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100+ Free GCC Factories Plant Engineering Exam Practice Questions
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Sample GCC Factories Plant Engineering Exam Practice Questions
Try these sample questions to test your GCC Factories Plant Engineering Exam exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A 500 kVA, 11 kV / 400 V distribution transformer has a full-load copper loss of 4.5 kW and an iron loss of 2.5 kW. What is the efficiency of the transformer at full load operating at a power factor of 0.8 lagging?
A.98.28%
B.97.56%
C.96.40%
D.98.90%
Explanation: Full load active output power P_out = S * cos(phi) = 500 kVA * 0.8 = 400 kW. Total losses at full load P_loss = P_cu + P_iron = 4.5 kW + 2.5 kW = 7.0 kW. Total input power P_in = P_out + P_loss = 400 + 7 = 407 kW. Efficiency eta = (P_out / P_in) * 100 = (400 / 407) * 100 = 98.28%.
2A 1000 kVA power transformer has a core (iron) loss of 3.0 kW and a full-load copper loss of 12.0 kW. At what load (in kVA) will the transformer achieve its maximum efficiency?
A.500 kVA
B.707 kVA
C.250 kVA
D.866 kVA
Explanation: Maximum efficiency occurs when variable copper loss equals constant iron loss: x^2 * P_cu_fl = P_iron, where x is the fraction of full load. x = sqrt(P_iron / P_cu_fl) = sqrt(3.0 / 12.0) = sqrt(0.25) = 0.5. Therefore, load for maximum efficiency = 0.5 * 1000 kVA = 500 kVA.
3Two 3-phase transformers A and B are connected in parallel to serve a total load of 700 kVA. Transformer A is rated at 500 kVA with 4% impedance, and Transformer B is rated at 250 kVA with 5% impedance. Assuming equal voltage ratios, how much load does Transformer A carry?
A.500 kVA
B.450 kVA
C.525 kVA
D.400 kVA
Explanation: Load sharing in parallel transformers: S_A = S_total * (S_A_rating / Z_A) / [(S_A_rating / Z_A) + (S_B_rating / Z_B)]. Here, S_A_rating / Z_A = 500 / 4 = 125. S_B_rating / Z_B = 250 / 5 = 50. Sum = 175. Thus, S_A = 700 * (125 / 175) = 500 kVA.
4A 4-pole, 3-phase induction motor operates from a 50 Hz supply. If the actual rotor speed is measured at 1440 rpm, what is the slip of the motor?
A.4.0%
B.2.5%
C.5.0%
D.3.3%
Explanation: Synchronous speed N_s = (120 * f) / P = (120 * 50) / 4 = 1500 rpm. Slip s = (N_s - N) / N_s = (1500 - 1440) / 1500 = 60 / 1500 = 0.04 or 4.0%.
5A 6-pole, 50 Hz, 3-phase induction motor runs at a slip of 3%. What is the frequency of the rotor currents under this operating condition?
A.1.5 Hz
B.3.0 Hz
C.50 Hz
D.48.5 Hz
Explanation: Rotor current frequency f_r = s * f, where s is slip and f is stator supply frequency. f_r = 0.03 * 50 Hz = 1.5 Hz.
6A 3-phase squirrel-cage induction motor develops a Direct-On-Line (DOL) starting torque equal to 2.2 times its full-load torque. If a Star-Delta starter is installed, what starting torque will the motor develop?
A.0.73 times full-load torque
B.1.10 times full-load torque
C.1.47 times full-load torque
D.2.20 times full-load torque
Explanation: In star connection, phase voltage is 1/sqrt(3) of line voltage. Since torque is proportional to the square of phase voltage, T_star = (1/3) * T_DOL. T_star = (1/3) * 2.2 T_FL = 0.733 times full-load torque.
7An industrial plant draws an active power load of 400 kW at a power factor of 0.75 lagging. To avoid utility penalties, the plant engineer decides to improve the power factor to 0.95 lagging using a static capacitor bank. Calculate the required rating of the capacitor bank in kVAR.
A.221.3 kVAR
B.180.5 kVAR
C.352.8 kVAR
D.131.5 kVAR
Explanation: Initial angle phi_1 = acos(0.75) = 41.41°, tan(phi_1) = 0.8819. Initial Q_1 = P * tan(phi_1) = 400 * 0.8819 = 352.76 kVAR. Target angle phi_2 = acos(0.95) = 18.19°, tan(phi_2) = 0.3287. Target Q_2 = P * tan(phi_2) = 400 * 0.3287 = 131.47 kVAR. Required capacitor rating Q_c = Q_1 - Q_2 = 352.76 - 131.47 = 221.3 kVAR.
8An 11 kV factory main busbar is supplied by a 10 MVA transformer with a per-unit percentage reactance of 8% (0.08 pu). Assuming the upstream grid capacity is infinite, what is the symmetrical short-circuit fault MVA level at the busbar?
A.125 MVA
B.80 MVA
C.100 MVA
D.150 MVA
Explanation: Short-circuit fault level S_sc = Rating / Z_pu = 10 MVA / 0.08 = 125 MVA.
9For the 11 kV busbar in Question 008 with a short-circuit fault level of 125 MVA, calculate the 3-phase symmetrical short-circuit fault current in kA.
A.6.56 kA
B.11.36 kA
C.3.79 kA
D.8.25 kA
Explanation: Fault current I_sc = S_sc / (sqrt(3) * V_line) = 125,000,000 VA / (1.73205 * 11,000 V) = 125,000,000 / 19,052.55 = 6,560.8 A = 6.56 kA.
10A 100 m long 3-phase 400 V cable supplies a motor drawing 80 A at 0.8 power factor lagging. The cable parameters per conductor are resistance R = 0.25 ohm/km and reactance X = 0.15 ohm/km. What is the line-to-line voltage drop across the cable?
A.4.02 V (1.01%)
B.6.96 V (1.74%)
C.2.32 V (0.58%)
D.8.04 V (2.01%)
Explanation: Cable length L = 0.1 km. R_line = 0.025 ohm, X_line = 0.015 ohm. cos(phi) = 0.8, sin(phi) = 0.6. Phase voltage drop V_drop_ph = I * (R_line * cos(phi) + X_line * sin(phi)) = 80 * (0.025 * 0.8 + 0.015 * 0.6) = 80 * (0.020 + 0.009) = 80 * 0.029 = 2.32 V. Line-to-line voltage drop = sqrt(3) * 2.32 V = 4.02 V. Percentage drop = (4.02 / 400) * 100 = 1.01%.
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