Question

Air flows over a heated plate at a velocity of 50 m/s. The local skin friction coefficient at a point on a plate is 0.004. The following properties for air are, $density=0.88kg/m^3,viscosity=2.286\times {10}^{-5}kg-m/s,specificheat=1.001kJ/kgK,$ thermal conductivity of air = 0.035 W/mK and use St.$P{r}^{1/3}=C_{fx}/2.$

The local heat transfer coefficient is

• A. $58.45W/m^{2}K$
• B. $116.9W/m^{2}K$
• C. $11.69W/m^{2}K$
• D. $29.225W/m^{2}K$

Answer 1

The correct option is D $29.225W/m^{2}K$


$u_{\infty }=50m/s$
$c_{fx}=0.004$
$\rho =0.88kg/m^3$
$\mu =2.286\times {10}^{-5}Pa-s$
$c_p=1001J/kg.K$
$k_{air}=0.035W/m-K$

According to correlation
$St\times P{r}^{2/3}=\frac{c_{fx}}{2}$

$\frac{N{u}_x}{Re.Pr}.P{r}^{2/3}=\frac{c_{fx}}{2}$

$N{u}_x=\frac{c_{fx}}{2}\times \frac{Re.Pr}{P{r}^{2/3}}$

$N{u}_x=\frac{c_{fx}}{2}\times Re.P{r}^{1/3}$

$\frac{h_x.L}{k}=\frac{c_{fx}}{2}\times \frac{\rho .u_{\infty }\times L}{u}\times P{r}^{1/3}$

$Pr=\frac{\mu c_p}{k}=\frac{2.286\times {10}^{-5}\times 1001}{0.035}=0.653796$

$h_x=\frac{c_{fx}}{2}\times \frac{\rho \times u_{\infty }}{\mu }\times P{r}^{1/3}\times k$

$=\frac{0.004}{2}\times \frac{0.88\times 50}{2.286\times {10}^{-5}}\times (0.6537{)}^{1/3}\times 0.035$

$h_x=116.89W/m^{2}K$