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​ = 400 VIL​ = ?M7.5 kW, η = 92%cos φ = 0.86

A three-phase induction motor delivers 7.5 kW of mechanical output power at full load. At this load its efficiency is 92% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.

Pout = 7.5 kWη = 92%cos φ = 0.86VL = 400 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 0.7 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 = 0.7pf2 = 0.98VL = 690 Vf = 60 Hz
Cph = µF
3.TransformersMedium
10 Ω21

A 10 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 2:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 2:1RL = 10 Ω
R1′ = Ω
4.TransformersHard
pf 0.85200 kVA, 50% load

A single-phase 200 kVA transformer has a core loss of 1600 W and a copper loss of 2400 W at rated load. It supplies 50% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.

S = 200 kVAPcore = 1600 WPcu,fl = 2400 Wx = 0.5pf = 0.85
Ploss = W
5.AC power & power factorMedium
P = 9 kWQ = ?Sφ

A single-phase motor takes 9 kW at a power factor of 0.9 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 9 kWpf = 0.9V = 230 V
Q = kvar
6.AC power & power factorHard
230 V1000 HzRs​ = 4 ΩXs​ = j30 ΩZL​ = ?

A 230 V (rms), 1000 Hz source has an internal impedance Zs = 4 + j30 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.

Vs = 230 V (rms)f = 1000 HzZs = 4 + j30 Ω
Pmax = W
7.AC impedance & phasorsMedium
24 V100 HzR = 10 ΩL = 250 mHC = 47 µFI

A series RLC circuit is connected to a 24 V (rms), 100 Hz supply. Find the rms current I.

V = 24 V (rms)f = 100 HzR = 10 ΩL = 250 mHC = 47 µF
I = A
8.AC impedance & phasorsHard
VfR = 20 ΩL = 50 mHC = 1 µF

A series RLC circuit has R = 20 Ω, L = 50 mH and C = 1 µF. Find its quality factor Q at resonance.

R = 20 ΩL = 50 mHC = 1 µF
Q =
9.DC motorsMedium
+−110 VRa​ = 0.4 ΩIa​ = 12 AM+−EωT

A DC motor draws 12 A from a 110 V supply. Its armature resistance is 0.4 Ω and its rotational (friction, windage and iron) losses are 50 W. Find its efficiency η.

V = 110 VIa = 12 ARa = 0.4 ΩProt = 50 W
η = %
10.DC motorsHard
+−120 VRs​ = ?Ra​ = 0.4 ΩM+−En = 800 rpmT = 15 N·m

A DC motor (k = 0.75 V·s/rad, Ra = 0.4 Ω) on a fixed 120 V supply drives a load of 15 N·m. Find the series resistance Rs that slows it down to 800 rpm.

V = 120 Vk = 0.75 V·s/radRa = 0.4 ΩT = 15 N·mn = 800 rpm
Rs = Ω

3Solutions

1.Three-phase power

Answer: IL = 13.68 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 = 7.5/0.92 = 8.152 kW

Line current

IL = Pin/(√3·VL·cos φ) = 8152/(√3·400·0.86) = 13.68 A

Answer

IL = 13.68 A
2.Three-phase power

Answer: Cph = 1138 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 0.7) = 250·1.02 = 255.1 kvar
Q2 = 250·tan(acos 0.98) = 250·0.2031 = 50.76 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 204.3 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) = 68095/(377·398.4²) = 1138 µF

Answer

Cph = 1138 µF
3.Transformers

Answer: R1′ = 40 Ω

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.

Referred resistance

The voltage goes up by a and the current down by a, so the ratio V/I goes up by a².

R1′ = a²·RL = 2²·10 = 40 Ω

Answer

R1′ = 40 Ω
4.Transformers

Answer: Ploss = 2200 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 = 0.25·2400 = 600 W

Total losses

The core loss is the same at any load.

Ploss = 1600 + 600 = 2200 W

Answer

Ploss = 2200 W

The efficiency at this load is η = 97.48%.

5.AC power & power factor

Answer: Q = 4.359 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.

Angle

φ = arccos(0.9) = 25.84°, tan φ = 0.4843

Reactive power

Q = P·tan φ = 9·0.4843 = 4.359 kvar

Answer

Q = 4.359 kvar
6.AC power & power factor

Answer: Pmax = 3306 W

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.

Conjugate match

The load reactance cancels the source reactance, and the load resistance equals the source resistance.

ZL = Zs* = 4 − j30 Ω

Maximum power

I = Vs/(2Rs) = 230/8 = 28.75 A
Pmax = I²·RL = Vs²/(4Rs) = 230²/(4·4) = 3306 W

Answer

Pmax = 3306 W
7.AC impedance & phasors

Answer: I = 0.1941 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|.

Reactances

ω = 2πf = 2π·100 = 628.3 rad/s
XL = ωL = 628.3·0.25 = 157.1 Ω
XC = 1/(ωC) = 1/(628.3·47·10⁻⁶) = 33.86 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(10² + (123.2)²) = 123.6 Ω

Current

The same current flows through every element of a series circuit.

I = V/|Z| = 24/123.6 = 0.1941 A

Answer

I = 0.1941 A
8.AC impedance & phasors

Answer: Q = 11.18

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|.

Resonant frequency

f0 = 1/(2π·√(0.05·1·10⁻⁶)) = 711.8 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/20)·√(0.05/(1·10⁻⁶)) = 11.18

Answer

Q = 11.18
9.DC motors

Answer: η = 91.85 %

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 = 110·12 = 1320 W
Pdev = 105.2·12 = 1262 W
Pout = 1262 − 50 = 1212 W

Efficiency

η = 1212/1320·100 = 91.85 %

Answer

η = 91.85 %
10.DC motors

Answer: Rs = 2.458 Ω

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.

Current and back-EMF

Ia = 15/0.75 = 20 A
E = 0.75·83.78 = 62.83 V

Series resistance

Ra + Rs = (120 − 62.83)/20 = 2.858 Ω
Rs = 2.458 Ω

Answer

Rs = 2.458 Ω

The resistor wastes 983.4 W; a chopper controls the speed with far less loss.

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