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​ = 690 VIL​ = 100 Acos φ = 0.85lagging

A balanced three-phase load draws a line current of 100 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 100 Acos φ = 0.85
P = kW
2.Three-phase powerHard
P = 120 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 120 kW at a power factor of 0.75 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 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 120 kWpf1 = 0.75pf2 = 0.95VL = 690 Vf = 50 Hz
Cph = µF
3.TransformersMedium
8 Ω121

An 8 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 = 8 Ω
R1′ = Ω
4.TransformersHard
pf 0.85500 kVA, 25% load

A single-phase 500 kVA transformer has a core loss of 2500 W and a copper loss of 10000 W at rated load. It supplies 25% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.

S = 500 kVAPcore = 2500 WPcu,fl = 10000 Wx = 0.25pf = 0.85
Ploss = W
5.AC power & power factorMedium
240 V18 kW, 0.8226 kW, 1I = ?

Two loads are connected in parallel to a 240 V single-phase supply: an 8 kW load at a power factor of 0.8 lagging and a 26 kW heater (unity power factor). Find the total current I drawn from the supply.

V = 240 VP1 = 8 kW, pf1 = 0.8 laggingP2 = 26 kW, pf2 = 1
I = A
6.AC power & power factorHard
P = 25 kWQ1​S1​φ1​QC​

A single-phase load takes 25 kW at a power factor of 0.8 lagging from a 400 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the new line current Inew.

P = 25 kWpf1 = 0.8pf2 = 0.92V = 400 Vf = 60 Hz
Inew = A
7.AC impedance & phasorsMedium
Vf0​ = 200 HzRL = 50 mHC = ?

A series RLC circuit with L = 50 mH must resonate at 200 Hz. Find the capacitance C needed.

f0 = 200 HzL = 50 mH
C = µF
8.AC impedance & phasorsHard
12 Vf0​R = 2 ΩL = 50 mHC = 10 µF

A series RLC circuit is driven by a 12 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 12 V (rms)R = 2 ΩL = 50 mHC = 10 µF
VL = V
9.DC motorsMedium
T (N·m)n (rpm)0.8 N·mn0​nFL​

A DC motor (k = 0.2 V·s/rad, Ra = 0.25 Ω) runs from 24 V. At full load it draws 4 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 24 Vk = 0.2 V·s/radRa = 0.25 ΩIFL = 4 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 = 101.6 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.

Apparent power

S = √3·VL·IL = √3·690·100 = 119.5 kVA

Real power

P = S·cos φ = 119.5·0.85 = 101.6 kW

Answer

P = 101.6 kW
2.Three-phase power

Answer: Cph = 148 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 = 120·tan(acos 0.75) = 120·0.8819 = 105.8 kvar
Q2 = 120·tan(acos 0.95) = 120·0.3287 = 39.44 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 66.39 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) = 22129/(314.2·690²) = 148 µF

Answer

Cph = 148 µF
3.Transformers

Answer: R1′ = 1152 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 = 12²·8 = 1152 Ω

Answer

R1′ = 1152 Ω
4.Transformers

Answer: Ploss = 3125 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 = 0.0625·10000 = 625 W

Total losses

The core loss is the same at any load.

Ploss = 2500 + 625 = 3125 W

Answer

Ploss = 3125 W

The efficiency at this load is η = 97.14%.

5.AC power & power factor

Answer: I = 143.9 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 of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 8·tan(arccos 0.8) = 6 kvar
Q2 = 0 = 0 kvar

Totals

P = 8 + 26 = 34 kW
Q = 6 + 0 = 6 kvar
S = √(34² + 6²) = 34.53 kVA

Total current

I = S/V = 34525/240 = 143.9 A

Answer

I = 143.9 A
6.AC power & power factor

Answer: Inew = 67.93 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 = 25·tan(36.87°) = 18.75 kvar
Q2 = P·tan φ2 = 25·tan(23.07°) = 10.65 kvar
QC = Q1 − Q2 = 8.1 kvar

New line current

Inew = P/(V·cos φ2) = 25000/(400·0.92) = 67.93 A

Answer

Inew = 67.93 A
7.AC impedance & phasors

Answer: C = 12.67 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·200 = 1257 rad/s
C = 1/(ω₀²·L) = 1/(1257²·0.05) = 12.67 µF

Answer

C = 12.67 µF
8.AC impedance & phasors

Answer: VL = 424.3 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|.

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.05·10·10⁻⁶) = 1414 rad/s
I = V/R = 12/2 = 6 A

Voltage across the inductor

XL = ω₀L = 1414·0.05 = 70.71 Ω
VL = I·XL = 6·70.71 = 424.3 V

Answer

VL = 424.3 V

This is Q = 35.36 times the supply voltage.

9.DC motors

Answer: SR = 4.348 %

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 = 60·24/(2π·0.2) = 1146 rpm
nFL = 60·(24 − 4·0.25)/(2π·0.2) = 1098 rpm

Speed regulation

SR = (1146 − 1098)/1098·100 = 4.348 %

Answer

SR = 4.348 %
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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