Condensation of Steam

Steam at atmospheric pressure condenses on a 0.25 m² vertical plate. The plate temperature is 96^oC. Find the heat transfer coefficient and the mass of steam condensed per hour. The length of the plate is 50 cm. At 97^oC, ρ_l = 960 kg/m³; k = 0.68 W/m.K; m_c = 2.82 x 10^-4 kg/m.s; h_fg = 2255 kJ/kg

Calculations:

The average value of condensing coefficient (h_m) can be calculated from the equation obtained by Nusselt's condensation theory.

${\displaystyle {ℎ}_{�}=0.943{\left(\frac{�{�}_{�}({�}_{�}-{�}_{�}){ℎ}_{��}{�}_{�}^3}{{�}_{�}({�}_{�}-{�}_{�})�}\right)}^{1/4}}$

Since condensation is at atmospheric pressure, steam temperature (T_v) is at 100^oC.

Film temperature T_f = (T_v + T_w)/2 = (100 + 96)/2 = 98^oC. Since there will be negligible change in the properties at 98^oC compared to 97^oC, we can utilize the properties given at 97^oC.

The data given are:

ρ_l = 960 kg/m³

k_l = 0.68 W/m.K

h_fg = 2255 kJ/kg = 2255 x 1000 J/kg

μ_l = 2.82 x 10^-4 kg/m.sec

T_w = 96^oC

L = 0.5 m

Compared to ρ_l, ρ_v can be taken as zero.

Therefore, h_m = 0.943 x [9.812 x 960² x 2255 x 1000 x 0.68³ / (2.82 x 10^-4 x 4 x 0.5) ]^1/4

= 9737.2 W/m².K

By adding a correction factor of 1.2 obtained from experiments to the Nusselt's equation, the average heat transfer coefficient is,

h_m = 1.2 x 9737.2 = 11685 W/m².K

Mass of steam condensed (M) is obtained by energy balance as:

M h_fg = h_mA(T_v - T_w)

i.e.,

M = 11685 x 0.25 x (100 - 96) / 2255000 = 5.1817 x 10^-3 kg/sec

Steam condensed per hour = 5.1817 x 10^-3 x 3600 = 18.654 kg