Your Federal Quarterly Tax Payments are due April 15th Get Help Now >>

PSE blade by mikesanye

VIEWS: 5 PAGES: 10

									                        BSE 422 Midterm, 4 November, 2010
                 Take Home – Due Tuesday, 9 November; 4:00 PM PST

This midterm is to be worked on by yourself with no assistance from anyone.
Failure to do so will be met with severe consequences.

This exam looks at lime mud filters used in the recovery area of a pulp mill. The model
presented here comes from "The Modeling Of Rotary Lime Mud Filters", by Kent R. Davey,
George Vachtsevanos, and Jim C. Cheng, Tappi Journal, Vol. 72, No.8, pp 150-156, August
1989.

Lime mud consists of CaCO3 and residual alkali from the causticizing process. The lime kiln
converts CaCO3 to CaO to be reused in the causticizing reaction to regenerate the kraft cooking
liquor. Before the lime mud can be ran through the lime kiln two things need to be done. The
lime mud is fed to the filter as slurry at an average solids content of 35%. The solids content
must be increased to around 70% prior to the kiln (30% moisture). The alkali concentration in the
mud also needs to be controlled. Too high an alkali concentration causes the drying lime to stick
together and form large balls. Too low and it won’t stick together at all and high dusting losses
occur. The job of the lime mud filter is to increase the solids and regulate the alkali concentration
of the lime mud going to the lime kiln. A diagram of a drum filter is shown in Figure 1.

Lime mud enters the vat and builds up on the vacuum drum. Wash water is applied to the
lime mud cake to reduce the alkali concentration. This also has the effect of increasing
the moisture content of the lime mud. A higher drum speed will give a lower moisture
content and lower alkali concentration. For the problems presented here the production
rate will be treated as a disturbance.


                        Lime Mud
                          Cake
                                                                      Wash
                                                                     Shower
                Scraper
                 Blade



                                             Vacuum
                                              Drum
                                                                     Vat
                                              Slurry
               Mud
             Conveyor


Figure 1. Drum mud filter.



                                                                                                   1
Table I lists the nominal operating conditions for the mud filter.

Table I. Nominal operating conditions for lime mud filter.
Mud moisture content (%)              30
Alkali concentration (%)              1.5
Blade position (cm)                   3.5
Drum vacuum (relative)                  1
Wash shower flow (L/s)                  3
Drum speed (rpm)                        5
Production rate (m3/s)                  1

Figures 2 and 3 represent block diagrams of the lime mud process.

                               W              P


                                9            36


     S           -8                                          M


Figure 2. Lime mud filter process model; drum speed (S), wash flow (W), production (P), moisture
content (M).


                                S             P


                              -0.2            2

    W         -0.125                                         A


Figure 3. Lime mud filter process model; wash flow (W), drum speed (S), production (P), alkali
concentration (A).




                                                                                                   2
Problem 1

The blade position (B) and the drum vacuum (V) affect the lime mud moisture content
(M) and alkali concentration (A). The data in table I show operational data for the mud
filter for these two variables. Note that the first row is the nominal operating point.


                            Table 1. Lime mud filter operational data.
          Blade Position, cm (B)   Vacuum, rel (V)   Moisture, % (M)     Alkali, % (A)
                   3.5                   1                 30                 1.5
                   3.0                   1                 26                 1.4
                   3.5                  1.25               27                 1.3



Derive a linear incremental model where the blade position (B) is the manipulated
variable, drum vacuum (V) is a disturbance, and the mud moisture (M) is the output.
Draw a block diagram for the model.

M / B  M / B  M / B  (26  30 ) /(3.0  3.5)  M / B  8 %/cm
M / V  M / V  M / V  (27  30 ) /(1.25  1)  M / V  12 %/rel
M  8 B  12V




                               V


                             -12

  B             8                        M




                                                                                          3
Problem 2

Design a feed forward control strategy (using the block diagrams and process data from
page 2) for the alkali concentration (A) that would result in perfect control. Use the wash
water (W) as the manipulated variable. The drum speed (S) and production rate (P) are
the disturbance variables. The solution should include a block diagram of the process
with the control strategy. Show the numerical values of the feed forward controller gains
in the block diagram. Calculate the change in alkali concentration (A) and the wash
water (W) when the alkali concentration setpoint is increased to 1.7% and the drum speed
is increased by 0.5 rpm.

The block diagram for this system is given by
                                                                                   P

                                                                      S


                            KFP           KFS                        -0.2          2

                                                W
 ASP         KSP                                       -0.125                                 A

The closed loop transfer function for the alkali concentration is given by

A  0.125 K SP ASP  (0.2  0.125 K FS ) S  (2  0.125 K FP ) P

For perfect control the closed loop gain for the alkali set point needs be one and the
closed loop gain for the production rate and drum speed should be zero.

G SP  0.125K SP  1  K SP  1 /(0.125)  K SP  8 %/%
G S  (0.2  0.125K FS )  0  K FS  0.2 / 0.125  K FS  1.6 L/s  rpm

G P  (2  0.125K FP )  0  K FP  2 / 0.125  K FP  16 L/m 3

The change in alkali concentration for a setpoint change of 0.2% (1.7%-1.5%) and
drum speed of 0.5 rpm will be 0.2% by design. The wash flow can be calculated as

W  8 ASP  1.6 S  W  8(0.2)  1.6(0.5)  W  2.4 L/s

We can use the open loop process gains to double check to make sure we achieved
the desired change of 0.2% for the alkali concentration.

A  0.125W  0.2 S  A  0.125 (2.4)  0.2(0.5)  A  0.3  0.1  A  0.2 %


                                                                                          4
The block diagram with the proper values for the gains is shown below.

                                                                         P

                                                               S


                         16          -1.6                     -0.2       2

                                                         W
ASP          -8                                 -0.125                           A




                                                                             5
Problem 3

Design a feedback control strategy (using the block diagrams and process data from page
2) for the mud moisture content (M) using the drum speed (S) as the manipulated
variable. The wash water (W) and production rate (P) are the disturbance variables. Use
a loop gain of 4 for the controller. The solution should include a block diagram of the
process with the controller. Calculate the change in mud moisture content (M) and the
drum speed (S) when the mud moisture content setpoint is reduced to 25%.

The block diagram for this system is given by
                                                     W         P


                                                     9         36

                               S
 MSP                  KC               -8                              M
              -
The closed loop transfer function for mud moisture is

         8K C                   9                36
M                  MSP                 W                P
     1  ( 8 K C )       1  ( 8 K C )    1  ( 8 K C )

where

loop gain  8 K C  K C  4 / 8  K C  0.5 rpm/%

so

             9    36
M  0.8 MSP  W     P
             5     5

For a -5% mud moisture setpoint change (25%-30%)

M  0.8(5)  M  4 %
S  K C ( MSP  M )  S  0.5(5  (4))  S  0.5 rpm

As was expected with a loop gain of 4 the process value came within 80% of the
setpoint.




                                                                                      6
Problem 4

a) The mud moisture needs to remain in the range of ± 5%, but the wash flow varies by ±
2 L/s. For the feedback only control strategy in Problem 3, determine the minimum
controller gain required to keep the mud moisture content in the specified range. What is
the loop gain for this situation?

With the wash flow varying by ± 2 L/s the mud moisture will vary by ± 18 %. We
want the mud moisture to only vary by ± 5 %.

        9            9
M          W M        (2)  5
     1  LG       1  LG

1  LG  18 / 5  LG  2.6
LG  8 K C  K C  2.6 / 8  K C  0.33 rpm/%

Another way to look at this is we need to reduce the variation due to the disturbance
to approximately 27% (5/18) of its open loop gain. This gives

       1
R          R(1  LG)  1  LG  (1  R) / R
    1  LG
R  reduction  5 / 18
LG  (1  0.278) / 0.278  LG  2.6

Either way you come up with the same answer for the loop gain.

The minimum controller gain to keep the mud moisture in the range of ± 5% with
wash water variations of ± 2 L/s is -0.33. This gives a loop gain of 2.6.




                                                                                        7
                      BSE 422 Midterm, 4 November, 2010
               Take Home – Due Tuesday, 9 November; 4:00 PM PST

This midterm is to be worked on by yourself with no assistance from anyone.
Failure to do so will be met with severe consequences.

Take Home Problem 1 – (50 points)


The following block diagram is a decoupled control system that uses the drum speed to
control the mud moisture content and the wash flow to control the alkali concentration.
The decouplers remove the influence of the drum speed controller output (SCO) on the
alkali concentration (A) and the wash flow controller output (WCO) on the mud moisture
content (M). The control engineer retuned the control system during a shutdown on a
Friday afternoon right before he was leaving on vacation for a remote island in the
Federated States of Micronesia. After the start up the system is very unstable and the
operators have to run the process in manual control. You get called in at midnight that
night to fix the problem. There appears to be some problems with the controller gains.
What changes would you make, if any, to the feedback controller gains (KCM and KCA)
and the decoupler gains (KDM and KDA) to make the process stable? Note that the 4 gains
on the right are the process gains and SCO and WCO are the feedback controller outputs for
the drum speed and wash flow, respectively.


           -                 SCO                       S
MSP                 -5.7                                         -7                          M
                    KCM                 KDA

                                       -2.14                     7


                                        -10                     -0.3
                    KCA                 KDM
ASP                 28.6                                       -0.14                         A
                             WCO                       W
           -
                 Feedback           Decouplers                Process

The first thing to check is the decouplers. The values for these are only based on the
process gains. First let’s look at the decoupler that removes the effect of drum speed
controller output on alkali concentration. The influence coefficient between the
alkali concentration and the drum speed controller output is given by

A  (0.14 K DA  0.3) S CO where   K DA  2.14 L/s  rpm

                                                                                        8
This influence coefficient should be zero. Let’s check this by calculating our own
value for KDA

 0.14K DA  0.3  0  K DA  0.3 / 0.14  K DA  2.14 L/s  rpm

This decoupler gain is good. Now we look at the decoupler that removes the effect
of wash flow controller output on mud moisture. The influence coefficient between
the mud moisture and the wash flow controller output is given by

M  (7 K DM  7)WCO where        K DM  10 s  rpm/L

Again the influence coefficient should be zero.

 7 K DM  7  0  K DM  7 / 7  K DM  1 s  rpm/L

Not only is the magnitude wrong it has the wrong sign as well. It made the
influence worse by going in the wrong direction and too much.

Dealing with the feedback controller gains is a little trickier since there are two
paths between the controller output and the controlled variable. You can see from
the block diagram the engineer ignored this. It looked like he was trying to tune the
loops for a loop gain of 4 (admittedly I did that making up the problem without
thinking about it). Right off you should at least see the sign of the feedback
controller gain on the alkali loop is wrong. It should be negative.

 If we write the closed loop transfer function for the mud moisture loop we get

        7 K CM K DA  7 K CM                 K CM (7 K DA  7)
M                               MSP  M                        MSP
     1  (7 K CM K DA  7 K CM )           1  K CM (7 K DA  7)

The decoupler eliminates any interaction of the wash flow on the mud moisture so
there are no terms from the alkali loop (i.e., ASP). First lets look at the current loop
gain.

LGM  K CM (7 K DA  7)  LGM  5.7(7(2.14)  7)
LGM  5.7(21.98)  LGM  125.3

We can see this is way too large. For a feedback controller a loop gain of 4 should
be reasonable so

LG M  K CM (7 K DA  7)  4  K CM  4 /(7 K DA  7)
K CM  4 /(7(2.14 )  7)  K CM  4 /(21 .98 )
K CM  0.18 %/%

                                                                                       9
Similarly for the alkali loop we write

         0.3 K CA K DM  0.14 K CA               K CA (0.3 K DM  0.14 )
A                                     ASP  A                              ASP
     1  (0.3 K CA K DM  0.14 K CA )           1  K CA (0.3 K DM  0.14 )

Calculating the loop gain for this loop (using the proper value for the decoupler gain
and the proper sign for KCA) gives

LG A  28.6(0.3(1)  0.14)  LG A  28.6(0.44)
LG A  12.6

Even with the correction on the decoupler (LGA=81.8 without correction) the loop
gain is still too high. Again specifying a loop gain of 4 and using the value we
calculated for KDM gives

LG  K CA (0.3 K DM  0.14 )  4  K CA  4 /(0.3 K DM  0.14 )
K CA  4 /(( 0.3)(1)  0.14 )  K CA  4 /(0.44 )
K CA  9.1 %/%

This confirms the negative sign for the feedback controller gain. This makes sense
physically. When the alkali concentration is above the setpoint the error will be
negative. With a negative controller gain this will cause the wash flow to increase
thus having the desired effect of lowering the alkali concentration towards the
setpoint.

To summarize

                   Gain                      Before           After
                   KCM                        -5.7           -0.18
                   KCA                        28.6            -9.1
                   KDM                         -10               1
                   KDA                       -2.14           -2.14




                                                                                   10

								
To top