Traffic calculation problem
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(Single-choice question) A differential pressure flow meter with on-site squaring functionality has a range of 40 kPa, with a corresponding maximum flow rate of 16 t/h. When the differential pressure is 10 kPa, what is the output current of the transmitter and the corresponding flow rate? A) 12 mA, 8 t/h B) 8 mA, 4 t/h C) 8 mA, 8 t/h D) 12 mA, 12 t/h The answer is A. This question came from a forum; shouldn’t the answer be B, since on-site squaring is involved? Option A refers to the situation where squaring is performed by a DCS system. Taking the square root on-site means that flow rate and differential pressure are compared directly, while taking the square root in the DCS corresponds to the ratio of the square of the flow rate to the differential pressure.
No matter which side is used for the square root, the flow rate should be the same, right? Which side performs the square root operation only affects the output current of the transmitter: if the square root is applied at the instrument side, the instrument’s output current is 12 milliamps; if it is applied at the system side, the instrument’s output current is 8 milliamps. The system receives the current signal, determines whether to take a square root, and then calculates the flow rate.
Taking the square root of the header value can be considered as yielding a flow signal. 4–20MA corresponds to 0–16 t/h. The range is 40 kPa; after taking the square root, this becomes 6.3. This means that a 4–20 mA signal corresponds to a differential pressure value of 0–6.3 after square root calculation. In practice, when the measured pressure is 10 kPa, the square root value is 3.16. This represents half of the full scale value of 6.3, which in turn corresponds to half of the 4–20 mA range—that is, 12 mA. It also corresponds to half of the 0–16 t/h range, or 8 t/h. If the DCS performs the square root calculation and the transmitter outputs a differential pressure signal, then with a range of 40 kPa and an actual measurement of 10 kPa (which is 1/4 of the full scale), the output signal would be 1/4 of the 4–20 mA range, i.e., 8 mA. Regardless of whether squaring is applied on that end, since the differential pressure is 10 kPa, the calculated flow rate remains unchanged at 8 t/h. In other words, it is the application of squaring that determines the unit and range of the variable received by the DCS. If squaring is applied to the gauge, then the 4–20 mA signal received by the DCS represents the flow rate, with a range corresponding to the flow rate scale; if squaring is applied to the DCS side, then the 4–20 mA signal received represents the differential pressure, with a range corresponding to the differential pressure scale.
The relationship between differential pressure and flow rate is given by differential pressure = k * (flow rate)². Since 40 = k * 16², it follows that k = 40 divided by 16². At a differential pressure of 10 kPa, the square of the flow rate equals 10 divided by 40 divided by 16², which is 64; therefore, the flow rate is 8. A range of 0–16 corresponds to 4–20 mA, so the output current is 12 mA
Building 4# is the correct choice; it’s equivalent to calculating the K value first. It’s easier to understand this way. In fact, taking the square root of the gauge reading means taking the square root of the differential pressure, and the percentage of current output by the gauge also involves taking a square root. In DCS, taking the square root is not applied to the current differential voltage on the primary side
According to the calculation formula for differential pressure flow meters, since square root operation is performed on-site, the current value is the one obtained after taking the square root of the differential pressure. Qmax/Q = (40/10)^1/2; (40/10)^1/2 = (20–4)/x. The final answer is A