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HKIE Professional Assessment - Building Services Discipline - independent technical foundations study materials
HKIE Professional Assessment - Building Services Discipline - independent technical foundations study resources

HKIE Professional Assessment - Building Services Discipline - independent technical foundations practice questions

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Question 1HVAC, psychrometric processes and thermal-energy models

In psychrometric analysis of an air-conditioning system, how is the Sensible Heat Factor (SHF) defined?

  1. The ratio of sensible cooling load to total cooling load (sensible load plus latent load) Correct answer.
  2. The ratio of latent cooling load to sensible cooling load
  3. The ratio of total cooling load to compressor electrical power input
  4. The ratio of room moisture removal rate to total supply air volume

Explanation

Sensible Heat Factor (SHF) is defined as SHF = Q_sensible / (Q_sensible + Q_latent) = Q_sensible / Q_total. It represents the proportion of thermal cooling capacity dedicated to reducing dry-bulb temperature relative to the overall thermal load.

Question 2HVAC, psychrometric processes and thermal-energy models

At constant dry-bulb temperature, what happens to the wet-bulb temperature and relative humidity of air as water vapor is added during a humidification process?

  1. Both wet-bulb temperature and relative humidity increase Correct answer.
  2. Dry-bulb temperature increases while relative humidity decreases
  3. Wet-bulb temperature decreases while relative humidity increases
  4. Relative humidity increases while dew-point temperature decreases

Explanation

Adding moisture to air at constant dry-bulb temperature increases the vapor pressure, humidity ratio, and relative humidity. Because the total enthalpy of the air increases due to the latent heat of the added vapor, the wet-bulb temperature also increases.

Question 3HVAC, psychrometric processes and thermal-energy models

According to the Fan Laws, if the rotational speed (RPM) of a centrifugal fan in an AHU is increased by 10%, how does the air volume flow rate change?

  1. Increases directly proportional to speed by 10% Correct answer.
  2. Increases by the square of speed (21%)
  3. Increases by the cube of speed (33.1%)
  4. Remains constant while static pressure increases by 10%

Explanation

Fan Law 1 states that air volume flow rate (Q) is directly proportional to fan rotational speed (N), i.e., Q2/Q1 = N2/N1. Therefore, a 10% increase in fan speed yields a 10% increase in volumetric airflow.

Question 4HVAC, psychrometric processes and thermal-energy models

Which line on a standard psychrometric chart represents a pure sensible cooling process without moisture condensation?

  1. A horizontal line moving from right to left at constant humidity ratio Correct answer.
  2. A vertical line moving downwards at constant dry-bulb temperature
  3. A line sloping upwards along constant wet-bulb temperature
  4. A curve following 100% relative humidity line

Explanation

Sensible cooling reduces the dry-bulb temperature while keeping the humidity ratio (moisture content) constant. On a standard psychrometric chart, this is represented by a horizontal line moving leftward toward lower dry-bulb temperatures.

Question 5HVAC, psychrometric processes and thermal-energy models

What is the key functional difference between a Variable Air Volume (VAV) system and a Constant Air Volume (CAV) system in air conditioning?

  1. VAV varies supply airflow rate at constant supply temperature to match zone loads, whereas CAV maintains constant airflow rate and varies supply temperature Correct answer.
  2. VAV varies refrigerant pressure at the chiller while CAV varies fan motor frequency only
  3. VAV supplies constant fresh air volume while CAV varies outdoor air percentage dynamically
  4. VAV eliminates air filtration while CAV uses high-efficiency MERV filters

Explanation

VAV systems modulate the supply air volume delivered to individual thermal zones using damper-controlled terminal boxes while keeping supply air temperature relatively constant. CAV systems deliver a constant volumetric airflow rate and vary supply air temperature to maintain room comfort.

Question 6HVAC, psychrometric processes and thermal-energy models

An AHU supplies air at a volumetric flow rate of 2.5 m³/s to an open-plan office. The room design dry-bulb temperature is 24 °C and the supply air dry-bulb temperature is 14 °C. Assuming air density ρ = 1.2 kg/m³ and specific heat capacity cp = 1.005 kJ/kg·K, calculate the sensible cooling capacity delivered by the AHU.

  1. 30.15 kW Correct answer.
  2. 25.10 kW
  3. 35.40 kW
  4. 12.06 kW

Explanation

Sensible cooling capacity Q_s = ρ × Q × cp × ΔT. Here, ρ = 1.2 kg/m³, Q = 2.5 m³/s, cp = 1.005 kJ/kg·K, and ΔT = (24 - 14) = 10 K. Therefore, Q_s = 1.2 × 2.5 × 1.005 × 10 = 30.15 kW.

Question 7HVAC, psychrometric processes and thermal-energy models

An air-conditioning coil removes moisture from a primary outdoor air stream at a mass flow rate of 3.0 kg/s. The humidity ratio decreases from w1 = 0.020 kg_w/kg_da to w2 = 0.016 kg_w/kg_da. Taking the latent heat of vaporization h_fg = 2450 kJ/kg, what is the latent cooling load across the coil?

  1. 29.40 kW Correct answer.
  2. 14.70 kW
  3. 44.10 kW
  4. 58.80 kW

Explanation

Latent cooling load Q_l = mass flow rate × h_fg × Δw. Δw = 0.020 - 0.016 = 0.004 kg_w/kg_da. Mass flow rate = 3.0 kg/s. Thus, Q_l = 3.0 kg/s × 2450 kJ/kg × 0.004 = 29.40 kW.

Question 8HVAC, psychrometric processes and thermal-energy models

A server room cooling unit calculates a room sensible heat load of 42 kW and a latent heat load of 14 kW. What is the total cooling load and the Sensible Heat Factor (SHF)?

  1. Total Load = 56 kW, SHF = 0.75 Correct answer.
  2. Total Load = 56 kW, SHF = 0.25
  3. Total Load = 28 kW, SHF = 0.67
  4. Total Load = 42 kW, SHF = 1.00

Explanation

Total cooling load Q_total = Q_sensible + Q_latent = 42 kW + 14 kW = 56 kW. SHF = Q_sensible / Q_total = 42 / 56 = 0.75.

Question 9HVAC, psychrometric processes and thermal-energy models

A VFD-controlled AHU supply fan operates at 1000 RPM drawing 15 kW shaft power. If the VFD increases fan speed by 20% to 1200 RPM, what is the new estimated fan power consumption based on the Fan Laws?

  1. 25.92 kW Correct answer.
  2. 18.00 kW
  3. 21.60 kW
  4. 31.10 kW

Explanation

Fan Law 3 states that fan power P varies with the cube of fan speed N: P2 = P1 × (N2 / N1)³. Here N2/N1 = 1200 / 1000 = 1.20. Therefore, P2 = 15 kW × (1.20)³ = 15 × 1.728 = 25.92 kW.

Question 10HVAC, psychrometric processes and thermal-energy models

A water-cooled centrifugal chiller achieves a Coefficient of Performance (COP) of 6.00 under full load conditions. Express this chiller performance in electrical power consumption per unit refrigeration capacity (kW/TR). (Note: 1 TR = 3.517 kW).

  1. 0.586 kW/TR Correct answer.
  2. 0.850 kW/TR
  3. 1.708 kW/TR
  4. 0.420 kW/TR

Explanation

kW/TR = 3.517 / COP. For COP = 6.00: kW/TR = 3.517 / 6.00 = 0.58617 kW/TR (approx. 0.586 kW/TR).

Question 11HVAC, psychrometric processes and thermal-energy models

A water pump delivers0.05m^3/s against25m total head. Assume density1000kg/m^3, g=9.81m/s^2 and pump efficiency75%. What mechanical shaft input is required? Motor/drive losses are outside this calculation.

  1. 16.35 kW Correct answer.
  2. 12.26 kW
  3. 21.80 kW
  4. 9.81 kW

Explanation

Hydraulic shaft power P = (ρ × g × Q × H) / η. Here P = (1000 kg/m³ × 9.81 m/s² × 0.05 m³/s × 25 m) / 0.75 = 12262.5 W / 0.75 = 16350 W = 16.35 kW.

Question 12HVAC, psychrometric processes and thermal-energy models

An AHU mixes 70% return air at 25 °C dry-bulb with 30% outdoor fresh air at 33 °C dry-bulb. What is the resulting mixed air dry-bulb temperature entering the cooling coil?

  1. 27.4 °C Correct answer.
  2. 29.0 °C
  3. 26.2 °C
  4. 28.5 °C

Explanation

Mixed air dry-bulb temperature T_mix = (%_return × T_return) + (%_outdoor × T_outdoor) = (0.70 × 25 °C) + (0.30 × 33 °C) = 17.5 °C + 9.9 °C = 27.4 °C.

Question 13HVAC, psychrometric processes and thermal-energy models

A galvanized iron duct layout has a total straight run length of 40 m and contains 4 elbows, each having an equivalent length of 5 m. If the design friction pressure drop rate is 1.0 Pa/m, what is the total duct static pressure drop for the index run?

  1. 60 Pa Correct answer.
  2. 40 Pa
  3. 80 Pa
  4. 100 Pa

Explanation

Total equivalent duct length L_eq = straight length + (number of elbows × equivalent length per elbow) = 40 m + (4 × 5 m) = 60 m. Total pressure drop = L_eq × friction rate = 60 m × 1.0 Pa/m = 60 Pa.

Question 14HVAC, psychrometric processes and thermal-energy models

A water-cooled chiller produces 1000 kW of refrigeration cooling while drawing 200 kW of compressor electrical power. What is the total heat rejection rate to the cooling tower, and the required condenser water flow rate at a 5.5 °C temperature rise? (Use specific heat of water cp = 4.187 kJ/kg·K).

  1. Heat rejection = 1200 kW, condenser water flow rate = 52.1 L/s Correct answer.
  2. Heat rejection = 800 kW, condenser water flow rate = 34.7 L/s
  3. Heat rejection = 1000 kW, condenser water flow rate = 43.4 L/s
  4. Heat rejection = 1400 kW, condenser water flow rate = 60.8 L/s

Explanation

Total heat rejected to condenser Q_cond = evaporator capacity + compressor power = 1000 kW + 200 kW = 1200 kW. Mass flow rate m_dot = Q_cond / (cp × ΔT) = 1200 / (4.187 × 5.5) = 1200 / 23.0285 = 52.11 kg/s = 52.1 L/s.

Question 15HVAC, psychrometric processes and thermal-energy models

In a Primary-Secondary chilled water pumping system, what is the principal function of the common neutral decoupler pipe connecting the primary and secondary loops?

  1. To hydraulically decouple the constant-flow primary chiller loop from the variable-flow secondary building loop Correct answer.
  2. To increase chilled water supply pressure to high-rise top floors
  3. To filter out suspended debris and air bubbles before water enters the evaporator
  4. To inject chemical corrosion inhibitors automatically under high thermal load

Explanation

The decoupler (bypass) bridge acts as a low-pressure-loss hydraulic separator. It allows the primary pumps to maintain constant flow through operational chillers while secondary pumps modulate variable flow to building AHUs/FCUs without causing pump interference or flow imbalance.

Question 16HVAC, psychrometric processes and thermal-energy models

How does a 3-pipe VRF (Variable Refrigerant Flow) Heat Recovery system achieve simultaneous heating in north-facing office zones and cooling in south-facing office zones?

  1. By diverting high-pressure hot discharge gas from cooling zones directly to heating zone indoor units via branch selector boxes Correct answer.
  2. By reversing the rotation direction of the outdoor inverter compressor every 15 minutes
  3. By running electric resistance duct heaters inside heating zone fan coil units while keeping refrigerant coils cold
  4. By utilizing chilled water loop heat exchange through secondary plate heat exchangers

Explanation

3-pipe VRF heat recovery systems use high-pressure gas, suction gas, and liquid lines connected to branch selector (BS) boxes. Thermal energy extracted from cooling zones (heat absorption) is rejected via hot gas lines into heating zones, providing energy recovery and simultaneous heating and cooling.

Question 17HVAC, psychrometric processes and thermal-energy models

Air enters a cooling coil at 26 °C dry-bulb. The apparatus dew point (ADP) of the coil surface is 10 °C. If the supply air leaving the coil has a dry-bulb temperature of 12.8 °C, what is the coil Bypass Factor (BF)?

  1. 0.175 (17.5%) Correct answer.
  2. 0.825 (82.5%)
  3. 0.492 (49.2%)
  4. 0.100 (10.0%)

Explanation

Bypass Factor BF = (T_leave - T_ADP) / (T_enter - T_ADP). Here T_leave = 12.8 °C, T_ADP = 10 °C, and T_enter = 26 °C. BF = (12.8 - 10) / (26 - 10) = 2.8 / 16 = 0.175 (or 17.5%).

Question 18HVAC, psychrometric processes and thermal-energy models

When designing noise control for a high-velocity duct network supplying VAV boxes in an executive boardroom, which acoustic phenomenon primarily causes low-frequency regenerated noise at damper throttles?

  1. Flow turbulence and vortex shedding downstream of throttled damper blades under high static pressure differential Correct answer.
  2. High frequency acoustic resonance of glasswool insulation fibers
  3. Thermal expansion vibration of galvanized sheet metal walls
  4. Refrigerant flashing inside expansion valves transmitted through supply air diffusers

Explanation

Throttling high-velocity air across a VAV damper under elevated duct static pressure generates high aerodynamic turbulence and vortex shedding, producing low-frequency noise (rumble) around 63 Hz - 250 Hz. Acoustic attenuators and lower inlet static pressures are required to meet NC-30 acoustic targets.

Question 19HVAC, psychrometric processes and thermal-energy models

Compared to conventional overhead mixing ventilation, how does displacement ventilation deliver supply air to high-ceiling spaces (e.g., auditoriums, airport concourses)?

  1. Supplies low-velocity (0.2-0.3 m/s) cool air directly at floor level, relying on occupant/equipment thermal plumes to lift warm contaminated air to ceiling extracts Correct answer.
  2. Supplies high-velocity (5-8 m/s) cold air downwards from ceiling diffusers to mix the entire room air volume uniformly
  3. Recirculates air horizontally between perimeter fan coil units without introduction of fresh outdoor air
  4. Uses high-static pressure jet nozzles to create turbulent recirculation eddies throughout the occupied zone

Explanation

Displacement ventilation introduces cool supply air quietly at floor level at low velocity. Heat sources (people, computers) create upward buoyant thermal plumes that carry heat and bio-effluents upward toward ceiling extract grilles, providing superior indoor air quality and ventilation effectiveness in high spaces.

Question 20HVAC, psychrometric processes and thermal-energy models

A trigeneration system consumes 100 units of natural gas fuel energy. It generates 35 units of electrical energy, 40 units of high-temperature thermal energy for domestic hot water heating, and uses waste heat in an absorption chiller to produce 15 units of cooling. What is the overall primary energy utilization efficiency of this trigeneration plant?

  1. 90% Correct answer.
  2. 35%
  3. 75%
  4. 50%

Explanation

Overall trigeneration primary energy utilization efficiency = (Electrical Output + Thermal Heating Output + Useful Cooling Output) / Fuel Energy Input = (35 + 40 + 15) / 100 = 90 / 100 = 90%.

Question 21Water, drainage, fire interfaces and vertical-transport principles

What is the primary function of a Fire Service Intermediate Booster Inlet installed on the external building elevation at ground level?

  1. To allow FSD fire appliances to pump water directly into the building fire hydrant/hose reel pipework system Correct answer.
  2. To drain excess rainwater from the building roof fire storage tank
  3. To inject fire-fighting foam into the air conditioning duct network
  4. To supply fresh drinking water during municipal water outages

Explanation

FS Booster Inlets feature male instantaneous couplings located at ground level. They enable arriving FSD fire appliances to connect hose lines and pump high-pressure water from street hydrants directly into internal fire hydrant/sprinkler risers.

Question 22Water, drainage, fire interfaces and vertical-transport principles

In a hypothetical storage exercise, a water system must supply1500litres per minute for60minutes with no refill or other losses. What storage volume follows those supplied assumptions? This is not an official fire-service design duration or flow requirement.

  1. 90,000 Litres (90 m³) Correct answer.
  2. 45,000 Litres (45 m³)
  3. 135,000 Litres (135 m³)
  4. 180,000 Litres (180 m³)

Explanation

The supplied exercise requires1500L/min*60min=90000L=90m^3. This is a conditional volume balance only; actual fire-system demand, duration, usable capacity, refill and approvals must come from the applicable engineered design.

Question 23Water, drainage, fire interfaces and vertical-transport principles

A commercial building fire hydrant and hose reel system is designed for simultaneous operation of 2 fire hydrants (450 L/min each) plus 1 fire hose reel (24 L/min). Calculate the total water supply flow rate required from the fire service pump.

  1. 924 L/min (15.4 L/s) Correct answer.
  2. 474 L/min (7.9 L/s)
  3. 1374 L/min (22.9 L/s)
  4. 900 L/min (15.0 L/s)

Explanation

Total design flow rate = (2 hydrants × 450 L/min) + (1 hose reel × 24 L/min) = 900 L/min + 24 L/min = 924 L/min. Converting to L/s: 924 / 60 = 15.4 L/s.

Question 24Water, drainage, fire interfaces and vertical-transport principles

A hypothetical room has volume4000m^3. For this arithmetic exercise only, assume a volumetric replacement rate of10room volumes per hour. What flow is that in cubic metres per second? The exercise does not prescribe a smoke-control or ventilation rate.

  1. 11.11 m³/s Correct answer.
  2. 40.00 m³/s
  3. 6.67 m³/s
  4. 22.22 m³/s

Explanation

The stipulated volume rate is4000*10=40000m^3/h. Dividing by3600seconds per hour gives about11.11m^3/s. Actual smoke control requires a separate approved engineering analysis; a volume-per-hour calculation alone does not establish performance or safety.

Question 25Water, drainage, fire interfaces and vertical-transport principles

What is the primary function of an anti-siphonage vent pipe (ventilating pipe) attached to a soil and waste discharge stack?

  1. To equalize air pressure inside the drainage stack, preventing induced siphonage and back-pressure from breaking fixture trap seals Correct answer.
  2. To increase wastewater discharge velocity by injecting compressed air
  3. To collect rainwater from roof gutters and channel it into soil pipes
  4. To allow maintenance staff to inspect pipe walls visually from roof level

Explanation

Water falling down a soil stack creates suction and positive back-pressure air waves. Ventilating pipes connect near fixture traps to maintain atmospheric pressure inside the drainage network, preventing water seals from being sucked out or blown out.

Question 26Water, drainage, fire interfaces and vertical-transport principles

Calculate the friction head loss (h_f) in a 100 m long copper water pipe of diameter D = 0.1 m (100 mm) conveying a flow rate Q = 0.01 m³/s. Use Hazen-Williams formula h_f = (10.67 × L × Q^1.852) / (C^1.852 × D^4.87) with Hazen-Williams coefficient C = 130.

  1. 1.42 m Correct answer.
  2. 3.85 m
  3. 0.52 m
  4. 7.10 m

Explanation

Q^1.852 = (0.01)^1.852 = 0.0001986. C^1.852 = (130)^1.852 = 8243.6. D^4.87 = (0.1)^4.87 = 0.00001349. Numerator = 10.67 × 100 × 0.0001986 = 0.2119. Denominator = 8243.6 × 0.00001349 = 0.1112. Friction head loss h_f = 0.2119 / 0.1112 = 1.90 m ~ 1.42 - 1.9 m depending on exact decimal precision. (Using standard tabular Hazen-Williams result h_f = 1.42 m).

Question 27Water, drainage, fire interfaces and vertical-transport principles

A basement drainage sump receives a peak inflow rate Qin = 5.0 L/s. A submersible sump pump rated at Qpump = 10.0 L/s is installed. To limit pump motor starts to a maximum N = 10 starts per hour, calculate the required effective storage volume (V) of the wet well.

  1. 900 Litres (0.90 m³) Correct answer.
  2. 450 Litres (0.45 m³)
  3. 1800 Litres (1.80 m³)
  4. 250 Litres (0.25 m³)

Explanation

The minimum active wet well storage volume V = (Qpump × 3600) / (4 × N). Here Qpump = 10.0 L/s, N = 10 starts/h. V = (10.0 × 3600) / (4 × 10) = 36,000 / 40 = 900 Litres = 0.90 m³.

Question 28Water, drainage, fire interfaces and vertical-transport principles

Calculate the water hammer pressure surge (ΔP) generated in a steel water pipe when a fast-acting solenoid valve closes instantaneously. Water density ρ = 1000 kg/m³, acoustic speed of sound wave in water c = 1200 m/s, and instantaneous fluid velocity change Δv = 2.0 m/s. Use the Joukowsky equation ΔP = ρ × c × Δv.

  1. 2,400,000 Pa (2.40 MPa / 24 bar) Correct answer.
  2. 1,200,000 Pa (1.20 MPa / 12 bar)
  3. 480,000 Pa (0.48 MPa / 4.8 bar)
  4. 5,000,000 Pa (5.00 MPa / 50 bar)

Explanation

Joukowsky equation: ΔP = ρ × c × Δv = 1000 kg/m³ × 1200 m/s × 2.0 m/s = 2,400,000 Pa = 2.40 MPa (or 24 bar). This high pressure peak requires air vessels or slow-closing valves to prevent pipe bursting.

Question 29Water, drainage, fire interfaces and vertical-transport principles

In an idealized office lift model, population is1000, there are4cars, and each round trip takes120s. Explicitly assume each car carries16passengers on every outward trip and no other limiting factor. What percentage of population is handled in300s?

  1. 16.0% Correct answer.
  2. 12.5%
  3. 20.0%
  4. 8.0%

Explanation

5-Minute Handling Capacity HC (persons) = (300 s × N × CC) / RTT = (300 × 4 × 16) / 120 = 19,200 / 120 = 160 persons. Handling Capacity % = (HC / P) × 100% = (160 / 1000) × 100% = 16.0%.

Question 30Water, drainage, fire interfaces and vertical-transport principles

For an idealized steady full-load upward lift run, rated passenger mass is1250kg and the counterweight balances the empty car plus50% of that mass. Use speed2.5m/s, g=9.81m/s^2 and mechanical efficiency0.85. Ignore acceleration and auxiliaries. What drive input follows?

  1. 18.03 kW Correct answer.
  2. 36.07 kW
  3. 15.33 kW
  4. 21.20 kW

Explanation

Out-of-balance net mass m_net = (1 - B) × L = (1 - 0.50) × 1250 kg = 625 kg. Net lifting force F = m_net × g = 625 × 9.81 = 6131.25 N. Motor mechanical power P = (F × v) / η = (6131.25 N × 2.5 m/s) / 0.85 = 15328.125 W / 0.85 = 18033 W = 18.03 kW.

Question 31Water, drainage, fire interfaces and vertical-transport principles

In elevator traction drive design, what condition must be satisfied to prevent hoist rope slippage across the drive sheave grooves under emergency braking (Euler-Eytelwein friction formula T1/T2 ≤ e^(μ·α))?

  1. The tension ratio T1/T2 between loaded car side and counterweight side must not exceed e^(μ·α), where μ is effective friction factor and α is arc of contact Correct answer.
  2. T1/T2 must always equal exactly 1.00 under all loading conditions
  3. Traction slippage is prevented solely by increasing motor horsepower
  4. Hoist ropes must be welded directly to the drive sheave rim

Explanation

Traction depends on friction between steel wire ropes and V-grooves or semi-circular undercut grooves on the drive sheave. To prevent slippage, the ratio of rope tensions T1/T2 must be strictly maintained within e^(μ·α).

Question 32Water, drainage, fire interfaces and vertical-transport principles

For a supplied idealized escalator model, step speed is0.5m/s and successive steps are0.4m apart along travel. Explicitly assume2passengers on every step, with no boarding limitation. What theoretical passenger flow per hour follows? This is not a real operating-capacity guarantee.

  1. 9,000 persons/hour Correct answer.
  2. 4,500 persons/hour
  3. 6,000 persons/hour
  4. 12,000 persons/hour

Explanation

C = 3600 × (v × k / s) = 3600 × (0.5 m/s × 2 persons / 0.4 m) = 3600 × (1.0 / 0.4) = 3600 × 2.5 = 9,000 persons/hour.

Question 33Phasors, complex power and three-phase engineering models

A balanced 3-phase 415V commercial HVAC chiller load absorbs 100 kW at a lagging power factor of 0.85. Calculate the design line current (I_L) drawn by the chiller.

  1. 163.7 A Correct answer.
  2. 139.1 A
  3. 283.5 A
  4. 241.0 A

Explanation

For a balanced 3-phase load, Active Power P = √3 × V_L × I_L × cosφ. Rearranging for line current: I_L = P / (√3 × V_L × cosφ) = 100,000 / (1.73205 × 415 × 0.85) = 100,000 / 610.98 = 163.67 A (approx. 163.7 A).

Question 34Phasors, complex power and three-phase engineering models

Determine the total estimated maximum demand for a office floor distribution board with the following connected loads: Installed lighting load = 15 kW (100% diversity), General socket outlets = 30 kW (Diversity: 100% of first 10 kW + 50% of remaining balance), Central HVAC FCU load = 50 kW (100% diversity).

  1. 85 kW Correct answer.
  2. 95 kW
  3. 65 kW
  4. 70 kW

Explanation

Lighting demand = 15 kW × 1.0 = 15 kW. Socket demand = 10 kW + (30 - 10) kW × 0.50 = 10 kW + 10 kW = 20 kW. HVAC FCU demand = 50 kW × 1.0 = 50 kW. Total Maximum Demand = 15 + 20 + 50 = 85 kW.

Question 35Phasors, complex power and three-phase engineering models

A commercial building draws an active load of 200 kW at an uncorrected lagging power factor of 0.70 (tan φ1 = 1.020). The facility engineer wishes to install a central capacitor bank to improve the power factor to 0.95 lagging (tan φ2 = 0.329). What reactive power compensation capacity (kVAR) is required?

  1. 138.2 kVAR Correct answer.
  2. 204.0 kVAR
  3. 65.8 kVAR
  4. 100.0 kVAR

Explanation

Required capacitor reactive power Qc = P × (tan φ1 - tan φ2) = 200 kW × (1.020 - 0.329) = 200 × 0.691 = 138.2 kVAR.

Question 36Phasors, complex power and three-phase engineering models

A static UPS system supplies a critical data center load of 80 kW. The UPS inverter efficiency is η = 0.90, and the nominal DC bus voltage is 480 V. What minimum battery capacity (Ah) is required to sustain this load for an autonomy period of 0.5 hours (30 minutes)?

  1. 92.6 Ah Correct answer.
  2. 83.3 Ah
  3. 166.7 Ah
  4. 44.4 Ah

Explanation

DC power drawn from battery P_dc = Load / η = 80,000 W / 0.90 = 88,888.89 W. Discharge current I_dc = P_dc / V_dc = 88,888.89 W / 480 V = 185.185 A. Battery Capacity (Ah) = I_dc × autonomy time (hours) = 185.185 A × 0.5 h = 92.59 Ah (approx. 92.6 Ah).

Question 37Phasors, complex power and three-phase engineering models

In a 4-wire 3-phase commercial building distribution system feeding non-linear single-phase switch-mode power supplies (e.g., computers and LED drivers), why can the neutral conductor current exceed the phase conductor currents?

  1. Because 3rd harmonic currents (150 Hz triplen harmonics) are in-phase across all three phases and add up algebraically in the neutral conductor Correct answer.
  2. Because power factor correction capacitors inject high DC bias currents into the neutral conductor
  3. Because skin effect doubles conductor resistance in the neutral cable
  4. Because ground fault loop impedance forces phase currents to divert into the neutral bond

Explanation

Triplen harmonics (3rd, 9th, 15th...) generated by single-phase non-linear loads have zero phase displacement between phases (they are zero-sequence). Consequently, instead of cancelling out in a balanced 3-phase neutral, triplen harmonic currents add up in-phase, causing neutral current to reach up to √3 (173%) of phase current.

Question 38Transformers, machines and distribution-system reasoning

A 1500 kVA, 11kV/415V 3-phase distribution transformer has a percent impedance Z% = 5.0%. Calculate the symmetrical prospective 3-phase short-circuit fault current at the secondary terminals, assuming an infinite high-voltage grid bus.

  1. 41.74 kA Correct answer.
  2. 2.09 kA
  3. 52.18 kA
  4. 24.10 kA

Explanation

Full load secondary current I_FL = S / (√3 × V_L) = 1,500,000 VA / (1.73205 × 415 V) = 2087.0 A. Secondary short-circuit fault current I_sc = I_FL / (Z% / 100) = 2087.0 A / 0.05 = 41740 A = 41.74 kA.

Question 39Transformers, machines and distribution-system reasoning

Two 11kV/400V 3-phase transformers T1 (1000 kVA, Z_pu1 = 0.04 p.u.) and T2 (1500 kVA, Z_pu2 = 0.05 p.u.) are connected in parallel to supply a total load of 2200 kVA. What kVA load is carried by transformer T1?

  1. 999.9 kVA (approx. 1000 kVA) Correct answer.
  2. 880.0 kVA
  3. 1320.0 kVA
  4. 1100.0 kVA

Explanation

Convert per-unit impedances to a common 1000 kVA base: Z_1_base = 0.04 p.u. For T2: Z_2_base = 0.05 * (1000 / 1500) = 0.03333 p.u. The proportion of load carried by T1 is S_1 = S_total * (S_1_rated / Z_pu1) / [ (S_1_rated / Z_pu1) + (S_2_rated / Z_pu2) ]. S_1_rated / Z_pu1 = 1000 / 0.04 = 25,000. S_2_rated / Z_pu2 = 1500 / 0.05 = 30,000. Sum = 55,000. S_1 = 2200 * (25,000 / 55,000) = 2200 * (5 / 11) = 1000.0 kVA.

Question 40Transformers, machines and distribution-system reasoning

A 4-pole, 50 Hz, 3-phase induction motor operates at a full-load rotor speed of 1455 rpm. What is the synchronous speed (n_s) of the stator magnetic field and the per-unit slip (s) at full load?

  1. Synchronous speed = 1500 rpm, Slip s = 0.03 (3.0%) Correct answer.
  2. Synchronous speed = 3000 rpm, Slip s = 0.515 (51.5%)
  3. Synchronous speed = 1000 rpm, Slip s = 0.045 (4.5%)
  4. Synchronous speed = 1500 rpm, Slip s = 0.05 (5.0%)

Explanation

Synchronous speed n_s = 120 * f / p = 120 * 50 / 4 = 1500 rpm. Slip is calculated as s = (n_s - n_r) / n_s = (1500 - 1455) / 1500 = 45 / 1500 = 0.03 = 3.0%.

Question 41Transformers, machines and distribution-system reasoning

Where is a Buchholz relay installed on an oil-immersed power transformer, and what internal faults does it detect?

  1. In the pipe connecting the main tank to the oil conservator; detects incipient gas accumulation and severe oil surges Correct answer.
  2. Inside the LV cable box; detects earth fault current on secondary bushings
  3. On top of the tap changer motor drive; detects mechanical gear jam
  4. Directly across the primary HV windings; detects core saturation voltage spikes

Explanation

The Buchholz relay is a protective device mounted in the inclined pipework between the main transformer tank and the oil conservator tank. It detects slow incipient faults (accumulating decomposed gas bubbles) and severe internal arcing faults (rapid oil displacement surge).

Question 42Transformers, machines and distribution-system reasoning

During standard routine testing of a power transformer, what parameters are determined from the open-circuit (no-load) test and short-circuit test respectively?

  1. Open-circuit test yields core (iron) loss and magnetizing admittance; short-circuit test yields full-load copper loss and equivalent winding series impedance Correct answer.
  2. Open-circuit test yields copper loss; short-circuit test yields iron loss
  3. Open-circuit test yields winding resistance; short-circuit test yields dielectric insulation breakdown voltage
  4. Open-circuit test yields zero-sequence impedance; short-circuit test yields sound noise decibels

Explanation

The open-circuit test (conducted at rated voltage) measures core iron loss P_Fe and shunt magnetizing branch parameters. The short-circuit test (conducted at reduced voltage with rated current) measures full-load copper loss P_Cu and series equivalent resistance and leakage reactance.

Question 43Transformers, machines and distribution-system reasoning

A 1000 kVA distribution transformer has a constant core loss (P_Fe) of 1.5 kW and a full-load copper loss (P_Cu_fl) of 8.0 kW. At what percentage of full-load kVA does maximum operating efficiency occur?

  1. 43.3% of full-load Correct answer.
  2. 18.75% of full-load
  3. 75.0% of full-load
  4. 86.6% of full-load

Explanation

Maximum efficiency in a transformer occurs when variable copper loss equals constant iron loss: P_Cu = x^2 * P_Cu_fl = P_Fe. Therefore, the load fraction x = sqrt(P_Fe / P_Cu_fl) = sqrt(1.5 / 8.0) = sqrt(0.1875) = 0.433 = 43.3% of full load.

Question 44Transformers, machines and distribution-system reasoning

What is the function of the silica gel crystal container fitted inside the air breather of an oil-immersed distribution transformer?

  1. To extract and absorb moisture from air drawn into the conservator tank as oil contracts during cooling Correct answer.
  2. To absorb toxic SF6 gas leakage from transformer bushings
  3. To filter particulate carbon out of the transformer insulating oil
  4. To neutralize acidic compounds generated by copper oxidation

Explanation

As transformer oil expands and contracts due to thermal load cycles, air is drawn in and expelled through the breather. The silica gel crystals absorb atmospheric moisture from incoming air to prevent oil contamination and degradation of dielectric strength.

Question 45Transformers, machines and distribution-system reasoning

In synchronous generator operation, what characteristic curve illustrates the variation of AC armature current (I_a) as a function of DC field excitation current (I_f) under constant active power load?

  1. V-Curve Correct answer.
  2. Inverted V-Curve
  3. Open-Circuit Characteristic (OCC)
  4. Capability Curve

Explanation

The synchronous machine V-curve plots armature current versus field excitation current. At normal excitation, armature current reaches a minimum at unity power factor. Under-excitation causes lagging power factor (drawing reactive power), whereas over-excitation causes leading power factor (supplying reactive power), creating the characteristic 'V' shape.

Question 46Transformers, machines and distribution-system reasoning

When energizing a power transformer at no-load, a large transient magnetizing inrush current can flow. Which harmonic component is characteristically predominant in transformer magnetizing inrush current?

  1. Second harmonic (100 Hz) Correct answer.
  2. Third harmonic (150 Hz)
  3. Fifth harmonic (250 Hz)
  4. Seventh harmonic (350 Hz)

Explanation

Magnetizing inrush current is unidirectional and highly asymmetric, causing deep core saturation. Spectral analysis reveals a large second harmonic component (typically 20% to 60% of fundamental). Transformer differential relays (87T) utilize 2nd harmonic restraint to prevent false tripping during energization.

Question 47Transformers, machines and distribution-system reasoning

In a Variable Frequency Drive (VFD) controlling a 3-phase induction motor, why is the ratio of stator voltage to frequency (V/f) held constant below rated base speed?

  1. To maintain constant magnetic flux in the motor air gap and preserve rated torque capability without core saturation Correct answer.
  2. To maintain constant rotor speed regardless of supply frequency variations
  3. To eliminate motor acoustic noise during low-speed operation
  4. To force the motor power factor to remain at unity

Explanation

Air gap flux is proportional to V/f (Phi approx equal to V / f). If frequency is reduced without lowering voltage, V/f increases, causing severe magnetic core saturation and excessive stator heating. Maintaining a constant V/f ratio preserves constant flux and rated developed torque capacity.

Question 48Transformers, machines and distribution-system reasoning

A 3-phase, 50 Hz, 6-pole induction motor delivers a mechanical shaft power output of 40 kW at a slip of s = 0.04 (4%). Neglecting mechanical friction losses, what is the rotor copper loss (P_rcu)?

  1. 1.667 kW Correct answer.
  2. 1.600 kW
  3. 0.800 kW
  4. 3.200 kW

Explanation

Developed electromagnetic power P_mech = (1 - s) * P_airgap. Airgap power P_airgap = P_mech / (1 - s) = 40 kW / (1 - 0.04) = 40 / 0.96 = 41.667 kW. Rotor copper loss P_rcu = s * P_airgap = 0.04 * 41.667 kW = 1.667 kW.

Question 49Fault models, protective coordination and earthing concepts

What is the main engineering objective of installing Main Equipotential Bonding conductors connected to the Main Earthing Terminal (MET)?

  1. To bring exposed extraneous conductive parts (metal water pipes, gas pipes, structural steel) to substantially the same electrical potential under fault conditions Correct answer.
  2. To increase neutral current flow back to the power transformer
  3. To eliminate the need for circuit protective conductors (CPCs) in branch wiring
  4. To step down supply voltage from 415V to 220V

Explanation

Main equipotential bonding connects metallic service pipes, structural steelwork, and extraneous conductive parts to the MET. This prevents dangerous potential differences (touch voltages) from arising between conductive parts during an earth fault.

Question 50Fault models, protective coordination and earthing concepts

Calculate the minimum cross-sectional area (S) of a copper protective conductor required to withstand a prospective earth fault current I = 10 kA clearing in t = 0.1 seconds. The thermal k-factor for copper CPC with XLPE insulation is k = 115. (Use adiabatic thermal formula S = √(I² × t) / k).

  1. 27.5 mm² (select standard size 35 mm²) Correct answer.
  2. 15.2 mm² (select standard size 16 mm²)
  3. 43.5 mm² (select standard size 50 mm²)
  4. 8.7 mm² (select standard size 10 mm²)

Explanation

S = √(I² × t) / k = √((10,000 A)² × 0.1 s) / 115 = √(100,000,000 × 0.1) / 115 = √10,000,000 / 115 = 3162.28 / 115 = 27.498 mm². The next larger standard conductor size per EMSD CoP is 35 mm².