# Reverse-acting proportional pressure controller

David J
Gold Member
Hello again, I completed the calculation for 1b which was a PB of 80% as shown in post 18.

I have attempted to plot this on a graph in the same way I plotted for question 1a but I cant seem to get it to make sense. I have taken a line from 8 mA on the Y axis. This works out at approx 16.2 bar on the x axis but it should be 14 bar. Below is a snapshot of the graph

Have I aligned the PB up correctly with the x axis or should the set point of 11 bar be inline with the 40% mark or, should the set point (desired value) be in line with the 50% measured value

Below is snap shot of the attachment, I thought I was right with this but I am missing something in the plotting or scaling of this graph I think

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jim hardy
Gold Member
2019 Award
Dearly Missed
I have attempted to plot this on a graph in the same way I plotted for question 1a but I cant seem to get it to make sense. I have taken a line from 8 mA on the Y axis. This works out at approx 16.2 bar on the x axis but it should be 14 bar. Below is a snapshot of the graph

Have I aligned the PB up correctly with the x axis or should the set point of 11 bar be inline with the 40% mark or, should the set point (desired value) be in line with the 50% measured value
remember
But there's a sneaky little fact not mentioned in the problem statement, namely,
....... in the absence of directions to the contrary you set up controllers so they have 50% output at zero error
and (b) the proportional band setting which will give an output of 8 mA when the measured value is 14 bar and the desired value is 11 bar.

you need to shift your output current line until 50% output (12 ma) intersects a vertical line going straight up from setpoint, 11 bar,
3 bar from setpoint is 3/15 = 20% of input span away from setpoint,
which multiplied by 1.25 = 25% of output span = 4 milliamps away from balance point of 12 ma
and that'll cross at 8 ma

i think the y = mx + b equation holds true,
practice the longhand arithmetic method until you trust it
and then it'll be obvious to the eye when your computer graphing program is fooling you..

old jim

just for reference so i didnt have to flip pages

okay,
y = 8
x is (bars - setpoint) = (14-11) = 3
b is 12 as before
solving y = mx + b for m
8 = m X 3 + 12
(8 -12 ) / 3 = m = -4/3
since m = munits X mgain
mgain = m / munits = ( -4/3 ) / (-16/15) = 5/4 ,

so pb = 1/mgain = 1/(5/4)= 0.8 = 80%

[ it's latei= --- i hope i didnt bungle my arithmetic. ]

David J
Gold Member
you need to shift your output current line until 50% output (12 ma) intersects a vertical line going straight up from setpoint, 11 bar,
3 bar from setpoint is 3/15 = 20% of input span away from setpoint,
which multiplied by 1.25 = 25% of output span = 4 milliamps away from balance point of 12 ma
and that'll cross at 8 ma
So from the 12mA mark on the Y axis I need to take a horizontal line till it meets the vertical line from the 11 bar mark on the PB scale??

Where you have stated 3 bar from the set point is ##\frac{3}{15}=20##% then multiply this by 1.5 to get 25%. Should this not be 30% ? What is the 1.5 ??

jim hardy
Gold Member
2019 Award
Dearly Missed
Where you have stated 3 bar from the set point is 315=20315=20\frac{3}{15}=20% then multiply this by 1.5 to get 25%. Should this not be 30% ? What is the 1.5 ??
Where do you see 1.5 ?

1.25 is 1/PB and PB is 0.8

So from the 12mA mark on the Y axis I need to take a horizontal line till it meets the vertical line from the 11 bar mark on the PB scale??
you need to move your output line left until it fits y=mx + b

milliamps = munits X mgain X (input - setpoint) + 12 milliamps
munits = -(16/15)
mgain = 1/PB = 1,25
m= munits X mgain = -(16/15) X 1.25 = -4/3
and B is 12 ma
so
ma = -(4/3) X (bars - 11) + 12

at 5 bars that's ma = -4/3 X (5-11) +12 = 20 ma
at 11 bars it's 12 ma
and at 17 bars it's 4 ma
if the controller is capable , at 20 bars it'd deliver zero ma.

this isn't right either

measured value is bars, 5 and 20 should align with 0 and 100 % .
So my line isn't quite right either
Fix that and replot your y = mx + b, see if it looks better.

this is how we learn dont get dejected
attention to detail is the key.

old jim

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David J
Gold Member
I think I am confusing myself with just where the PB needs to be positioned on the graph. I am placing the PB on the x axis and associating it with the input which is bars.
measured value is bars, 5 and 20 should align with 0 and 100 % .
I think the PB should be relating to the output which is mA in this question.

I have not been aligning 5 - 20 bar with 0 - 100% measured value, I have been shrinking it as a percentage of the MV which is why the graph is not making any sense

A snap shot from my notes below shows how the "output" is squeezed as a percentage of the measured value. In this case that is the mA output.

I will work on what you have said and see if I can grasp this

Thanks once again

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jim hardy
Gold Member
2019 Award
Dearly Missed
I think the PB should be relating to the output which is mA in this question.
PB is the fraction of full scale input required to cause full scale output
Δin/Δout where Δout is 100%

80% of your 15 bar full scale input is 12 bar

jim hardy
Gold Member
2019 Award
Dearly Missed

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David J
Gold Member
I think, finally, after looking at this for hours it just hit home. Its been staring me in the face to be honest but I finally got it. It made me question my answer to the first part though, even though I had been told my answer was correct my graph wasn't so I ended up changing it. We are not even required to create a graph, I just did it to get a better understanding and I think given this problem in the future being able to plot a graph will be a big help

Thanks again for your help with this problem

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jim hardy
jim hardy
Gold Member
2019 Award
Dearly Missed
I think, finally, after looking at this for hours it just hit home. Its been staring me in the face to be honest but I finally got it.
Isn't that a great feeling ?
We old guys enjoy them too , if only vicariously.

In my plant we made such drawings of our analog reactor control and protection system for our own use.We called them "Scaling Diagrams".
Actually one young engineer made them to teach himself how it all worked and we all recognized their value. ( In case he's reading this , his initials were jpm , nice work Jorge !)
Ten years later our engineering department stumbled across them, cleaned them up and made them into official plant drawings.

As you have found a picture is worth a thousand words .

I'm glad to hear 'the light came on' for you, thanks for the feedback.
Now - Share the light !

old jim

David J