What is the gain and time constant in the process transfer function? If at another time the engineer changes the power input linearly at a rate of 0.5 kW/min, what can you say about the maximum rate of change of process temperature: When will it occur? Show your calculations in a table form.

After 30 minutes, the temperature stabilizes at $T_{ss} = 500 \text{ °C}$.
The steady-state deviation in temperature is $T'_{ss} = T_{ss} - T_0 = 500 \text{ °C} - 100 \text{ °C} = 400 \text{ °C}$.
For a first-order system, the steady-state gain is given by:

$$K = \frac{\text{Change in Output}}{\text{Change in Input}} = \frac{T'_{ss}}{\Delta P}$$ $$K = \frac{400 \text{ °C}}{1 \text{ kW}} = \boxed{400 \text{ °C/kW}}$$

For a first-order system subjected to a step change, the output response in deviation variables is:

$$T'(t) = K \Delta P (1 - e^{-t/\tau})$$

At $t = 4 \text{ minutes}$, the temperature rises to $T(4) = 400 \text{ °C}$.
The deviation temperature at $t=4$ min is $T'(4) = T(4) - T_0 = 400 \text{ °C} - 100 \text{ °C} = 300 \text{ °C}$.
Substitute the known values into the equation:

$$300 \text{ °C} = (400 \text{ °C/kW}) (1 \text{ kW}) (1 - e^{-4/\tau})$$ $$300 = 400 (1 - e^{-4/\tau})$$ $$0.75 = 1 - e^{-4/\tau}$$ $$e^{-4/\tau} = 1 - 0.75$$ $$e^{-4/\tau} = 0.25$$

Take the natural logarithm of both sides:

$$\ln(e^{-4/\tau}) = \ln(0.25)$$ $$-4/\tau = -1.38629$$ $$\tau = \frac{-4}{-1.38629} = \boxed{2.885 \text{ min}}$$

2.2 If at another time the engineer changes the power input linearly at a rate of 0.5 kW/min, what can you say about the maximum rate of change of process temperature: When will it occur? Show your calculations in a table form.

The power input changes linearly, which is a ramp input. Let the rate of change be $R = 0.5 \text{ kW/min}$.
In deviation variables, the input is $P'(t) = R t$. In the Laplace domain, $P'(s) = \frac{R}{s^2}$.

The inverse Laplace transform of $T'(s)$ for a ramp input is:

$$T'(t) = KR \left( t - \tau + \tau e^{-t/\tau} \right)$$

To find the rate of change of process temperature, we differentiate $T'(t)$ with respect to time:

$$\frac{dT'(t)}{dt} = \frac{d}{dt} \left[ KR \left( t - \tau + \tau e^{-t/\tau} \right) \right]$$ $$\frac{dT'(t)}{dt} = KR \left( 1 - 0 + \tau \left( -\frac{1}{\tau} \right) e^{-t/\tau} \right)$$ $$\frac{dT'(t)}{dt} = KR (1 - e^{-t/\tau})$$

The term $(1 - e^{-t/\tau})$ starts at 0 when $t=0$ and increases, asymptotically approaching 1 as $t \rightarrow \infty$.
Therefore, the maximum value of $\frac{dT'(t)}{dt}$ occurs as $t \rightarrow \infty$.
The maximum rate of change is $KR(1) = KR$.
Using the values $K = 400 \text{ °C/kW}$ and $R = 0.5 \text{ kW/min}$:

$$\text{Maximum rate of change} = (400 \text{ °C/kW}) \times (0.5 \text{ kW/min}) = \boxed{200 \text{ °C/min}}$$

This maximum rate of change occurs as $\boxed{t \rightarrow \infty}$.

We calculate the rate of change $\frac{dT'(t)}{dt} = 200 (1 - e^{-t/2.885})$ for various time intervals.

t (min)	$e^{-t/\tau}$	$1 - e^{-t/\tau}$	$\frac{dT'(t)}{dt}$ (°C/min)
0	1.000	0.000	0.0
1	0.707	0.293	58.6
2	0.500	0.500	100.0
3	0.353	0.647	129.4
4	0.250	0.750	150.0
5	0.177	0.823	164.6
10	0.031	0.969	193.8
15	0.005	0.995	199.0
20	0.001	0.999	199.8
$\infty$	0.000	1.000	200.0

The table clearly shows that the rate of change of process temperature asymptotically approaches $200 \text{ °C/min}$ as time increases.