Feedwater flow loop reading low after a second indicator is added in series
Feedwater flow of 0-100 t/h is measured by an orifice plate and a 0-250 kPa differential-pressure transmitter. The transmitter is 2-wire, 4-20 mA, loop-powered from a 24 V DC supply, and needs at least 12 V at its terminals at 20 mA. The loop contains a 250 ohm input resistor at the PLC analogue card, 42 ohm of total cable resistance and a 100 ohm local indicator, all in series. The square root is extracted in the PLC, which scales 4 mA to 0 t/h and 20 mA to 100 t/h. During a plant trial the PLC showed 11.2 mA while the operators insisted the flow was 'about 45% of the expected 100 t/h', and an instrument technician replaced the transmitter once with no change in the reading. Later, a second recorder with a 250 ohm input was added in the same series loop, and at high flow the indicated value stuck at about 96% and would not go higher.
1. Calculate the actual flow at 11.2 mA and decide whether the transmitter is faulty.
Read the worked answer
Convert the current to a fraction of span: (11.2 - 4)/16 = 0.45. That is the fraction of differential pressure, not of flow: DP = 0.45 x 250 = 112.5 kPa.
Because flow is proportional to the square root of DP, flow = 100 x sqrt(0.45) = 67.1 t/h. The operators' '45%' is the linear current fraction (45 t/h); the true flow is 67.1 t/h, which is the correct reading for a properly scaled square-root loop.
The transmitter is not faulty - it is delivering a healthy 11.2 mA, which is 45% of its DP span, and replacing it changed nothing, exactly as expected. The fault is in the scaling: either the PLC's square-root extraction has been switched to linear (so the operator sees 45 t/h), or the square root is being applied twice (transmitter configured as square-root output and the PLC also taking a square root, which would give 100 x sqrt(0.45) once more and under-read badly). Check the transmitter's configuration for a built-in square-root output and the PLC's function block; verify with a 12.00 mA injected signal, which must display 50.0 t/h, and at 4-20 mA check the endpoints.
2. Explain why the indication stuck at about 96% after the second recorder was added, and give the fix with numbers.
Read the worked answer
Sum the loop resistance: 250 ohm PLC input + 100 ohm local indicator + 250 ohm new recorder + 42 ohm cable = 642 ohm.
At 20 mA the voltage dropped across this resistance is 0.020 x 642 = 12.84 V, leaving 24 - 12.84 = 11.16 V at the transmitter terminals, which is below its 12 V minimum. The transmitter therefore saturates: the maximum current it can drive is (24 - 12)/642 = 18.7 mA. That corresponds to (18.7 - 4)/16 = 0.919 of DP span, so the displayed flow caps at 100 x sqrt(0.919) = 95.9%, which matches the observed 96%.
The maximum permissible loop resistance is (24 - 12)/0.02 = 600 ohm, so the loop as built is 42 ohm over the limit. Fixes, in order of preference:
- Remove the recorder from the current loop and feed it from a separate signal isolator or a second analogue output, restoring the loop to 392 ohm.
- If the recorder must stay in series, use a lower-burden input (for example a 50 ohm or 1-5 V input) so the total returns below 600 ohm.
- Raise the supply to 30 V only if the transmitter is rated for it and the loop is not in a hazardous area with an associated apparatus limit.
- After any change, prove it by driving the loop to 20 mA with a calibrator and confirming at least 12 V at the transmitter and 100% at the PLC and recorder.
3. The transmitter reads 4.3 mA with the orifice isolated and the manifold equalised. Quantify the effect on the indicated flow at 60% and at 20% flow, and state whether a zero trim is the right action.
Read the worked answer
A 4.3 mA output is (4.3 - 4)/16 = 0.01875 of span, i.e. a zero offset of 0.01875 x 250 = 4.69 kPa of DP. This is a fixed offset in DP, but its effect on flow is not fixed, because flow follows the square root.
At 60% flow the true DP is 0.36 x 250 = 90 kPa. The measured DP becomes 94.69 kPa, giving an indicated flow of 100 x sqrt(94.69/250) = 61.6%, i.e. about +1.6 percentage points, or +2.6% of reading.
At 20% flow the true DP is 0.04 x 250 = 10 kPa. The measured DP becomes 14.69 kPa, giving 100 x sqrt(14.69/250) = 24.2%, i.e. +4.2 percentage points, or +21% of reading. The zero error is therefore far more damaging at low flow, which matters for leak detection and low-flow alarms.
Before trimming, establish the cause. If the transmitter is mounted below the taps with condensate-filled impulse lines, a constant static-head offset is expected and a zero trim is legitimate - but only with the manifold truly equalised at operating pressure and temperature, and only if the condensate level is stable. If the lines are dry or partly filled, or if the offset drifts, the head correction must be calculated and the installation corrected rather than trimmed out, because an unstable offset cannot be zeroed away. After trimming, verify at 4.0 mA at zero DP and recheck one mid-scale point against the flow computer.