Laminar forced convection over a heated flat plate

Film temperature $T_f = (100+20)/2 = 60$ °C, the temperature at which the given properties apply.

(a) Reynolds number

$$Re_L = \frac{u_\infty L}{\nu} = \frac{3\times1}{18.97\times10^{-6}} = 1.58\times10^5$$

This is less than $5\times10^5$, so the flow over the whole plate is laminar and the laminar correlation is valid.

(b) Average heat transfer coefficient

$$\begin{aligned} Nu_L &= 0.664\,Re_L^{0.5}\,Pr^{1/3}\\ &= 0.664\,(1.58\times10^5)^{0.5}\,(0.7202)^{1/3}\\ &= 0.664\,(397.7)(0.8962) = 236.7 \end{aligned}$$ $$\bar h = \frac{Nu_L\,k}{L} = \frac{236.7\times0.02896}{1} = 6.85\ \text{W/m}^2\text{K}$$

(c) Heat loss

$$Q = \bar h A (T_s-T_\infty) = 6.85\times(1\times0.6)\times(100-20) = 329\ \text{W}$$

(Heat is lost from one side. For both faces the value doubles.)

(d) Boundary layer thickness at $x = 1$ m

$$\delta = \frac{5x}{\sqrt{Re_x}} = \frac{5(1)}{397.7} = 0.01257\ \text{m} = 12.6\ \text{mm}$$ $$\delta_t = \frac{\delta}{Pr^{1/3}} = \frac{12.57}{0.8962} = 14.0\ \text{mm}$$

Since $Pr<1$ the thermal layer is slightly thicker than the velocity layer.

Answer: $Re_L = 1.58\times10^5$ (laminar); $\bar h = 6.85\ \text{W/m}^2$K; $Q = 329$ W; $\delta = 12.6$ mm, $\delta_t = 14.0$ mm.