Write the mass balance equation for a plugflow system stream

Write the mass balance equation for a plug-flow system stream with a second order decaying substance, C.

Find the steady state concentration of C 5km downstream. The rate constant is 0.3 L mg^-1 day^-1, the mean velocity of the river is 0.5 m/s, and the concentration of C at x=0 is 100mg/l.

Sketch a plot of 1/C as a function of x.

Solution

since in a plug flow reaction, concentration varies along the length of reactor, we need to write a differential balance

considering a differential element

FAO= Molar Flow rate of A into the reactor, FA + dFA= Molar flow rate out of reactor

FA = FAO*(1 - XA), XA= conversion

dFA= FAO*-dXA

General mass balance for steady state ( Rate of accumulation = 0 for steady state)

Rate of input= Rate of output + rate of loss due to chemical reaction

ler -rA= rate of reaction = KCA²   and dV= differential Volume of reactor

FA = FA +dFA+(-rA)* KCA²dV

-dFA= KCA²dV

but -dFA= FAO*dXA

FAO*dXA= KCA²dV

but T= VCAO/FAO, CAO= initial concentration and FAO= Initial molar flow rate , T = space time

dV= FAO*dXA/KCA²

integrating the above equation gives

K= rate constant = 0.3 L/mg.day, , space time, T = disance/ Velocity= 5*1000/0.5= seconds=10000/(60*60*24)=0.11574 days

KCAOT= 0.11574*0.3*100 =XA/(1-XA), 3.47= XA/(1-XA)

3.47-3.47XA=XA

3.47/4.47= XA

XA= 0.78

1-XA= 0.22

CA/CAO=0.22, CA= 100*0.22= 22 mg/L

for each 1 km T= needs to be calculated.

Since KCAOT= XA/(1-XA), XA= needs to be calculaed and bases on XA ,CA= CAO*(1-XA) needs to be calculated.

XA =KCAOT*(1-XA)

XA*(1+KCAOT)= KCAOT

XA= KCAOT/ (1+KCAOT)

CA=CAO*(1-XA)

the calculations along with the plot is shown below

0.3* 100mg/L* 0.011574= XA/(1-XA)

0.35 = XA/(1-XA)

0.35(1-XA)= XA

0.35 = 1.35XA

XA= 0.35/1.35=0.26

1-XA= 1-0.26= 0.74

CA/CAO= 0.74

CA= 0.74*100 mg/L= 74 mg/L

with disance x, the space time changes and this changes XA which in turn changes C