Controller Principle Question

CONTROL PRINCIPLE - EXAMPLE 1

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

            ii)         Temperature for pressure output at  100 kN/m²    

    

       

        a)Pressure at 20KN/m

 P=20KN/m² ,  P= 0%

     P(t)= kpEp + P(0) ,  kp=100/PB

          = 100/15 EP+50       100/15=6.67

      0= 100/15 EP+50

           100/15 EP= -50

                 EP= -7.5

    

     E= SP- MV/Range x 100

    -7.5= 750-MV/1000-0 x 100

    -0.075= 750-mv/1000

     -75= 750- mv

     -825= -mv

     MV= 825°C

  b) Pressure at 100KN/m²

 

 P=100% , kp=100/15=6.67

 P(t)= kp Ep + P(0)

100=100/15 EP + 50

100-50=100/15 EP

7.5=EP

7.5= 750-MV/1000-0 x 100

 MV=6750°C

  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.

      PB=100/Kp
      Kp=100/20
           = 5

P(t)=Kp Ep + P(0)
  0= 5 EP +P(0)
  0= 5 EP  + 50
 0-50= 5 EP
  EP= -50/5
  EP= -10% 



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

       
P(t)=10m current=?
E=Sp-mv/range x 100%
   =8-10/14-1 x 100
    = -15.4
P(t)= Kp Ep + P(0)
      =100/20(-15.4) +50
     = -27
P(%)=Pk-Pmin/Pmax-Pmin x 100
        = -27 x 15/100 +5
        = 1

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
                 =1.25
         
             E=Po/Kp
            E=30/1.25
               = 24/100
               = 0.24

ii)                   40%.

              
          PB=100/proportional gain,Kp
            80=100/Kp
             Kp=100/40 
                  =2.5
             E=Po/Kp
                = 30/2.5
                =12/100
                =0.12                                                                                                        

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

                            
                                              PB=100/Kp
                                            
                                             PB=input(E)/output x 100%
                                                   =   3-(-3)/100-0 x 100%
                                                    = 6 
                                              Kp= 100/6
                                                   = 16.67
                                              
               
                                    

                                              

                                                                                        

                              ii.            The controller Proportional Gain