Agricultural and Biosystems Engineers Licensure Examination (ABELE) · Philippines · Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC
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
How our questions for Agricultural and Biosystems Engineers Licensure Examination spread across the units of Electrical Machines & Power Utilisation. Difficulty: 2 easy, 19 medium, 20 hard.
| Rural power distribution and tariffs | 10 | |
| Selecting motors for pumps, threshers and processing machines | 7 | |
| Transformers in practice | 4 | |
| Power factor correction | 4 | |
| Wiring and protection: fuses, MCB, ELCB and earthing | 4 | |
| DC machines | 3 | |
| Motor starters | 3 | |
| Variable-frequency drives | 3 | |
| Single-phase motors, including capacitor-start | 2 | |
| Three-phase induction motors: slip and torque-speed curve | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
An 11 kV/433 V transformer at the tail end of a rural feeder sees only 10.2 kV on its HV side. With the tap on the nominal position, what is the no-load LV voltage?
B. 401.5 V
The ratio is fixed by turns: V_LV = 433 × 10.2/11 = 401.5 V. 467.0 V inverts the ratio. 433 V assumes the LV is unaffected by HV voltage. 432.2 V subtracts the 0.8 kV shortfall as if it were 0.8 V. The remedy is a tap with fewer HV turns, which raises LV voltage.
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.
Which motor and starting arrangement best suits a large thresher with a heavy drum that must be run up to speed before crop is fed?
B. High-starting-torque motor started direct-on-line or by a drive
A heavy drum has large inertia; the motor needs enough accelerating torque through the whole run-up without overheating, which favours a high-starting-torque design (such as NEMA Design C or D) on DOL or a VFD. Star-delta and a low-voltage soft start cut torque to a fraction; a shaded-pole motor is far too weak.
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.
A 30 kW load is corrected from 0.7 to 0.95 power factor at constant voltage. By what percentage does the supply current fall?
B. 26.3 %
Current ∝ kVA = P/pf: 30/0.7 = 42.86 kVA, 30/0.95 = 31.58 kVA. Reduction = 1 − 0.7/0.95 = 26.3 %. 25.0 % is the change in power factor. 45.7 % is the reduction in copper loss (∝ I²). 35.7 % divides the change by 0.7.
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.
What does a two-part electricity tariff consist of?
C. A fixed or demand charge plus a charge for energy used
A two-part tariff combines a fixed or maximum-demand charge (recovering capacity cost) with an energy charge per kWh. Different day and night rates describe a time-of-use tariff; horsepower charges are flat-rate; kVAh billing is a separate method.
A solar pump controller in an open field fails repeatedly after thunderstorms, though no direct strike is seen. What should be added?
C. Surge protective devices on both DC and AC sides with good bonding
Induced lightning surges on long DC and motor cables destroy electronics; SPDs on the array and motor sides, bonded to a low-impedance earth, divert them. Array size does not limit surges; a 30 mA RCD gives shock protection, not surge protection; star-delta cannot be used after a drive.
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