I am working with a DCS control system, and I need to extract the square root of a flow transmitter. It's a very simple procedure, but I have some doubts.

1) According to Bernulli law, the output from the flow transmitter is a square root, right?

2) In the control system, the function I have to use do the following calculation:

sqrt (input/100) * (Span high-Span low) + Span low

If the output from the flow transmitter is a square root value, using this function I am applying another square root to the signal.

Hope someone can help me.

Thanks.

1) According to Bernulli law, the output from the flow transmitter is a square root, right?

2) In the control system, the function I have to use do the following calculation:

sqrt (input/100) * (Span high-Span low) + Span low

If the output from the flow transmitter is a square root value, using this function I am applying another square root to the signal.

Hope someone can help me.

Thanks.

This may sound silly, but are you sure the signal out from the transmitter is a square root? This is usually something a transmitter does, to extract a square root and linearize the signal and produce a 4-20 ma signal proportional to flow. Check your documentation.

Walt Boyes

Editor in Chief

Control and ControlGlobal.com

555 W. Pierce Rd Suite 301

Itasca, IL 60143

wboyes [at] putman.net

www.controlglobal.com

Walt Boyes

Editor in Chief

Control and ControlGlobal.com

555 W. Pierce Rd Suite 301

Itasca, IL 60143

wboyes [at] putman.net

www.controlglobal.com

You will have to determine if the output of the flow transmitter is actually flow or differential pressure. Some smart transmitters can do the square root extraction internally. You don't say what the process fluid is. If it is a liquid, you may not need any temperature and pressure compensation. If it is a gas, you would need pressure and temperature compensation (unless those parameters are essentially constant). If compensation is required, it should be applied before the square root is taken, so if done within the transmitter, those signals would also have to be available within the transmitter.

there are some smart transmitters which can be configured to give square root output. put any way you can make sure by calibrating the transmitter and check whether the output linear or square

I don't have the instruments specifications. I only have the P & ID's for analysis. But considering the transmitter extract the square root to linearize the signal, why should I have a square root extraction calculation (sqrt (input/100) * (Span high-Span low) + Span low) in the DCS?

"why should I have a square root extraction calculation (sqrt (input/100) * (Span high-Span low)+ Span low) in the DCS?

This could well be a mistake, I know I have done that myself, square root in two places. It was very easy to miss because the flow was only out by a few percent.

Do as Almonster suggests and put some DP on your transmitter e.g. 50%, that will show you for sure.

Roy

This could well be a mistake, I know I have done that myself, square root in two places. It was very easy to miss because the flow was only out by a few percent.

Do as Almonster suggests and put some DP on your transmitter e.g. 50%, that will show you for sure.

Roy

> Do as Almonster suggests and put some DP on your transmitter e.g. 50%, that

> will show you for sure.

Where the Transmitter extracts the Square Root, the loop will function as follows:

JRS

> will show you for sure.

Where the Transmitter extracts the Square Root, the loop will function as follows:

4 mA shows 0% at the DCS

5 mA shows 25% at the DCS

8 mA shows 50% at the DCS

13 mA shows 75% at the DCS

20 mA shows 100% at the DCS

Where the DCS extracts the Square Root, the loop will function as follows:

4 mA shows 0% at the DCS

8 mA shows 25% at the DCS

12 mA shows 50% at the DCS

16 mA shows 75% at the DCS

20 mA shows 100% at the DCS

JRS

> Do as Almonster suggests and put some DP on your transmitter e.g. 50%, that will show you for sure.

> Where the Transmitter extracts the Square Root, the loop will function as follows:

What JRS said is right but it also depends upon the constant. As flow is proportional to the difference in pressure but not equal. In our case We are using ABB supplied DP for flow measurement, where the constant is 10.

> Where the Transmitter extracts the Square Root, the loop will function as follows:

> 4 mA shows 0% at the DCS

> 5 mA shows 25% at the DCS

> 8 mA shows 50% at the DCS

> 13 mA shows 75% at the DCS

> 20 mA shows 100% at the DCS

> Where the DCS extracts the Square Root, the loop will function as follows:

> 4 mA shows 0% at the DCS

> 8 mA shows 25% at the DCS

> 12 mA shows 50% at the DCS

> 16 mA shows 75% at the DCS

> 20 mA shows 100% at the DCS

What JRS said is right but it also depends upon the constant. As flow is proportional to the difference in pressure but not equal. In our case We are using ABB supplied DP for flow measurement, where the constant is 10.

If the transmitter output is already a square root of the differential pressure, then the DCS should just scale the signal directly, without taking the square root again. If the signal from the transmitter has no pressure &/or temperature compensation, and if this is required, the DCS will have to provide it.

I suggest that you find the specs on the device. The vendor will be delighted to help you. But assuming you have done the SQRT in the transmitter, you don't need to use the SQRT function block in the DCS.

normally the default configuration for the flow transmitters is non-linear output, and you have to extract this signal to converted to linear before it enters to your control system

You only do the square root ONCE; EITHER at the transmitter OR at the DCS.

If you are dealing with an SIS, those logic solvers usually only accept linear signals, so in that case set the DCS to sqrt and the xmtr to linear.

Remember that differential pressure is a linear function, but flow rate is derived from volume. Assuming a normal orfice plate-type DP sensor and a circular pipe, it's easy to see the need for square root since area = pi * Radius squared.

If you set the DP Transmitter to square root conformity and then also do a square root extraction at the DCS, you will either give away all your gasoline (if outbound flowmeter) or blow up your refinery (if process-control flow meter).

Your P&ID will indicate the type of sensor and transmitter; your application and Plant practice (DCS Programmer/Engineer) will dictate where sqrt is taken.

Good Luck!

JRS

If you are dealing with an SIS, those logic solvers usually only accept linear signals, so in that case set the DCS to sqrt and the xmtr to linear.

Remember that differential pressure is a linear function, but flow rate is derived from volume. Assuming a normal orfice plate-type DP sensor and a circular pipe, it's easy to see the need for square root since area = pi * Radius squared.

If you set the DP Transmitter to square root conformity and then also do a square root extraction at the DCS, you will either give away all your gasoline (if outbound flowmeter) or blow up your refinery (if process-control flow meter).

Your P&ID will indicate the type of sensor and transmitter; your application and Plant practice (DCS Programmer/Engineer) will dictate where sqrt is taken.

Good Luck!

JRS

You always do the square root extraction in the DCS for the very simple reason that if you have to replace the Transmitter you only calibrate it for the correct DP and you are good to go again.

In the transmitter you have the feature to select if you want to do squire root extraction or not, make sure this is switched off and then do the extraction in the DCS or PLC.

In the transmitter you have the feature to select if you want to do squire root extraction or not, make sure this is switched off and then do the extraction in the DCS or PLC.

if the transmitter is doing square root there is no need to put another logic in DCS.

Can anybody tell me if i am using an orifice of 10,000mmWC flow capacity and using transmitter of range 6000 mmWC. will it show correct flow (tph) using a square root extractor type DP??? plz suggest.

Can anybody tell me if i am using an orifice of 10,000mmWC flow capacity and using transmitter of range 6000 mmWC. will it show correct flow (tph) using a square root extractor type DP??? plz suggest.

yes it will show correct reading.

you need not to use the full range of DP.

If your requirement is of 6000mmWc, you configure your tx in 0~6000mmwc range. you will get better accuracy (instead of feeding 0~10000mmwc).

you need not to use the full range of DP.

If your requirement is of 6000mmWc, you configure your tx in 0~6000mmwc range. you will get better accuracy (instead of feeding 0~10000mmwc).

Okay I am not your ones right or not.. but I have my own .. Let me explain it..

Example:

if you reverse this calculation it will give from dcs to field..

i have chkd.. hloveslol@gmail.com

Example:

DCS Range: 0-1000 Kg/hr

field in ft. : 0-2500 mm h20

first find constant..

constant = 1000/square root of 2500 =1000/50=20 constant

now for 25 % = square root of field 625 * 20(constant)

=25*20

=500 kg /hr

so when in field we are giving 25 % it shows here 50 % 500 kg/hr

now we check for 50% Means for 1250 field.\

square root of 1250 * 20 (constant)

=35.35*20

=707 kg/hr

if you reverse this calculation it will give from dcs to field..

i have chkd.. hloveslol@gmail.com

Your use of this site is subject to the terms and conditions set forth under Legal Notices and the Privacy Policy. Please read those terms and conditions carefully. Subject to the rights expressly reserved to others under Legal Notices, the content of this site and the compilation thereof is © 1999-2014 Nerds in Control, LLC. All rights reserved.

Users of this site are benefiting from open source technologies, including PHP, MySQL and Apache. Be happy.

**Fortune**

Whenever you find that you are on the side of the majority, it is time

to reform.

-- Mark Twain