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 …

Question

asked Jun 22, 2020 173k views

1 Answer

Asaf Bartov · Answer 1 · 2020-06-29T04:03:14+0000

Answer:

Steam condensation rate = 0.642 kg/s

Step-by-step explanation:

Given data:

P_(sat) = 1 atm

L = H = 2m

b = 10 m

T_s = 85 degree celcius

film temperature
= ( 100+85)/(2) = 92.5 degree/ celcius

$\rho = 965 kg/m^3$

$\\u = 3.19* 10^(-7) m^2/s$

$\mu = \rho \\u = 965* 10^(-4) kg/m - s$

Pr = 1.92

Cp= 4208.12 J/kg K

K = 0.676

Hfg = 2257* 10^3 J/kg

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

h = 0.946[(0.676^3* 965^2* 2257* 10^3)/(3.07* 10^(-4) * 2* (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* 10) (100-85)

= 1450.74 kw

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

= (1450.74* 10^3)/(2257* 10^3)

= 0.642 kg/s