NCEES FE Other Disciplines · USA · NCEES
Original practice questions written to the published syllabus · not past papersMeasuring temperature, pressure, flow, force and moisture, logging the data, and controlling processes automatically.
In the exams: ICAR AICE subjects 38 (Unit 7), 39 (Unit 8) and 41 (Unit 9); NCEES PE sensors and data acquisition; NCEES FE 4–6 questions
NCEES FE Other Disciplines — Instrumentation and Controls (4–6 questions)
Sensors and transducers, bridge circuits, sampling and ADC resolution, block diagrams, closed-loop transfer functions, steady-state error, stability and first- and second-order response.
How our questions for NCEES FE Other Disciplines spread across the units of Instrumentation, Sensors & Process Control. Difficulty: 6 easy, 24 medium, 5 hard.
| Control modes: on-off, P, PI and PID | 8 | |
| Signal conditioning | 6 | |
| Pressure, flow and level measurement | 5 | |
| Static and dynamic characteristics of instruments | 4 | |
| The generalised measurement system | 3 | |
| Temperature transducers | 3 | |
| Data acquisition and data loggers | 3 | |
| Force, torque and strain: strain gauges and load cells | 2 | |
| Programmable logic controllers | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A thermometer with a first-order time constant of 10 s is moved from air at 20 °C into hot water at 80 °C. What does it read 15 s later?
D. 66.6 °C
y = 80 − (80 − 20)e^(−15/10) = 80 − 60 × 0.2231 = 66.6 °C. 57.9 °C is the reading at one time constant (10 s), 62.1 °C ignores the 20 °C starting point, and 33.4 °C adds the remaining fraction instead of the completed fraction.
Dryer plenum air warms steadily at 2 °C per minute. The air-temperature sensor is first-order with a time constant of 30 s. Once the ramp is established, how far does the reading lag behind the true temperature?
B. 1.0 °C
For a ramp, a first-order sensor lags by rate × τ = (2 °C/60 s) × 30 s = 1.0 °C. 60 °C multiplies 2 by 30 without converting minutes to seconds; 2.0 °C would correspond to a 60 s time constant; 0.033 °C is the rate per second, not the lag.
A Pt100 sensor (α = 0.00385 /°C, linear model) in a pasteuriser hot-water line reads 119.4 Ω. What is the temperature?
C. 50.4 °C
R = R0(1 + αT), so T = (R/R0 − 1)/α = (1.194 − 1)/0.00385 = 50.4 °C. 310.1 °C forgets to subtract 1; 19.4 °C takes the ohms change as degrees; 7.5 °C multiplies the 19.4 Ω change by α (with a factor-of-100 slip) instead of dividing.
An orifice meter gives 16 kPa differential at 40 L/s. If flow falls to 20 L/s, what differential will it show?
C. 4 kPa
Q ∝ √Δp, so Δp ∝ Q². Halving flow divides Δp by four: 16/4 = 4 kPa. 8 kPa assumes a linear relation, which is why obstruction meters have limited turndown. 2 kPa assumes Δp ∝ Q³; 32 kPa doubles instead of quartering.
What is the size of one least significant bit (LSB) for a 12-bit ADC with a 0–5 V input range?
C. 1.22 mV
LSB = FS/2ⁿ = 5/4096 = 1.22 mV. 4.88 mV is the 10-bit value; 417 mV divides by the number of bits instead of 2ⁿ; 2.44 mV assumes a 10 V range.
In an ultimate-cycle test on a dryer burner loop, sustained oscillation occurs at Ku = 8 with period Pu = 120 s. What are the Ziegler–Nichols PID settings?
D. Kp = 4.8, Ti = 60 s, Td = 15 s
Ziegler–Nichols PID: Kp = 0.6Ku = 4.8, Ti = Pu/2 = 60 s, Td = Pu/8 = 15 s. Kp = 3.6 with Ti = 100 s is the PI rule; 0.5Ku is the P-only rule. These settings give quarter-amplitude decay and are usually softened in practice.
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