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square root extraction
kindly furnish the difference between a transmitter configured with square root extraction and linear...

kindly furnish the difference between a transmitter configured with square root extraction and linear.

By Thomas Hergenhahn on 6 October, 2002 - 1:37 pm

If the measured quntity doubles, the linear confidured sensor delivers the double output,the square root configuration gives 1.41 time the output signal.
This is useful in situations where you are not interested in the measured quntity, but in a derived one, which depends on the square root of the first. E.g. you want to know the flow of gas and measure pressure difference. Then you can use a differential pressure transmitter which calculates the square root, so the output signal is proportional to the flow.

By narendrakumar on 7 October, 2002 - 2:08 pm

all draught and pressure transmitter are used only linear output. If you are using the tranmitter for flow then squre root is used.Based on the requirements linear or sqare root is selected

By elavazhagan_s on 14 October, 2002 - 11:19 am

Well,

The square root extractor is used for All Differential pressure Flow measurement Transmitters (for eg. Orifice, Flow Nozzle etc.,) For all other applications like pressure and Level the linear transmitters are used.

With best regards,

S. Elavazhagan,
Engineer - I&C,
TCE Consulting Engineers Ltd.,
Bangalore, India.

By Manoj Topiwala on 17 October, 2002 - 12:48 pm

Square root extraction facility generally used with Differential Pressure transmitter used to measure flow with orifice. You might be knowing that the flow is proportional to square root of DP across the orifice. The
output of the DP transmitter is generally linear with respect to the DP. If the transmitter is equipped with square root extraction facility, the output will be proportional to square root of the DP, i.e. it will be
DIRECTLY proportional to FLOW.

See the table below for electronics DP transmitter.

`|---------+-----------+------------------------+----------------------|| % of DP |% of FLOW  |Linear o/p              |Square root o/p       ||         |           |(proportional to DP) mA |(proportional to      ||         |           |                        |FLOW), mA             ||---------+-----------+------------------------+----------------------||    0    |     0     |           4            |          4           ||---------+-----------+------------------------+----------------------||   25    |    50     |           8            |          8           ||---------+-----------+------------------------+----------------------||   50    |   70.71   |           12           |        15.31         ||---------+-----------+------------------------+----------------------||   75    |   86.60   |           16           |        17.85         ||---------+-----------+------------------------+----------------------||   100   |    100    |          100           |         100          ||---------+-----------+------------------------+----------------------|`

Regards,
Manoj Topiwala,
Reliance Industries Ltd., Hazira-India

Mr. Topiwala,
I am calculating the output signals for square root o/p (proportional to flow)indicated in your table below. Please show me how you arrived at these values. I have tried every possible method that I know but I have not arrived with the same values as yours.

By Bob Peterson on 14 March, 2003 - 11:37 am

<P>Keep in mind that percent relates to some portion of 1. </P>
<PRE>
% DP DP(0-1) SQR DP (0-1)
0 0 0
25 0.25 0.5
50 0.5 0.707106781
75 0.75 0.866025404
100 1 1
</PRE>
<BR>Bob Peterson

But how can I arrive at the series of output signal values of 4, 8, 15.31, 17.85 and 100 mA which are said to be square root o/p proportional to flow?

By Bob Peterson on 17 March, 2003 - 2:05 pm

> But how can I arrive at the series of output signal values of 4, 8, 15.31,
> 17.85 and 100 mA which are said to be square root o/p proportional to flow?
>

4 mA = 0%
8 mA = 25%
15.31 mA = 70.68%
17.85 mA = 86.56 %

My guess is that the 100 mA figure is a type and probably should be 20 mA = 100%

Bob Peterson

By mike kozak on 14 July, 2014 - 6:58 pm

>> But how can I arrive at the series of output signal values of 4, 8, 15.31,
>> 17.85 and 100 mA which are said to be square root o/p proportional to flow?

> 4 mA = 0%
> 8 mA = 25%
> 15.31 mA = 70.68%
> 17.85 mA = 86.56 %
>
> My guess is that the 100 mA figure is a type and probably should be 20 mA = 100%

Don't worry about 4mA & 20mA they stay exact. You are working w/a span of 16mA starting at 4 increasing through 20. Take your 16mA x 0.866025404 = 13.85 then add 4 thats 17.85 also called 75%.16 mA x 0.707106781 = 11.31 + 4mA = 15.31! Hope this helps

By John Ruback on 1 July, 2003 - 11:32 pm

4-20 mA signal

out mA = out minimum (4mA) + (4* sq root of input-input min(4mA))

By mike kozak on 14 July, 2014 - 6:42 pm

Thanks,
This was the first unimpressive, yet clear explanation of how to calculate the DP, or mA output for a sq.rt. curve, perfect!

> Keep in mind that percent relates to some portion of 1.

`> % DP     DP(0-1)     SQR DP (0-1)  > 0         0            0> 25        0.25        0.5> 50        0.5         0.707106781> 75        0.75        0.866025404> 100     1             1`

> Bob Peterson

There appears to be some mistakes in your table. I think the correct values should be:

`|---------+-----------+------------------------+----------------------|| % of DP |% of FLOW  |Linear o/p              |Square root o/p       ||         |           |(proportional to DP) mA |(proportional to      ||         |           |                        |FLOW), mA             ||---------+-----------+------------------------+----------------------||    0    |     0     |            4           |         4            ||---------+-----------+------------------------+----------------------||   25    |    50     |            8           |        12            ||---------+-----------+------------------------+----------------------||   50    |   70.71   |           12           |        15.31         ||---------+-----------+------------------------+----------------------||   75    |   86.60   |           16           |        17.85         ||---------+-----------+------------------------+----------------------||   100   |    100    |           20           |        20            ||---------+-----------+------------------------+----------------------|`

dear

what reason for applying Squre root in dp transmitter in flow measurement.

By Sushil Kumar on 16 October, 2013 - 11:31 pm

Dear Friend,

From where it come the formula

Flow = (sqrt (flow percentage*10)) /100 * span (TPH)

i think you have ignored the orifice data,
>

`>|---------+-----------+------------------------+----------------------|>| % of DP |% of FLOW  |Linear o/p      >       |Square root o/p       |>|         |           |(proportional to>DP) mA |(proportional to      |>|         |           |                >       |FLOW), mA             |>|---------+-----------+------------------------+----------------------|>|    0    |     0     |           4    >       |          4           |>|---------+-----------+------------------------+----------------------|>|   25    |    50     |           8    >       |          8           |>|---------+-----------+------------------------+----------------------|>|   50    |   70.71   |           12   >       |        15.31         |>|---------+-----------+------------------------+----------------------|>|   75    |   86.60   |           16   >       |        17.85         |>|---------+-----------+------------------------+----------------------|>|   100   |    100    |          100   >       |         100          |>|---------+-----------+------------------------+----------------------|>`