Upload
nicole
View
103
Download
1
Embed Size (px)
DESCRIPTION
Acid Neutralization Reactor. Module 4: Acid neutralization reactor Lecture 2: Analysis of the feed tank and the reactor for the case of no reaction. Mark J. McCready Chemical Engineering. Outline for today. Quick review of mass balance equations - PowerPoint PPT Presentation
Citation preview
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- 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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Reactor, what will the exit
concentration be?
Control volume
Acid in, 1
Base in, 2
Flow out, 3
3
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Feed Tank, how fast does it
drain?
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Feed Tank, how fast does it
drain?
h
We will use this control volume
for the tank
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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.
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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))
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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!!
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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.
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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….
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Feed Tank, how fast does it
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.
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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!
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
White board
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
White board
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
White board
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
White Board
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
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