CHE 354 Chemical Reactor Design

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CHE 354 Chemical Reactor Design. The General Mole Balance & Ideal Reactors. F A0. F A. What do we do with this term?. Example of Reactor Design. In – Out + Generation = Accumulation. F A0. F A. As D V goes to zero and M goes to infinity:. - PowerPoint PPT Presentation

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CHE 354 CHE 354 Chemical Reactor Chemical Reactor

DesignDesign

The General Mole The General Mole BalanceBalance

&&Ideal ReactorsIdeal Reactors

Example of Reactor Example of Reactor DesignDesign

In – Out + Generation = Accumulation

FA0 FA

dt

dNGFF AAAA 0

What do we do with this term?

The Generation TermThe Generation Term

iAi

M

iAi

M

iA VrGG

11

FA0 FA

Divide reactor into little regions, small enough that intensive variables (CA, T, P) do not vary within.

dVrG A

V

A

As V goes to zero and M goes to infinity:

The General Mole The General Mole BalanceBalance

dt

dNdVrFF AA

V

AA 0

Let’s apply it to the three ideal reactors!

Batch ReactorBatch Reactor

Perfectly mixed

No flow in or out

dt

dNdVrFF AA

V

AA 0

Batch ReactorBatch Reactor

No flow in

No flow out

Perfectly mixedVrdVrdVr A

V

AA

V

Batch ReactorBatch Reactor

Vrdt

dNA

A

And if the volume is not a function of time (constant)

AAA r

dt

dC

dt

VNd

)/(

Team ExerciseTeam Exercise

What does design equation (mole balance) look like for a semi-batch

reactor - continuous flow of Reactant A, Reactant B, or

Product P?

Continuous Stirred Tank Continuous Stirred Tank ReactorReactorCSTRCSTR

Perfectly mixed

Steady state

dt

dNdVrFF AA

V

AA 0

CSTRCSTR

Perfectly mixed

VrdVrdVr A

V

AA

V

Steady state

CSTRCSTR

00 VrFF AAA

What would the design equation look like at

startup and with some spatial variations?

dt

dNdVrFF AA

V

AA 0

Plug Flow ReactorPlug Flow ReactorPFRPFR

No radial gradients – only axial

Steady state

PFRPFR

dt

dNdVrFF AA

V

AA 0

Steady state

This is the integral form. Often the differential form is more useful. Take the derivative with respect to volume of each term.

PFRPFR

00

dV

dVrd

dV

dF

dV

dF A

V

AA

Derivative of a constant is

just 0 rA

PFRPFR

AA r

dV

dF

)],,([)( PTCfTkrdV

dFAAA

A

Here is where some complication sets in:

Example 1-1Example 1-1

Assume isomerization is first order with k

= 0.23 min-1

What volume PFR and CSTR are required for 90% conversion and 10

L/min?