Unsteady State Material Balance Assignment 2022

Fig 1 Tank system
1. Unsteady state material balance
Accumulation of NaOH in the tank = Total input of NaOH - Total output of
NaOH
d(c.V)/dt = (c1F1 +c2F2) – c3F3
or, V.(dc/dt) + c.dv/dt= (c1F1 +c2F2) – c3F3
again, dV/dt = A.dh/dt = (F1+F2)-F3
Substituting we get, V.(dc/dt) + c.[ (F1+F2)-F3]= (c1F1 +c2F2) – c3F3,
Now the tank is well mixed,
So, c = c3 and, c2 = 0 as it is only water no NaOH
V.(dc/dt) = c1F1-cF1-cF2
This is the equation of the unsteady state material balance
Now, for steady state material balance,

Accumulation of NaOH in the tank = Total input of NaOH - Total output of
NaOH
Here accumulation = 0
So, Total input of NaOH = Total output of NaOH
V.(dc/dt) = c1F1-cF1-cF2 = 0
Or, c1F1 = c(F1+F2)
2. Linearization of the equation,
V.(dc/dt) = c1F1-cF1-cF2
And, A.dh/dt = (F1+F2)-F3
Now, F3 = a.h where a is the constant
Then, A.dh/dt +a.h = (F1+F2)
3. Now, deviation variable, when t = 0, c3s = 0
There is no out flow as there is no concentration of the NaOH
Deviation variable, C3 = c3-c3s
Or, C3=c3
Again, H = h – hs
So, the equation reduces to,
V. dC/dt + C. (F1+F2) = C1F1
4. In the equation,
V. dC/dt + C. (F1+F2) = C1F1
Now, F1+F2 = F
So the equation reduces to,
V. dC/dt + C.F = C1F1
Taking the laplace transform of both side,
C(s).s.V – C(0) + F.C(s) = c1.F1(s)
Or, C(s).[V.s+F] = c1.F1(s)
Or, C(s)/F1(s) = c1/[ V.s+F] = (c1/F)/[(V/F).s + 1]

Where, time constant is V/F = V/[F1+F2] and steady state gain is, c1/F
5.
Fig. 2. Block diagram of the transfer function
6. Under the steady state condition,
F1 = 60 ml/min, F2 = 40 ml/min, c1 = 0.1M
The concentration of the NaOH after the mixing is c,
Then, c1.F1 = c(F1+F2)
Or, c = [F1/(F1+F2)].c1 = (60/(60+40)).0.1 = 0.06M
7. If it is maintained the ratio of the inlet feed stream a constant value then only
it is solely depends on the inlet concentration,
In this problem, let assume, F2/F1 = m, or, F2 = m.F1,
Then,
Unsteady state material balance
Accumulation of NaOH in the tank = Total input of NaOH - Total output of
NaOH
d(c.V)/dt = (c1F1 +c2F2) – c3F3
or, V.(dc/dt) + c.dv/dt= (c1F1 +c2F2) – c3F3
again, dV/dt = A.dh/dt = (F1+m.F1)-F3
Substituting we get, V.(dc/dt) + c.[ (F1+mF1)-F3]= (c1F1 +c2mF1) – c3F3,
Now the tank is well mixed,
So, c = c3 and, c2 = 0 as it is only water no NaOH