### Condensation of Steam

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

Calculations:

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

$$\displaystyle h_m = 0.943\left(\frac{g\rho_l(\rho_l-\rho_v)h_{fg}k_l^3}{\mu_l(T_v-T_w)L}\right)^{1/4}$$

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

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

The data given are:

ρl = 960 kg/m3

kl = 0.68 W/m.K

hfg = 2255 kJ/kg = 2255 x 1000 J/kg

μl = 2.82 x 10-4 kg/m.sec

Tw = 96oC

L = 0.5 m

Compared to ρl, ρv can be taken as zero.

Therefore, hm = 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,

hm = 1.2 x 9737.2 = 11685 W/m2.K

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

M hfg = hmA(Tv - Tw)

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