#### 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.

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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:

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

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