Answer:
(a) 45.17×10^-14 kg/s
(b) since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.
Explanation:
Helium gas at temperature T=293k
Helium gas at pressure P= 500kPa
The inner diameter of spherical tank is [tex]D_1 = 2m[/tex]
The inner radius of spherical tank is : [tex]r_1 = \frac{D_1}{2}[/tex]
= [tex]\frac{2}{2}[/tex]
=1m
Thickness of the container r = 1cm =0.01m
Outer radius of the spherical tank is ;
t = [tex]r_2 - r_1[/tex]
[tex]-r_2 = -t - r-1[/tex]
multiplying through with (-) we have ;
[tex]r_2 = t + r_1[/tex]
[tex]r_2 = 1 + 0.01m[/tex]
[tex]r_2 = 1.01m[/tex]
From table of binary diffusion coefficients of solids, the diffusion coefficients of helium in silica is noted as
[tex]D_A_B =4.0 ×10^-14 \frac{m^2}{s}[/tex]
From table molar mass and gas constant, the molecular weight of helium is:
M = 4.003kg/kmol
The solubility of helium in fused silica is determined from Table of Solubility of selected gases and soilids.
[tex]S_He[/tex] = 0.00045 kmol/m³. bar
Considering total molar concentration as constant, the molar concentration of helium inside the container is determined as
[tex]C_B_I = S_H_e×P[/tex]
= 0.00045kmol/m³. bar × (5)
[tex]C_B_I = 2.25×10^-3 kmol/m³[/tex]
From one dimensional mass transfer through spherical layers is expressed as:
[tex]N_di_f_f= 4πr_1 r_2 D_A_B \frac{C_B_I - C_B_2}{r_2 - r_1}[/tex]
substituting all the values in the above relation, we have;
[tex]M_di_f_f= 4π(1) (1.01) (4.0×10^-14) \frac{2.25 × 10^-3 -0}{1.01-1}[/tex]
[tex]M_di_f_f=11.42×10^-14kmol/s[/tex]
(a) The mass flow rate is expressed as
[tex]M_di_f_f = MN_diff[/tex]
[tex]M_di_f_f=4.003×11.42×10^-14[/tex]
[tex]M_di_f_f=45.71×10^-14kg/s[/tex]
(b) The pressure drop in the tank after a week;
For one week the mass flow rate of helium is
[tex]N_di_ff =11.42×10^-14kmol/s[/tex]
[tex]N_di_ff= 11.42×10^-14×7×24×3600 kmol/week[/tex]
[tex]N_di_f_f=6.9×10^-8kmol/week[/tex]
The volume of the spherical tank is [tex]V=\frac{4}{3} πr_1^3[/tex]
[tex]V=\frac{4}{3}π×1^3[/tex]
V = 4.189m³
The initial mass of helium in the sphere is determined from the ideal gas equation:
PV=NRT
where R is the universal gas constant and its value is R = 8.314 KJ/Kmol.k
N= PV/RT
N= 500 × 4.189/ 8.314 × 293
N= 0.86kmol
The number of moles of helium gas remaining in the tank after one week is:
[tex]N_di_f_f-final = 0.86 - 6.9 × 10^-8 kmol/week[/tex]
[tex]N_di_f_f-final ≅ 0.86[/tex]
therefore, since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.
