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 VIL​ = 15 Acos φ = 0.75lagging

A balanced three-phase load draws a line current of 15 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.75 lagging. Find the total real power P drawn by the load in kW.

VL = 208 VIL = 15 Acos φ = 0.75
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 0.75 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 Y. Find the capacitance Cph of each capacitor in µF.

P = 60 kWpf1 = 0.75pf2 = 0.92VL = 480 Vf = 60 Hz
Cph = µF
3.TransformersMedium
7200 V240 VIHV​ = ?ILV​100 kVA

A single-phase 100 kVA, 7200/240 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 100 kVAV1 = 7200 VV2 = 240 V
IHV = A
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 5000 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 = 5000 Wx = 0.25pf = 0.85
Ploss = W
5.AC power & power factorMedium
240 V113 kW, 0.7525 kW, 0.7I = ?

Two loads are connected in parallel to a 240 V single-phase supply: a 13 kW load at a power factor of 0.75 lagging and a 5 kW load at a power factor of 0.7 lagging. Find the total current I drawn from the supply.

V = 240 VP1 = 13 kW, pf1 = 0.75 laggingP2 = 5 kW, pf2 = 0.7 lagging
I = A
6.AC power & power factorHard
P1​ = 79 kWQ1​S1​φ

A 150 kVA single-phase transformer supplies a load of 79 kW at a power factor of 0.7 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?

Smax = 150 kVAP1 = 79 kWpf1 = 0.7
ΔP = kW
7.AC impedance & phasorsMedium
230 V200 HzR = 60 ΩL = 100 mHC = 47 µF

A series RLC circuit is connected to a 230 V (rms), 200 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 230 V (rms)f = 200 HzR = 60 ΩL = 100 mHC = 47 µF
φ = °
8.AC impedance & phasorsHard
VfR = 5 ΩL = 10 mHC = 0.22 µF

A series RLC circuit has R = 5 Ω, L = 10 mH and C = 0.22 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 5 ΩL = 10 mHC = 0.22 µF
B = Hz
9.DC motorsMedium
T (N·m)n (rpm)n0​ = ?

A DC motor with k = 2 V·s/rad runs from 220 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.

V = 220 Vk = 2 V·s/rad
n0 = rpm
10.DC motorsHard
+−48 VRa​ = 0.8 ΩM+−EωT

A small DC motor has an armature resistance of 0.8 Ω and runs from 48 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.

V = 48 VRa = 0.8 Ω
Pmax = W

3Solutions

1.Three-phase power

Answer: P = 4.053 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 = √3·208·15 = 5.404 kVA

Real power

P = S·cos φ = 5.404·0.75 = 4.053 kW

Answer

P = 4.053 kW
2.Three-phase power

Answer: Cph = 314.9 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 0.75) = 60·0.8819 = 52.92 kvar
Q2 = 60·tan(acos 0.92) = 60·0.426 = 25.56 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 27.36 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 480/√3 = 277.1 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 9118/(377·277.1²) = 314.9 µF

Answer

Cph = 314.9 µF
3.Transformers

Answer: IHV = 13.89 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 = 100000/7200 = 13.89 A

Answer

IHV = 13.89 A
4.Transformers

Answer: Ploss = 2813 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 = 0.0625·5000 = 312.5 W

Total losses

The core loss is the same at any load.

Ploss = 2500 + 312.5 = 2813 W

Answer

Ploss = 2813 W

The efficiency at this load is η = 97.42%.

5.AC power & power factor

Answer: I = 101.9 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 = 13·tan(arccos 0.75) = 11.46 kvar
Q2 = 5·tan(arccos 0.7) = 5.101 kvar

Totals

P = 13 + 5 = 18 kW
Q = 11.46 + 5.101 = 16.57 kvar
S = √(18² + 16.57²) = 24.46 kVA

Total current

I = S/V = 24463/240 = 101.9 A

Answer

I = 101.9 A
6.AC power & power factor

Answer: ΔP = 47.51 kW

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 the existing load

Q1 = 79·tan(arccos 0.7) = 80.6 kvar

Largest total real power

At full rating S = Smax, and Q is unchanged by the resistive load.

P = √(Smax² − Q1²) = √(150² − 80.6²) = 126.5 kW

Extra real power

ΔP = 126.5 − 79 = 47.51 kW

Answer

ΔP = 47.51 kW
7.AC impedance & phasors

Answer: φ = 61.11 °

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·200 = 1257 rad/s
XL = ωL = 1257·0.1 = 125.7 Ω
XC = 1/(ωC) = 1/(1257·47·10⁻⁶) = 16.93 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(60² + (108.7)²) = 124.2 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(108.7/60) = 61.11°

Answer

φ = 61.11°
8.AC impedance & phasors

Answer: B = 79.58 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|.

Resonant frequency

f0 = 1/(2π·√(0.01·0.22·10⁻⁶)) = 3393 Hz

Quality factor

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

Q = (1/R)·√(L/C) = (1/5)·√(0.01/(0.22·10⁻⁶)) = 42.64

Bandwidth

B = f0/Q = 3393/42.64 = 79.58 Hz

Answer

B = 79.58 Hz
9.DC motors

Answer: n0 = 1050 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 = 220/2 = 110 rad/s
n0 = 60·110/(2π) = 1050 rpm

Answer

n0 = 1050 rpm
10.DC motors

Answer: Pmax = 720 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 = 24 V

Maximum power

Pmax = (V/2)²/Ra = V²/(4Ra) = 48²/(4·0.8) = 720 W

Answer

Pmax = 720 W

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

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