Question

The figure below shows three reactors connected by pipes. As indicated, chemical pollutants rate of transfer through each pipe is equal to a volumetric flow rate Q in m/s multiply by the concentration from the prior system where the chemical originates. If the system is at a steady-State, the transfer into each reactor will balance the transfer out. Develop mass balance equations for reactors and leave answers in Ax=b, where x shall be (C1, C2, C3) Show all steps. Do not solve. 9139 03393 850 400 021C2 should be Q21 c1 922021 Q33 = 350 Q13 = 30 Q21 = 40 Q23 = 50 Q12 = 20 Q22 = 5

Answer 1

Here, we need to balance the mass flow rate of chemical pollutant inside each reactors. Note that the system is in steady state. Hence, the mass coming out of reactor equals Mass leaving the reactor.

Note: As you have instructed in the question, I have taken Q₂₁C₂ as Q₂₁C₁

Thus, for Reactor 1,

Mass flow inside = Mass flow outside = 850 = Q13C1 +Q12C1 + Q21C1 Q13C1 + Q12C1 + Q21C1 = 850

For reactor 2,

Mass flow inside = Mass flow outside =Q12C1 +Q21C1 + 400 = Q23C2 + Q22C2 Q23 C2 + Q22C2 - Q12C1 - Q21C1 = 400

For reactor 3,

Mass flow inside = Mass flow outside =Q13C1 + Q23C2 = 400+ Q33C3 Q13C1 + Q23C2 - Q33 C3 = 400

Thus, we have obtained a set of three equations,

Q13C1 +212C1 +Q21C1 = 850

Q23C2+Q22C2-Q12C - Q21C1 = 400

Q13C1 + Q23C2-33 C3 = 400

Converting this to matrix form, we will have,

850 Q13 + Q12 + Q210 -Q12 - Q21 Q23 +Q220 Q13 Q23 –Q33] [C3 400 400

Putting in the values, we have,

850 C2 400 C3 400 30+20 + 40 0 0 -20 – 40 50 + 5 0 30 5 0 -350 † 90 0 0 ][c] [850 -60 55 0 = 400 30 50 -350 = 400

This is the final answer.