Question

Determine the settings of KP, TI and TD required for a three-mode controller which gave a process reaction curve with a lag L of 200 s and a gradient R of 0.010%/s when the test signal was a 5% change in the control valve position.

Answer 1

Answer:

$\mathbf{K_p =2.857}$

$\mathbf{\tau_1= 400 \ s}$

$\mathbf{ \tau_d = 100 \ s}$

Explanation:

We can determine each parameter by using the first method of Zeigler Nichols

$K_p = 1.2 \times \dfrac{\tau}{KL}$

$\tau_1= 2L$

$\tau_d = 0.5 L$

For this process;

the initial state of the output = $A_o$

$\tau = constant$

For calculating the gradient, we use the equation:

$y-y_o = G(t-t_o) \\ \\ y-A_o = 0.01 \% A_o (t-L)$

where.

Time constant relates to the time $y = K = 1.05 A_o$

∴

$1.05A_o -A_o = 0.01\% A_o (t-200)$

$0.05 A_o = 0.0001 A_o (t-200)$

$t = 500A_o + 200$

To find time t

$\tau = (500A_o +200) -200$

$\tau = 500 \ A_o$

Recall that:

Using the first method of Zeigler Nichols

$K_p = 1.2 \times \dfrac{\tau}{KL}$

$\tau_1= 2L$

$\tau_d = 0.5 L$

$K_p = 1.2 \times \dfrac{500 \ A_o}{1.05\ A_o \times 200}$

$K_p = 1.2 \times 2.38095$

$\mathbf{K_p =2.857}$

$\tau_1= 2L$

$\tau_1= 2(200)$

$\mathbf{\tau_1= 400 \ s}$

$\tau_d = 0.5 \ L$

$\tau_d = 0.5 \times 200$

$\mathbf{ \tau_d = 100 \ s}$