Acid Neutralization Reactor

Introduction to Engineering Systems

Copyright ©2001, University of Notre Dame

Module 4- Acid Neutralization Reactor

Module 4:Acid neutralization

reactorLecture 2:

Analysis of the feed tank and thereactor for the case of no

reaction

Mark J. McCreadyChemical Engineering

Acid Neutralization

Reactor




Outline for today

Quick review of mass balance equations Analysis of the reactor for 2 feeds but no

reaction Expectation of a Steady State ...

Analysis of the feed tank that is draining by gravity How does the depth of liquid affect the flow

rate? Bernoulli equation to relate effects of gravity,

pressure and velocity within a fluid




Reactor, what will the exit

concentration be?

Control volume

Acid in, 1

Base in, 2

Flow out, 3

3




Feed Tank, how fast does it

drain?





drain?

h

We will use this control volume

for the tank




Start of new material

First we will analyze the reactor using the mass balance equations.

Today, there will be no reaction. But, we will allow for inlet streams of

different concentration. We will see that if the inlet concentrations

and flows are constant, a steady - state is expected where there is no change in the concentration with time in the tank.




Today, we will not do reaction,just

use the tank as a mixer

Control volume

Salt solution in, 1

Another salt solution in, 2

Flow out, 3

3




RecallRecall Mass Balance

General mass balance equation for a fixed control volume Rate of Accumulation = Rate In - Rate Out + Production by reaction- Consumption by reaction

Overall

Component mass (mole) balance

dρVdt

=∑jqj ρj

masstime

⎛

⎝ ⎜

⎞

⎠ ⎟

dciVdt

=∑jqjc j i −riV

molestime

⎛

⎝ ⎜

⎞

⎠ ⎟

j - density of stream j, (mass/length3 )qj -- volumetric flow rate of stream j, (length3 /time)V -- active volume of reactor,(length3)

cji-- molar concentration of species i in stream j, (moles/ length3 )ri -- molar reaction rate per volume (moles/ (length3 -time))




Today, we will not do reaction,just use the

tank as a mixer

Control volume

Salt solution in, 1

Another salt solution in, 2

Flow out, 3




We see two inputs and one outputOverall mass balance

Component mass balance, for salt

qj -- volumetric flow rate of stream j (m3/s)V -- active volume of reactor (m3)cjsalt-- molar concentration of salt in stream j (moles/m3)

Mass Balance

dVdt

=q1 +q2 −q3volume

time

⎛

⎝ ⎜

⎞

⎠ ⎟

21

3

V

dc3salt

dt=q1c1salt +q2c2salt −q3c3salt

A sketch of our problem looks like:

3

The reaction term is 0!!




Simplifications

Flowrates in are not changing in time Reactor is filled at the beginning Thus, overall mass balance tells us

nothing we don’t find obvious.

What about the salt balance? We expect that it will tell us what comes out, if we know what goes in.

q3=q1 +q2




Balance equation

The salt balance equation#

Can be solved to give…

You can solve this equation by numerical integration

V

dc3

dt=q1c1+q2c2 −q3c3

c3 =

(c1q1 +c2q2) 1−e−

tq3V

⎛

⎝ ⎜

⎞

⎠ ⎟

q3

#A green background slide means that we don’t expect you to get the answer, because we used mathematics you may not yet understand. But, the answer will be insightful.




Plot of the concentration

We see that there is an initial transient (exponential) that depends on reactor volume and then a steady state is reached after which there is no further time variation. (If the inlets remain

constant!) Steady state

answer:

c3 =

(c1q1 +c2q2)q3

Initial concentration =0

Initial concentration =0

Note different volumes andabscissa scales




Steady state concentration

For this example we have q1 = 10 m3/s, c1 = 2 moles/

m3

q2 = 5 m3/s, c2 = 3 moles/m3

Thus: q3 = (10 + 5)= 15 m3/s

c3 =

(c1q1 +c2q2)q3

c3 =(2*10+3*5)

15=2.33moles /m3

Note different volumes and

abscissa scales




Steady state behavior?

Is there always a steady state if we have steady inputs to a reactor? Maybe this is obvious ??

Should we have even bothered to integrate?

Think of some examples….





drain?

h

Now let’s examine a feed tank

We need a new control volume

This tank has an exit stream, but

no inlet streams.




We have just one outputOverall mass balance:

since

we can use the chain rule to get

But…, how do we get u3?

Draining tank

dVdt

=−q3

u3,A3

area of exit pipe, velocity of fluid leaving in stream 3.

A sketch of our problem looks like:

ATank

dhdt

=−q3

Control valve

ATank

dhdt

=−u3A3

h

V =ATankh

q3 =u3A3

we know the flowrate and velocity are related by

thus




ATank --not really h-- yup!A3 -- yes, consider a

4” pipe versus a hypodermic needle

How open the valve is (as denoted by K)g, gravity -- well of course

• can’t drain a tank on the space station with gravity!

Draining tank

u3,A3

area and velocity

Factors that affect exit liquid flowrate

Control valve, K ATank

dhdt

=−u3A3

h

h=0




Pressure-depth relation

Common occurrence in the summer

Basic equation of hydrostatics:

Wow,my ears hurt

ΔP =ρgΔh

=densityg=gravitation constantP=pressureh=depth of liquid




Effect of depth

So we expect that if the depth is greater, the flow rate will be faster

Can we quantify this? Recall from Physics,

Consider conversion of potential to kinetic energy for a fluid blob.

First we take the case of no “friction” or drag




KE--PE relation,we get velocity

Consider a blob of fluid in our tank. It will follow the path shown with no friction

u3,A3

area and velocity

Control valve, K

m g h = PE

KE = 1/2 m u2

KE+ PE = 01/2 m (ub

2 - ua2 ) +mg (hb-ha)

= 0 ub

2 = 2 g h

ub = 2gΔh

a

b

h

h

h=0

h=0




Draining tank with control

valve We see that the velocity will not depend

on the area of exit pipe. Now for the real system we have a

control valve that can open and close, the easiest way to deal with this is to consider that it causes a “loss” of energy.

KE + PE + “losses” = 0 1/2 m (ub

2 - ua2 ) +mg(hb-ha) + K/2 ub

2 = 0 (1+K) ub

2 = 2 g h

ub =

2gΔh1+K

As K increases, velocitydecreases.As the valve is closed,K increases!




WHITE BOARD STUFF

Notre Dame law of wind direction How momentum of fluid is converted

to an increase in pressure as it impinges on a wall? Student--University paradox How the pressure must increase if

the fluid is to be slowed down. Work--Energy Principle from Physics Bernoulli Equation




Bernoulli equation

The relation between the pressure, the velocity, the change in height and frictional losses:

For our draining tank, there is no pressure change, and the relation between u and h is

ΔPρ

+Δu2

2+gΔh +K

u2

2=0

Now we can go back to the mass balance and finish solving the problem

ub =

2gΔh1+K




White board




White board




White board




White Board




You can solve this numerically to find how h changes in time.

Draining tank

u3,A3

area and velocity

RecallRecall the mass balance

Control valve, K ATank

dhdt

=−u3A3

ATank

dhdt

=−A32gh1+K

We use our relation, note that the “b” subscript is now

“3”

u3 =

2gΔh1+K

To get a final equation that canbe solved ...

h

h=0




Draining tank

h(t)=

This one has a rather ugly analytical solution…

K varies from 0 to

12

Here is a plot of some results

K=0

K=12




Filling/draining tank (for homework)

What do the equations for this tank look like?

This last equation can be easily solved numerically to get height versus time.

1 2

33

dVdt

=q1 +q2 −q3

ATank

dhdt

=q1 +q2 −u3A3

ATank

dhdt

=q1 +q2 −2gh1+K

A3

Substitute for the unknown flow rate and the liquid depth

Now use the Bernoulli equation




Recap (mixing tank)

Component mass balance for “mixing tank”

The behavior is:

21

3

V

dc3salt

dt=q1c1salt +q2c2salt −q3c3salt

c3 =

(c1q1 +c2q2)q3

Steady-state answer




Recap (draining tank)

Overall mass balance for draining tank

u3,A3

area and velocity

Control valve, K

ATank

dhdt

=−u3A3

u3 =

2gΔh1+K

ATank

dhdt

=−A32gh1+K K varies

from 0 to 12




RecapBernoulli equation

Bernoulli equation

Useful engineering equation to describe large-scale fluid flows. It relates changes in pressure, height and velocity and accounts for frictional losses.

ΔPρ

+Δu2

2+gΔh +K

u2

2=0

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Acid Neutralization Reactor