Heat Transfer Miscellaneous
- Consider a two-dimensional laminar flow over a long cylinder as shown in the figure below.
The free stream velocity is U∞ and the freestream temperature T∞ is lower than the cylinder surface temperature T∞. The local heat transfer coefficient is minimum at point
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For laminar flow, the heat transfer coefficient is minimum where the boundary layer thickness is maximum and vice versa. For turbulentregion boundary layer thickness is maximum at 3 but for laminar boundary layer thickness is maximum at 2 so minimum heat transfer coefficient.
Correct Option: B
For laminar flow, the heat transfer coefficient is minimum where the boundary layer thickness is maximum and vice versa. For turbulentregion boundary layer thickness is maximum at 3 but for laminar boundary layer thickness is maximum at 2 so minimum heat transfer coefficient.
- A spherical steel ball of 12 mm diameter is initially at 1000 K. It is slowly cooled in a surrounding of 300 K. The heat transfer coefficient between the steel ball and the surroundings is 5 W/m²K. The thermal conductivity of steel is 20 W/mK. The ternperature difference between the centre and the surface of the steel ball is
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Bi = hL = 5 ×0.002 = 0.0005 K 20
For the given condition, Biot number tends to zero, i.e. conduction resistance is far less than convection resistance. Therefore temperature between centre and surface is very small.Correct Option: D
Bi = hL = 5 ×0.002 = 0.0005 K 20
For the given condition, Biot number tends to zero, i.e. conduction resistance is far less than convection resistance. Therefore temperature between centre and surface is very small.
- Water flows through a tube of diameter 25 mm at an average velocity of 1.0 m/s. The properties of water are ρ = 1000 kg/m³, μ= 7.25 × 104 Ns/m², k = 0.625 W/mK, Pr = 4.85. Using Nu = 0.023 Re0.8 Pr0.4, the convective heat transfer coefficient (inW/m²K) is _______
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Re = ρVD μ = 1000 × 1 × 25 × 10-3 7.25 × 10-4
= 34482.75
Pr = 4.85
Nu = 0.023 Re0.8 Pr0.4.
Nu = .0.23 × (34482.758)0.8 (4.85)0.4.
= 184.546& Nu = hd k h = Nu.k d = 184.546 × 0.625 25 × 10-3 = 4613.66 W m²K Correct Option: B
Re = ρVD μ = 1000 × 1 × 25 × 10-3 7.25 × 10-4
= 34482.75
Pr = 4.85
Nu = 0.023 Re0.8 Pr0.4.
Nu = .0.23 × (34482.758)0.8 (4.85)0.4.
= 184.546& Nu = hd k h = Nu.k d = 184.546 × 0.625 25 × 10-3 = 4613.66 W m²K
- A 0.2 m thick infinite black plate having a thermal conductivity of 3.96 W/m-K is exposed to two infinite black surfaces at 300 K and 400 K as shown in the figure. At steady state the surface temperature of the plate facing the cold side is 350 K. The value of Stefan-Boltzmann constant a is 5.67 × 10–8 W/m² K4. Assuming 1- D heat conduction, the magnitude of heat flux through the plate (in W/m²) is ______(correct to two decimal places).
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Heat flux through the wall} = heat flux from wall to surface at 300 k
= × σ(3504 - 3004) 1 + 1 - 1 1 1 = 391.612 w (∵ 6 = 5.67 × 10-8 w m² m²k4 Correct Option: C
Heat flux through the wall} = heat flux from wall to surface at 300 k
= × σ(3504 - 3004) 1 + 1 - 1 1 1 = 391.612 w (∵ 6 = 5.67 × 10-8 w m² m²k4
- The shape factors with themselves of two infinitely long black body concentric cylinders with a dia ratio of 3 are _____ for the Inner and _____ for the outer
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d2 = 3 d1 
For Body – 1
F11 + F12 = 1
F12 = 1 (∵ F11 = 0)For Body - 2 F21 + F22 = 1 - (A)
By applying reciprocity theorem,
A1 F12 = A2 F21
A1 = A2 F21or F21 = A1 = πd1l = d1 = 1 A2 πd2l d2 3 F21 = 1 3 Substituting, F21 = 1 in(A),weget 3 1 F22 = 1 3 F22 = 2 3 Correct Option: B
d2 = 3 d1
For Body – 1
F11 + F12 = 1
F12 = 1 (∵ F11 = 0)For Body - 2 F21 + F22 = 1 - (A)
By applying reciprocity theorem,
A1 F12 = A2 F21
A1 = A2 F21or F21 = A1 = πd1l = d1 = 1 A2 πd2l d2 3 F21 = 1 3 Substituting, F21 = 1 in(A),weget 3 1 F22 = 1 3 F22 = 2 3
