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​ = 480 VIL​ = ?M37 kW, η = 92%cos φ = 0.84

A three-phase induction motor delivers 37 kW of mechanical output power at full load. At this load its efficiency is 92% and its power factor is 0.84 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 37 kWη = 92%cos φ = 0.84VL = 480 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 0.65 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 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.65pf2 = 0.98VL = 690 Vf = 50 Hz
Cph = µF
3.TransformersMedium
10000 V400 VIHV​ = ?ILV​500 kVA

A single-phase 500 kVA, 10000/400 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 500 kVAV1 = 10000 VV2 = 400 V
IHV = A
4.TransformersHard
250 kVA

A single-phase 250 kVA transformer has a core loss of 1000 W and a copper loss of 2500 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 250 kVAPcore = 1000 WPcu,fl = 2500 W
Sη,max = kVA
5.AC power & power factorMedium
120 VZ8 + j6 ΩI

A load with impedance Z = 8 + j6 Ω is connected to a 120 V (rms) single-phase supply. Find the apparent power S.

V = 120 VZ = 8 + j6 Ω
S = VA
6.AC power & power factorHard
P = 40 kWQ1​S1​φ1​QC​

A single-phase load takes 40 kW at a power factor of 0.7 lagging from a 230 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the capacitance C needed.

P = 40 kWpf1 = 0.7pf2 = 0.92V = 230 Vf = 50 Hz
C = µF
7.AC impedance & phasorsMedium
Vf0​ = 2000 HzRL = 1 mHC = ?

A series RLC circuit with L = 1 mH must resonate at 2000 Hz. Find the capacitance C needed.

f0 = 2000 HzL = 1 mH
C = µF
8.AC impedance & phasorsHard
120 V500 HzR = 30 ΩL = 150 mHC = 100 µF

A series RLC circuit is connected to a 120 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.

V = 120 V (rms)f = 500 HzR = 30 ΩL = 150 mHC = 100 µF
f0 = Hz
9.DC motorsMedium
T (N·m)n (rpm)n0​ = ?

A DC motor with k = 1 V·s/rad runs from 120 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.

V = 120 Vk = 1 V·s/rad
n0 = rpm
10.DC motorsHard
+−110 VRa​ = 1 ΩM+−En = ?T = 4 N·m

A separately excited DC motor (k = 1 V·s/rad at full field, Ra = 1 Ω, V = 110 V) drives a constant load torque of 4 N·m. The field flux is reduced to 85 % of its full value. Find the new speed n.

V = 110 Vk = 1 V·s/radRa = 1 ΩT = 4 N·mΦ = 85 %
n = rpm

3Solutions

1.Three-phase power

Answer: IL = 57.59 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 = 37/0.92 = 40.22 kW

Line current

IL = Pin/(√3·VL·cos φ) = 40217/(√3·480·0.84) = 57.59 A

Answer

IL = 57.59 A
2.Three-phase power

Answer: Cph = 538.2 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 = 250·tan(acos 0.65) = 250·1.169 = 292.3 kvar
Q2 = 250·tan(acos 0.98) = 250·0.2031 = 50.76 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 241.5 kvar

Capacitance per phase

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

V = 690 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 80506/(314.2·690²) = 538.2 µF

Answer

Cph = 538.2 µF
3.Transformers

Answer: IHV = 50 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 = 500000/10000 = 50 A

Answer

IHV = 50 A
4.Transformers

Answer: Sη,max = 158.1 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

x²·2500 = 1000 → x = √(1000/2500) = 0.6325

Load at maximum efficiency

Sη,max = x·S = 0.6325·250 = 158.1 kVA

Answer

Sη,max = 158.1 kVA
5.AC power & power factor

Answer: S = 1440 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| = √(8² + 6²) = 10 Ω
I = V/|Z| = 120/10 = 12 A

Apparent power

S = V·I = 120·12 = 1440 VA

Answer

S = 1440 VA
6.AC power & power factor

Answer: C = 1430 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 = 40·tan(45.57°) = 40.81 kvar
Q2 = P·tan φ2 = 40·tan(23.07°) = 17.04 kvar
QC = Q1 − Q2 = 23.77 kvar

Capacitance

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

C = QC/(2πf·V²) = 23768/(2π·50·230²) = 1430 µF

Answer

C = 1430 µF
7.AC impedance & phasors

Answer: C = 6.333 µF

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
C = 1/(ω₀²·L) = 1/(12566²·0.001) = 6.333 µF

Answer

C = 6.333 µF
8.AC impedance & phasors

Answer: f0 = 41.09 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·√(0.15·100·10⁻⁶)) = 41.09 Hz

Answer

f0 = 41.09 Hz
9.DC motors

Answer: n0 = 1146 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.

No-load speed

ω₀ = V/k = 120/1 = 120 rad/s
n0 = 60·120/(2π) = 1146 rpm

Answer

n0 = 1146 rpm
10.DC motors

Answer: n = 1183 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 = 0.85·1 = 0.85 V·s/rad

New armature current

Ia = 4/0.85 = 4.706 A

New speed

ω = (110 − 4.706·1)/0.85 = 123.9 rad/s → n = 1183 rpm

Answer

n = 1183 rpm

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

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