A fluid (Prandtl number $P r = 1$ ) at $500 K$ flows over a flat plate of $1.5 m$ length, maintained at $300 K$ . The velocity of the fluid is $10 m / s .$ Assuming kinematic viscosity, $v = 30 \times 10^{- 6} m^{2} / s,$ the thermal boundary layer thickness $(i n m m)$ at $0.5 m$ from the leading edge is

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Solution

Given,
$P r = 1$
$T_{\infty} = 500 K$
$L e n g t h = 1.5 m$
$T_{p l a t e} = 300 K$
$U_{(} \infty) = 10 m / s$
$v = 30 \times 10^{- 6} m^{2} / s$
Thermal boundary layer at $0.5 m$ from is leading edge $= δ_{t}$
$R e_{x = 0.5} = \frac{U x}{v} = \frac{10 \times 0.5}{30 \times 10^{- 6}} = 166, 666.67$

Since $R e_{x = 0.5} < 5 \times 10^{5} .$ Hence flow is laminar
$∴$ $δ = \frac{5 x}{\sqrt{R e_{x}}}$
$= \frac{5 \times 0.5}{\sqrt{166, 666.67}}$
$= 6.1237 \times 10^{- 3} m$
$\frac{δ_{t}}{δ} = \frac{1}{1.026} (P r)^{- 1 / 3}$
$δ_{t} = 5.97 m m$

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