Assignment 7

This is the diagram of a jacketed CSTR in which a first order reaction is taking place. The reaction is

exothermic and hence, it is cooled using cooling water in its jacket to maintain its temperature at Ty.

Assume:-

1) No energy balance for jacket.

2) Inside is perfectly mixed.

3) Constant volume and constant parameters.

Use the values as given:-

CAf = 10 kgmol/m3

UA/V = 150 Kcal/m3 oC hr

Tf = 25 oC

Ty = 25 oC

ρCP = 500 Kcal/m3 oC

F/V = 1 hr -1

Ko = 9700 x 3600 hr -1

(-ΔH) = 5960 Kcal / Kgmol

Ea = 11843 Kcal/Kgmol

R = 1.987 Kcal K−1 Kmol−1

Find:

(1) Steady State solution for above values?

(2) Dynamic behavior for 3 initial conditions:

[CA , T]i1 = [9 , 300]

[CA , T]i2 = [5 , 350]

[CA , T]i3 = [1 , 450]

(3) Perform stability analysis for 3 steady state temperatures you arrive based on above 3 initial

conditions.

CA, T, V

F, CAf, Tf

Fy, Ty

Fy, Tyin

F, CA, T

T

Ty

Solution:

Overall material balance

𝑑(𝑉𝜌)

𝑑𝑡= 𝐹𝑖𝑛𝜌 − 𝐹𝜌

i.e., 𝐹𝑖𝑛 = 𝐹

Material balance of Component A

𝑉𝑑𝐶𝐴

𝑑𝑡= 𝐹𝐶𝐴𝑓 − 𝐹𝐶𝐴 − 𝑟𝑉

𝑑𝐶𝐴

𝑑𝑡= (𝐹/𝑉)(𝐶𝐴𝑓 − 𝐶𝐴) − 𝑟

Energy Balance on Reactor

𝑉𝜌𝐶𝑝𝑑𝑇

𝑑𝑡= 𝐹𝜌𝐶𝑝𝑇𝑓 − 𝐹𝜌𝐶𝑝𝑇 + (−∆𝐻)𝑟𝑉 − 𝑈𝐴(𝑇 − 𝑇𝑦)

𝑑𝑇

𝑑𝑡= (

𝐹

𝑉) (𝑇𝑓 − 𝑇) + (−

∆𝐻

𝜌𝐶𝑝) 𝑟 − (𝑈𝐴/𝑉𝜌𝐶𝑝)(𝑇 − 𝑇𝑦)

Steady State Solution:

The steady state solution is obtained when 𝑑𝐶𝐴

𝑑𝑡= 0 and

𝑑𝑇

𝑑𝑡= 0

The final steady state values depend upon initial values of Ca, T in the reactor.

Dynamic Behaviour:

Case 1: ( CA = 9 and T =300)

Final Steady state Ts : 311.16 K

Final Steady state CAs : 8.56 Kmol/m3

Case 2: ( CA = 5 and T =350)

Final Steady state Ts : 368.04 K

Final Steady state CAs : 2.36 Kmol/m3

Case 3: ( CA = 1 and T =450)

Final Steady state Ts : 368.05 K Final Steady state CAs : 2.36 Kmol/m3

Stability Analysis:

The Eigenvalues of the A matrix determine the stability of the steady state operating points.

- Case 1: λ1 = -0.8953 and λ2 = -0.5197 As both values are negative, steady state point is Stable (node)

- Case 2: λ1 = -0.7653+0.9568i and λ2 = -0.7653-0.9568i As real part of lambda is negative , the steady state point is stable (focus)

- Case 3: λ1 = -0.7652+0.9507i and λ2 = -0.7652-0.9507i As real part of lambda is negative , the steady state point is stable (focus)

Eigen value method involves linearization of system around a point. But here the system assumes non-linear behaviour due to the exponential term exp(-E/RT). Therefore Eigen value method is not much effective.

Alternate method for stability analysis:

The steady-state concentration solution (dCA/dt) = 0) for concentration is

From this equation we get:

The steady-state temperature solution (dT/dt = 0) is

The terms in above equation can be related to the energy removed and generated.

In the above expression,

and Qgen can be written in the form of:

A steady-state solution exists when there is an intersection of the Qrem and Qgen curves. Also note that

from the above eqn, Heat Removal Qrem is linear curve(Straight line).

A slight perturbation at point 1 brings back to the original steady state and so is stable. But at point 2 it is

unstable.

Therefore the stability criterions for operating points of CSTRs is that the slope of the straight line of heat

removal must be larger than the slope of the heat generation curve ie.

𝑑𝑄𝑔𝑒𝑛

𝑑𝑇>

𝑑𝑄𝑟𝑒𝑚

𝑑𝑇

Algorithm:

We defined Qgen as a function of Ts (Reactor temperature) and determined 𝑑𝑄𝑔𝑒𝑛

𝑑𝑇 and

𝑑𝑄𝑟𝑒𝑚

𝑑𝑇 at the

precalcutaed steady state temparature value.

i. If the above inequality condition satisfies then the steady state is stable and the condition does not

satisfy the steady state is unstable.

Using this method we got same stability analysis results as the previous one.