Power Systems and Electrical Machines: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Three-phase power

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

Transformers

Turns ratio
a = N1/N2 = V1/V2
Current ratio
I1/I2 = N2/N1 = 1/a
Rated current
I = S/V
Referred impedance
Z1′ = a²·ZL
Copper loss at part load
Pcu = x²·Pcu,fl
Efficiency
η = x·S·pf/(x·S·pf + Pcore + x²·Pcu,fl)
Maximum efficiency
x = √(Pcore/Pcu,fl)

AC power & power factor

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

AC impedance & phasors

Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z

Method

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

DC motors

Armature circuit
V = E + IaRa
Back-EMF and torque
E = k·ω, T = k·Ia
Speed
ω = 2π·n/60, ω = (V − IaRa)/k
No-load speed and stall torque
ω0 = V/k, Tstall = k·V/Ra
Powers
Pin = V·Ia, Pdev = E·Ia = T·ω, Pout = Pdev − Prot
Starting
Istart = V/(Ra + Rst)
Speed regulation
SR = (n0 − nFL)/nFL
Maximum developed power
Pmax = V²/(4Ra)
Chopper drive
Va = D·Vdc

Method

  1. Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
  2. Use V = E + Ia·Ra to link voltage, current and back-EMF.
  3. Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
  4. For power and efficiency, start from V·Ia, subtract the copper loss Ia²·Ra and then the rotational losses.

2Practice problems

1.Three-phase powerMedium
ABCVL​ = 415 VIL​ = ?M110 kW, η = 88%cos φ = 0.8

A three-phase induction motor delivers 110 kW of mechanical output power at full load. At this load its efficiency is 88% and its power factor is 0.8 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.

Pout = 110 kWη = 88%cos φ = 0.8VL = 415 V
IL = A
2.Three-phase powerHard
P = 250 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 250 kW at a power factor of 0.78 lagging from a 690 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.78pf2 = 0.98VL = 690 Vf = 60 Hz
Cph = µF
3.TransformersMedium
4160 V480 VIHV​ILV​ = ?100 kVA

A single-phase 100 kVA, 4160/480 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 100 kVAV1 = 4160 VV2 = 480 V
ILV = A
4.TransformersHard
pf 0.975 kVA, 75% load

A single-phase 75 kVA transformer has a core loss of 225 W and a copper loss of 562.5 W at rated load. It supplies 75% of its rated kVA at a power factor of 0.9 lagging. Find the total losses Ploss in watts at this load.

S = 75 kVAPcore = 225 WPcu,fl = 562.5 Wx = 0.75pf = 0.9
Ploss = W
5.AC power & power factorMedium
P = 12 kWQSφ

A single-phase motor takes 12 kW at a power factor of 0.9 lagging from a 240 V supply. Find the line current I.

P = 12 kWpf = 0.9V = 240 V
I = A
6.AC power & power factorHard
400 V50 Hzload9 kW, 0.7C33 µFI

A single-phase load takes 9 kW at a power factor of 0.7 lagging from a 400 V, 50 Hz supply. A 33 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 9 kWpf1 = 0.7V = 400 Vf = 50 HzC = 33 µF
cos φ =
7.AC impedance & phasorsMedium
230 V1000 HzR = 100 ΩL = 150 mHC = 2.2 µF

A series RLC circuit is connected to a 230 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 230 V (rms)f = 1000 HzR = 100 ΩL = 150 mHC = 2.2 µF
φ = °
8.AC impedance & phasorsHard
5 Vf0​R = 15 ΩL = 20 mHC = 0.22 µF

A series RLC circuit is driven by a 5 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 5 V (rms)R = 15 ΩL = 20 mHC = 0.22 µF
VL = V
9.DC motorsMedium
+−110 VRa​ = 0.8 ΩIa​ = 8 AM+−EωT

A DC motor draws 8 A from a 110 V supply. Its armature resistance is 0.8 Ω and its rotational (friction, windage and iron) losses are 25 W. Find its efficiency η.

V = 110 VIa = 8 ARa = 0.8 ΩProt = 25 W
η = %
10.DC motorsHard
+−110 VRa​ = 1.2 ΩIa​: 8 → 16 AM+−En = ?T

A separately excited DC motor on 110 V (Ra = 1.2 Ω) runs at 1000 rpm with an armature current of 8 A. The load changes and the current becomes 16 A, with the field unchanged. Find the new speed n2.

V = 110 VRa = 1.2 Ωn1 = 1000 rpmIa1 = 8 AIa2 = 16 A
n2 = rpm

3Solutions

1.Three-phase power

Answer: IL = 217.4 A

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Electrical input power

Pin = Pout/η = 110/0.88 = 125 kW

Line current

IL = Pin/(√3·VL·cos φ) = 125000/(√3·415·0.8) = 217.4 A

Answer

IL = 217.4 A
2.Three-phase power

Answer: Cph = 834.6 µF

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 250·tan(acos 0.78) = 250·0.8023 = 200.6 kvar
Q2 = 250·tan(acos 0.98) = 250·0.2031 = 50.76 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 149.8 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 690/√3 = 398.4 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 49935/(377·398.4²) = 834.6 µF

Answer

Cph = 834.6 µF
3.Transformers

Answer: ILV = 208.3 A

Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.

Rated current

Both windings carry the same apparent power.

ILV = S/V = 100000/480 = 208.3 A

Answer

ILV = 208.3 A
4.Transformers

Answer: Ploss = 541.4 W

Transformer losses: the core loss (hysteresis and eddy currents) depends on the voltage and is the same at any load. The copper loss in the windings goes with the current squared, so at a fraction x of rated load it is x²·Pcu,fl. Efficiency η = Pout/(Pout + Pcore + Pcu), with Pout = x·S·pf. The efficiency is highest at the load where the copper loss equals the core loss.

Copper loss at this load

Pcu = x²·Pcu,fl = 0.5625·562.5 = 316.4 W

Total losses

The core loss is the same at any load.

Ploss = 225 + 316.4 = 541.4 W

Answer

Ploss = 541.4 W

The efficiency at this load is η = 98.94%.

5.AC power & power factor

Answer: I = 55.56 A

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Line current

Only the real part of the power does work, so divide by the power factor.

I = P/(V·cos φ) = 12000/(240·0.9) = 55.56 A

Answer

I = 55.56 A
6.AC power & power factor

Answer: cos φ = 0.7673

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = 9·tan(arccos 0.7) = 9.182 kvar
QC = V²·2πf·C = 400²·2π·50·33·10⁻⁶ = 1659 var = 1.659 kvar
Q2 = 9.182 − 1.659 = 7.523 kvar

New apparent power

S2 = √(9² + 7.523²) = 11.73 kVA

New power factor

cos φ = P/S2 = 9/11.73 = 0.7673

Answer

cos φ = 0.7673
7.AC impedance & phasors

Answer: φ = 83.44 °

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Reactances

ω = 2πf = 2π·1000 = 6283 rad/s
XL = ωL = 6283·0.15 = 942.5 Ω
XC = 1/(ωC) = 1/(6283·2.2·10⁻⁶) = 72.34 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(XL − XC).

|Z| = √(R² + (XL − XC)²) = √(100² + (870.1)²) = 875.9 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(870.1/100) = 83.44°

Answer

φ = 83.44°
8.AC impedance & phasors

Answer: VL = 100.5 V

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.02·0.22·10⁻⁶) = 15076 rad/s
I = V/R = 5/15 = 0.3333 A

Voltage across the inductor

XL = ω₀L = 15076·0.02 = 301.5 Ω
VL = I·XL = 0.3333·301.5 = 100.5 V

Answer

VL = 100.5 V

This is Q = 20.1 times the supply voltage.

9.DC motors

Answer: η = 91.34 %

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Powers

Pin = 110·8 = 880 W
Pdev = 103.6·8 = 828.8 W
Pout = 828.8 − 25 = 803.8 W

Efficiency

η = 803.8/880·100 = 91.34 %

Answer

η = 91.34 %
10.DC motors

Answer: n2 = 904.4 rpm

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Back-EMF before and after

E1 = 110 − 8·1.2 = 100.4 V
E2 = 110 − 16·1.2 = 90.8 V

New speed

n2 = 1000·90.8/100.4 = 904.4 rpm

Answer

n2 = 904.4 rpm

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