NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersMotors and power supply on farms and in processing plants: choosing, starting, protecting and paying for electric drives.
In the exams: ICAR AIEEA-PG Unit II (explicit; not in the GATE AG syllabus); NCEES PE electric motors; rural electrification in PRC ABE
NCEES PE Agricultural and Biological Engineering
Electric motors under Common System Applications: motor input power and current from pump and fan duty, selecting a standard motor size, efficiency and wire-to-water efficiency, energy cost, VFD savings with the affinity laws, capacitor sizing for power-factor correction, and conductor sizing for voltage drop on long farm runs. Equipotential planes (ASABE EP473) appear under facilities.
How our questions for NCEES PE Agricultural and Biological Engineering spread across the units of Electrical Machines & Power Utilisation. Difficulty: 1 easy, 8 medium, 20 hard.
| Selecting motors for pumps, threshers and processing machines | 5 | |
| Three-phase induction motors: slip and torque-speed curve | 4 | |
| Variable-frequency drives | 4 | |
| Power factor correction | 4 | |
| Transformers in practice | 3 | |
| Rural power distribution and tariffs | 3 | |
| Wiring and protection: fuses, MCB, ELCB and earthing | 3 | |
| Motor starters | 2 | |
| DC machines | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A three-phase, 400 V, 7.5 kW (output) pump motor has full-load efficiency 0.88 and power factor 0.85. What is its full-load line current?
C. 14.5 A
I = P_out/(√3 V η pf) = 7500/(1.732 × 400 × 0.88 × 0.85) = 14.5 A. 10.8 A ignores efficiency and power factor. 25.1 A omits √3. 8.1 A multiplies by η and pf instead of dividing.
A star-delta starter for a 15 kW centrifugal pump is set to change over to delta after only 1 s, before the motor has accelerated. What problem results?
D. A second current surge at changeover, nearly as large as a direct-on-line start
If the timer switches to delta while the motor is still at low speed, it receives full voltage at high slip and draws close to DOL current, losing the benefit of star starting. Changeover does not leave it in star, does not reverse the phase sequence, and does not bypass the overload relay.
An irrigation pump lifts 15 L/s through a total head of 20 m. Pump efficiency is 0.70 and motor efficiency 0.85. What electrical input power does the set draw?
D. 4.95 kW
Water power = 1000 × 9.81 × 0.015 × 20 = 2943 W. Wire-to-water efficiency = 0.70 × 0.85 = 0.595. Input = 2.943/0.595 = 4.95 kW. 4.20 kW is the pump shaft power (motor efficiency omitted). 2.94 kW is water power. 3.46 kW applies only the motor efficiency.
A borewell pump lifts water through a large static head with little pipe friction. Why does slowing it with a drive save much less energy than the cube law predicts?
A. Head falls with speed squared, so a small slowdown leaves too little head to lift water
The affinity laws hold for friction-only systems. With a large static head, the pump's head (∝ N²) soon falls to the static lift and flow collapses, so the usable speed range is narrow and savings must be read from the pump curve against the system curve. The cube law is for centrifugal machines; drive losses are only a few per cent; lower speed does not raise motor current.
Line copper losses supplying an irrigation load are 2 kW at 0.7 power factor. If power factor is raised to 0.95 with the same real power and voltage, what do the losses become?
D. 1.09 kW
Losses ∝ I² ∝ 1/pf², so new losses = 2 × (0.7/0.95)² = 1.09 kW. 1.47 kW scales linearly with current. 3.68 kW inverts the ratio. 2.00 kW assumes power factor has no effect on losses.
A 15 kVAR, 400 V, 50 Hz capacitor bank is delta-connected. What capacitance is needed in each phase?
A. 99.5 µF
Each delta capacitor sees 400 V and supplies 5 kVAR: C = Q/(3ωV²) = 15000/(3 × 314.16 × 400²) = 99.5 µF. 298.4 µF is the star-connected value (only 231 V per capacitor). 33.2 µF divides by 3 twice. 625.0 µF uses f = 50 instead of ω = 2πf.
A rice mill's maximum demand is 100 kW at 0.75 power factor. Its tariff charges 300 currency units per kVA of maximum demand per month. If the power factor is raised to 0.95 at the same kW, what is the monthly saving in demand charges?
D. 8,421 units
kVA before = 100/0.75 = 133.33; after = 100/0.95 = 105.26. Saving = (133.33 − 105.26) × 300 = 8,421 units. 6,000 multiplies kW by the change in power factor. 31,579 is the new bill, not the saving. 16,597 charges the reactive power removed (kVAR) instead of the change in kVA.
A processing plant adds several VFDs and soon afterwards its fixed power-factor capacitor bank fails repeatedly with blown fuses. What is the most likely reason and fix?
B. Harmonic resonance with the capacitors; replace them with a detuned, reactor-protected bank
Drive harmonics can excite a parallel resonance between capacitors and supply inductance, overloading the capacitors. Detuned banks with series reactors shift the resonance below the dominant harmonics. Drives keep displacement power factor near unity, do not change supply frequency, and capacitors must not be put on drive outputs.
A three-phase 400 V borewell motor draws 14.47 A at 0.85 power factor through 100 m of copper cable (resistivity 0.0175 Ω·mm²/m). Reactance is neglected and voltage drop is limited to 3 % (12 V). What minimum conductor cross-section is needed?
B. 3.11 mm²
ΔV = √3·I·(ρL/A)·cos φ, so A = √3 × 14.47 × 100 × 0.0175 × 0.85/12 = 3.11 mm²; choose the next standard size (4 mm²) and check current capacity. 3.59 mm² uses the single-phase factor 2. 3.66 mm² ignores power factor. 6.21 mm² doubles the length, which is only right for single-phase loop length.
Dairy cows in a milking parlour hesitate to enter and let down milk poorly, and small voltages are measured between the feed rail and the wet floor. What is the recommended remedy?
B. Install an equipotential plane bonded to the metal fittings
Stray voltage between metalwork and floor drives current through animals, which are more sensitive than people. A bonded equipotential plane (ASABE EP473) keeps everything the animal touches at the same potential. Overload relays and motor phase do not address it, and mats wear and do not equalise potential.
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