The purification of hydrogen gas is possible by diffusion through a thin palladium sheet Calculate the number of kilograms of hydrogen that pass per hour (in kg/h) through a 4.6 mm thick sheet of palladium having an area of 0.25 m at 500C. Assume a diffusion coefficient of 6.0 x 10^-8 m^2/s, that the concentrations at the high and low-pressure sides of the plate are 5.2 and 0.32 kg/m^3 (kilogram of hydrogen per cubic meter of palladium), and that steady-state conditions have been attained.

The purification of hydrogen gas is possible by diffusion through a thin palladium sheet Calculate the number of kilograms of hydrogen that pass per hour (in kg/h) through a 4.6 mm thick sheet of palladium having an area of 0.25 m² at 500°C. Assume a diffusion coefficient of 6.0 x 10^-8 m^2/s, that the concentrations at the high and low-pressure sides of the plate are 5.2 and 0.32 kg/m^3 (kilogram of hydrogen per cubic meter of palladium), and that steady-state conditions have been attained.

Answers

Answer 1

The number of kilograms of hydrogen that pass per hour (in kg/h) through is equal to 0.0573 kg/h.

What is Diffusion?

The diffusion of material is the movement of action atoms for the region from high to low concentration. In other words, it is the transportation of material from atomic movement dependent on time.

There are different types of mechanism diffusion: interstitial diffusion and substitutional diffusion.

The diffusion can be explained by the Fick Law (Fick's 1st Law and Fick's 2nd Law). For solving you should apply Fick's 1st Law, which says:

M=JAt=[tex]-DAt\frac{ \Delta C }{\Delta X}[/tex]., where:

D= diffusion coefficientA = area t= time ΔC/ΔX = the concentration of the gradient x= thickness sheet

The exercise asks the number of kilograms of hydrogen that pass per hour (in kg/h), and it gives:

D= diffusion coefficient= 6.0 x 10^-8 m²/sA = area = 0.25 m² ΔC= 0.32 - 5.2 kg/m^3 x= [tex]4.6*10^{-3} m[/tex]

Thus,

[tex]-(6.0* 10^{-8} m^2/s)*0.25 m^2*(\frac{3600*s}{h} )*(\frac{0.32-5.2 kg/m^3}{4.6*10^-3 m} )= 0,0573 \frac{kg}{h}[/tex]

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The diffusion time that is needed for the junction depth of 1.21 m is given as 13875 seconds

How to solve for the diffusion time

background concentration of As in Si is 2.5 x 10²⁰ atoms/m

tp = 1000°c

tp = 45 minutes

= 45 x 60 seconds

surface concentration = 8x 10^26

Td = 1100°c

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qd = 4.10ev

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We have to use the equation

[tex]Dp = Doexp(-\frac{qd}{kTP} )\\\\= (2.23 * 10^-^1^3 m^2/s) exp [-\frac{4.10}{8.62*10^-^5 *(1000+273)} ][/tex]

= 1.36 x 10⁻¹³

From the relation that we have here

[tex]Qo = 2cs\sqrt{\frac{Dpte}{pi} } \\\\= 2 * 8* 10^2^6*\sqrt{\frac{1.36*10^-^1^3(45*60)}{pi} }[/tex]

Qo = 1.73 x 10¹³

Dd at 1100

[tex]Dd = 2.29*10^-^3exp(\frac{-4.10}{8.62x10^-^5)(1100 + 273)}[/tex]

Dd = 2.06 x 10⁻¹⁸m²/s

From the relation we have

[tex]exp[\frac{(1.2 * 10^-^6)^2}{4*(2.06x10^-^1^8)+d} ]=\frac{1.73*10^1^9}{2.5x10^2^0*\sqrt{pi*2.06*10^-^1^8x+d} }[/tex]

[tex]ln\frac{2.72*10^7}{\sqrt{+d} } =\frac{1.75*10^5}{+d}[/tex]

Using heat and trial, we would have the value of +d = 13875 seconds

The diffusion time is 13875 seconds

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Explanation:

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Answer:

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Explanation:

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