[Class 1] Power Generation, Transmission, Substations and Electrical Applications

P = 9.8 × Q × H × η = 9.8 × 15 × 80 × 0.8 = 9,408 kW. 960 kW forgets the gravitational acceleration 9.8, 7,526 kW applies the efficiency twice, and 11,760 kW leaves out the efficiency.

A pumped-storage plant uses surplus electricity to pump water up to the upper reservoir and generates with that water when demand is high, storing energy as height of water. Using river water directly is run-of-river (流込み式), using underground steam is geothermal, and using tides is tidal power; none of these is pumped storage.

The Pelton turbine (an impulse turbine), turned by jets from nozzles striking its buckets, suits high heads. Propeller and Kaplan turbines suit low heads with large flows, and the Francis turbine is a reaction turbine widely used for medium heads.

The condenser cools the steam leaving the turbine back into water and keeps the pressure low so that more energy can be taken out in the turbine. Making steam by burning fuel is the boiler's job, raising water pressure is the feed pump's job, and further heating steam is the superheater's job.

Electrical energy = 200,000 kW × 1 h = 200,000 kWh; since 1 kWh = 3,600 kJ, that is 7.2 × 10⁸ kJ. Fuel heat = 40,000 kg × 44,000 kJ/kg = 1.76 × 10⁹ kJ. η = 7.2 × 10⁸ ÷ 1.76 × 10⁹ ≈ 0.409, about 40.9%. 59.1% is the share of losses (100 − 40.9), 11.4% wrongly takes 1 kWh as 1,000 kJ, and 244.4% divides in the wrong order (fuel heat ÷ electrical energy = 1.76 × 10⁹ ÷ 7.2 × 10⁸); efficiency can never exceed 100%.

In combined-cycle generation, the hot exhaust from the gas turbine produces steam in a heat-recovery boiler, which drives a steam turbine as well, giving high thermal efficiency. A diesel engine is internal-combustion generation, reactor steam drives a steam turbine in nuclear generation, and co-locating solar and wind is simply a mix of renewables; none of these is combined cycle.

A BWR boils water inside the reactor and sends that steam to the turbine. Using a steam generator to boil water in a separate loop is the pressurised water reactor (PWR, 加圧水型). Present nuclear plants use fission heat, not fusion, and both types drive turbines with steam.

Wind power is proportional to the cube of the wind speed, so 2³ = 8 times. 2 times assumes proportionality to speed, 4 times assumes the square (looking only at ½mv² and forgetting that the mass of air passing each second also grows with speed), and 6 times mistakes 2³ for 2 × 3 (treating the power as multiplication).

A fuel cell takes DC electricity directly from the chemical reaction of hydrogen and oxygen, and the product is water. Burning fuel to turn a turbine is thermal power, converting light is a solar cell (which gives DC), and only storing and releasing electricity is a storage battery, not a fuel cell.

N = 120f ÷ p = 120 × 60 ÷ 4 = 1,800 min⁻¹. 3,600 min⁻¹ is for 2 poles, 900 min⁻¹ is for 8 poles, and 1,500 min⁻¹ uses 50 Hz.

Doubling the voltage halves the current, and the loss 3I²r is proportional to the square of the current, so (1/2)² = 1/4. 1/2 assumes loss proportional to current, 1/8 assumes the cube, and 'no change' overlooks that the current falls.

The overhead ground wire is an earthed wire strung above the conductors to keep lightning from striking them directly (shielding). Suppressing vibration is the damper's job, moving the arc away from insulators is the arcing horn's job, and reducing corona is the role of bundled conductors.

Arcing horns are metal fittings at both ends of an insulator string that move the flashover arc off the insulator surface so the insulator is not destroyed by heat. Absorbing vibration is the damper's job, lowering voltage is the transformer's job, and converting current for meters is the current transformer's job.

Dampers are weights attached to the conductor that suppress wind-induced vibration and prevent strand breakage near the support points. Carrying lightning current to earth is done by the ground wire and earthing, keeping conductor spacing is the spacer's job, and dampers do not cool the conductor.

I = P ÷ (√3 × V × cosθ) = 1,000 ÷ (√3 × 6.6 × 0.8) ≈ 109.3 A. Loss = 3 × I² × r = 3 × 109.3² × 0.5 ≈ 17,900 W ≈ 17.9 kW. 6.0 kW counts only one conductor, 12.0 kW counts two conductors as for single-phase two-wire, and 11.5 kW leaves out the power factor (current 87.5 A).

e = √3 × I × (r cosθ + x sinθ). sinθ = √(1 − 0.8²) = 0.6, so e = √3 × 100 × (0.4 × 0.8 + 0.3 × 0.6) = √3 × 100 × 0.5 ≈ 86.6 V, about 87 V. 50 V forgets the √3, 83 V swaps cosθ and sinθ, and 121 V simply adds r and x (√3 × 100 × 0.7).

GIS encloses circuit breakers, disconnectors, busbars and so on in earthed metal enclosures insulated with highly insulating SF₆ (sulfur hexafluoride) gas, making the equipment compact for urban substations. Live parts exposed in air is air insulation, extinguishing the arc in oil is an oil circuit breaker, and interrupting in a vacuum is a vacuum circuit breaker; none of these describes GIS.

Article 37(1) requires arresters on high-voltage and extra-high-voltage circuits at the lead-in (excluding customers' lead-ins) and lead-out of overhead lines at power stations, storage stations, substations and similar places. Because it covers high and extra-high voltage circuits, low-voltage branch circuits and low-voltage customers' service entrances are not included, and the listed points relate to overhead lines, not supply by underground lines only.

At light load the line capacitance tends to raise the voltage, so a shunt reactor (分路リアクトル) absorbs leading reactive power to hold the voltage down. A power capacitor supplies lagging reactive power at heavy load to stop the voltage falling, and would raise the voltage further at light load. An instrument voltage transformer steps voltage down for measurement and an arrester protects against overvoltages; neither is phase-modifying equipment.

Article 38 (Table 38-1) requires the fence height plus the distance from the fence to live parts to be at least 6 m when the voltage is over 35,000 V and up to 160,000 V; 66,000 V falls in this range. 5 m is the value for 35,000 V or less, and 3 m and 4 m are not in the table.

The distance to the point is r = √(3² + 4²) = 5 m, and cosθ = 3 ÷ 5 = 0.6. Eh = (1,000 ÷ 5²) × 0.6 = 40 × 0.6 = 24 lx. 40 lx is the illuminance on a surface facing the light (cosθ forgotten), 32 lx wrongly uses cosθ = 4 ÷ 5, and 111 lx uses only the 3 m height, giving the illuminance directly below.

N = E × A ÷ (F × U × M) = 500 × 200 ÷ (5,000 × 0.6 × 0.7) = 100,000 ÷ 2,100 ≈ 47.6, rounded up to 48. 34 forgets the maintenance factor (33.3), 29 forgets the utilisation factor (28.6), and 20 forgets both.

Total flux is the intensity times the solid angle of a whole sphere, 4π sr: F = 4πI = 4 × 3.14 × 100 ≈ 1,257 lm. 628 lm uses 2π (a hemisphere), 314 lm uses π, and 8 lm divides 100 by 4π instead of multiplying.

Heat needed = 20 × 4.2 × (85 − 15) = 5,880 kJ. With 80% efficiency the heater must supply 5,880 ÷ 0.8 = 7,350 kJ. 2 kW = 2 kJ/s, so 7,350 ÷ 2 = 3,675 s ≈ 61 min. 49 min ignores the efficiency, 39 min multiplies by the efficiency instead of dividing, and 77 min divides by the efficiency twice.

Induction heating uses the alternating magnetic field of a coil to induce eddy currents in a metal (conductive object) and heats it with their Joule heat; it is used in IH cookers. Heating an insulator by dielectric loss is dielectric heating, heating with an arc is arc heating, and shining infrared rays is infrared heating.

Dielectric heating (誘電加熱) heats an object from the inside by the dielectric loss that arises in an insulator in a high-frequency electric field (microwaves in a microwave oven). Resistance heating uses Joule heat from current, induction heating uses eddy currents in metal, and arc heating uses the heat of an arc; none heats an insulator from inside.

Heat output = 1.2 kW × 4.0 = 4.8 kW, i.e. 4.8 kWh in one hour; with 1 kWh = 3,600 kJ that is 17,280 kJ. 4,320 kJ converts only the electrical input (1.2 kWh) and forgets the COP, 1,080 kJ divides by the COP, and 4.8 kJ forgets the 3,600 kJ/kWh conversion.

Q = 1.2 ÷ 60 = 0.02 m³/s. P = 9.8 × Q × H × K ÷ η = 9.8 × 0.02 × 25 × 1.1 ÷ 0.75 ≈ 7.19 kW, about 7.2 kW. 6.5 kW forgets the margin factor, 4.9 kW forgets both the margin factor and the efficiency, and 431 kW does not divide m³/min by 60.

v = 30 ÷ 60 = 0.5 m/s. P = 9.8 × m × v ÷ η = 9.8 × 1.5 × 0.5 ÷ 0.7 = 10.5 kW. 7.4 kW forgets the efficiency (7.35), 5.1 kW multiplies by the efficiency instead of dividing, and 630 kW does not divide m/min by 60.

From P = 2π × (n ÷ 60) × T, T = 60P ÷ (2πn) = 60 × 15,000 ÷ (2 × 3.14 × 1,450) ≈ 98.8 N·m. 621 N·m forgets to divide by 2π, 10.3 N·m just divides power by speed, and 0.099 N·m leaves the power in kW (15).