1. **Problem Statement:** Find the washout function for both a Continuous Stirred Tank Reactor (CSTR) and a pipe using Laplace transforms. 2. **Background:** The washout function describes the response of a system to an impulse input, often used in chemical reactor analysis to characterize residence time distribution. 3. **CSTR Washout Function:** - The governing equation for concentration $C(t)$ in a CSTR with flow rate $Q$, volume $V$, and input concentration $C_0(t)$ is: $$V\frac{dC(t)}{dt} = Q(C_0(t) - C(t))$$ - Rearranged: $$\frac{dC(t)}{dt} + \frac{Q}{V}C(t) = \frac{Q}{V}C_0(t)$$ - Define $\tau = \frac{V}{Q}$ (mean residence time). 4. **Laplace Transform of CSTR Equation:** - Taking Laplace transform (denote $\mathcal{L}\{f(t)\} = F(s)$): $$sC(s) - C(0) + \frac{1}{\tau}C(s) = \frac{1}{\tau}C_0(s)$$ - Assuming zero initial concentration $C(0) = 0$: $$C(s)\left(s + \frac{1}{\tau}\right) = \frac{1}{\tau}C_0(s)$$ - Transfer function: $$\frac{C(s)}{C_0(s)} = \frac{1/\tau}{s + 1/\tau}$$ 5. **Impulse Response (Washout Function) for CSTR:** - For an impulse input $C_0(t) = \delta(t)$, $C_0(s) = 1$. - Thus: $$C(s) = \frac{1/\tau}{s + 1/\tau}$$ - Taking inverse Laplace transform: $$C(t) = \frac{1}{\tau}e^{-t/\tau}$$ - This is the washout function for the CSTR. 6. **Pipe Washout Function:** - Model the pipe as a plug flow reactor with velocity $u$ and length $L$. - The residence time is $\tau = \frac{L}{u}$. - The washout function is a delayed impulse: $$E(t) = \delta(t - \tau)$$ - Laplace transform: $$E(s) = e^{-s\tau}$$ 7. **Summary:** - CSTR washout function: $$E(t) = \frac{1}{\tau}e^{-t/\tau}$$ - Pipe washout function: $$E(t) = \delta(t - \tau)$$ These functions describe the residence time distribution for each system.