer

1. 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 104. The Prandtl and Nusselt numbers for P are 1/8 and 35 respectively. The Prandlt and Nusselt numbers for Q are respectively

1.  δt = 1 × Pr-1/3 δ 1.026

 For fluid Q: 1/2 = δt = 1 × Pr-1/3 δ 1.026

⇒ Pr = 8
For fluid P: Laminar flow over flat plate Nu = 0.664 ReL1/2 Pr1/3 = 35
Similarly for fluid Q: Nu = 0.664 ReL1/2 Pr1/3
= 0.664 (104)1/2 81/3 ≃ 140

##### Correct Option: A

 δt = 1 × Pr-1/3 δ 1.026

 For fluid Q: 1/2 = δt = 1 × Pr-1/3 δ 1.026

⇒ Pr = 8
For fluid P: Laminar flow over flat plate Nu = 0.664 ReL1/2 Pr1/3 = 35
Similarly for fluid Q: Nu = 0.664 ReL1/2 Pr1/3
= 0.664 (104)1/2 81/3 ≃ 140

1. The total e-missive power of a surface is 500 W/ m2 at temperature T1 and 1200 W/m2 at temperature T2, where the temperature are in Kelvin. Assuming the e-missivity of the surface to be constant, the ratio of the temperature T1 /T2 is

1. E-missive power ∝ T4

 ⇒ 500 = T1 4 1200 T2

 ∴ T1 = 0.8034 T2

##### Correct Option: C

E-missive power ∝ T4

 ⇒ 500 = T1 4 1200 T2

 ∴ T1 = 0.8034 T2

1. 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 T1 = T3 = 927°C while the side walls are attemperature T2 = T4 = 527°C. The view factor, F1 - 2 is 0.26.The net radiation heat loss or gain on side 1 is ______ W/m.
Stefan-Boltzmann constant = 5.67 × 10-8 W/ m2K4.

1. T1 = 927°C = 1200 K,
T2 = 527°C = 800 K,
F12 = F14 = 0.26
F11 + F12 + F13 + F14 = 1
F13 = 0.48
Q = Q12 + Q13 + Q14
Q13 = 0 since the temperatures are same
Q = Q12 + Q14 = 2 × σb × A × F12 ×(T14 - T24)
Q = 2 × 5.67 × 10-8 × (0.5 × 1) × 0.26 × (12004 – 8004)
= 24530.688

##### Correct Option: A

T1 = 927°C = 1200 K,
T2 = 527°C = 800 K,
F12 = F14 = 0.26
F11 + F12 + F13 + F14 = 1
F13 = 0.48
Q = Q12 + Q13 + Q14
Q13 = 0 since the temperatures are same
Q = Q12 + Q14 = 2 × σb × A × F12 ×(T14 - T24)
Q = 2 × 5.67 × 10-8 × (0.5 × 1) × 0.26 × (12004 – 8004)
= 24530.688

1. Two large parallel plates having a gap of 10 mm in between them are maintained at temperatures T1, = 1000 K and T2 = 400 K. Given emissivity values, ε1 = 0.5, ε2 = 0.25 and Stefan-Boltzmann constant σ = 5.67 × 10-8 W/ m2K4, the heat transfer between the plates (in kW/m2) is ______

1.  Q + 2 = σ(T14 - T24) 1 + 1 - 1 ε1 ε2

 = 5.76 ×10-8(10004 - 4004) 1 + 1 - 1 0.5 0.25

= 11049.696 W/m2
= 11.049 kW/m2

##### Correct Option: A

 Q + 2 = σ(T14 - T24) 1 + 1 - 1 ε1 ε2

 = 5.76 ×10-8(10004 - 4004) 1 + 1 - 1 0.5 0.25

= 11049.696 W/m2
= 11.049 kW/m2

1. 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 F12 = 0.5. T1 = 800 K, T2 = 600 K, Tsurrounding = 300 K and Stefan Boltzmann constant, a = 5.67 × 10-8 W/ (m2K4), the heat transfer rate from Surface 2 to the surrounding environment is _____kW.

1. F12 = 0.5

T1 = 800 K, T2 = 600 K
Tsurrounding = 300 K
Let (3) denote the surrounding environment (Assume unit width perpendicular to plane of Fig.)
From summation rule:
F21 + F22 + F23 = 1
F21 + F23 = 1
∴ F22 = 0
or F23 = 1 – F21
A1 F12 = A2 F21 = 6 × 1 × 0.5
= (5 × 1) × F21

 F21 = 6 × 0.5 = 0.6 5

∴ F23 = 1 – F21 = 0.4
Since surrounding environment being large. No surface has a surface resistance

Heat transfer rate from (2) and (3)
 = Eb2 - Eb3 1 A2F23

 = σ(T24 - T34) 1 A2F23

 = 10-8(6004 - 3004) 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

F12 = 0.5

T1 = 800 K, T2 = 600 K
Tsurrounding = 300 K
Let (3) denote the surrounding environment (Assume unit width perpendicular to plane of Fig.)
From summation rule:
F21 + F22 + F23 = 1
F21 + F23 = 1
∴ F22 = 0
or F23 = 1 – F21
A1 F12 = A2 F21 = 6 × 1 × 0.5
= (5 × 1) × F21

 F21 = 6 × 0.5 = 0.6 5

∴ F23 = 1 – F21 = 0.4
Since surrounding environment being large. No surface has a surface resistance

Heat transfer rate from (2) and (3)
 = Eb2 - Eb3 1 A2F23

 = σ(T24 - T34) 1 A2F23

 = 10-8(6004 - 3004) 1 5 × 1 × 0.4

= 13.778 kW/metre
Assume that heat exchange is per unit width perpendicular to plane of figure.