CONTROL SYSTEM ENG: August 2013

1) A Proportional Controller is used for controlling temperature in melting process. Temperature set point is750°C and temperature measuring tool range is 0 - 1000°C . Proportional space is set at 15%. Pressure output range from controller is 20 –100 kN/m² and pressure output rises when temperature increases. If pressure output is set at 60 kN/m² for temperature set point, find

i) Temperature for pressure output at 20 kN/m²

Pb = 15 % Kp = 100

Pb(0) = 50% 50

P = 20kN/m²

^{Kp = 100}

¹⁵

E = Sp – Mv (100)
Range P(t) = KpEp + P(0)

-7.5= 750 - Mv (100) 0 = 100 Ep + 50
1000 -0 15

Mv = 825 ⁰ C 100Ep = 50
15

Ep = 50 (15)
100
Ep = -7.5

ii) Temperature for pressure output at 100kN/m²

P = 100 kN/m²E= Sp – Mv (100)

                   P = 100%                                          Range
                   Kp = 100                                    7.5 = 750 –Mv   (100)
                             15                                                 1000

P(t) = KpEp + P(0) Mv = (7.5 x 1000) - 750

                    100 = 100 Ep + 50                       Mv = -6750⁰C #
                               15
                  100 Ep = 100- 50

                   15
                         Ep= 50    (15)
                                 100

Ep = 7.5

2) In a process, a Proportional Controller is used to control the liquid level in the boiler. The

level is set at 8 meters and the level range is 1 to 14 meters. Proportional band is at

20%. The output current has a range of 5 – 20mA. Determine:-

i. The output level when current is 5 mA.

Ep = Sp – Mv ( 100 ) P(t) = KpEp + P(0)

range

Ep = 8 – 10 x 100 P = 5(-15.38) + 50

14 – 1

Ep = -15.38 p = -26.9 %

= current – min. x 100

Max - min

-26.9 = Mv – 1 x 100

14 – 1

-26.9(13) = Mv – 1
100

Mv = - 2.5

ii. The output current at a level of 10 meters.

controller output = current – Min out x 100

max out – min out

= 10 – 5 x 100
20 – 5

= 5 x 20%
15

= 7 meters

3) An air to open valve on the inflow controls level in a tank. When the process is at the set point the valve opening is 50%. An increase in outflow results in the valve opening increasing to a new steady state value of 80%. What is the resulting offset if the controller PB is:

i)                    80%

Pb = 100
  proportional gain, Kp

80 = 100
Kp
Kp = 100
  80

ii) 40%.

Pb = 100

Proportional gain, Kp

40 = 100
Kp

Kp = 100
40

4. A temperature controller has the following characteristic curve below:

The controller has a range of 100 ⁰C to 400 ⁰C. The set‐point is 250 ⁰C. Calculate:

i. The controller Proportional Band

Positive error band                                                               negative error band
=        3 ⁰C                x 100                                                     = - 5⁰C
    (400 – 100)⁰C                                                                          (400 – 100)⁰C
= 1.7 %                                                                                         = - 1.7 %

= 1.7 – ( - 1.7 )
= 3.4 %

ii. The controller Proportional Gain

Proportional band, Pb = 100

Proportional gain, Kp

                             Pb = 100
  Kp

3.4 = 100
Kp

Kp = 100
3.4

Kp = 29. 41