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​ = 690 VIL​ = 60 Acos φ = 0.88lagging

A balanced three-phase load draws a line current of 60 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.88 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 60 Acos φ = 0.88
P = kW
2.Three-phase powerHard
P = 80 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 80 kW at a power factor of 0.8 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. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 80 kWpf1 = 0.8pf2 = 0.98VL = 690 Vf = 60 Hz
QC = kvar
3.TransformersMedium
2400 V240 VIHV​ = ?ILV​10 kVA

A single-phase 10 kVA, 2400/240 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 10 kVAV1 = 2400 VV2 = 240 V
IHV = A
4.TransformersHard
75 kVA

A single-phase 75 kVA transformer has a core loss of 450 W and a copper loss of 1350 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 75 kVAPcore = 450 WPcu,fl = 1350 W
Sη,max = kVA
5.AC power & power factorMedium
120 VR + jXP = 792 W11 A

An inductive load takes 11 A and 792 W from a 120 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find X.

V = 120 VI = 11 AP = 792 W
X = Ω
6.AC power & power factorHard
P = 50 kWQ1​S1​φ1​QC​

A single-phase load takes 50 kW at a power factor of 0.7 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the reactive power QC the capacitor must supply.

P = 50 kWpf1 = 0.7pf2 = 0.9V = 230 Vf = 60 Hz
QC = kvar
7.AC impedance & phasorsMedium
100 V200 HzR = 20 ΩL = 20 mHI

A resistor and an inductor are connected in parallel across a 100 V (rms), 200 Hz supply. Find the rms current I.

V = 100 V (rms)f = 200 HzR = 20 ΩL = 20 mH
I = A
8.AC impedance & phasorsHard
24 Vf0​R = 20 ΩL = 10 mHC = 4.7 µF

A series RLC circuit is driven by a 24 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 24 V (rms)R = 20 ΩL = 10 mHC = 4.7 µF
VL = V
9.DC motorsMedium
+−220 VRa​ = 0.4 ΩIa​ = 25 AM+−EωT

A DC motor draws 25 A from a 220 V supply. Its armature resistance is 0.4 Ω and its rotational (friction, windage and iron) losses are 420 W. Find its efficiency η.

V = 220 VIa = 25 ARa = 0.4 ΩProt = 420 W
η = %
10.DC motorsHard
T (N·m)n (rpm)3.6 N·m600 rpm

A DC motor with k = 0.6 V·s/rad and Ra = 2 Ω must drive a load of 3.6 N·m at 600 rpm. Find the armature voltage V it needs.

k = 0.6 V·s/radRa = 2 ΩT = 3.6 N·mn = 600 rpm
V = V

3Solutions

1.Three-phase power

Answer: P = 63.1 kW

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.

Apparent power

S = √3·VL·IL = √3·690·60 = 71.71 kVA

Real power

P = S·cos φ = 71.71·0.88 = 63.1 kW

Answer

P = 63.1 kW
2.Three-phase power

Answer: QC = 43.76 kvar

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 = 80·tan(acos 0.8) = 80·0.75 = 60 kvar
Q2 = 80·tan(acos 0.98) = 80·0.2031 = 16.24 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 43.76 kvar

Answer

QC = 43.76 kvar
3.Transformers

Answer: IHV = 4.167 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.

IHV = S/V = 10000/2400 = 4.167 A

Answer

IHV = 4.167 A
4.Transformers

Answer: Sη,max = 43.3 kVA

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 equals core loss

x²·1350 = 450 → x = √(450/1350) = 0.5774

Load at maximum efficiency

Sη,max = x·S = 0.5774·75 = 43.3 kVA

Answer

Sη,max = 43.3 kVA
5.AC power & power factor

Answer: X = 8.727 Ω

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.

Resistance from the real power

R = P/I² = 792/11² = 6.545 Ω

Reactance from the impedance

|Z| = V/I = 120/11 = 10.91 Ω
X = √(|Z|² − R²) = √(10.91² − 6.545²) = 8.727 Ω

Answer

X = 8.727 Ω
6.AC power & power factor

Answer: QC = 26.79 kvar

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 = P·tan φ1 = 50·tan(45.57°) = 51.01 kvar
Q2 = P·tan φ2 = 50·tan(25.84°) = 24.22 kvar
QC = Q1 − Q2 = 26.79 kvar

Answer

QC = 26.79 kvar
7.AC impedance & phasors

Answer: I = 6.39 A

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|.

Reactance

XL = ωL = 2π·200·0.02 = 25.13 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

IR = V/R = 100/20 = 5 A
IX = V/XL = 100/25.13 = 3.979 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(IR² + IX²) = √(5² + 3.979²) = 6.39 A

Answer

I = 6.39 A
8.AC impedance & phasors

Answer: VL = 55.35 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/√(0.01·4.7·10⁻⁶) = 4613 rad/s
I = V/R = 24/20 = 1.2 A

Voltage across the inductor

XL = ω₀L = 4613·0.01 = 46.13 Ω
VL = I·XL = 1.2·46.13 = 55.35 V

Answer

VL = 55.35 V

This is Q = 2.306 times the supply voltage.

9.DC motors

Answer: η = 87.82 %

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 = 220·25 = 5500 W
Pdev = 210·25 = 5250 W
Pout = 5250 − 420 = 4830 W

Efficiency

η = 4830/5500·100 = 87.82 %

Answer

η = 87.82 %
10.DC motors

Answer: V = 49.7 V

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

E = 0.6·62.83 = 37.7 V

Armature current

Ia = 3.6/0.6 = 6 A

Armature voltage

V = 37.7 + 6·2 = 49.7 V

Answer

V = 49.7 V

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