Title Chemistry in Chemical Engineering Reynolds Number Fluid Dynamics Heat Exchanger Turbulence Heat Transfer 8.9 MB 40
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Plot agraph of t(yaxis) against In _a- (x axis).
a- x

e Note the time when the blue colour
reappears, and add afurther 1.0 cm3 of

thiosulphate solution to the reaction mixture.

f  Repeat until a total of 12.0 cm3 of
thiosulphate solution has been added, noting
the total time from the start of the experiment
as the blue colour reappears after each

1.0 cm3 addition of thiosulphate. Record

Volume of

thiosulphate
x cm3

Figure 4.4

Time

t/min

a

a- x

a- x
a

a- x
In

a
a- x

Treatment of results
If the reaction is first order with respect to
hydrogen peroxide, then:

The design equation for the batch reactor is:

a is the volume of thiosulphate solutiQ,n
equivalent to the initial number of moles of

H202•

x is the volume of thiosulphate solution

Thus the design equation becomes:

Q2

Confirm that the reaction is first order with
respect to hydrogen peroxide.

Q3

Calculate the rate constant (k) for the
reaction under these conditions from the

1 In[H202]o

k [H202]

[H2°2] 0 is the initial hydrogen peroxide
concentration, and

[H202] is the hydrogen peroxide
concentration at time t.

Expressing hydrogen peroxide concentration
in terms of the volume of thiosulphate
solution used:

=!-In a
k  a-x

If the reaction is first order, agraph of

tagainst In _a_ should bea straight line with
a- x

d
" I

gra lent -
k

Draw up a table of results as shown in

figure 4.4 above.

Q4
Why can the effect of iodide concentration
on reaction rate beignored in this experiment?

Look carefully at Equations [4.1] and [4.2].

Q5

Using the design equation, calculate the time

taken for 10, 20, and 30% conversion of the
initial hydrogen peroxide in your batch

reactor.

Figure 4.5

Vk[ A]

[A]o - [A]

1.0 X 0.03 X 0.0060

0.0015

then [A] =0.0060 mol dm-3

•'. if [A]o =0.0075 mol dm-3

k  =0.03 min-1

Target conversion =20 %

Rearranging the design equation gives:

u = total volume flow rate in dm3 min-1

u

[A ]O - [A ]

k[A]

O.02M acidified
potassium iodide
solution
+starch

own solution using the results of Experiment
4.2a.

--

d Decide upon the degree of conversion of
peroxide for which you will design (between
10 % and 30 %). Each group in the class
should aim for a different target conversion.

e Use these conditions in the design equation
for a continuous stirred tank reactor to
calculate the flow rate of reagents required.

u

Example:

V  =1.0dm3

E xperiment 4.2b
T he cont in uou s-fl ow stir r ed tank r eactor
In this experiment you will design a

continuous stirred tank reactor (CSTR) to
produce a certain percentage conversion of

reactants to products. You wil l then construct
the reactor to your own specifications and
compare its operating performance with your
design calculations. The reagents will be

'I volume' hydrogen peroxide solution and
acidifi ed potassium iodide solution (0.02M) as
used in the batch reactor experiment.

a Design areactor vessel of capacity between
0.5 and 1 dm3 which will enable reactants to
be added continuously and products to be
withdrawn at the same flow rate. It must be

possible to agitate the contents of the reactor
mechanically so that they are thoroughly
mixed at all times. (Check the design with

b Determine the working capacity of your
reactor by filling it with water and switching
on the stirrer. Water will overflow until a

steady state i sreached. Switch off and
measure the volume of water left in the vessel.

This is the working volume of the reactor
(V dm3).

c For the sake of comparison, aim to work at
the same initial concentrations as in the batch

reactor experiment, so the inlet stream
should be 'I volume' ('" 0.083M) hydrogen
peroxide mixed with 0.02M acidified

potassium iodide solution in a volume ratio of
1: 1O.Allowing for dilution, this would make
the initial hydrogen peroxide concentration

0.083 x - - l I = 0.0075 mol dm-3
Note. As the concentration of hydrogen
peroxide solution may change significantly
during storage, you should standardize your

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= 0.12 dIn3min-I

So the required total flow rate is 120 cm3

min-I.

To giveperoxide/iodide flow rates in the ratio
of 1:10, the peroxide flow rate should be

120 x ..!- = 11 cm3min-I
11

And the iodide flow rate should be

120 x 10 = 109 cm3min-I
11

f  Put about 9 dm3 of 0.02M acidifi ed
potassium iodide solution (containing 10 cm3

of 1% starch solution) into a constant head

reservoir. Position the reservoir above the
reactor vessel and adjust the flow rate to the
desired value using ameasuring cylinder and
stopclock. (109 ±S cm3 min-I in the above
example.)

g Set up asimilar reservoir containing about
2 dIn3 of '1 volume' hydrogen peroxide

solution and adjust the flow rate to the
calculated value. (11 ± 1 cm3 min-I in the

above example.)

h Al low the reactor vessel to fill up and
reach equilibrium. This will take approximately
four times the mean residence time (7) after

the reactor is full. Since

7=~
u

the residence time in the above example is

1000 = 8.3 minutes.
120

Thus at least 30 minutes should be allowed if

possible.

i  While the system iscoming to equilibrium,
drops of saturated sodium thiosulphate
solution should be added to remove the blue

colour of the iodine each time it appears. This

will ensure that the iodide concentration in
the reaction mixture remains constant.

i  Once the reactor has reached a steady state,
then O.IM sodium thiosulphate from a
constant head device should be carefully run
into the reaction mixture at such a rate that

the colour of the reactor contents appears to
'hover' between blue and colourless. It may
take a few minutes to determine this
equilibrium flow rate. Measure the rate of flow

of thiosulphate solution required using a
measuring cylinder and stopwatch.

Treatment of results
When the reaction mixture 'hovers' between
blue and colourless:

rate of production of iodine from hydrogen

peroxide
= rate of removal of iodine by thiosulphate

Assuming that the reaction between iodine
and thiosulphate is instantaneous, use your

results to calculate the percentage conversion
in the reactor.

The method is as follows:
i  Calculate the number of moles of

ii Hence calculate the number of moles of
iodine being produced per minute, using
Equation [4.2].
iii  Hence calculate the number of moles
of hydrogen peroxide reacting per minute in

your reactor, volume V  dIn3, using Equation

[4.1].
iv Calculate the rate of reaction in moles
dm-3 min-I.

v Using your value fOIthe rate constant and
the rate expression for the reaction

calculate the concentration of hydrogen
peroxide in the reaction mixture and hence

also in the exit stream.

Q6

Compare the actual percentage conversion
with your design conversion. Try to account
for any discrepancies which exist.

Q7

Consider the l ikely effect of the following

changes of conditions on the percentage
conversion within the reactor:

a increased reactant concentration in feed
b increased total flow rate through reactor

c increased reactor volume
d increased temperature.

The average residence time 7is

For a first order reaction in a batch reactor the design

equation is

Then

= 13.3 minutes

= 2 . . - In [Alo
k  [A]

t

v  = 22
u 1.65

[A]o = 1 mol dm-3
k  = 0.122 min-1

t = 13.3 minutes (same time interval as CSTR)

If:

v = 7 = [A]o - [A]
u k[A]

4 . 3
B A T C H O R C O NT IN U O U S O P E R A T IO N ?

In the design of any chemical reactor, two factors - the

kinetics of the reaction and the required output of product -

are normally fIXed from the outset. Using all the available

information, the chemical engineer must make decisions

concerning the type of reactor to be used, its physical

dimensions, and the optimum conditions under which it is to

operate.

The design equations developed earlier in this section

enable comparisons to be made between theoretical yields of

product from a continuous stirred tank reactor and a batch

reactor during the same time interval.

For a fust order reaction in a continuous stirred tank

reactor, the design equation is

... [A]

If:
Volume (V) = 22 m3
Flow rate (u) = 1.65 m3 min-I

Rate constant (k) = 0.122 min-I

[A] 0 = 1mol dm-3

In 1

[A]
= 0.122 x 13.3

= 0.20 mol dm-3

Then using the design equation

[A] =0.38 mol dm-I

This represents a 62 % yield of products.

Since [A]o was 1 mol dm-3, this represents an 80 % yield of

products.

The batch reactor gives a larger percentage conversion

than the continuous stirred tank reactor, using the same size

vessel over the same period of time. Normally, a manufac-

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