Question

A polymeric extruder is turned on and immediately begins producing a product at a rate of 10 kg/min. An operator realizes 20 minutes later that the production rate is too low, and increases the production rate to an immediate 15 kg/min. An hour later, an emergency causes the outlet valve to rapidly adjust to 1 kg/min. One minute later, the emergency is resolved, and the outlet valve is allowed to rapidly readjust to 10 kg/min. Plot production rate m(t) in kg/min against time t in min. Determine the production rate function m(t) in the time domain, and then determine the Laplace transform of m(t).

Answer 1

Answer:

The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.

The production rate function M(t) is:

$m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min$ (1)

The Laplace transform of this function is:

$\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min$    (2)

Explanation:

The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.

For the Laplace transform we use the following rules:

$\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)$    (3)

$\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}$    (4)