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​ = 15 Acos φ = 0.95lagging

A balanced three-phase load draws a line current of 15 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.95 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 15 Acos φ = 0.95
P = kW
2.Three-phase powerHard
P = 200 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 200 kW at a power factor of 0.72 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 200 kWpf1 = 0.72pf2 = 0.9VL = 690 Vf = 50 Hz
Cph = µF
3.TransformersMedium
7200 V240 VIHV​ILV​ = ?10 kVA

A single-phase 10 kVA, 7200/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 10 kVAV1 = 7200 VV2 = 240 V
ILV = A
4.TransformersHard
pf 0.85500 kVA, η = 97.5%

A single-phase 500 kVA transformer has a core loss of 3000 W. At rated load and a power factor of 0.85 lagging its efficiency is 97.5%. Find the full-load copper loss Pcu,fl in watts.

S = 500 kVAPcore = 3000 Wpf = 0.85η = 97.5%
Pcu,fl = W
5.AC power & power factorMedium
P = 8 kWQSφ

A single-phase motor takes 8 kW at a power factor of 0.9 lagging from a 230 V supply. Find the line current I.

P = 8 kWpf = 0.9V = 230 V
I = A
6.AC power & power factorHard
230 V50 Hzload13 kW, 0.7C220 µFI = ?

A single-phase load takes 13 kW at a power factor of 0.7 lagging from a 230 V, 50 Hz supply. A 220 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 13 kWpf1 = 0.7V = 230 Vf = 50 HzC = 220 µF
I = A
7.AC impedance & phasorsMedium
Vf0​ = 2000 HzRL = ?C = 4.7 µF

A series RLC circuit with C = 4.7 µF must resonate at 2000 Hz. Find the inductance L needed.

f0 = 2000 HzC = 4.7 µF
L = mH
8.AC impedance & phasorsHard
2 Vf0​R = 15 ΩL = 20 mHC = 1 µF

A series RLC circuit is driven by a 2 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 2 V (rms)R = 15 ΩL = 20 mHC = 1 µF
VL = V
9.DC motorsMedium
+−240 VRa​ = 1.5 ΩIa​ = 6 AM+−EωT

A DC motor draws 6 A from a 240 V supply. Its armature resistance is 1.5 Ω and its rotational (friction, windage and iron) losses are 42 W. Find its efficiency η.

V = 240 VIa = 6 ARa = 1.5 ΩProt = 42 W
η = %
10.DC motorsHard
+−120 VRa​ = 1.2 ΩM+−En = ?T = 4.5 N·m

A separately excited DC motor (k = 0.75 V·s/rad at full field, Ra = 1.2 Ω, V = 120 V) drives a constant load torque of 4.5 N·m. The field flux is reduced to 70 % of its full value. Find the new speed n.

V = 120 Vk = 0.75 V·s/radRa = 1.2 ΩT = 4.5 N·mΦ = 70 %
n = rpm

3Solutions

1.Three-phase power

Answer: P = 17.03 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 = √3·690·15 = 17.93 kVA

Real power

P = S·cos φ = 17.93·0.95 = 17.03 kW

Answer

P = 17.03 kW
2.Three-phase power

Answer: Cph = 641.2 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 = 200·tan(acos 0.72) = 200·0.9639 = 192.8 kvar
Q2 = 200·tan(acos 0.9) = 200·0.4843 = 96.86 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 95.91 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·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 31969/(314.2·398.4²) = 641.2 µF

Answer

Cph = 641.2 µF
3.Transformers

Answer: ILV = 41.67 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 = 10000/240 = 41.67 A

Answer

ILV = 41.67 A
4.Transformers

Answer: Pcu,fl = 7897 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.

Output and input power

Pout = 500·0.85 = 425 kW
Pin = 425000/0.975 = 435897 W

Total and copper losses

Ploss = 435897 − 425000 = 10897 W
Pcu,fl = 10897 − 3000 = 7897 W

Answer

Pcu,fl = 7897 W
5.AC power & power factor

Answer: I = 38.65 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) = 8000/(230·0.9) = 38.65 A

Answer

I = 38.65 A
6.AC power & power factor

Answer: I = 70.28 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.

Reactive power before and after

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

Q1 = 13·tan(arccos 0.7) = 13.26 kvar
QC = V²·2πf·C = 230²·2π·50·220·10⁻⁶ = 3656 var = 3.656 kvar
Q2 = 13.26 − 3.656 = 9.606 kvar

New apparent power

S2 = √(13² + 9.606²) = 16.16 kVA

New line current

I = S2/V = 16164/230 = 70.28 A

Answer

I = 70.28 A
7.AC impedance & phasors

Answer: L = 1.347 mH

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·2000 = 12566 rad/s
L = 1/(ω₀²·C) = 1/(12566²·4.7·10⁻⁶) = 1.347 mH

Answer

L = 1.347 mH
8.AC impedance & phasors

Answer: VL = 18.86 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/√(0.02·1·10⁻⁶) = 7071 rad/s
I = V/R = 2/15 = 0.1333 A

Voltage across the inductor

XL = ω₀L = 7071·0.02 = 141.4 Ω
VL = I·XL = 0.1333·141.4 = 18.86 V

Answer

VL = 18.86 V

This is Q = 9.428 times the supply voltage.

9.DC motors

Answer: η = 93.33 %

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 = 240·6 = 1440 W
Pdev = 231·6 = 1386 W
Pout = 1386 − 42 = 1344 W

Efficiency

η = 1344/1440·100 = 93.33 %

Answer

η = 93.33 %
10.DC motors

Answer: n = 1996 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.

Weakened motor constant

k₂ = 0.7·0.75 = 0.525 V·s/rad

New armature current

Ia = 4.5/0.525 = 8.571 A

New speed

ω = (120 − 8.571·1.2)/0.525 = 209 rad/s → n = 1996 rpm

Answer

n = 1996 rpm

Field weakening is how a DC drive runs above its base speed.

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