AEE Electrical Engineering Core · Chapter 7

Power Systems I: Generation, Transmission Lines, Cables, Insulators, Corona and Distribution

What to remember

Load factor = average load / maximum demand. Diversity factor = sum of individual maximum demands / maximum demand of the system (always at least 1). Plant capacity factor = average demand / installed capacity.
Transmission line constants: short line A = D = 1, B = Z, C = 0. Nominal-π: A = D = 1 + YZ/2. Always AD − BC = 1. Surge impedance Zc = √(L/C); SIL = V²/Zc.
Corona starts when the conductor surface stress exceeds the breakdown stress of air (about 30 kV/cm peak). Insulator string efficiency is below 100 percent because the unit nearest the conductor carries the most voltage.

1. Power generation

Thermal station: coal burns in a boiler to make steam, which turns a turbine coupled to the alternator. It works on the Rankine cycle. Main parts: coal handling, boiler, superheater, turbine, condenser, cooling tower, economiser, air pre-heater, ash handling, and alternator. Efficiency is limited by condenser heat rejection.

Hydro station: water power P = ρ g Q H η watts (ρ = 1000 kg/m³, Q in m³/s, H = head in metres). With Q = 10 m³/s, H = 100 m and η = 0.9, P = 1000 × 9.81 × 10 × 100 × 0.9 = 8.83 MW. Turbine types by head: Pelton (high head, impulse), Francis (medium head, reaction), Kaplan (low head, high flow). Hydro stations start quickly and suit peak load. Pumped-storage stations store energy by pumping water uphill off-peak.

Nuclear station: heat comes from fission of uranium; it uses a moderator (slows neutrons), control rods (absorb neutrons) and a coolant. Fuel cost is low and capital cost is high; suitable for base load.

Gas turbine and combined-cycle: gas turbines follow the Brayton cycle, start fast and suit peaking duty. A combined cycle uses the hot exhaust to make steam for a steam turbine, giving higher efficiency.

Diesel stations are used for standby and small remote supply. Renewable sources (solar PV, wind) are intermittent and are integrated with the grid through converters.

In Andhra Pradesh, generation is by APGENCO (thermal and hydro stations, including large hydro schemes on the Krishna such as Srisailam), transmission by APTRANSCO, and retail distribution by the regional discoms. Station names and capacities change, so check the latest official release.

2. Load curve and factors

Term	Definition
Connected load	Sum of ratings of all connected appliances
Maximum demand	Greatest demand in a period
Demand factor	Maximum demand / connected load (at most 1)
Load factor	Average load / maximum demand
Diversity factor	Sum of individual maximum demands / maximum demand of the group (at least 1)
Plant capacity factor	Actual energy produced / (installed capacity × time)
Plant use factor	Energy produced / (capacity in use × time operated)
Reserve capacity	Installed capacity minus maximum demand

Higher load factor and higher diversity factor reduce the cost per unit. Units generated = average load × hours. Example: a 100 kW average load for 24 hours = 2400 kWh. A system with average load 40 MW and maximum demand 80 MW has a load factor of 0.5.

Base load plants (nuclear, large thermal) run constantly. Peak load plants (hydro, gas) meet the daily peak. The load duration curve shows load against the number of hours it is equalled or exceeded; its area is the energy generated.

3. Transmission line parameters

Resistance increases with frequency because of skin effect (current crowds toward the surface). Proximity effect also increases it.
Inductance of a conductor with an effective radius r′ = 0.7788 r: L = 2 × 10⁻⁷ ln(D/r′) H/m. Using bundled conductors reduces inductance and corona.
Capacitance per unit length of a single-phase line: C = π ε0 / ln(D/r) per metre between conductors. Transposition balances the phase inductances and capacitances.
A line is short if up to about 80 km, medium from about 80 to 250 km, and long above about 250 km. These are working limits.

4. Line models and ABCD constants

Sending-end relations: Vs = A Vr + B Ir, Is = C Vr + D Ir.

Model	A	B	C	D
Short line (Z only)	1	Z	0	1
Nominal π (medium)	1 + YZ/2	Z	Y(1 + YZ/4)	1 + YZ/2
Nominal T (medium)	1 + YZ/2	Z(1 + YZ/4)	Y	1 + YZ/2
Long line (rigorous)	cosh γl	Zc sinh γl	sinh γl / Zc	cosh γl

For a symmetrical, reciprocal network A = D and AD − BC = 1. B has the unit of impedance (ohm); C of admittance (siemens).

Voltage regulation = (|Vr,no-load| − |Vr,full-load|)/|Vr,full-load|. For a short line this is (|Vs| − |Vr|)/|Vr|. Transmission efficiency = Pr/Ps (receiving-end power over sending-end power).

Ferranti effect: at no load or light load, the receiving-end voltage of a long line is higher than the sending-end voltage, because of charging current flowing through line inductance.

Surge impedance Zc = √(L/C), about 400 Ω for overhead lines (much lower for cables). Surge impedance loading SIL = V²/Zc. For V = 400 kV and Zc = 400 Ω, SIL = 160000/400 = 400 MW. At SIL the line neither supplies nor absorbs reactive power.

5. Underground cables

Parts: conductor (copper or aluminium), insulation (paper, XLPE, PVC), metallic sheath (lead or aluminium, against moisture), bedding, armouring (steel wire or tape) and serving.

Belted cable: used up to about 11 kV. Screened cables (H-type, SL-type): each core is separately screened, which removes tangential stress; used from about 22 to 66 kV. Pressure cables (oil-filled, gas-filled) for 66 kV and above.
Insulation resistance of a cable: R = (ρ / 2πl) ln(r2/r1); it is inversely proportional to cable length.
Capacitance of a single-core cable: C = 2π ε0 εr l / ln(r2/r1).
Grading: capacitance grading (layers of different permittivity) and intersheath grading make the electric stress more uniform. The stress is greatest at the conductor surface.
Cables have higher capacitance and lower inductance than overhead lines, so they have a large charging current. They cost more but are safer and neater.

6. Insulators

Pin type: used up to about 33 kV.
Suspension (disc) type: for higher voltages; units are added to suit the voltage. Flexible.
Strain type: at dead ends and sharp bends.
Shackle type: distribution lines at low voltage.
Materials: porcelain, glass, composite polymer.

String efficiency = (voltage across the whole string)/(n × voltage across the unit nearest the conductor). Voltage across the unit nearest the conductor is highest. This is due to the shunt capacitance between the units and the tower. Example: three discs with 20 V, 30 V and 50 V (total 100 V): efficiency = 100/(3 × 50) = 66.7 percent. Methods to improve it: grading ring (guard ring), using longer cross-arms, and capacitance grading of discs. The efficiency is lower for longer strings.

7. Sag and corona

Sag for supports at the same level: S = w l² / (8 T) (w = weight per unit length, l = span, T = tension). For w = 2 kg/m, l = 200 m, T = 1000 kg, S = 2 × 40000/8000 = 10 m. Length of conductor = l + 8 S²/(3 l). Extra length for S = 10 m, l = 200 m: 8 × 100/600 = 1.33 m.

Corona is the ionisation of air around a conductor when the voltage gradient exceeds the breakdown value of air. Breakdown stress is about 30 kV/cm peak (21.1 kV/cm rms) at normal temperature and pressure. A faint violet glow and a hissing noise appear.

Critical disruptive voltage: the voltage at which corona starts; given by Vc = g0 δ m r ln(D/r) (per phase, rms, with g0 about 21.1 kV/cm).
Visual critical voltage is higher than the disruptive voltage.
Factors: line voltage, conductor radius (larger means less corona), spacing (larger means less), air density factor δ, surface condition (irregularity increases corona), weather (rain and humidity increase).
Effects: power loss, interference with radio and communication, ozone, corrosion of conductors, non-sinusoidal line current.
Methods of reduction: larger conductor diameter, hollow or bundled conductors, increased spacing.

8. Distribution systems

Radial: simple and cheap, but supply fails when the feeder fails and regulation is poor.
Ring main: two paths to each load; more reliable.
Interconnected: several generating stations or feeders joined; most reliable.
Feeder: carries power from the substation to the distribution area with no tapping. Distributor: has tappings for consumers. Service mains connect the distributor to the consumer's premises.
Kelvin's law: the most economical conductor size is where the annual cost of energy loss equals the annual interest and depreciation on the capital for the conductor.
Voltage drop in a two-wire feeder = 2 I R (R per conductor). Three-phase loss = 3 I² R. A 3-phase, 4-wire system with a phase voltage of 230 V has a line voltage of about 400 V.
A DC distributor fed at both ends has its lowest voltage at a point set by load distribution. An AC distributor needs power factor for voltage drop: V drop ≈ I (R cosφ + X sinφ).
Primary distribution is at 11 kV; secondary distribution is at 415/240 V.

Exam traps

1. Load factor uses average load; demand factor uses connected load. Diversity factor is at least 1; load factor is at most 1.
2. A = D = 1 for a short line, C = 0, but C is not zero for medium lines.
3. Nominal-π and nominal-T have the same A and D but different B and C.
4. String efficiency is lowest in long strings; the unit next to the conductor (the lowest disc of the string) is most stressed.
5. Corona loss is reduced by larger radius and larger spacing, not smaller.
6. Skin effect rises with frequency and conductor size.
7. Surge impedance of a cable is lower than that of an overhead line.
8. Ferranti rise happens at no load, not at full load.

One-liners

1. Load factor of a system is at most 1.
2. Pelton for high head, Kaplan for low head.
3. AD − BC = 1 for a passive reciprocal network.
4. Skin effect is the uneven distribution of AC current over the conductor section.
5. Pin insulators are used up to about 33 kV.
6. SIL = V²/Zc.
7. Sag S = wl²/(8T).
8. Bundled conductors reduce corona and line inductance.
9. Kelvin's law decides the economical conductor cross-section.
10. Ring main gives better reliability than radial.
11. A grading ring improves string efficiency.
12. XLPE is a common cable insulation.

Practice questions

Load factor is defined as
1. average load / maximum demand
2. maximum demand / connected load
3. maximum demand / average load
4. connected load / maximum demand
Answer
A. average load / maximum demand
Load factor = average load ÷ maximum demand (at most 1).
A system has an average load of 40 MW and a maximum demand of 80 MW. Its load factor is
1. 2
2. 0.4
3. 0.8
4. 0.5
Answer
D. 0.5
40/80 = 0.5.
The demand factor of a consumer with a connected load of 10 kW and a maximum demand of 6 kW is
1. 0.6
2. 1.67
3. 0.4
4. 4
Answer
A. 0.6
6/10 = 0.6.
The sum of individual maximum demands of consumers is 100 kW and their combined maximum demand is 80 kW. The diversity factor is
1. 20
2. 1.25
3. 0.8
4. 1.8
Answer
B. 1.25
100/80 = 1.25; it is always at least 1.
An average load of 100 kW for 24 hours means energy supplied is
1. 4.17 kWh
2. 2400 kWh
3. 240 kWh
4. 24 kWh
Answer
B. 2400 kWh
100 × 24 = 2400 kWh.
The area under a load duration curve represents
1. the energy generated
2. the maximum demand
3. the plant factor only
4. the connected load
Answer
A. the energy generated
Load × time = energy.
The turbine type best suited to a low-head, high-flow hydro site is
1. Steam
2. Francis (high head)
3. Kaplan
4. Pelton
Answer
C. Kaplan
Kaplan is an axial-flow reaction turbine for low head.
A hydro station has Q = 10 m³/s, head 100 m and overall efficiency 0.9. The power output is about (g = 9.81)
1. 88.3 MW
2. 0.88 MW
3. 9.81 MW
4. 8.83 MW
Answer
D. 8.83 MW
P = 1000×9.81×10×100×0.9 = 8.83 MW.
A Pelton wheel is an
1. reaction turbine for low head
2. axial-flow turbine for low head
3. impulse turbine for high head
4. steam turbine
Answer
C. impulse turbine for high head
Water jets strike buckets at high head.
Gas turbine power stations work on the
1. Rankine cycle
2. Carnot cycle only
3. Otto cycle
4. Brayton cycle
Answer
D. Brayton cycle
Thermal steam stations use the Rankine cycle.
Control rods in a nuclear reactor are used to
1. carry heat
2. absorb neutrons and control the chain reaction
3. shield radiation
4. slow down neutrons
Answer
B. absorb neutrons and control the chain reaction
A moderator slows neutrons; the coolant carries heat.
The A, B, C, D constants of a short transmission line are
1. A = D = 0, B = Z, C = 1
2. A = Z, B = 1, C = 0, D = 1
3. A = D = 1, B = Z, C = 0
4. A = 1, B = 0, C = Y, D = 1
Answer
C. A = D = 1, B = Z, C = 0
Shunt admittance is neglected in a short line.
For a nominal-π model, A equals
1. 1 + YZ
2. 1 + YZ/2
3. Z(1 + YZ/4)
4. 1 − YZ/2
Answer
B. 1 + YZ/2
The nominal-π has A = D = 1 + YZ/2.
For a nominal-T model, the constant C equals
1. Y
2. Y(1 + YZ/4)
3. 1 + YZ/2
4. Z
Answer
A. Y
T model: B = Z(1 + YZ/4), C = Y.
For a passive, reciprocal two-port network the relation among the constants is
1. AD − BC = 1
2. AB − CD = 1
3. AD + BC = 1
4. A + D = 1
Answer
A. AD − BC = 1
The determinant of the transmission matrix is unity.
The constant B of a transmission line has the unit of
1. siemens
2. henry
3. ohm
4. dimensionless
Answer
C. ohm
Vs = A Vr + B Ir, so B is an impedance.
The Ferranti effect is the
1. fall of voltage at full load
2. rise of receiving-end voltage over sending-end voltage on a lightly loaded long line
3. loss due to corona
4. skin effect in cables
Answer
B. rise of receiving-end voltage over sending-end voltage on a lightly loaded long line
Charging current flowing through the line inductance causes it.
A line has a sending-end voltage of 10.5 kV and a full-load receiving-end voltage of 10 kV, with a no-load receiving voltage equal to the sending voltage. The regulation is
1. 4.76%
2. 10%
3. 0.5%
4. 5%
Answer
D. 5%
(10.5 − 10)/10 = 5%.
A line receives 95 MW when 100 MW is sent. The transmission efficiency is
1. 5%
2. 95%
3. 105%
4. 90%
Answer
B. 95%
Pr/Ps = 95/100.
A line has L = 1.6 mH/km and C = 10 nF/km. Its surge impedance is
1. 400 Ω
2. 160 Ω
3. 4000 Ω
4. 40 Ω
Answer
A. 400 Ω
Zc = √(L/C) = √(1.6×10⁻³/10⁻⁸) = √160000 = 400 Ω.
The surge impedance loading of a 400 kV line with Zc = 400 Ω is
1. 40 MW
2. 160 MW
3. 400 MW
4. 1000 MW
Answer
C. 400 MW
SIL = V²/Zc = (400×10³)²/400 = 400 MW.
Skin effect in a conductor
1. has no effect on resistance
2. decreases the AC resistance
3. raises DC resistance
4. increases the AC resistance with frequency
Answer
D. increases the AC resistance with frequency
Current crowds near the surface, so the effective area falls.
Bundled conductors are used in EHV lines to
1. increase corona
2. reduce the capacitance only
3. increase skin effect
4. reduce corona loss and line inductance
Answer
D. reduce corona loss and line inductance
A bundle behaves like a larger-radius conductor.
Statements about overhead line parameters: 1. Transposition balances the line inductances and capacitances of the phases. 2. Larger conductor spacing increases the inductance.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are correct standard facts.
Corona is the ionisation of air around a conductor when
1. the line carries DC only
2. the voltage gradient exceeds the breakdown stress of air
3. the current is too small
4. the line is underground
Answer
B. the voltage gradient exceeds the breakdown stress of air
Breakdown stress is about 30 kV/cm peak.
Which of the following reduces corona loss?
1. A larger conductor diameter
2. A rough conductor surface
3. A smaller spacing between conductors
4. Operating in heavy rain
Answer
A. A larger conductor diameter
A bigger radius lowers the surface stress.
The critical disruptive voltage of a line increases with
1. increase in humidity
2. surface irregularities
3. decrease in air density
4. increase in conductor radius and spacing
Answer
D. increase in conductor radius and spacing
Vc = g0 δ m r ln(D/r) rises with r and D.
Cable insulation resistance is
1. inversely proportional to the cable length
2. directly proportional to the cable length
3. proportional to the square of the length
4. independent of length
Answer
A. inversely proportional to the cable length
Leakage paths are in parallel along the length.
Screened cables (H-type, SL-type) are used to
1. increase armouring
2. reduce the conductor size only
3. remove tangential stress in the insulation
4. increase charging current
Answer
C. remove tangential stress in the insulation
Metallic screens make the electric field radial.
The electric stress in a single-core cable is highest at
1. the sheath
2. the conductor surface
3. the armouring
4. the middle of the insulation
Answer
B. the conductor surface
Stress varies inversely with radius.
Cable grading is done to
1. increase the cable length
2. make the electric stress more uniform in the insulation
3. reduce the temperature rise
4. reduce armouring
Answer
B. make the electric stress more uniform in the insulation
Layers of different permittivity share the stress.
Pin insulators are generally used up to about
1. 400 kV
2. 33 kV
3. 765 kV
4. 1 kV only
Answer
B. 33 kV
For higher voltages suspension insulators are used.
Strain insulators are used
1. only for 400 V lines
2. at dead ends and sharp corners of lines
3. inside substations only
4. only in cables
Answer
B. at dead ends and sharp corners of lines
They withstand the full tension of the line.
In a string of three suspension insulator discs, the voltages across the discs are 20 V, 30 V and 50 V (nearest the conductor). The string efficiency is
1. 66.7%
2. 33.3%
3. 50%
4. 100%
Answer
A. 66.7%
100/(3 × 50) = 0.667.
The unit nearest the conductor in a suspension string has the highest voltage because of
1. lower voltage of that unit
2. higher resistance of porcelain
3. shunt capacitance to earth of the intermediate units
4. the weight of the conductor
Answer
C. shunt capacitance to earth of the intermediate units
Leakage through the capacitance to the tower reduces the voltage across upper units.
A grading ring is used to
1. reduce the span
2. increase conductor sag
3. reduce the line weight
4. improve the string efficiency
Answer
D. improve the string efficiency
It equalises the voltage distribution across units.
Statements about suspension insulators: 1. String efficiency decreases as the number of discs increases. 2. A larger string efficiency means more uniform voltage distribution.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are correct.
The sag of a line with supports at the same level is
1. wl²/(8T)
2. w²l/(8T)
3. wl/(8T)
4. wl²/(4T)
Answer
A. wl²/(8T)
S = w l² / (8T).
A span is 200 m, conductor weight 2 kg/m and tension 1000 kg. The sag is
1. 20 m
2. 5 m
3. 2.5 m
4. 10 m
Answer
D. 10 m
S = 2×200²/(8×1000) = 80000/8000 = 10 m.
Kelvin's law states that the most economical conductor size is where
1. the current density is maximum
2. voltage drop is zero
3. annual energy-loss cost equals annual interest and depreciation on conductor cost
4. the conductor is thinnest
Answer
C. annual energy-loss cost equals annual interest and depreciation on conductor cost
It balances running and capital costs.
Which distribution system is the most reliable?
1. Interconnected system
2. Radial system
3. Open feeder with a single source
4. Single-wire earth return
Answer
A. Interconnected system
Several sources and paths give continuity.
A 3-phase, 4-wire system has a phase voltage of 230 V. The line voltage is about
1. 132 V
2. 230 V
3. 690 V
4. 400 V
Answer
D. 400 V
230 × √3 = 398 V ≈ 400 V.
A two-wire feeder has a resistance of 0.1 Ω per conductor and carries 100 A. The voltage drop in the feeder is
1. 20 V
2. 10 V
3. 40 V
4. 5 V
Answer
A. 20 V
Drop = 2 × 100 × 0.1 = 20 V.
A three-phase line carries 100 A per phase with a resistance of 0.1 Ω per phase. The total I²R loss is
1. 30 kW
2. 300 W
3. 1 kW
4. 3 kW
Answer
D. 3 kW
3 × 100² × 0.1 = 3000 W.
The part of a distribution system with no tappings between the substation and the distribution area is the
1. bus-bar
2. service main
3. feeder
4. distributor
Answer
C. feeder
A feeder carries bulk power; a distributor has tappings.

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1.Which of the following principle is related to solar power systems?
2.The central nervous system that receives information from all parts of the body and integrates it is constituted by:
3.The fundamental unit which is common in FPS and MKS systems is _________.
4.The formula to find the power is-
5.The generation of heat in a conductor due to electric current is known as:
6.The efficiency of a pump is 50%. It lifts 2 tonnes of water by 60 m in 20 minutes. Find the input power (in kW) of the pump.
7.The electric power consumed by a resistor of resistance R, carrying current I, is _______
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