# Evaluation Question

An analog electronic d/p transmitter has an output of 4-20 mA which represents 0 to 100 gpm (0 to 378.5 lpm). On a properly calibrated transmitter, what is the output in gpm (lpm), if the transmitter output is 12 mA?

a.    25                 (94 lpm)

b.   33.3             (126 lpm)

c.   70.7             (267.6 lpm)

d.   50                 (189.3 lpm)

This question has to be read carefully because a lot of people including myself , tend to say 50 because 12 mA would normally would represent 50% of your span. However, the question states that you are using a differential pressure transmitter to measure gallons per minute which is a volumetric flow unit. Therefore you have to use the square root because flow is directly proportional to the square root of the differential pressure. This can be done in the transmitter or the DCS, but not both.

The correct formula for the above question is as follows:

In other words,  take your output signal which was 12 mA and subtract 4, giving you 8. Divide 8 by16 which gives you 0.5. Then you take the square root of 0.5 which is 0.7071 and multiply it by the maximum flow rate of 100.

The answer is c. 70.7 gpm.

• Benjamin Ncube

why do you subtract the four from the output signal?

• Mark Weisner

The reason that you take the 4 mA off is because we are solving for the percentage of span at this part of the equation. We have a span of 16 mA since zero starts a 4mA..

• Nishanth N

In other words, since the LRV and URV (i.e. 0 and 100 gpm respectively) are a result of square root extraction of the differential pressure measured, the minimum and maximum values of DP are calculated as square(0) and square (100) which is 0 and 10000. 50% of this span works out to 5000, the square root of which is 70.7 gpm.

• Very good Nishanth. Sometimes 1 problem hides many ways to be solved. This interchange board shows some of those ways.

• Anonymous

Thanks nice informations

• Renatoalexpa

good informations, thanks