Material Balance on a Reactor example

(chapter 4, A Reaction) Example:

.7044 mol/s Hexane and 9.287 mol/s Methane and 100 % excess air are fed into a reactor to produce only CO₂ and H₂O as products of the reaction. Determine all inlet and outlet flow rates, and the extents of reaction.

Flow Diagram:

Our Reactions

Here are the reactions that occur in our reactor.

C₆H₁₄ + O₂ ® CO₂ + H₂O

CH₄ + O₂ ® CO₂ + H₂O

or, in getting the stoichimetry right, we have,

1C₆H₁₄ + 9.5O₂ ® 6CO₂ + 7H₂O, and call this reaction #1

1CH₄ + 2O₂ ® 1CO₂ + 2H₂O, and call this reaction #2
And, the useful formulas are,

n_{i, leaving the reactor} = n_{i,into the reactor} + S_{of all reactions} (v_i*x)

And, rearranging this to show that we can calculate x if we just know how many moles of any of the components that reacted (because we would already know v_i for that species from the reaction),

x = n_{i, that reacted} / |v_i|

and finally, our percent excess formula,

%xs = (feed - stoich) / stoich * 100%, or = amount left / stoich

where, "stoich" referes to the number of moles that react.
Here now, we need not write out the first equation given for each component, because we can calculate the extents of reaction with the second equation, but we proceed to write out the first equation for each component, for illustrative purposes. Here, out referes to the outlet stream, and in referes to the respective inlet stream.

ndot_out,Hex = .7044mol/s + (-1)*x₁ + (0)*x₂ = 0
ndot_out,Me = 9.287mol/s + (0)*x₁ + (-1)*x₂ = 0
ndot_out,O2 = ndot_in,O2 + (-9.5)*x₁ + (-2)*x₂
ndot_out,H2O = 0 + (+7)*x₁ + (+2)*x₂
ndot_out,CO2 = 0 + (+6)*x₁ + (+1)*x₂ = 0
ndot_out,N2 = ndot_in,N2 + (0) + (0) = 0 The hexane and methane balances give both the extents of reaction, this is the first thing we would calculate, after this, we could sub in the extents of reaction into the other equations, thus, each of the above 6 equations readily give,

x₁ = .7044 mol/s
x₂ = 9.287 mol/s
ndot_out,O2 = ndot_in,O2 + -25.26 mol/s
ndot_out,H2O = 23.50 mol/s
ndot_out,CO2 = 13.51 mol/s
ndot_out,N2 = ndot_in,N2

Now, we can go no farther, and we find that we can still use the %xs relation, so, given air in 100% xs, this gives,

%xs air = [ndot_{O2 fed} - ndot_{O2 stoich}] / ndot_{O2 stoich} * 100% = 100%
or,
[ndot_{O2 fed} - ndot_{O2 stoich}] / ndot_{O2 stoich} = 1

and since ndot_{O2 stoich} = 9.5x₁ + 2x₂, and we have found the extents of reaction, thus we get ndot_{O2 stoich} = 25.26 mol/s, now, if we plug this into our equation, we get ndot_{O2 fed} = 50.52 mol/s.
Now, we can readily get the fed N₂, and we get .79/.21 * 50.52 mol/s = 189.0 mol/s.
Now, going back to our O₂ balance, with the feed O₂ now determined for the %xs relation, we can now sub in this value to get the O₂ in the outlet, to get, ndot_out,O2 = 25.26 mol/s. Now, all flows have been determined as have both extents of reaction.

x₁ = .7044 mol/s
x₂ = 9.287 mol/s

	Hexane	Methane	Oxygen	Nitrogen	CO₂	Water
Inlets	.7044	9.207	50.52	189.0	X	X
Outlets	9.992	.7044	25.26	189.0	13.51	23.50

All flowrates in mol/s.

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