Linear ma to Square Root ma Conversion Formula | DP Transmitter Square Root Extractor
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- Опубліковано 14 лип 2024
- In this video, I have given brief explanation about square root extractor in differential pressure transmitter(DPT).In addition to this, I have given examples for linear mA to square root mA conversions, square root mA to linear mA conversions, and linear percentage to square root percentage conversion. If you have any questions regarding conversions, do ask me in the comment box. And please let us know if you need videos on some specific topics related to instrumentation such as level measurement ,flow measurement , pressure measurement and lot more .
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Video Content:
1. What is Square Root Extractor in DPT?
2. Linear mA To Square Root mA Conversion Formula for DP Transmitter.
3. Square Root mA To Linear mA Conversion Formula for DP Transmitter.
4. Linear to Square Root Percentage Conversion Formula for DP Transmitter.
1. What is Square Root Extractor in DPT?
- The relationship between flow rate and differential pressure is non linear.
- Flow rate is directly proportional to square root of Differential Pressure. Or we can say that, differential pressure developed by orifice plate or any other acceleration based flow element is proportional to square of the flow rate.
-So, square root extractor is an electronic or pneumatic device that takes the square root of the signal from the flow transmitter, and produce outputs a corresponding to linear flow signal.
In other words, it takes square root input milli ampere signal from dp transmitter, and converts it into linear milli ampere signal.
The square root function was performed in a separate device called a square root extractor in the days of pneumatic instrumentation.
Nowadays, all the smart dp transmitter comes with square root extractor function.
In other words, user can select square root extractor function from transmitter.
2. Linear mA To Square Root mA Conversion Formula for DP Transmitter.
Example- If linear analog output is 5 mA, What is Square Root output for this Input?
Square Root Output = 4 mA + [4 * Square Root Of {Output Linear}-4 mA}]
Sqrt Output = 4 mA + [4 * Square Root Of(5-4)]
Sqrt Output = 4 mA + [4*1]
Sqrt Output = 8 mA
3. Square Root mA To Linear mA Conversion Formula for DP Transmitter.
Linear Output = 4 mA + [([Output Sq Rt] - 4mA)^2/16]
Example: Square Root Output is 6 mA. What is Linear Output?
Linear Output = 4 mA + [6-4]^2/16
Linear Output = 4 mA + [2]^2/16
= 4 mA + 4/16
= 4 mA + 0.25
Linear Output = 4.25 mA.
4. Linear to Square Root Percentage Conversion Formula for DP Transmitter.
Formula:
Square Root = Sqrt(X).(10)
Where X is Linear Percentage Value.
Example:
Convert 75% Linear to Square Root
Percentage Output.
Square Root % = √x *10
Square Root % = √75 *10
Square Root % = 8.660*10
Square Root % = 86.60%.
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Dear Sir,
Good Explanation sir... everything understood but Why we multiply by 10 in the %
Dear Srinivas...
In this example, 50 is in percentage form. This is a reason we have multiplied with 10. There is other way too. Just convert percentage in to normal form. For example, 50%/100 =0.5 and sqrt root of 0.5 is 0.7071. Now convert 0.7071 in to percentage
0.7071*100 = 70.71%
I hope you understood.
@@InstrumentCalibration Thank you for your support marvelous answer...
@@srinivasv3214 You are welcome... please do ask me question whenever you have confusion.
@@InstrumentCalibration thank you & from which country yours sir
@@srinivasv3214 You are welcome
Thanks A lot for sharing informative calculation about flow
Thanks
Dear sir
Really very good example for understanding
Thanks ..Keep supporting
Please define in detail what is linear non linear and why we convert into linear value... difference between square root linear and linear value...
Hello Khan
Thanks for asking this question. Please watch following video to get ur answer.
ua-cam.com/video/3NSL_2CYAh4/v-deo.html
dear sir
please the calibration range 0-- 500 m bar for differentiation transmitter
how to calculate square root the out 4 to 20 mA
flow rat 21 bopod as max
pipeline 8"
Flow is directly proportional to differential pressure but what about constant of proportionality which is considered based c and beta etc, is it mean all these factors are also incorporated in the transmitter
Simple q is approximately equal to dp what if exact flow is to be inquired secondly what other factors needs to be considered as well
Hi sir! Could you please make a video on dp type ft calibration
Hi Shabab
I have already made video on Dp Type FM Calibration.
ua-cam.com/video/nDGVIH_sMDA/v-deo.html
Dear Sir,
I wanna ask some questions
1. Linear Output is signal from transmitters delta P. converting 4-20mA?
2. Flow rate is proportional to squareroot delta P.
So linear output needs to be converted to squareroot output?
1. Yes, But you need to select square root function from transmitter to convert into linear mA.
2. In order to cancel square root, you have to select square root function at transmitter side or at PLC or DCS side.
Hope you will get your answer
Dear Sir
I am having Siemens DP transmitter, I want to calibration flow with respect to pressure, and why we take square root for flow calculation WRT pressure transmitter pls explain sir.
It is because flow rate is directly proportional to square root of differential pressure. And in order to convert this differential pressure to linear, we take square root in DP Transmitter. Please feel free to ask me, if you have any further queries.
@@InstrumentCalibration Thank you sir👍. Do you have any WhatsApp group or blog?
@@samuelsolomon2028 At Present, we do not have any. But we will let yo know when we will make it in future.
How we can convert square root % to linear %
Linear(%)= (x)^2*100
For example, if you want to convert 86.66% to linear %.
First Convert Percentage to normal number
86.66/100 =0.866
Linear(%)= (0.866)^2*100= 0.74.99*100
=74.99% =75%
Hope you understand.
Please feel free to ask me your question in comment box.
In linear percentage to square root percentage conversion formula why we multiply with 10 . What is the reason behind it. Plz and me one person asked me in an interview.
Hello Ajay
Thanks for asking this question. Linear Value is in percentage. And we need to convert this percentage value into normal Value. This is reason we are multiplying with 10.
For Example.=50% This is a percentage value.
There is also other way. You can also divide 50 by 100 then take square root.
50/100 = 0.5
Square Root of 0.5 is 0.707
If you convert this value into percentage.
0.707*100=70.7 %.
I hope you get your answer.
@@InstrumentCalibration thanks for information. your explanation very well.
Can I ask questions in different topics also? Like thermocouple or radar level tx.
why square root extractor is not used in level and temperature measurement?
Square root extractor is only used when we measure flow by using differential pressure. Since flow is proportional to square root of differential pressure. All other variable such as temp, level does not have such relationship
@@InstrumentCalibration THANK YOU SIR YOUR VALUABLE ANSWER.
Please dont show subtitles it disturbing the screen
Ok...we will do it.
@@InstrumentCalibration Thank you
Ok sure..we will make changes
Remove the caption not able to see properly
Thank you for feedback. I will make sure to remove caption in next videos