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​ = 208 VNZ = 6 + j8 Ωper phase

A balanced Y-connected load with an impedance of 6 + j8 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.

VL = 208 VZ = 6 + j8 Ω per phase
P = kW
2.Three-phase powerHard
P = 60 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 60 kW at a power factor of 0.72 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 60 kWpf1 = 0.72pf2 = 0.92VL = 480 Vf = 60 Hz
Cph = µF
3.TransformersMedium
50 Ω121

A 50 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 12:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 12:1RL = 50 Ω
R1′ = Ω
4.TransformersHard
100 kVA

A single-phase 100 kVA transformer has a core loss of 600 W and a copper loss of 2400 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 100 kVAPcore = 600 WPcu,fl = 2400 W
Sη,max = kVA
5.AC power & power factorMedium
230 VZ12 + j16 ΩI

A load with impedance Z = 12 + j16 Ω is connected to a 230 V (rms) single-phase supply. Find the apparent power S.

V = 230 VZ = 12 + j16 Ω
S = VA
6.AC power & power factorHard
P = 5 kWQ1​S1​φ1​QC​

A single-phase load takes 5 kW at a power factor of 0.65 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the capacitance C needed.

P = 5 kWpf1 = 0.65pf2 = 0.95V = 240 Vf = 60 Hz
C = µF
7.AC impedance & phasorsMedium
230 V50 HzR = 30 ΩL = 20 mHC = 100 µFI

A series RLC circuit is connected to a 230 V (rms), 50 Hz supply. Find the power factor cos φ of the circuit.

V = 230 V (rms)f = 50 HzR = 30 ΩL = 20 mHC = 100 µF
cos φ =
8.AC impedance & phasorsHard
12 V60 HzR = 15 ΩL = 100 mHC = 100 µF+−VC​

A series RLC circuit is connected to a 12 V (rms), 60 Hz supply. Find the rms voltage VC across the capacitor.

V = 12 V (rms)f = 60 HzR = 15 ΩL = 100 mHC = 100 µF
VC = V
9.DC motorsMedium
T (N·m)n (rpm)60 N·mn0​nFL​

A DC motor (k = 1.5 V·s/rad, Ra = 0.4 Ω) runs from 240 V. At full load it draws 40 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 240 Vk = 1.5 V·s/radRa = 0.4 ΩIFL = 40 A
SR = %
10.DC motorsHard
+−110 VRa​ = 3 ΩM+−EωT

A small DC motor has an armature resistance of 3 Ω and runs from 110 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.

V = 110 VRa = 3 Ω
Pmax = W

3Solutions

1.Three-phase power

Answer: P = 2.596 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.

Phase voltage

Vph = 208/√3 = 120.1 V

Phase current

|Z| = √(6² + 8²) = 10 Ω
Iph = 120.1/10 = 12.01 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·12.01²·6 = 2.596 kW

Answer

P = 2.596 kW

Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.

2.Three-phase power

Answer: Cph = 123.8 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 = 60·tan(acos 0.72) = 60·0.9639 = 57.83 kvar
Q2 = 60·tan(acos 0.92) = 60·0.426 = 25.56 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 32.27 kvar

Capacitance per phase

In a Δ bank each capacitor is connected between two lines.

V = 480 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 10757/(377·480²) = 123.8 µF

Answer

Cph = 123.8 µF
3.Transformers

Answer: R1′ = 7200 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 = 12²·50 = 7200 Ω

Answer

R1′ = 7200 Ω
4.Transformers

Answer: Sη,max = 50 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

x²·2400 = 600 → x = √(600/2400) = 0.5

Load at maximum efficiency

Sη,max = x·S = 0.5·100 = 50 kVA

Answer

Sη,max = 50 kVA
5.AC power & power factor

Answer: S = 2645 VA

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.

Current

|Z| = √(12² + 16²) = 20 Ω
I = V/|Z| = 230/20 = 11.5 A

Apparent power

S = V·I = 230·11.5 = 2645 VA

Answer

S = 2645 VA
6.AC power & power factor

Answer: C = 193.5 µF

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 = 5·tan(49.46°) = 5.846 kvar
Q2 = P·tan φ2 = 5·tan(18.19°) = 1.643 kvar
QC = Q1 − Q2 = 4.202 kvar

Capacitance

A capacitor across the supply delivers Q = V²·ωC.

C = QC/(2πf·V²) = 4202/(2π·60·240²) = 193.5 µF

Answer

C = 193.5 µF
7.AC impedance & phasors

Answer: cos φ = 0.7613

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

XL = 6.283 Ω, XC = 31.83 Ω

Impedance

|Z| = √(30² + (−25.55)²) = 39.4 Ω

Power factor

cos φ = R/|Z| = 30/39.4 = 0.7613
The circuit is capacitive, so the power factor is leading.

Answer

cos φ = 0.7613
8.AC impedance & phasors

Answer: VC = 17.02 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|.

Reactances

ω = 2πf = 2π·60 = 377 rad/s
XL = ωL = 377·0.1 = 37.7 Ω
XC = 1/(ωC) = 1/(377·100·10⁻⁶) = 26.53 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (11.17)²) = 18.7 Ω

Current

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

I = V/|Z| = 12/18.7 = 0.6416 A

Voltage across the capacitor

VC = I·XC = 0.6416·26.53 = 17.02 V

Answer

VC = 17.02 V
9.DC motors

Answer: SR = 7.143 %

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.

No-load and full-load speed

n₀ = 60·240/(2π·1.5) = 1528 rpm
nFL = 60·(240 − 40·0.4)/(2π·1.5) = 1426 rpm

Speed regulation

SR = (1528 − 1426)/1426·100 = 7.143 %

Answer

SR = 7.143 %
10.DC motors

Answer: Pmax = 1008 W

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.

Power as a function of the back-EMF

P(E) = E·(V − E)/Ra
dP/dE = 0 → E = V/2 = 55 V

Maximum power

Pmax = (V/2)²/Ra = V²/(4Ra) = 110²/(4·3) = 1008 W

Answer

Pmax = 1008 W

At this point the efficiency is only 50 %, so real motors run well above half speed.

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