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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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    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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