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Reverse-acting proportional pressure controller

  • Thread starter Antonio76
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Antonio76

Homework Statement


[/B]
A 5 to 20 bar reverse acting proportional pressure controller has an output of 4 to 20 mA. The set point is 11 bar. Determine:
(a) the measured value pressure which gives an output of 15 mA when the proportional band setting of the controller is 40%
(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.


Homework Equations




The Attempt at a Solution


Can anyone help me with my findings whether i am in the right direction?
Part (b) should be a figure or a %? Thanks

(a)

PB = 40% = 0.4
Controller output (Co) = initial control valve + (error x gain)
Gain Comax - Comin / PB = -(20-4) / 0.4= - 40

error e(t) = SP - MP
Controller output = Comax + Gain (SP - MP) 15 = 20 - 40 (11 - MP) MP = 10.9 bar


(b)
Controller output = Comax + Gain (SP - MP) 8 = 20 + Gain (11 - 14) Gain = 4

4 = 16 / PB = 4
 

Answers and Replies

  • #2
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Hi, have you resolved your problem yet?
 
  • #3
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I'm struggling with this.

So far I've got 4ma = 5 bar and 20ma=20bar..... So, 15ma is equal to 4ma +11ma which is an increase of 68.75%. 68.75% of 5-20bar is 10.3125bar.

So, a 15mA signal will give 10.3125 bar if the PB were 100%???? I'm not sure how to work in the PB of the controller being 40%
 
  • #4
Tom.G
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So far I've got 4ma = 5 bar and 20ma=20bar
A 5 to 20 bar reverse acting proportional pressure controller has an output of 4 to 20 mA.
 
  • #5
David J
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Does 4mA = 20 bar and 20mA = 5bar ?? (Reverse acting)
 
  • #6
Tom.G
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Does 4mA = 20 bar and 20mA = 5bar ?? (Reverse acting)
Yes.
 
  • #7
David J
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Thanks for that, I am trying to work this out based on examples given in lessons but they are not clear.

Can you look at the attachment. This contains 2 examples from the lesson I am following.

There are 2 diagrams each with 3 proportional bands, 30%, 50% and 100%. They are all centered at 50% of the measured value.

I understand from this that the proportional band is always associated with the measured value, the input which in the case of this question, is the pressure, 5 - 20 bar. This will be the x axis

The controller output is 4 - 20 mA. This is the y axis

So for this question I think that I could create a graph similar to the attached examples with the 5 - 20 bar input on the x axis and the 4 - 20 mA as the Y axis.

The scaling of the pressure would have to be reduced (squeezed) to 40% on the x axis, centered around 50% of the measured value.

From this I should be able to strike a line from the 15 mA mark on the y axis and find the approx pressure required to generate a value 15 mA from the x axis

Does this make sense ??

Or should the PB be centered around the set-point or desired value, which in the case of this question, is 11 bar
 

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  • #8
Tom.G
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I think it will be easier to see if you start by drawing a graph similiar to the first one in your attachment, but with the axes labelled with the actual pressure and the actual current output of the controller. Start both axes at Zero and initially draw the graphed line for the full working range of the pressure with the controller covering 100% of the pressure range.
 
  • #9
David J
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Hello again, I worked on this yesterday, trying different methods and I have attached my examples. I seem to get a value of approx 9.75 bar in all cases no matter where I position the PB. I am not sure if this is the correct approach or not and I am not able to mathematically verify this value of 9.75 bar. I am not sure how to do that.
In the first instance could you look at my graphs and advise if this is the correct approach or not please. My "first attempt" is in answer to your advice in the last post
 
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  • #10
Tom.G
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Hmmm. Your results work ONLY IF YOU CAN CHANGE THE OFFSET to get the (a) and (b) answers.

Given the problem statement, I would design on the basis of 20 bar pressure always gives 4mA output, and then, if possible, to use one PB that could cover both parts (a) and (b). If more than one PB is needed for the two questions, then state/show the two different values.

Does the original problem, or its context, say anything about offsets? The problem statement given in this thread refers only to the "proportional band setting" with no mention of even the concept of "offset."

It would be a big help if someone can copy and paste the original problem.

Optional:
If you want to generate a formula for a final answer rather than a graph, I suggest that the graph 'Y' axis start at Zero current. This will make it more obvious when generating the formula that there is an offset. (it's easy to forget otherwise :biggrin:)

Cheers,
Tom

p.s. This has got me thinking some too; it has been a few decades since I've used these!
 
  • #11
David J
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Hi

The original problem is copied below. It does not mention offsets so I have to assume tat the control loop is stable and the controller is set up properly. The 3 graphs I posted were my attempts, varying the position of the PB with respect to the measured value. I am wondering if I am looking to deeply into this because part B of the question seems to be written in a way that can be calculated if I knew of the equation

A 5 to 20 bar reverse acting proportional pressure controller has an output of 4 – 20 mA. The set point is 11 bar. Determine:

(a) The measured value pressure which gives an output of 15 mA when the proportional band setting of the controller is 40%
(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


In the first post of this thread there is an equation (see below) which I think assumes there is an error but the result of that calculation (10.9) does not agree with my graphs so that has kind of thrown me of the route a bit

PB = 40% = 0.4
Controller output (Co) = initial control valve + (error x gain)
Gain Comax - Comin / PB = -(20-4) / 0.4= - 40

error e(t) = SP - MP
Controller output = Comax + Gain (SP - MP) 15 = 20 - 40 (11 - MP) MP = 10.9 bar
 
  • #12
Tom.G
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Perhaps we are interpreting the problem statement differently.
A 5 to 20 bar reverse acting proportional pressure controller has an output of 4 to 20 mA.
I am reading that as there being a self-contained pressure controller that reports the measured outlet pressure to the 4-20mA loop.
Another interpretation could be a proportional pressure valve that is driven by a controller... which seems to be the interpretation you are using.

The apparent ambiguity of the problem statement in the First Post is why I asked for the ORIGINAL statement, as in a link to where it appeared or at least a copy-and-paste.

I have put in a request for other any other Advisor to join this discussion to help sort things out.

Cheers,
Tom
 
  • #13
David J
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Ok thanks for the update, hopefully we can solve this
 
  • #14
David J
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I was given a small amount of information regarding this, an example which could help.
As we have said the
"controller is reverse acting hence 20mA corresponds to 8 bar and 4mA to 14 bar. 6 bar pressure change = 16mA then calculate for output of 15mA (i.e a change of 11mA from 4 to 15)"
I cant work out how the above statement can be incorporated in this problem but it obviously can be
 
  • #15
Tom.G
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It may help to realize that a straight-line graph can be characterized by its Slope-Intercept form, Y=mX+b. You made use of that in your earlier calculations.
Y= Dependent variable
m= Slope of the line
X= Independent variable
b= Y Intercept

Cheers,
Tom
 
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  • #16
David J
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I have been unofficially told that one of my earlier graphs was actually correct although there are other methods of calculating this question. What I am trying to do now is mathematically work out the answer. I know the answer is approx. 9.75 bar for a 15mA signal when the PB is 40%. I am just trying to prove the math now. I need to read up on Y=mX+b as I think there maybe some differential in there which is not my strong point
 
  • #17
jim hardy
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I would venture that
as @Tom.G suggested
y = mx + b is the right approach.
I'd make y be output milliamps
and x be input bars.
so as to make the independent variable, pressure, plot on the X axis
and the dependent one , milliamps, on the y axis.

x and m and b are not simple single term constants though.
Look at the units - we have to get from psi to ma,
so m will be the product of at least two individual m's -
one for units conversion let's call it munits
and another for controller gain which is 1/(proportional band) let's call it mgain
and m's sign will reflect direct or reverse acting..

Next, since controllers operate on an error signal, namely difference between input and setpoint,
input x will be that difference namely (pressure - setpoint)
so x = (bars - setpoint)

lastly b will take care of the 4ma offset in controller output.
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
that way they'll control somewhere around mid output and not bang against a limit.
so our b term also takes care of the fact that at zero error, ie x = zero,. the controller's output will be 50% or 12 ma.
meaning b is 12 ma not the 4 you'd likely think. .

y = munits X mgain X (input - setpoint) + 12 milliamps

so let's step through what we know
A 5 to 20 bar reverse acting proportional pressure controller has an output of 4 – 20 mA.
..
He didn't state at what proportional band that relation holds so let us assume it's at pb of 100%, meaning it take a fullscale change at input to cause a fullscale change at output.
So, munits = Δoutput / Δinput = 16 ma / 15 bar
Furthermore, since it's reverse acting, ie increasing input causes decreasing output,
Δ's in numerator and denominator have opposite signs
meaning munits must be negative.
So munits = -(16/15) ma / bar .




The set point is 11 bar. Determine:
aha ! Error then will be (pressure -11) bars and that's x .
x = (bars - 11)

(a) The measured value pressure which gives an output of 15 mA when the proportional band setting of the controller is 40%

Great ! He's told us pb is 0.4 which is 1/(controller gain),
That makes mgain =2.5 ,
and m = munits X mgain = -16/15 X 2.5 = = -8/3
and x = pressure - 11
Now y = mx + b
and y = 15
so
15 = -8/3 X (bars-11) + 12
(15-12) X 3/8 = -(bars - 11)
9/8 = 11 - bars
bars = 11 - 9/8 = 9.875



(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
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. ]


Does above make sense ?

There's no step that is complicated. Just you have to be meticulous.

let me know if you find a mistake.

old jim
 
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  • #18
jim hardy
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Perhaps we are interpreting the problem statement differently.
I am reading that as there being a self-contained pressure controller that reports the measured outlet pressure to the 4-20mA loop.
Another interpretation could be a proportional pressure valve that is driven by a controller... which seems to be the interpretation you are using.
Wow those would be two different problems with different answers.
Old adage "A question well stated is half answered... "
I sure missed that ambiguity. (red faced chagrin icon)

Seems to me that were this gizmo reporting measured process pressure its setpoint and PB would be superfluous .
it would be acting as a simple pressure transmitter with simple span and zero adjustments.
Asking about PB and Setpoint would be a misleading "dirty trick" on students.

old jim
 
  • #19
David J
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Gents
Post 18 is correct. It makes a lot of sense to me now. In past lessons I have seen the use of y = mx + b but never connected it with this problem. I had confirmation that my graph was correct and your answer in post 18 mathematically proves that. All I need to do now is revisit the lessons using this and understand how it calculates in this case. I dont think I am finished with this just yet but very close now. Thanks to you both (Tom and Jim) for helping out with this so far
 
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  • #20
David J
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Good Morning, just a recap, last night I followed up on the advice given in posts 16 and 18. I found a website (link below) that focuses on this area

http://www.math.com/school/subject2/lessons/S2U4L2GL.html

Just one final question.

....... in the absence of directions to the contrary you set up controllers so they have 50% output at zero error
that way they'll control somewhere around mid output and not bang against a limit.
so our b term also takes care of the fact that at zero error, ie x = zero,. the controller's output will be 50% or 12 ma.
meaning b is 12 ma not the 4 you'd likely think.
.

Do you mean that unless there is an error stated we always assume that x = zero? In the case of this question there is an error (x = (bars - 11) What kind of example would show an error? How would we know if there was an error in the question? I understand `b` but if you had not explained this I would never have assumed b = 12 mA and not 4 mA

I have attached the graph I keep talking about and marked on it the exact center of the output which equals 12 mA
 

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  • #21
jim hardy
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Do you mean that unless there is an error stated we always assume that x = zero? In the case of this question there is an error (x = (bars - 11) What kind of example would show an error? How would we know if there was an error in the question?
i am not sure i understand your question "What kind of example would show an error?"
To me, the term "controller" defines that we're dealing with some sort of automatic control system.

The basic idea behind Automatic Control is to measure something and take some physical action action to hold it at or near a desired value.
Call that something "the process" , it might be a temperature, a flow, a pressure, or anything else in our modern world.
To accomplish that we measure the process and compare its value to a desired value.
In my day we just called measured value "process" and desired value "setpoint".
Error is always the difference between those two..

There are lots of turorials out there just search 'basic automatic control'
my first hit was http://www.ent.mrt.ac.lk/~rohan/teaching/EN5001/Reading/DORFCH1.pdf
try a few and find one that suits your taste
upload_2018-11-14_8-41-14.png


in this one your reverse acting controller includes the error summer (circle) and control device that i highlighted in red ..
Actuator will be something that moves according to the controller's 4-20 milliamp output.
In my plant we controlled pressure with 400 kilowatts of electric heaters that boiled water into steam to maintain pressure at 2235 psi.
The 4 to 20 ma signal controlled heater power over a narrow range of pressure , i forget the exact value perhaps 50psi.
That would mean Setpoint of 2235 , at 2210 heaters would be full on (400 kw) and at 2260 fully off and at 2235 50% 200 kw.
That's oversimplifyimg a little because our proportional controller had a second function called "integral"
but should convey the idea.

Do you mean that unless there is an error stated we always assume that x = zero?
not at all, read on.
How would we know if there was an error in the question?
Error is always process minus setpoint ,or if your author prefers setpoint minus error, but always the difference.
Sometimes setpoint is calculated externally and applied as a second input to the controller.
You always arrange something downstream of the controller that enables it to drive process toward setpoint *hence error toward zero)
just like the feedback in an operational amplifier circuit.

If it's an automatic control system there's a process and there's a setpoint and their difference is error and in the problem you presented that's x.


any help ?

old jim
 

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  • #22
jim hardy
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those attached files ask to install on my computer and i dont want them on my disk, i just want to view in the webpage.

so i didnt look at them.

Does adobe offer you an option to make them not self-install ?

i just wont have self installing unknowns writing on my disk.

old jim
 
  • #23
David J
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Hello Jim

The attachments are just graph paper. They are taken from a www.mathsphere.co.uk.
I use the graph paper online as I can get a more detailed result using the adobe system, apologies, I didn't know they try to self install.
I am more or less finished with this question now. Your help has been really appreciated.
I have snipped my graph below, not sure if you will be able to see it or not. You may need to enlarge it to see the detail. I didn't actually start this post so I cant mark it as solved but thanks for your help with this. I will no doubt come up against other problems like this as I am only half way through my control and automation module
View attachment 234006
 
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  • #24
jim hardy
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Thanks for the feedback.

I see our snip and paste although it is a little blurry wnen i enlarge it to 300%.

You might try the UPLOAD button

But i think you're flying now on your pwn wings, so - soar my friend !
Usually one just needs a small boost to get started up the learning curve.

Automatic control theory used to be called "cybernetics"
the concepts apply to everyday life too, not just hard science
there's an interesting old pop-psychology book by one Maxwell Maltz called "PsychoCybernetics"
i enjoyed it fifty years ago
should you spot it in a used bookstore for cheap, pick it up ,. You might enjoy it too and i'd be curious to know if it has cross generational appeal...

old jim
 
  • #25
David J
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will certainly keep an eye out for that yes, thanks again
 

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