material balance

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555gag
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material balance

Post by 555gag »

100 kg / h H2 is to be produced. The conversion rate reached is 85%. The reaction proceeds at
temperature 800 ° C and pressure 202,6 kPa. The feed gases should be preheated from 200 ° C. O2 contains
7% inert impurities and CH4 are dosed with a 5% molar excess. Draw the scheme
with substance and energy flows. Perform mass and volume balance a
energy balance process. (Neglect heat to heat dirt). Specify the conditions for
which process is used in industry (temperature, catalyst, pressure,).


my solution:

CH4 +1/2 O2 = CO + 2H2

we have a node where the current 1 - ch4 and the current 2 - o2, inert and the node outputs the current 3 - ch4, inert, o2, CO, H2
nH2 = mH2/MH2 = 100000/2,009 = 47 846 mol
nCO = nH2/2* = 23 923 mol
nCH4(surplus)= nH2/2 *100/85 * 1,05 = 29 551 mol
nO2=nH2/4 = 11961 mol

Current composition 3.
n(inert) = no2*5/85 = 11961*5/85 = 703,58 mol
nH2 = 47846 mol
nCO=23 923 mol
nCH4 = nch4(surplus) - 1/2*nH2 = 29551 - 0,5*47846 = 5628 mol
nO2 = 11 961 mol
I could ask for a check, I'm not sure that the conversion is counted correctly

I only deal with this part, I will calculate the rest then, and I do not want to count unnecessarily if I do not know the correct composition ...

Is it correct? If not, could I ask for an explanation? Thank you
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ChenBeier
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Post by ChenBeier »

I think the calculation of oxygen has also considered by 85%, like you did for methane. Also the amount has to be increased, because of 7% impurities.
555gag
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Post by 555gag »

thank you so much, so now it's right?

Current composition 1
nO2=(nH2/4)*100/85= 14 072 mol

Current composition 3.
nO2 = 14 072 mol
n(inert) = no2*5/85 = 14 072*5/85 = 828 mol
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ChenBeier
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Post by ChenBeier »

The inert I dont understand.

First I would calculate all at 100%. You did partly already.

100 kg H2 is 47846 mol
The reaction yield is 85%
So it will be 56289 mol in consideration.

The CO will be the half 28147 mol

The CH4 is the same 28147 plus 5% excess = 31099 mol
O2 will be 14072 mol, but 7% impurities its only 93% = so will need 15131 mol
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