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## Q. 14.4

Consider flow of air at atmospheric pressure and 300 K parallel to a flat plate 2m long. The velocity of air far away from the plate is 10 m/s. The plate surface is held at a constant temperature of 400 K. Determine the heat transfer coefficient at the trailing edge of the plate using the formulae based on the various analogies presented in the text. Comment on the results.

## Verified Solution

Step 1 Given data is written down first.

Plate length: L = 2m

Plate temperature: $T_{w}=400 K$

Free stream velocity: $U_{\infty}=10 m / s$

Free stream temperature: $T_{\infty}=300 K$

Step 2 Air properties are taken from tables of properties at the film temperature of $T_{f}=\frac{400+300}{2}=350 K$.

Density: $\rho_{f}=0.995 kg / m ^{3}$

Kinematic viscosity:$\nu_{f}=20.92 \times 10^{-6} m ^{2} / s$

Thermal conductivity: $k_{f}=0.030 W / m K$

Prandtl number: $\operatorname{Pr}_{f}=0.7$

Step 3 Then, the Reynolds number at the trailing edge of the plate is

$R e_{L}=\frac{U_{\infty} L}{\nu_{f}}=\frac{10 \times 2}{20.92 \times 10^{-6}}=9.56 \times 10^{5}$

The flow is turbulent since the Reynolds number is greater than the critical Reynolds number $R e_{c}=5 \times 10^{5}$ . The heat transfer coefficient at the trailing edge of the plate is now estimated using the three analogies given in the text.

Step 4 (a) Colburn analogy: The Nusselt number is obtained using Eq. 14.43 as

$N u_{x}=0.0296 R e_{x}^{0.8} \operatorname{Pr}_{\infty}^{\frac{1}{3}}$ (14.43)

$N u_{x}=0.0296 \times\left(9.56 \times 10^{5}\right)^{0.8} \times 0.7^{\frac{1}{3}}=1599.7$

The heat transfer coefficient is then given by

$h_{L}=\frac{N u_{L} k_{f}}{L}=\frac{1599.7 \times 0.030}{2}=24 W / m ^{2} K$

Step 5 (b) Prandtl analogy: The friction factor at x = L is calculated using Eq. 14.41 with the constant modified to 0.0592, as mentioned in the text, as

$C_{f, x}=0.045\left[\frac{\nu_{\infty} R e_{x}^{\frac{1}{5}}}{0.371 U_{\infty} x}\right]=0.0583 R e_{x}^{-\frac{1}{5}}$  (14.41)

$C_{f, L}=0.0592 \times\left(9.56 \times 10^{5}\right)^{-\frac{1}{5}}=0.003769$

The Stanton number is then obtained using Eq. 14.49 as

$S t_{x}=\frac{\frac{C_{f x}}{2}}{1+5 \sqrt{\frac{C_{f x}}{2}}\left(\operatorname{Pr}_{\infty}-1\right)}$  (14.49)

$S t_{L}=\frac{\frac{0.003769}{2}}{1+5 \sqrt{\frac{0.003769}{2}(0.7-1)}}=0.002016$

Noting that $S t_{L}=\frac{N u_{L}}{R e_{L} P r_{f}}$ , we have

$N u_{L}=S t_{L} R e_{L} P r_{f}=0.002016 \times 9.56 \times 10^{5} \times 0.7=1349$

The heat transfer coefficient is then given by

$h_{L}=\frac{N u_{L} k_{f}}{L}=\frac{1349 \times 0.030}{2}=20.2 W / m ^{2} K$

Step 6 (c) von Karman analogy: The friction factor calculated above may now be used in Eq. 14.50 to get the Stanton number as

$S t_{x}=\frac{\frac{C_{f x}}{2}}{1+5 \sqrt{\frac{C_{f x}}{2}}\left[\left(\operatorname{Pr}_{\infty}-1\right)+\ln \left(1+\frac{5}{6}\left(\operatorname{Pr}_{\infty}-1\right)\right)\right]}$ (14.50)

$S t_{L}=\frac{\frac{0.003769}{2}}{1+5 \sqrt{\frac{0.003769}{2}}\left[(0.7-1)+\ln \left(1+\frac{5}{6}(0.7-1)\right)\right]}=0.00216$

$N u_{L}$ is then calculated as

$N u_{L}=S t_{L} \operatorname{Re}_{L} P r_{f}=0.00216 \times 9.56 \times 10^{5} \times 0.7=1445.5$

The heat transfer coefficient is then given by

$h_{L}=\frac{N u_{L} k_{f}}{L}=\frac{1445.5 \times 0.030}{2}=21.7 W / m ^{2} K$

All three analogies give heat transfer coefficient values within a 9% band around a mean value of $22 W / m ^{2} K$.