1. The problem is to find the washout function for both a Continuous Stirred Tank Reactor (CSTR) and a pipe using Laplace transforms. 2. The washout function describes the system's response to an input, often modeled by differential equations representing concentration changes over time. 3. For a CSTR, the governing equation is $$\tau \frac{dC(t)}{dt} + C(t) = C_{in}(t)$$ where $\tau$ is the residence time, $C(t)$ is the concentration inside the reactor, and $C_{in}(t)$ is the input concentration. 4. Taking the Laplace transform of both sides, assuming zero initial conditions, we get $$\tau s C(s) + C(s) = C_{in}(s)$$ 5. Factor out $C(s)$: $$C(s)(\tau s + 1) = C_{in}(s)$$ 6. The transfer function (washout function) for the CSTR is $$G_{CSTR}(s) = \frac{C(s)}{C_{in}(s)} = \frac{1}{\tau s + 1}$$ 7. For a pipe, assuming plug flow with no dispersion, the concentration at the outlet is a delayed version of the input: $$C(t) = C_{in}(t - \tau)$$ where $\tau$ is the transit time. 8. Taking the Laplace transform, $$C(s) = e^{-\tau s} C_{in}(s)$$ 9. Thus, the washout function for the pipe is $$G_{pipe}(s) = \frac{C(s)}{C_{in}(s)} = e^{-\tau s}$$ 10. Summary: - CSTR washout function: $$G_{CSTR}(s) = \frac{1}{\tau s + 1}$$ - Pipe washout function: $$G_{pipe}(s) = e^{-\tau s}$$ These functions describe how the output concentration relates to the input concentration in the Laplace domain, capturing the dynamics of washout in each system.