Saturated water vapor at 25°C (Psat = 3.17 kPa) flows in a pipe that passes through air at 25°C with a relative humidity of 40 percent. The vapor is vented to the atmosphere through a 9-mm internal-diameter tube that extends 10 m into the air. The diffusion coefficient of vapor through air is 2.5 × 10?5 m2/s. The amount of water vapor lost to the atmosphere through this individual tube by diffusion is

Question

(a) 1.7?×?10?6 kg
(b) 2.3?×?10?6 kg
(c) 3.8?×?10?6 kg
(d) 5.0?×?10?6 kg
(e) 7.1?×?10?6 kg

Answer 1

(b) 2.3?×?10?6 kg
T=25 [C]
phi=0.40
D=0.009 [m]
L=10 [m]
D_AB=2.5E-5 [m^2/s] "Table 14-2"
P_A_0=pressure(steam_iapws, x=1, T=T) "pressure of vapor at x=0"
P_A_L=phi*P_A_0 "pressure of vapor at x=L=10 m"
A=pi*D^2/4
R_u=8.314[kPa-m^3/kmol-K]
N_dot_vapor=(D_AB*A)/(R_u*T)*(P_A_0-P_A_L)/L
MM=MolarMass(H2O)
m_dot_vapor=N_dot_vapor*MM
time=24*3600 [s]
m_vapor=m_dot_vapor*time
"Some Wrong Solutions with Common Mistakes"
W_P_A_L=0 "Taking the vapor pressure at air side zero"
W_N_dot_vapor=(D_AB*A)/(R_u*T)*(P_A_0-W_P_A_L)/L
W_m_dot_vapor=W_N_dot_vapor*MM
W_m_vapor=W_m_dot_vapor*time