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 VZ = 7 + j24 Ωper phase

A balanced Δ-connected load with an impedance of 7 + j24 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 = 7 + j24 Ω per phase
P = kW
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 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 0.98 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 = 0.65pf2 = 0.98VL = 400 Vf = 60 Hz
Cph = µF
3.TransformersMedium
RL​ = 8 Ω41960 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 4:1 has 960 V applied to its primary winding. An 8 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 4:1V1 = 960 VRL = 8 Ω
I1 = A
4.TransformersHard
pf 0.8250 kVA, η = 98%

A single-phase 250 kVA transformer has a core loss of 1250 W. At rated load and a power factor of 0.8 lagging its efficiency is 98%. Find the full-load copper loss Pcu,fl in watts.

S = 250 kVAPcore = 1250 Wpf = 0.8η = 98%
Pcu,fl = W
5.AC power & power factorMedium
P = 14 kWQ = ?Sφ

A single-phase motor takes 14 kW at a power factor of 0.8 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 14 kWpf = 0.8V = 230 V
Q = kvar
6.AC power & power factorHard
P = 20 kWQ1​S1​φ1​QC​

A single-phase load takes 20 kW at a power factor of 0.75 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.

P = 20 kWpf1 = 0.75pf2 = 0.98V = 230 Vf = 60 Hz
Inew = A
7.AC impedance & phasorsMedium
Vf0​ = 1000 HzRL = ?C = 10 µF

A series RLC circuit with C = 10 µF must resonate at 1000 Hz. Find the inductance L needed.

f0 = 1000 HzC = 10 µF
L = mH
8.AC impedance & phasorsHard
VfR = 100 ΩL = 200 mHC = 0.1 µF

A series RLC circuit has R = 100 Ω, L = 200 mH and C = 0.1 µF. Find its quality factor Q at resonance.

R = 100 ΩL = 200 mHC = 0.1 µF
Q =
9.DC motorsMedium
T (N·m)n (rpm)3.6 N·mn0​n = ?

A DC motor (k = 0.3 N·m/A, Ra = 0.2 Ω) runs from 48 V and drives a load torque of 3.6 N·m. Find its speed n in rpm.

V = 48 Vk = 0.3 N·m/ARa = 0.2 ΩT = 3.6 N·m
n = rpm
10.DC motorsHard
+−110 VRs​ = ?Ra​ = 0.25 ΩM+−En = 600 rpmT = 15 N·m

A DC motor (k = 0.75 V·s/rad, Ra = 0.25 Ω) on a fixed 110 V supply drives a load of 15 N·m. Find the series resistance Rs that slows it down to 600 rpm.

V = 110 Vk = 0.75 V·s/radRa = 0.25 ΩT = 15 N·mn = 600 rpm
Rs = Ω

3Solutions

1.Three-phase power

Answer: P = 1.454 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 = VL = 208 V

Phase current

|Z| = √(7² + 24²) = 25 Ω
Iph = 208/25 = 8.32 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·8.32²·7 = 1.454 kW

Answer

P = 1.454 kW

Check: cos φ = R/|Z| = 0.28, and √3·VL·IL·cos φ gives the same power.

2.Three-phase power

Answer: Cph = 4004 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 Y bank each capacitor is connected between a line and the star point.

V = 400/√3 = 230.9 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 80506/(377·230.9²) = 4004 µF

Answer

Cph = 4004 µF
3.Transformers

Answer: I1 = 7.5 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.

Secondary voltage and current

V2 = 960/4 = 240 V
I2 = 240/8 = 30 A

Primary current

An ideal transformer has no losses: V₁·I₁ = V₂·I₂.

I1 = I2/4 = 7.5 A

Answer

I1 = 7.5 A

Check: the load seen from the primary is a²·RL = 128 Ω, and 960/128 gives the same current.

4.Transformers

Answer: Pcu,fl = 2832 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 = 250·0.8 = 200 kW
Pin = 200000/0.98 = 204082 W

Total and copper losses

Ploss = 204082 − 200000 = 4082 W
Pcu,fl = 4082 − 1250 = 2832 W

Answer

Pcu,fl = 2832 W
5.AC power & power factor

Answer: Q = 10.5 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(0.8) = 36.87°, tan φ = 0.75

Reactive power

Q = P·tan φ = 14·0.75 = 10.5 kvar

Answer

Q = 10.5 kvar
6.AC power & power factor

Answer: Inew = 88.73 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 = 20·tan(41.41°) = 17.64 kvar
Q2 = P·tan φ2 = 20·tan(11.48°) = 4.061 kvar
QC = Q1 − Q2 = 13.58 kvar

New line current

Inew = P/(V·cos φ2) = 20000/(230·0.98) = 88.73 A

Answer

Inew = 88.73 A
7.AC impedance & phasors

Answer: L = 2.533 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·1000 = 6283 rad/s
L = 1/(ω₀²·C) = 1/(6283²·10·10⁻⁶) = 2.533 mH

Answer

L = 2.533 mH
8.AC impedance & phasors

Answer: Q = 14.14

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·√(0.2·0.1·10⁻⁶)) = 1125 Hz

Quality factor

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

Q = (1/R)·√(L/C) = (1/100)·√(0.2/(0.1·10⁻⁶)) = 14.14

Answer

Q = 14.14
9.DC motors

Answer: n = 1451 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 = 3.6/0.3 = 12 A

Back-EMF and speed

E = 48 − 12·0.2 = 45.6 V
ω = E/k = 152 rad/s → n = 1451 rpm

Answer

n = 1451 rpm
10.DC motors

Answer: Rs = 2.894 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 = 0.75·62.83 = 47.12 V

Series resistance

Ra + Rs = (110 − 47.12)/20 = 3.144 Ω
Rs = 2.894 Ω

Answer

Rs = 2.894 Ω

The resistor wastes 1158 W; a chopper controls the speed with far less loss.

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