Traffic calculation
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There is an electrical differential pressure transmitter with a range of 25 KPa; the corresponding maximum flow rate is 30 t/h. What is the differential pressure corresponding to an output signal of 8 mA? What is the corresponding traffic volume? (There are two cases: with and without square root of the sub-transmitter value)
1. When the transmitter applies a square root operation, the output signal represents the flow rate; for a flow rate of 30*0.25=7.5 t/h, the corresponding pressure difference is (7.5/30)²×25=1.56 KPA. 2. If no square root operation is applied by the transmitter, the output pressure difference signal is 25×0.25=6.25 KPA; in this case, the flow rate is √6.25/25 × 30=15 t/h. Is this the correct way to calculate it? Sea friends
Without taking the square root: 30/sqrt(25)Xsqrt([[8-4]/16]X25) = 15 T/H. With the square root taken: 30/sqrt(25)Xsqrt({[[8-4]/16]X[[8-4]/16]]}X25) = 7.5 T/H. Absolutely correct
Is it the case that for marine professionals, the header does not take the square root – that is, the differential pressure is proportional to the milliampere signal?； The square root of the header value is proportional to the current and milliampere signals