Question

Saturated steam at 1 atm condenses on a 2 m high, 10 m wide vertical plate that is maintained at 85 °C. Determine the rate of heat transfer by condensation to the plate and the rate at which the condensate drips off the plate at the bottom

Answer 1

Answer:

Steam condensation rate = 0.642 kg/s  

Explanation:

Given data:

$P_{sat} = 1 atm$

L = H = 2m

b = 10 m

$T_s = 85 degree celcius$

film temperature $= \frac{ 100+85}{2} = 92.5 degree/ celcius$

$\rho = 965 kg/m^3$

$\nu = 3.19\times 10^{-7} m^2/s$

$\mu = \rho \nu = 965\times 10^{-4} kg/m - s$

Pr = 1.92

Cp= 4208.12 J/kg K

K = 0.676

$Hfg = 2257\times 10^3 J/kg$

$h = 0.949[\frac{k^3\rho^2 g Hfg}{\mu l (T_{sat}_T_{s}}]^{1/4}$

$h = 0.946[\frac{0.676^3\times 965^2\times 2257\times 10^3}{3.07\times 10^{-4} \times 2\times (100-85)}]^{1/4}$

solving for h we get

h = 4835.32 W/m^2 K

Heat transfer rate $Q = h A(t_{sat} - T_{s})$

                                 $= 4835.8(2\times 10) (100-85)$

                                   = 1450.74 kw

Steam condensation rate $m = \frac[Q}{Hfg}$

$= \frac{1450.74\times 10^3}{2257\times 10^3}$

                                               = 0.642 kg/s