# Reactor design chemical engineering question ? Relevance
• A=ammonia, B=oxygen, C=nitrogen, D=water vapor

Basis; 100 moles/s of dry air (FD0=0 moles/s)is fed to the CSTR having FB0=21moles/s of B and FC0=79moles/s of C

Reaction is A+0.75B->0.5C+1.5D

Delta= 0.5+1.5-0.75-1=0.25

This means 21moles/s of B will stoichiometrically need 21x4/3 =28moles/s of A=FA0 moles/s

So total molar flow rate of gas fed to the CSTR is 128moles/s

Mole fraction of A in feed gas is 28/128=0.21875=yA0

Mole fraction of B in feed gas is 21/128=0.16406=yB0

Mole fraction of C in feed gas is 79/128=0.61719=yC0

Mole fraction of D in feed gas is 1-yA0-yB0-yC0=0=yD0

epsilon=yA0 x delta=0.0546875=0.055

The feed gas enters the CSTR at 1200K and 3atm absolute pressure

Assuming ideal gas law at the above temperature and pressure conditions and for gases A, B and C, Volumetric flow rate of the feed gas is v0=(128 x 8.314 x 1200)/(3x101325)=4201103.4cm3/s

Molar concentration of A in feed gas is CA0=28/4201103.4=6.664916e-6moles/cm3

X=conversion of reactant A, x=multiplication

v=v0(1+epsilonxX)=cm3/s of gas leaving the reactor

CA=FA/v

CB=FB/v

CA0=FA0/v0

At T=1200K, k=19830268.4141 (moles/cm3)^1.95/s

Mole Balance on reactant A using well-mixed steady state model gives FA0=-rAxV+FA0x(1-X)

FA0xX=-rAxV

reactor volume V cm3= FA0xX/-rA, V=v0xCA0xX/-rA

residence time tau seconds = V/v0=CA0xX/-rA

Moles/s of A out from the reactor =FA0x(1-X)=FA=28x(1-X)

Moles/s of B out from the reactor=FB0-0.75xFA0xX=FB=0.75xFA0x(1-X)=21x(1-X)

Moles/s of C out from the reactor=FC0+0.5xFA0xX=FC=79+14xX

Moles/s of D out from the reactor=FD0+1.5xFA0xX=FD=0+42xX

CA^0.75=(FA0x(1-X)/v0/(1+epsilonxX))^0.75

CB^1.2=(0.75xFA0x(1-X)/v0/(1+epsilonxX))^1.2

using the value of k =19830268.411and above two expressions for CA and CB find -rA

tau= CA0xXx(1/k)x[v0x(1+epsilonxX)/FA0/(1-X)]^1.95x(1/0.75)^1.2

tau=6.664916e-6x5.043e-8x1.4123x12405104065.84x[(1+0.055xX)/(1-X)]^1.95

tau=0.0059x[(1+0.055xX)/(1-X)]^1.95

For X=0.95, tau = 1.83s

(FA0xCpA+FB0xCpB+FC0xCpC)x(1200-300)+316600xFA0xX-(FAxCpA+FBxCpB+FCXCpC+FDxCpD)x(1200-300)=Heat removal rate.

CpA=37.2J/mole/K, CpB=29.4J/mole/K, CpC=29.1J/mole/K, CpD=43.7J/mole/K

(37.2x28+29.4x21+29.1x79+43.7x0)x900=3562110J/s

For X=0.95 heat of reaction is 316600x28x0.95=8421560J/s

FA=28x(1-0.95)=1.4moles/s

FB=21x(1-0.95)=1.05moles/s

FC=79+14x0.95=92.3moles/s

FD=42x0.95=39.9moles/s

(37.2x1.4+29.4x1.05+29.1x92.3+43.7x39.9)x900=4061259J/s

Heat removal rate=7922411J/s =282943.25 J/mole of A fed

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• Anonymous
1 month ago

It's not a question though.

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