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 = 50 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 50 kW at a power factor of 0.72 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 50 kWpf1 = 0.72pf2 = 0.95VL = 400 Vf = 60 Hz
Cph = µF
3.TransformersMedium
4160 V480 VIHV​ILV​ = ?500 kVA

A single-phase 500 kVA, 4160/480 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 500 kVAV1 = 4160 VV2 = 480 V
ILV = A
4.TransformersHard
150 kVA

A single-phase 150 kVA transformer has a core loss of 900 W and a copper loss of 2700 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 150 kVAPcore = 900 WPcu,fl = 2700 W
Sη,max = kVA
5.AC power & power factorMedium
P = 33 kWQ = ?Sφ

A single-phase motor takes 33 kW at a power factor of 0.85 lagging from a 400 V supply. Find the reactive power Q it draws.

P = 33 kWpf = 0.85V = 400 V
Q = kvar
6.AC power & power factorHard
P = 35 kWQ1​S1​φ1​QC​

A single-phase load takes 35 kW at a power factor of 0.7 lagging from a 240 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.

P = 35 kWpf1 = 0.7pf2 = 0.98V = 240 Vf = 50 Hz
Inew = A
7.AC impedance & phasorsMedium
12 V100 HzR = 30 ΩL = 200 mHC = 100 µFI

A series RLC circuit is connected to a 12 V (rms), 100 Hz supply. Find the real power P taken from the supply.

V = 12 V (rms)f = 100 HzR = 30 ΩL = 200 mHC = 100 µF
P = W
8.AC impedance & phasorsHard
230 V200 HzR = 100 ΩL = 20 mHC = 2.2 µF

A series RLC circuit is connected to a 230 V (rms), 200 Hz supply. Find the resonant frequency f0 of the circuit.

V = 230 V (rms)f = 200 HzR = 100 ΩL = 20 mHC = 2.2 µF
f0 = Hz
9.DC motorsMedium
T (N·m)n (rpm)0.24 N·mn0​n = ?

A DC motor (k = 0.12 N·m/A, Ra = 1.5 Ω) runs from 24 V and drives a load torque of 0.24 N·m. Find its speed n in rpm.

V = 24 Vk = 0.12 N·m/ARa = 1.5 ΩT = 0.24 N·m
n = rpm
10.DC motorsHard
+−120 VRa​ = 0.4 ΩIa​: 10 → 15 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 0.4 Ω) runs at 1440 rpm with an armature current of 10 A. The load changes and the current becomes 15 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 0.4 Ωn1 = 1440 rpmIa1 = 10 AIa2 = 15 A
n2 = rpm

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 = 175.5 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 = 50·tan(acos 0.72) = 50·0.9639 = 48.19 kvar
Q2 = 50·tan(acos 0.95) = 50·0.3287 = 16.43 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 31.76 kvar

Capacitance per phase

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

V = 400 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 10586/(377·400²) = 175.5 µF

Answer

Cph = 175.5 µF
3.Transformers

Answer: ILV = 1042 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 = 500000/480 = 1042 A

Answer

ILV = 1042 A
4.Transformers

Answer: Sη,max = 86.6 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

x²·2700 = 900 → x = √(900/2700) = 0.5774

Load at maximum efficiency

Sη,max = x·S = 0.5774·150 = 86.6 kVA

Answer

Sη,max = 86.6 kVA
5.AC power & power factor

Answer: Q = 20.45 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(0.85) = 31.79°, tan φ = 0.6197

Reactive power

Q = P·tan φ = 33·0.6197 = 20.45 kvar

Answer

Q = 20.45 kvar
6.AC power & power factor

Answer: Inew = 148.8 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 = P·tan φ1 = 35·tan(45.57°) = 35.71 kvar
Q2 = P·tan φ2 = 35·tan(11.48°) = 7.107 kvar
QC = Q1 − Q2 = 28.6 kvar

New line current

Inew = P/(V·cos φ2) = 35000/(240·0.98) = 148.8 A

Answer

Inew = 148.8 A
7.AC impedance & phasors

Answer: P = 0.3337 W

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 = 125.7 Ω, XC = 15.92 Ω

Impedance

|Z| = √(30² + (109.7)²) = 113.8 Ω

Real power

Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.

I = 12/113.8 = 0.1055 A
P = I²·R = 0.1055²·30 = 0.3337 W

Answer

P = 0.3337 W
8.AC impedance & phasors

Answer: f0 = 758.7 Hz

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 inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.02·2.2·10⁻⁶)) = 758.7 Hz

Answer

f0 = 758.7 Hz
9.DC motors

Answer: n = 1671 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.

Armature current

Ia = T/k = 0.24/0.12 = 2 A

Back-EMF and speed

E = 24 − 2·1.5 = 21 V
ω = E/k = 175 rad/s → n = 1671 rpm

Answer

n = 1671 rpm
10.DC motors

Answer: n2 = 1415 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 = 120 − 10·0.4 = 116 V
E2 = 120 − 15·0.4 = 114 V

New speed

n2 = 1440·114/116 = 1415 rpm

Answer

n2 = 1415 rpm

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