Summaries

Reactor Design Equations Summary Table

Reactor Type	First-Order Reaction ($-r_A = kC_A$)	Second-Order Reaction ($-r_A = kC_A^2$)
Batch Reactor	Concentration: $C_A = C_{A0} e^{-kt}$ Conversion: $t = \frac{1}{k}\ln\left(\frac{1}{1-X}\right)$	Concentration: $\frac{1}{C_A} - \frac{1}{C_{A0}} = kt$ Conversion: $t = \frac{X}{k C_{A0}(1-X)}$
Plug Flow Reactor (PFR)	Concentration: $C_A = C_{A0} e^{-k\tau}$ Conversion: $\tau = \frac{1}{k}\ln\left(\frac{1}{1-X}\right)$	Concentration: $\tau = \frac{1}{k}\left(\frac{1}{C_A} - \frac{1}{C_{A0}}\right)$ Conversion: $\tau = \frac{X}{k C_{A0}(1-X)}$
Single CSTR (Microfluid)	Concentration: $C_A = \frac{C_{A0}}{1 + k\tau}$ Conversion: $\tau = \frac{X}{k(1-X)}$	Concentration: $\tau = \frac{C_{A0} - C_A}{k C_A^2}$ Conversion: $\tau = \frac{X}{k C_{A0}(1-X)^2}$
Multiple Equal CSTRs in Series N tanks, $\tau_i = \frac{\tau_{\text{total}}}{N}$	Concentration: $C_N = \frac{C_{A0}}{(1 + k\tau_i)^N}$ Conversion: $X_N = 1 - \frac{1}{(1 + k\tau_i)^N}$	Concentration (Recursive): $C_i = \frac{\sqrt{1 + 4 k \tau_i C_{i-1}} - 1}{2 k \tau_i}$ Conversion (per stage $i$): $\tau_i = \frac{X_i - X_{i-1}}{k C_{A0} (1 - X_i)^2}$
Single CSTR (Macrofluid)	Concentration: $C_A = \frac{C_{A0}}{1 + k\tau}$ (Identical to microfluid) Conversion: $X = \frac{k\tau}{1 + k\tau}$ (Identical to microfluid)	Concentration: $C_A = \int_0^\infty \left(\frac{C_{A0}}{1 + k C_{A0} t}\right) \frac{e^{-t/\tau}}{\tau} dt$ Conversion: $X = 1 - \int_0^\infty \left(\frac{1}{1 + k C_{A0} t}\right) \frac{e^{-t/\tau}}{\tau} dt$

Reactor RTD and Moments Summary Table

Reactor Configuration	Residence Time Distribution, $E(t)$	First Moment (Mean, $\mu_1 = \tau$) & Second Moment (Variance, $\sigma^2 = \mu_2$)
Batch Reactor	Not strictly applicable (Closed system with no continuous flow)	Not applicable (All molecules remain inside for the exact batch time $t$)
Plug Flow Reactor (PFR)	$E(t) = \delta(t - \tau)$	Mean: $\mu_1 = \tau$ Variance: $\sigma^2 = 0$
Single Ideal CSTR	$E(t) = \frac{1}{\tau} e^{-t/\tau}$	Mean: $\mu_1 = \tau$ Variance: $\sigma^2 = \tau^2$
$N$ Identical CSTRs in Series	$E(t) = \frac{t^{N-1}}{(N-1)! \, \tau_i^N} e^{-t/\tau_i}$ where $\tau_i = \tau/N$	Mean: $\mu_1 = N\tau_i = \tau$ Variance: $\sigma^2 = N\tau_i^2 = \frac{\tau^2}{N}$
2 Non-Identical CSTRs in Series	$E(t) = \frac{1}{\tau_1 - \tau_2} \left( e^{-t/\tau_1} - e^{-t/\tau_2} \right)$ assuming $\tau_1 \neq \tau_2$	Mean: $\mu_1 = \tau_1 + \tau_2 = \tau_{\text{total}}$ Variance: $\sigma^2 = \tau_1^2 + \tau_2^2$
2 Non-Identical CSTRs in Parallel	$E(t) = \left(\frac{v_1}{v_0}\right) \frac{1}{\tau_1} e^{-t/\tau_1} + \left(\frac{v_2}{v_0}\right) \frac{1}{\tau_2} e^{-t/\tau_2}$ where $\tau_1 = \frac{V_1}{v_1}$, $\tau_2 = \frac{V_2}{v_2}$, $v_0 = v_1 + v_2$	Mean: $\mu_1 = \left(\frac{v_1}{v_0}\right)\tau_1 + \left(\frac{v_2}{v_0}\right)\tau_2 = \frac{V_{\text{total}}}{v_0}$ Variance: $\sigma^2 = 2\left[\frac{v_1}{v_0}\tau_1^2 + \frac{v_2}{v_0}\tau_2^2\right] - \mu_1^2$

The Gamma Integral Identity

When working with Residence Time Distributions (RTD), the raw moments are calculated using integrals of the form $\int_0^\infty t^n e^{-at} dt$. To completely avoid tedious integration by parts, we use the standard Gamma function identity for any integer $n \geq 0$ and constant $a > 0$:

\[\int_0^\infty t^n e^{-at} dt = \frac{n!}{a^{n+1}}\]

Specific Cases for Exponents $n = 0, 1, 2$:

For $n = 0$:
$\int_0^\infty e^{-at} dt = \frac{0!}{a^1} = \frac{1}{a}$
For $n = 1$:
$\int_0^\infty t e^{-at} dt = \frac{1!}{a^2} = \frac{1}{a^2}$
For $n = 2$:
$\int_0^\infty t^2 e^{-at} dt = \frac{2!}{a^3} = \frac{2}{a^3}$

Application 1: $N$ Identical CSTRs in Series

The RTD for $N$ identical CSTRs, each with a space time of $\tau_i$, is given by: $E(t) = \frac{t^{N-1}}{(N-1)! \, \tau_i^N} e^{-t/\tau_i}$

The raw second moment $\mu_2’$ is defined as: $\mu_2' = \int_0^\infty t^2 E(t) dt = \int_0^\infty t^2 \left[ \frac{t^{N-1}}{(N-1)! \, \tau_i^N} e^{-t/\tau_i} \right] dt$

Step-by-Step Resolution using the Identity:

Pull the constant terms out of the integral: $\mu_2' = \frac{1}{(N-1)! \, \tau_i^N} \int_0^\infty t^{N+1} e^{-t/\tau_i} dt$
Identify the parameters for our Gamma identity: $n = N+1$ and $a = \frac{1}{\tau_i}$. Apply the identity: $\int_0^\infty t^{N+1} e^{-t/\tau_i} dt = \frac{(N+1)!}{\left(\frac{1}{\tau_i}\right)^{N+2}} = (N+1)! \, \tau_i^{N+2}$
Substitute this back into the expression for $\mu_2’$ and simplify the factorials: $\mu_2' = \frac{(N+1)! \, \tau_i^{N+2}}{(N-1)! \, \tau_i^N} = \frac{(N+1)(N)(N-1)!}{(N-1)!} \tau_i^2$ $\mu_2' = N(N+1)\tau_i^2$

Note: Since $\mu_1 = N\tau_i$, this easily yields the variance: $\sigma^2 = \mu_2’ - \mu_1^2 = N(N+1)\tau_i^2 - (N\tau_i)^2 = N\tau_i^2$.

Application 2: 2 Unequal CSTRs in Series

The RTD for two non-identical CSTRs in series ($\tau_1 \neq \tau_2$) is given by: $E(t) = \frac{1}{\tau_1 - \tau_2} \left( e^{-t/\tau_1} - e^{-t/\tau_2} \right)$

The raw second moment $\mu_2’$ is defined as: $\mu_2' = \int_0^\infty t^2 E(t) dt = \frac{1}{\tau_1 - \tau_2} \left[ \int_0^\infty t^2 e^{-t/\tau_1} dt - \int_0^\infty t^2 e^{-t/\tau_2} dt \right]$

Step-by-Step Resolution using the Identity:

Apply the specific case for $n = 2$ to both terms individually:
- For the first term ($a = \frac{1}{\tau_1}$): $\int_0^\infty t^2 e^{-t/\tau_1} dt = \frac{2}{\left(\frac{1}{\tau_1}\right)^3} = 2\tau_1^3$
- For the second term ($a = \frac{1}{\tau_2}$): $\int_0^\infty t^2 e^{-t/\tau_2} dt = \frac{2}{\left(\frac{1}{\tau_2}\right)^3} = 2\tau_2^3$
Substitute these solved terms back into the bracket: $\mu_2' = \frac{1}{\tau_1 - \tau_2} \left( 2\tau_1^3 - 2\tau_2^3 \right) = \frac{2}{\tau_1 - \tau_2} \left( \tau_1^3 - \tau_2^3 \right)$
Use the algebraic identity for the difference of cubes $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ to simplify: $\mu_2' = \frac{2(\tau_1 - \tau_2)(\tau_1^2 + \tau_1\tau_2 + \tau_2^2)}{\tau_1 - \tau_2}$ $\mu_2' = 2(\tau_1^2 + \tau_1\tau_2 + \tau_2^2)$

Note: Since $\mu_1 = \tau_1 + \tau_2$, this simplifies straight to the variance: $\sigma^2 = \mu_2’ - \mu_1^2 = 2\tau_1^2 + 2\tau_1\tau_2 + 2\tau_2^2 - (\tau_1^2 + 2\tau_1\tau_2 + \tau_2^2) = \tau_1^2 + \tau_2^2$.

Tank Configuration	Mean Residence Time ($\tau_{\text{sys}}$)	Second Central Moment / Variance ($\sigma^2$)
Two Equal Sized Tanks in Series ($\tau_1 = \tau_2$)	$2\tau_1$	$\sigma^2 = 2\tau_1^2$
Two Unequal Sized Tanks in Series ($\tau_1 \neq \tau_2$)	$\tau_1 + \tau_2$	$\sigma^2 = \tau_1^2 + \tau_2^2$
Two Equal Sized Tanks in Parallel ($V_1 = V_2$, but $f_1 \neq f_2 \implies \tau_1 \neq \tau_2$)	$f_1\tau_1 + f_2\tau_2$	$\sigma^2 = f_1(2\tau_1^2) + f_2(2\tau_2^2) - (f_1\tau_1 + f_2\tau_2)^2$
Two Unequal Sized Tanks in Parallel ($V_1 \neq V_2$, split fractions $f_1, f_2$)	$f_1\tau_1 + f_2\tau_2$	$\sigma^2 = f_1(2\tau_1^2) + f_2(2\tau_2^2) - (f_1\tau_1 + f_2\tau_2)^2$

Reactor Type	First-Order Reaction ($-r_A = kC_A$)	Second-Order Reaction ($-r_A = kC_A^2$)
Batch Reactor	Concentration: \(C_A = C_{A0} e^{-kt}\) Conversion: \(t = \frac{1}{k}\ln\left(\frac{1}{1-X}\right)\)	Concentration: \(\frac{1}{C_A} - \frac{1}{C_{A0}} = kt\) Conversion: \(t = \frac{X}{k C_{A0}(1-X)}\)
Plug Flow Reactor (PFR)	Concentration: \(C_A = C_{A0} e^{-k\tau}\) Conversion: \(\tau = \frac{1}{k}\ln\left(\frac{1}{1-X}\right)\)	Concentration: \(\tau = \frac{1}{k}\left(\frac{1}{C_A} - \frac{1}{C_{A0}}\right)\) Conversion: \(\tau = \frac{X}{k C_{A0}(1-X)}\)
Single CSTR (Microfluid)	Concentration: \(C_A = \frac{C_{A0}}{1 + k\tau}\) Conversion: \(\tau = \frac{X}{k(1-X)}\)	Concentration: \(\tau = \frac{C_{A0} - C_A}{k C_A^2}\) Conversion: \(\tau = \frac{X}{k C_{A0}(1-X)^2}\)
Multiple Equal CSTRs in Series N tanks, $\tau_i = \frac{\tau_{\text{total}}}{N}$	Concentration: \(C_N = \frac{C_{A0}}{(1 + k\tau_i)^N}\) Conversion: \(X_N = 1 - \frac{1}{(1 + k\tau_i)^N}\)	Concentration (Recursive): \(C_i = \frac{\sqrt{1 + 4 k \tau_i C_{i-1}} - 1}{2 k \tau_i}\) Conversion (per stage $i$): \(\tau_i = \frac{X_i - X_{i-1}}{k C_{A0} (1 - X_i)^2}\)
Single CSTR (Macrofluid)	Concentration: \(C_A = \frac{C_{A0}}{1 + k\tau}\) (Identical to microfluid) Conversion: \(X = \frac{k\tau}{1 + k\tau}\) (Identical to microfluid)	Concentration: \(C_A = \int_0^\infty \left(\frac{C_{A0}}{1 + k C_{A0} t}\right) \frac{e^{-t/\tau}}{\tau} dt\) Conversion: \(X = 1 - \int_0^\infty \left(\frac{1}{1 + k C_{A0} t}\right) \frac{e^{-t/\tau}}{\tau} dt\)