Question

Sectionl Proportional or Integral Control of First Order Process 2) A first-order gas pressure control process with a time constant of 100 seconds is being controlled with a proportional-only controller with a gain of 1.5. At time - 0, the setpoint was instantly changed from 100 to 110 psig. Please determine the expected change of the controlled (pressure) variable after: a) 15 seconds b) 30 seconds c) 2 minutes d) 10 minutes Express your answer in psig. 3) In the pressure process of problem 2, the control is changed from P-only to I-only control. a) Determine the integral control gain which will result in a critically damped system. b) Determine the effective natural system of this controlled process.

Answer 1

2):

Transfer funtion of the process is given by

$G(s) = \frac{1}{\tau s+1}= \frac{1}{100s+1}$

where $\tau$ is the time constant.

Controller transfer function is C(s) = 1.5.

Closed loop transfer funciton is:

$T(s) = \frac{G(s)C(s)}{1+G(s)C(s)} = \frac{1.5}{100s+1+1.5} = \frac{1.5}{100s+2.5}$

For a linear system, the output when the set point is changed from 100 to 110 is the same as steady output for 100 plus the output for a step input of 110-100 = 10.

Output for a step input of 10 is:

$\mathcal{L}^{-1} \left (T(s)*\frac{10}{s} \right )= \mathcal{L}^{-1} \left (\frac{15}{100s^2+2.5s} \right )= \mathcal{L}^{-1} \left ( \frac{6}{s}-\frac{600}{100s+2.5} \right )$

$=6(1-e^{-0.025t})$

Therefore, change in the output is

$6(1-e^{-0.025t})psig$

a:

After 15 seconds, it is

b:

After 30 seconds, it is

c:

After 2 minutes, it is

d:

After 10 minutes, it is

3):

The closed loop transfer function is:

$T(s) = \frac{G(s)C(s)}{1+G(s)C(s)} = \frac{\frac{1}{100s+1}*\frac{k}{s}}{1+\frac{1}{100s+1}*\frac{k}{s}}= \frac{k}{100s^2+s+k} = \frac{0.01k}{s^2+0.01s+0.01k}$

a:

Comparing the denominator with the standard second order equation  $s^2+2\zeta \omega_n s+ \omega_n^2$, we get

$\omega_n = 0.1\sqrt{k}$

$\zeta = \frac{0.01}{2*0.1\sqrt{k}}=\frac{0.05}{\sqrt{k}}$

for a critically damped case,

$\zeta = \frac{0.05}{\sqrt{k}}=1$

=> k = 0.05² = 0.0025.

b:

The effective natural frequency of the system is:

$\omega_n = 0.1\sqrt{k} = 0.1*0.05 = 0.005 rad/s$