Question

Steam reforming of natural gas is the most common method of producing commercial hydrogen. A stream with a flow rate of 1150 mol/h containing 85.0 mol% CH4 and 15.0 mol% of water is combined with additional water steam and fed to a steam reforming reactor to produce hydrogen. The stream coming out is in chemical equilibrium. The fractional conversion for both the water and methane are 0.600. Balance the chemical equation and calculate how much additional water steam is fed to the steam reforming reactor and the flow rate of the outlet hydrogen.CH4 + H2O <--> CO + 3H2 (balanced)

Answer 1

Answer:

The flow rate of the outlet hydrogen is 1,759.5 mol H2/h

The additional water steam fed is 414 mol H2O/h

Explanation:

First, we calculate the flow rate for each reactant:

$Reactant Flow = Entry Flow * Molar Fraction$

Therefore:

$FCH4 = 1150 mol/h * 0.85 = 977.5 mol/h\\FH2O = 1150 mol/h * 0.15 =  172.5 mol/h$

Next, we calculate how much H2 (in mol/h) would be generated according to the balanced chemical equation:

977.5 mol CH4/h * 3 mol H2 / 1 mol CH4 = 2,932.5 mol H2/h

But the conversion rate for the CH4 is 60%, therefore the actual H2 generated is:

2,932.5 mol H2/h * 0.60 = 1,759.5 mol H2/h

According to the balanced chemical equation, we can calculate the amount of H2O needed for the reaction:

977.5 mol CH4/h * 1 mol H2O / 1 mol CH4 = 977.5 mol H2O/h

The conversion rate for the H2O is also 60%, so the actual H2O used is:

977.5 mol H2O/h * 0.60 = 586.5 mol H2O/h

Therefore the additional water stream would be:

F2 H2O = F H2O used - F1 H2O = 586.5 mol H2O/h - 172.5 mol H2O/h = 414 mol H2O/h