EXPERIMENT NUMBER 9

RTD Analysis of Three CSTRs Connected in Series

Aim: To study the RTD due to a pulse input for a system of 3 CSTRs in series.

INTRODUCTION: In practical situations, we often encounter non-ideal conditions when using

ideal Continuous Stirred Tank Reactors (CSTRs) and Plug Flow Reactors (PFRs) to determine
conversion and other performance metrics. To address this non-ideality, Real-Time Delay (RTD)
analysis, and more precisely, the examination of the E(t) curve, serves as a valuable tool for
making necessary adjustments. Non-ideal flow in a PFR is primarily attributed to the following
factors:
1. The occurrence of inactive regions, which diminish the usable reactor volume.

2. The presence of axial dispersion or mixing along the reactor's length.

3. Non-uniform mixing in the radial direction.

Theory

Let’s consider that 10 atoms are injected in a PFR in a short period of time, since there is no
axial mixing in PFRs, ideally all these atoms should come out of the reactor at the same time.
However, this is not the case with a real system.

Different atoms take different times to come out of the reactor. The time the atoms spend in the
reactor is known as residence time and the distribution of various atoms coming out of the
reactor w.r.t time is called residence time distribution.

RTD can be determined experimentally by injecting an inert chemical called tracer into the
reactor at t=0. Pulse and step inputs are two methods of injection.

Pulse Input

In a pulse input, an amount of tracer ◻◻ is suddenly injected in one shot into the feed stream
entering the reactor in as short time as possible. The outlet concentration is then measured as a
function of time. The effluent concentration-time curve is referred to as the C(t) curve in the
RTD analysis. If we select an increment of time Δt sufficiently small that the concentration of
tracer C(t), exiting between time t and t + Δt is essentially constant, then the amount of tracer
material, ΔN, leaving the reactor between time t and t + Δt is
 ΔN = ϑC(t)Δt (1)

where, ϑ is the effluent volumetric flow rate.

In other words, Δ◻ is the amount of material that has spent an amount of time between t and
t+Δt in the reactor. If we now divide by the total amount of material that was injected into the
reactor (◻◻), we obtain:

(2)
This represents the fraction of the material that has a residence time in the reactor between time t
and t + Δt.

For a pulse injection, we define:

(3)
Thus, Eq. (2) can be written as

(4)

Re-writing the above equations in the differential form,

(5)

Integrating Eq. (5), we obtain

(6)

The volumetric flow rate ϑ is constant, and so we can define E(t) as

(7)

The integral in the denominator is the area under the C(t) curve.

The mean residence time of all particles is given by:

(8a)
For theoretical calculation of E-curve for N equal sized CSTRs in series, the dimensionless E
curve as a function of N is given by

(9)

Step Input (Additional information)

Now, consider a constant rate of tracer addition to a feed that is initiated at time t = 0. Before this
time, no tracer was added to the feed. Thus, we have

C0(t) = 0 t<0
C0(t) = C0 t>0

The concentration of the tracer in the feed to the reactor is kept at this level until the
concentration in the effluent is indistinguishable from that in the feed. The test may then be
discontinued.

The output concentration from a vessel is related to the input concentration by the convolution
integral:

(9)

Since the inlet concentration is a constant with time, we can take it outside the integral sign.
Thus,

(10)

Dividing by ◻◻ yields

(11)

Thus, the normalized concentration vs. time profile gives the F curve. The relation between E(t)
and F(t) is:

(12)
For a laminar flow reactor, the velocity profile is parabolic with the fluid in the center of the tube
spending the shortest time. By using a similar analysis as shown above, we obtain the complete
RTD function for a laminar flow reactor as
We can also determine the dispersion number in the following manner:
The variance is defined as:

(13)
where ◻◻ is defined by Eq. (8)

(14)
The dispersion number can be calculated using Eq. (15).

(15)

APPARATUS:

Procedure:
1. Begin by filling the reservoir with water.

2. Determine the volume of the reactors in the specified configuration.

3. Activate all the switches and initiate the water flow through the reactors,
allowing the system to reach a stable condition.

4. Subsequently, measure the volumetric flow rate from the effluent stream.

5. Prepare a 10 ml solution of 5 M NaOH for a pulse input study and a 0.1 M

HCl solution for titration.
6. At a specific time (t=0), introduce the NaOH solution into the initial CSTR
all at once and start the stopwatch.
7. Collect samples from each CSTR at designated time intervals (at t=1, 4, 7…
minutes from tank 1; at t=2, 5, 8… minutes from tank 2; and at t=3, 6, 9…
minutes from tank 3). Acquire 15 readings from each tank.

8. Perform titration on the samples to determine the tracer's concentration in

each tank as a function of time.
Calculations
At any time, t= 600sec =10min.

Volume of HCL consumed in titration, VHCL = 1.4ml

Concentration of NaOH in sample taken, C = (0.1*1.4)/10 = 0.014 M

Theoretical E(t) = { 0, if t < 𝜏/2 and 𝜏2/2𝑡3 if t > 𝜏/2 }

𝜏 = ∑𝐶i𝑡i /∑𝐶i

∑𝐶i𝑡I = 135.36, ∑𝐶i = 0.198

𝜏 = ∑𝐶i𝑡i /∑𝐶i = 135.36/0.198 = 683.6364

𝜏/2 = 341.818182
Since for t = 600 sec, t > 341.818182 ,

Theoretical E(t) = 𝜏2/2𝑡3 = (683.6364*683.6364)/(2*(600)3) = 0.001082

∞
Experimental E(t) = C(t)/ ∫0 C(t)dt

Area under curve, at any point area is calculated using trapezoidal rule
at any ti area ai = (1/2)*(Ci + C(i+1))*(ti+1 – ti) = (1/2)*(0.013+0.014)*(780 – 600) = 2.43
∞
thus, ∫0 C(t)dt = ∑ ai = 30.87
∞
Experimental E(t) = C(t)/ ∫0 C(t)dt = 0.014/30.87 = 0.000454

Theta = t/𝜏 = 600/683.6364 = 0.877659528

𝐸𝜃(θ) = (𝑁N exp(−𝑁θ)/(𝑁−1)!)θN-1

For three CSTRs, N=3 , Theoretical 𝐸𝜃(θ) = 0.249101641

𝑡 ∞
Mean residence time , 𝑡m = ∫ tC(t)dt / ∫ C(t)dt = ∑ tiCiΔti /∑ CiΔti = 24004.8/135.36
0 0
=809.8785 sec

Variance, σ2 = ∑ 𝑡i2𝐶𝑖Δt/ ∑ 𝐶𝑖Δti − 𝑡m2

∑ 𝑡i2𝐶𝑖Δt = 32692032
∑ CiΔti = 29.64
𝜎2 = ∑ 𝑡i2𝐶𝑖Δt/ ∑ 𝐶𝑖Δti − 𝑡 2
= (32692032/29.64) – (177.34*177.34) = 1102970.04 –
𝑡 m
31449.4756 = 447066.855
𝜎2 = 𝜎2/ 𝑡 2 = 447066.855/655903.185 = 0.6816
𝜃 𝑡 m
Dispersion number (𝐷/𝑢𝐿) can be found as,
𝜎2 = 𝜎2/ 𝑡 2 = (2𝐷/𝑢𝐿) − 2 (𝐷/𝑢𝐿)2(1 – 𝑒xp(− 𝑢𝐿/𝐷)) = 0.6816
𝜃 𝑡 m

Number of theoretical CSTRs, N = 1/𝜎𝜃2 = 1.4459

𝐷/𝑢𝐿 = 0.7825

Calculation table
• For tank 1
tank 1 Column1 Column2 Column3 Column4 Column5 Column6 Column7 Column8 Column9 Column10 Column11
Volume of hcl used (ml) time (sec.) conc of NaOH E(t) exp E(t) theo theta E(theta) theo ai Ci*ti*del_ti Ci*ti Ci*del_ti Ci*ti*ti*del_ti
5 60 0.05 0.0016197 0 0.087765953 0.026638873 7.65 180 3 3 10800
3.5 240 0.035 0.0011338 0 0.351063811 0.193459211 5.4 1512 8.4 6.3 362880
2.5 420 0.025 0.0008098 0.0031541 0.614361669 0.268917515 3.51 1890 10.5 4.5 793800
1.4 600 0.014 0.0004535 0.0010818 0.877659528 0.249101641 2.43 1512 8.4 2.52 907200
1.3 780 0.013 0.0004211 0.0004924 1.140957386 0.191080721 2.16 1825.2 10.14 2.34 1423656
1.1 960 0.011 0.0003563 0.0002641 1.404255244 0.131378322 1.8 1900.8 10.56 1.98 1824768
0.9 1140 0.009 0.0002915 0.0001577 1.667553103 0.084090034 1.53 1846.8 10.26 1.62 2105352
0.8 1320 0.008 0.0002592 0.0001016 1.930850961 0.051172457 1.35 1900.8 10.56 1.44 2509056
0.7 1500 0.007 0.0002268 6.924E-05 2.194148819 0.029993317 1.26 1890 10.5 1.26 2835000
0.7 1680 0.007 0.0002268 4.928E-05 2.457446678 0.0170771 1.08 2116.8 11.76 1.26 3556224
0.5 1860 0.005 0.000162 3.631E-05 2.720744536 0.009501125 0.81 1674 9.3 0.9 3113640
0.4 2040 0.004 0.0001296 2.753E-05 2.984042394 0.005187559 0.72 1468.8 8.16 0.72 2996352
0.4 2220 0.004 0.0001296 2.136E-05 3.247340253 0.002788446 0.63 1598.4 8.88 0.72 3548448
0.3 2400 0.003 9.718E-05 1.69E-05 3.510638111 0.001479219 0.54 1296 7.2 0.54 3110400
0.3 2580 0.003 9.718E-05 1.361E-05 3.773935969 0.000775895 1393.2 7.74 0.54 3594456
summation 0.198 0.006414 30.87 24004.8 135.36 29.64 32692032
D/uL = 0.7825
tau 683.6364 tm 809.878 variance 0.6816 N = 1.4459

• For tank 2
Volume of hcl used (ml) time (sec.) conc of NaOH E(t) exp E(t) theo theta E(theta) theo ai Ci*ti*del_ti Ci*ti Ci*del_ti Ci*ti*ti*del_ti
1.6 120 0.016 0.0005487 0 0.126666667 0.049374794 3.33 230.4 1.92 1.92 27648
2.1 300 0.021 0.0007202 0 0.316666667 0.174516887 3.69 1134 6.3 3.78 340200
2 480 0.02 0.0006859 0.0040577 0.506666667 0.252655972 3.51 1728 9.6 3.6 829440
1.9 660 0.019 0.0006516 0.0015609 0.696666667 0.270138889 3.06 2257.2 12.54 3.42 1489752
1.5 840 0.015 0.0005144 0.0007571 0.886666667 0.247462819 3.24 2268 12.6 2.7 1905120
2.1 1020 0.021 0.0007202 0.0004229 1.076666667 0.206349715 3.24 3855.6 21.42 3.78 3932712
1.5 1200 0.015 0.0005144 0.0002597 1.266666667 0.161516973 2.34 3240 18 2.7 3888000
1.1 1380 0.011 0.0003772 0.0001708 1.456666667 0.120799738 1.62 2732.4 15.18 1.98 3770712
0.7 1560 0.007 0.0002401 0.0001182 1.646666667 0.087298978 1.26 1965.6 10.92 1.26 3066336
0.7 1740 0.007 0.0002401 8.518E-05 1.836666667 0.061420112 1.17 2192.4 12.18 1.26 3814776
0.6 1920 0.006 0.0002058 6.34E-05 2.026666667 0.042292827 0.99 2073.6 11.52 1.08 3981312
0.5 2100 0.005 0.0001715 4.846E-05 2.216666667 0.028612446 0.81 1890 10.5 0.9 3969000
0.4 2280 0.004 0.0001372 3.786E-05 2.406666667 0.019073845 0.54 1641.6 9.12 0.72 3742848
0.2 2460 0.002 6.859E-05 3.014E-05 2.596666667 0.012557145 0.36 885.6 4.92 0.36 2178576
0.2 2640 0.002 6.859E-05 2.439E-05 2.786666667 0.008178632 950.4 5.28 0.36 2509056
summation 0.171 29.16 29044.8 162 29.82 39445488
D/uL = 0.2662
tau 947.368421 tm 974.004024 variance 0.394 N = 2.5189
 • For tank 3
tank 3 Column1 Column2 Column3 Column4 Column5 Column6 Column7 Column8 Column9 Column10 Column11
Volume of hcl used (ml) time (sec.) conc of NaOH E(t) exp E(t) theo theta E(theta) theo ai Ci*ti*del_ti Ci*ti Ci*del_ti Ci*ti*ti*del_ti
1.5 180 0.015 0.0006083 0 0.182160804 0.092769285 2.79 486 2.7 2.7 87480
1.6 360 0.016 0.0006488 0 0.364321608 0.230539863 3.15 1036.8 5.76 2.88 373248
1.9 540 0.019 0.0007705 0.0031004 0.546482412 0.322262949 3.33 1846.8 10.26 3.42 997272
1.8 720 0.018 0.0007299 0.001308 0.728643216 0.355934166 2.97 2332.8 12.96 3.24 1679616
1.5 900 0.015 0.0006083 0.0006697 0.91080402 0.34551868 2.52 2430 13.5 2.7 2187000
1.3 1080 0.013 0.0005272 0.0003876 1.092964824 0.309111991 2.07 2527.2 14.04 2.34 2729376
1 1260 0.01 0.0004055 0.0002441 1.275125628 0.261391379 1.62 2268 12.6 1.8 2857680
0.8 1440 0.008 0.0003244 0.0001635 1.457286432 0.212107967 1.53 2073.6 11.52 1.44 2985984
0.9 1620 0.009 0.000365 0.0001148 1.639447236 0.166779956 1.44 2624.4 14.58 1.62 4251528
0.7 1800 0.007 0.0002839 8.371E-05 1.82160804 0.127920652 1.17 2268 12.6 1.26 4082400
0.6 1980 0.006 0.0002433 6.289E-05 2.003768844 0.096162969 0.9 2138.4 11.88 1.08 4234032
0.4 2160 0.004 0.0001622 4.844E-05 2.185929648 0.071099544 0.63 1555.2 8.64 0.72 3359232
0.3 2340 0.003 0.0001217 3.81E-05 2.368090452 0.051840938 0.36 1263.6 7.02 0.54 2956824
0.1 2520 0.001 4.055E-05 3.051E-05 2.550251256 0.037352876 0.18 453.6 2.52 0.18 1143072
0.1 2700 0.001 4.055E-05 2.48E-05 2.73241206 0.026639884 486 2.7 0.18 1312200
summation 0.145 24.66 25790.4 143.28 26.1 35236944
D/uL = 0.2554
tau 988.137931 tm 988.137 variance 0.382685 N =2.613

Graphs
For CSTR tank 1

CNaOH vs time t
0.06

0.05

0.04
CNaOH

0.03

0.02

0.01

0
0 500 1000 1500 2000 2500 3000
time t (sec)
 0.0035

0.003

0.0025
E(t) exp /E() theo

0.002

0.0015

0.001

0.0005

0
0 500 100 0 150 0 200 0 250 0 3000
-0.0005
time t (sec)
E(t) exp E(t) theo

𝐸_𝜃(θ) theo. vs time t

0.3

0.25

0.2
𝐸_𝜃(θ)

0.15

0.1

0.05

0
0 0.5 1 1.5 2 2.5 3 3.5 4
θ
For CSTR tank 2

CNaOH vs time t
0.025

0.02

0.015
CNaOH

0.01

0.005

0
0 500 1000 1500 2000 2500 3000
time t (sec)

0.0045
0.004
0.0035
0.003
E(t) exp /E() theo

0.0025
0.002
0.0015
0.001
0.0005
0
0 500 100 0 150 0 200 0 250 0 3000
-0.0005
-0.001
time t (sec)

E(t) exp E(t) theo

 𝐸_𝜃(θ) theo. vs time t
0.3

0.25

0.2
𝐸_𝜃(θ)

0.15

0.1

0.05

0
0 0.5 1 1.5 2 2.5 3
θ

For CSTR tank 3

CNaOH vs time t
0.02
0.018
0.016
0.014
0.012
CNaOH

0.01
0.008
0.006
0.004
0.002
0
0 500 1000 1500 2000 2500 3000
time t (sec)
 0.0035

0.003

0.0025
E(t) exp /E() theo

0.002

0.0015

0.001

0.0005

0
0 500 100 0 150 0 200 0 250 0 3000
-0.0005
time t (sec)

E(t) exp E(t) theo

𝐸_𝜃(θ) theo. vs time t

0.4
0.35

0.3

0.25
𝐸_𝜃(θ)

0.2

0.15

0.1

0.05

0
0 0.5 1 1.5 2 2.5 3
θ

Results
Cases Exp. Mean theo. Mean Exp. No. of variance Dispersion
residence residence tanks number
time time D/uL
Tank 1 683.636 809.878 1.44 0.6816 0.7825
Tank 2 947.368 974.004 2.52 0.394 0.2662
Tank 3 988.137 988.137 2.61 0.3827 0.2554
average 873.047 924.006 2.19 0.4861 0.4347
total 2619.141 2772.019
Error analysis
( theoretical value – experimental value )
Percentage error = x 100
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒

Error % in total mean residence time = (2772.019–2619.141)*100/2772.019= 5.515 %

Error % in total number of tanks = ( 3 – 2.19 )*100/3 = 27%

Discussion analysis
From error analysis we are getting error in total number of tanks less than 30% and error in
total mean residence time near to 5%. From graphs of concentration of NaOH against time
we found its concentration in tank 1 is decreases down with time , in tank 2 it goes through
maxima point and in tank 3 we saw erratic behaviour of concentration that is it shows
increases then decreases continuously. We used trapezoidal rule for area calculation which
has its effect on E(t) calculated value. We found highest E(t) theoretical and experimental
value for tank 3 at same time but not in other tanks, initially both differs in magnitude by
large margin but as time passes their difference goes down. For dimensionless E(theta), its
graph against dimensionless time theta similar for all three tanks but differs in magnitude.
E(t) theoretical against time show similar behaviour like concentration with time as it is
reduced version of it. We found experimental mean residence time highest in tank 3 followed
by in tank 2 than in tank 1 similar to theoretical prediction. This happens because tracer
spends more time in tank 2 and 3 than in tank 1 because of axial mixing. Dispersion value
decreases as number of tank increases, we found lowest dispersion in tank 3 than in other
tanks indicates not much spreading of tracer and suggest plug flow behaviour if number of
tanks increases in accordance with theory.

Sources of error
1. Solution of NaOH may contain impurities while dissolving with distilled water.
2. Parallax error while taking burette reading for titration volume of HCL.
3. Lag in taking 10ml solution out of tanks may give error in calculated values.
4. Humar error or experimental error.

Precautions
1. Apparatus must be cleaned before using
2. Only Distilled water should be used for NaOH solution
3. Flow rate should remain constant throughout experiment
 4. Do titration slowly and observe colour change.

Scope improvement
1. More different types of input may be used
2. Different combinations of CSTRs may be used
3. Different method may be used for area calculation other than trapezoidal rule.