Question

Pure gaseous hydrogen at 10 atm is stored in a 10-cm-diameter, spherical steel pressure vessel having a wall thickness of 3 mm. The solubility of H2 in steel under these conditions is 0.085 m3 (STP)/(m3 of solid . atm) and the diffusivity of H2 in steel at 300oC is 3.8x10-10 m2 /s (independent of pressure). You may assume that the H2 partial pressure outside the vessel is zero.

a) Calculate the steady–state rate of leakage (in kmol/h) of H2 from the vessel.

b) Calculate the time in hours for the pressure inside the vessel to decrease to 9 atm, assuming the temperature stays constant at 300oC.

Answer 1

1). ANSWER :

GIVENTHAT :

Given=

p1 = 10 atm ( inside hydrogen pressure)

diameter of sphere = 10 cm =0.1 m

Thickness of the sphere =0.003 m.

r1 inside radius =0.05 m

r2=Outside radius =0.05+0.003=0.053 m.

surface area of sphere =4*π*r1*r2

= 4*π*0.05*0.053=0.0333 m².

Diffusivity of H2 = 3.8*10^-10 m²/s

T = 300°C =573K and P2 =outside pressure of H2 =0 atm

solubility of H2=S =0.085 kmol/m³ steel atm

C1 =inside H2 conc =S*inside pressure of H2

C1 = 10 (atm)*(0.085 kmol/m³ atm)

C1 = 0.85 kmol /m3

C2 = 0 kmol /m3

Rate of Diffusion NA through sphere is given by Ficks Law ...

NA =4πr1r2*D*(C1-C2)/(r2-r1) ..........kmol/sec.

NA = 0.0333*3.8*10^-10*(0.85-0)/(0.053-0.05)

Rate of leakage of H2 is = 3.5854 *10^-9 kmol/sec...

In Kg / sec = 7.17*10^-9 kg /sec..

................

2) now time required to pressure 9 atm inside sphere is to be calculated......

we know that n = PV /RT

for this case

Now total H2 moles transferred when H2 pressure changes from 10 atm to 9 atm

n =(P1-P2)*V/(RT)

Volume of sphere =( π/6) *d³= 5.236*10^-4 m³

n = (1*5.2359*10^-4)/0.08206*573

=1.1136*10^-5 kmol

Total time required= Molar flow/ moles transferred

time = 3.5854 * 10^ -9/(1.1136* 10^ -5)

= ​3.2196*10^-4 sec .

=8.9434 *10 ^ (-8) hrs.

Plz make note here solubility unit is given as m3(STP)/m³*atm

which make no sense because its have unit of

S= Conc / pressure .= kmol / m³ atm.