Material balance- purge cycle help!

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Main Question or Discussion Point

Hydrogen is used to reduce 2.3 tonne per hour, Fe2O3 to metallic iron according to the reaction

[tex] Fe_{2}O_{3} + 3H_{2} = 2Fe + 3H_{2}O [/tex]

The water is condensed and the unreacted hydrogen is recycled (see diagram attached). Because the hydrogen in the fresh feed contains 1.1% CO2 as an impurity, some of the unreacted hydrogen must be purged. Calculate the flow rate and the composition of the purge stream required to limit the CO2 in the reactor feed to 2.8% if the ratio of recycle to fresh feed is 11:2 on a molar basis.
MW(Fe) = 55.85, MW(H) = 1.01, MW(O) = 16.00.

I am really struggling to get to grips with this question, ( struggling with material balance overall!)

I can do dimple material balance however for things with a recycle stream or purge stream, really confuses me.

I have the solution which I will post up, but I still dont get 'why' it is solved in this manner. I would really appreciated it, if someone explains to me why. Or if you have simpler way of working the flow rate of the purge stream?

thank you,
 

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SOLUTION;

CO2- Balance;


[tex] 0.011x + y = 6.5x (0.028) [/tex]

[tex] y = 0.0171x [/tex] in the recycle stream.

[tex] \frac{0.0171x}{5.5x} \times 100 = 3.1 \% [/tex] of CO2 in purge stream,
100 - 3.1 = 96.9 % H2 in purge stream.


Flow rate of purge:

CO2 BALANCE-

[tex] 0.011x = 0.031p , x = \frac{0.031p}{0.011} [/tex]

H2 Balance-
[tex] 0.989x = 43.204 + 0.969p [/tex]

[tex] 2.787p = 43.204 + 0.969p [/tex]

[tex] p = 23.76 kmol h^{-1} [/tex]




Firstly, I dont get why the flow rate of the purge stream is equal to the flow rate of the fresh feed stream?

Surely it wont be the same, as some CO2 is being recycled? Your also loosing water and Fe iron, from the fresh fee, you have two outputs, so surely the flow rate of the fresh feed would be more than the flow rate of the purge stream, so you can't set them equal to each other?
 

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