Condensation of steam on a vertical plate

(a) Average coefficient (Nusselt)

$\Delta T = 100 - 80 = 20$ K, $L = 0.5$ m, $g = 9.81$ m/s$^2$.

$$\begin{aligned} \bar h &= 0.943\left[\frac{\rho_l(\rho_l-\rho_v)\,g\,h_{fg}\,k_l^3}{\mu_l\,L\,\Delta T}\right]^{1/4}\\ &= 0.943\left[\frac{6.34\times10^{12}}{3.148\times10^{-3}}\right]^{1/4}\\ &= 0.943\,[2.014\times10^{15}]^{1/4} = 0.943\times6699\\ &= 6317\ \text{W/m}^2\text{K} \end{aligned}$$

(b) Heat transfer rate

$$Q = \bar h A\,\Delta T = 6317\times(0.5\times1)\times20 = 63\,175\ \text{W} = 63.2\ \text{kW}$$

(c) Condensate rate

$$\dot m = \frac{Q}{h_{fg}} = \frac{63\,175}{2.257\times10^6} = 0.0280\ \text{kg/s} = 100.8\ \text{kg/h}$$

(d) Film Reynolds number

$$Re_f = \frac{4\dot m}{b\,\mu_l} = \frac{4\times0.0280}{1\times314.8\times10^{-6}} = 356$$

$Re_f$ lies between 30 and 1800, so the film is wavy laminar. Nusselt's equation is then conservative: ripples usually raise $\bar h$ by about 20 percent (Kutateladze correction), so the true value may be near 7500 W/m$^2$K.

Answer: $\bar h \approx 6320\ \text{W/m}^2$K; $Q \approx 63.2$ kW; $\dot m \approx 101$ kg/h; $Re_f \approx 356$ (wavy).