LearnChemE

#### Multiple Steady States in a Non-Isothermal CSTR: Screencasts

Analyzes a continuous stirred tank reactor in terms of heat generated and heat removed to predict which solutions to the mass and energy balances are stable.

We suggest you list the important points in this screencast as a way to increase retention.

Explain multiple steady states and how they arise for a CSTR operating at steady state.

We suggest you list the important points in this screencast as a way to increase retention.

Explain multiple steady states and how they arise for a CSTR operating at steady state, part 2.

We suggest you list the important points in this screencast as a way to increase retention.

##### Important Equations:

Mass balances on transient CSTR for the irreversible reaction $$A \rightarrow 2B$$ whose rate is nth order in $$C_A$$. The reactor volume does not change.

$V\frac{dC_A}{dt} = F_{A0} – F_A – kC^n _AV$

Accumulation = in – out + rate of generation

$V \frac{dC_B}{dt} = F_{B0} – F_B – 2kC^n _AV$

where $$F_A$$ and $$F_B$$ are the molar flow rates (mol/s) of A and B, respectively, $$V$$ is the volume of the reactor contents (L), $$k$$ is the rate constant, $$C_A$$ is the molar concentration of A (mol/L), $$C_B$$ is the molar concentration of B (mol/L), and $$t$$ is time.

These equations use the relations: $$F_A = \nu C_A$$ and $$F_B = \nu C_B$$, where $$\nu$$ is the volumetric flow rate (L/s). For a liquid-phase system $$\nu$$ can often be assumed constant.

Energy balance for transient CSTR:

$V\rho C_{P,mass} \frac{dT}{dt} = – \Delta HkC^n _AV – \nu \rho C_{P,mass} (T – T_0) – UA(T – T_c)$

where $$\Delta H$$ is the heat of reaction, $$\rho$$ is the density of the feed, $$C_{P,mass}$$ is the mass heat capacity of the feed, $$T_0$$ is the feed temperature, $$U$$ is the heat transfer coefficient, $$A$$ is the heat transfer area, and $$T_c$$ is the cooling temperature.

Initial conditions:

$at \,\, t = 0, \hspace{1cm} N_A = N_{A0}, \hspace{1cm} N_B = N_{B0}$

where $$N_A$$ is the number of moles of A in the reactor at time $$t \hspace{5mm} N_A = C_AV$$, $$N_B$$ is the number of moles of B in the reactor at time $$t \hspace{5mm} N_B = C_BV$$, $$N_{A0}$$ is the number of moles of A in the reactor at $$t$$ = 0, and $$N_{B0}$$ is the number of moles of B in the reactor at $$t$$ = 0.

$X = \frac{F_{A0} – F_A}{F_{A0}}$

where $$X$$ is the fractional conversion.

Percent conversion = 100$$X$$

Residence time in CSTR:

$\tau = \frac{V}{\nu}$

where $$\nu$$ is the outlet volumetric flow rate (approximately the same as inlet volumetric flow rate for liquid-phase system)

Separating steady-state energy balance into heat generated and heat removed:

$Heat \,\, generated \,\,=-\Delta HkC^n _AV$

$Heat \,\, removed \,\, = \nu \rho C_{P,mass}(T – T_0) + UA(T – T_c)$