A fluid (Prandtl number, Pr = 1) at 500 K flows over a flat plate of 1.5 m length, maintained at 300 K. The velocity of the fluid is 10 m/s. Asuming kinematic viscosity, v = 30 x 10-6 m2/s, the thermal boundary layer thickness (in mm) at 0.5 m from the leading edge is ____(Important - Enter only the numerical value in the answer)Correct answer is '6'. Can you explain this answer?

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Calculation of Thermal Boundary Layer Thickness

Given:
Prandtl number, Pr = 1
Temperature of fluid, T∞ = 500 K
Temperature of plate, Tw = 300 K
Length of plate, L = 1.5 m
Velocity of fluid, V = 10 m/s
Kinematic viscosity, ν = 30 x 10-6 m2/s
Distance from leading edge, x = 0.5 m

Reynolds number (Re) can be calculated as:

Re = (ρVx)/μ
where ρ is the density of the fluid, and μ is the dynamic viscosity.

Using the ideal gas law, ρ can be calculated as:

ρ = P/(R*T∞)
where P is the pressure of the fluid, and R is the gas constant.

Assuming the pressure and gas constant to be constant, ρ can be simplified as:

ρ = constant/T∞

Using the value of ρ, Re can be further simplified as:

Re = constant*Vx/ν

Using the calculated value of Re, the thermal boundary layer thickness (δ) can be calculated using the relation:

δ/x = 0.664/√Re*Pr1/2

Substituting the given values, we get:

Re = 2.5 x 105
δ/x = 0.664/√(2.5 x 105) = 0.006 m

Converting δ to mm, we get:

δ = 6 mm

Therefore, the thermal boundary layer thickness at 0.5 m from the leading edge is 6 mm.

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