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A differential nitrogen pressure exists across a 2-mm-thick steel furnace wall. After some time, steady-state diffusion...

A differential nitrogen pressure exists across a 2-mm-thick steel furnace wall. After some time, steady-state diffusion of the nitrogen is established across the wall. Given that the nitrogen concentration on the high-pressure surface of the wall is 5 kg/m3kg/m3 and on the low-pressure surface is 0.3 kg/m3 , calculate the flow of nitrogen through the wall (in kg/m2.h) if the diffusion coefficient for nitrogen in this steel is 1.0 ×10−10m2/s at the furnace operating temperature.

Express your answer to three significant figures.

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Answer #1

According to Fick's first law of diffusion the rate of diffusion per unit area of the surface is directly proportional to the concentration difference and is inversely proportional to the distance between the two points. Following mathematical equation is obtained from the Fick's law for steady state flow:

J​= -D​​​​​AB

Where J is the flux of species A

D​​​​​​AB is the diffusivity constant having S.I. unit of m​​​​​​2​​​​​/s

Ca is the concentration of species A

And x represent the length between the two points

   is known as concentration gradient.

The unit of flux is kg/m​​​​​​2​​​​​.h

The negative sign represents that the concentration of species A decreases with increasing length with reference to concentration of A i.e. as we move farther from species A its concentration decreases.

In the given problem we are given the following data

Diffusivity constant D = 1×10-10 m​​​​​​2​​​​​/s

The distance between the high pressure surface and low pressure suface dx= 2 mm = 2× 10-3 m

Concentration on the high pressure surface is 5 kg/m​​​​​​3 and at low pressure suface is 0.3 kg/m​​​​​​3​​​. We can calculate dCa by subtracting the above values.

dCa=( 0.3 - 5 )kg/m³ = - 4.7 kg/m³

Now we are in a position to calculate the required value of flow by using the Fick's law

J= - (1×10-10 m²/s) ( - 4.7 kg/m³)/(2×10-³ m)

J= 2.35× 10-​​​​​7 kg/ m².s

Since we are asked the value of flow in kg/m².h we can convert it into required units

1 hr =60×60 seconds. So the conversion takes place as follows.

Finally J = 8.46 ×10-4 kg /m².h

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