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Condensation of Steam

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Steam at atmospheric pressure condenses on a 0.25 m^{2} vertical plate. The plate temperature is 96^{o}C. Find the heat transfer coefficient and the mass of steam condensed per hour. The length of the plate is 50 cm. At 97^{o}C, ρ_{l} = 960 kg/m^{3}; 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.

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

Since condensation is at atmospheric pressure, steam temperature (T_{v}) is at 100^{o}C.

Film temperature T_{f} = (T_{v} + T_{w})/2 = (100 + 96)/2 = 98^{o}C. Since there will be negligible change in the properties at 98^{o}C compared to 97^{o}C, we can utilize the properties given at 97^{o}C.

The data given are:

ρ_{l} = 960 kg/m^{3}

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^{o}C

L = 0.5 m

Compared to ρ_{l}, ρ_{v} can be taken as zero.

Therefore, h_{m} = 0.943 x [9.812 x 960^{2} x 2255 x 1000 x 0.68^{3} / (2.82 x 10^{-4} x 4 x 0.5) ]^{1/4}

= 9737.2 W/m^{2}.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/m^{2}.K

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

M h_{fg} = h_{m}A(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

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Last Modified on: 04-Feb-2022

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