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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.

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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:

\[Heat \,\, generated \,\,=-\Delta HkC^n _AV\]

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