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​ = 150 Acos φ = 0.82lagging

A balanced three-phase load draws a line current of 150 A from a supply with a line-to-line voltage of 480 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.

VL = 480 VIL = 150 Acos φ = 0.82
P = kW
2.Three-phase powerHard
P = 200 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 200 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 Δ. Find the capacitance Cph of each capacitor in µF.

P = 200 kWpf1 = 0.65pf2 = 0.98VL = 400 Vf = 60 Hz
Cph = µF
3.TransformersMedium
6600 V230 VIHV​ILV​ = ?15 kVA

A single-phase 15 kVA, 6600/230 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 15 kVAV1 = 6600 VV2 = 230 V
ILV = A
4.TransformersHard
pf 0.85150 kVA, 75% load

A single-phase 150 kVA transformer has a core loss of 600 W and a copper loss of 1200 W at rated load. It supplies 75% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.

S = 150 kVAPcore = 600 WPcu,fl = 1200 Wx = 0.75pf = 0.85
Ploss = W
5.AC power & power factorMedium
120 VZ4 + j3 ΩI

A load with impedance Z = 4 + j3 Ω is connected to a 120 V (rms) single-phase supply. Find the apparent power S.

V = 120 VZ = 4 + j3 Ω
S = VA
6.AC power & power factorHard
240 V50 Hzload23 kW, 0.8C100 µFI

A single-phase load takes 23 kW at a power factor of 0.8 lagging from a 240 V, 50 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 23 kWpf1 = 0.8V = 240 Vf = 50 HzC = 100 µF
cos φ =
7.AC impedance & phasorsMedium
100 V500 HzR = 50 ΩL = 150 mHC = 10 µF

A series RLC circuit is connected to a 100 V (rms), 500 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 100 V (rms)f = 500 HzR = 50 ΩL = 150 mHC = 10 µF
φ = °
8.AC impedance & phasorsHard
100 V200 HzR1​ = 25 ΩR2​ = 80 ΩC = 10 µF

A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 100 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.

V = 100 V (rms)f = 200 HzR1 = 25 ΩR2 = 80 ΩC = 10 µF
|Z| = Ω
9.DC motorsMedium
T (N·m)n (rpm)0.16 N·mn0​n = ?

A DC motor (k = 0.08 N·m/A, Ra = 1.2 Ω) runs from 24 V and drives a load torque of 0.16 N·m. Find its speed n in rpm.

V = 24 Vk = 0.08 N·m/ARa = 1.2 ΩT = 0.16 N·m
n = rpm
10.DC motorsHard
+−120 VD = ?Ra​ = 1 ΩM+−En = 400 rpmT = 0.6 N·m

A DC chopper feeds a motor (k = 0.12 V·s/rad, Ra = 1 Ω) from a 120 V DC bus. The average armature voltage is D·120 V. Find the duty cycle D that runs the motor at 400 rpm with a load of 0.6 N·m.

Vdc = 120 Vk = 0.12 V·s/radRa = 1 ΩT = 0.6 N·mn = 400 rpm
D = %

3Solutions

1.Three-phase power

Answer: P = 102.3 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 = √3·480·150 = 124.7 kVA

Real power

P = S·cos φ = 124.7·0.82 = 102.3 kW

Answer

P = 102.3 kW
2.Three-phase power

Answer: Cph = 1068 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 = 200·tan(acos 0.65) = 200·1.169 = 233.8 kvar
Q2 = 200·tan(acos 0.98) = 200·0.2031 = 40.61 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 193.2 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) = 64405/(377·400²) = 1068 µF

Answer

Cph = 1068 µF
3.Transformers

Answer: ILV = 65.22 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 = 15000/230 = 65.22 A

Answer

ILV = 65.22 A
4.Transformers

Answer: Ploss = 1275 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 = 0.5625·1200 = 675 W

Total losses

The core loss is the same at any load.

Ploss = 600 + 675 = 1275 W

Answer

Ploss = 1275 W

The efficiency at this load is η = 98.68%.

5.AC power & power factor

Answer: S = 2880 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| = √(4² + 3²) = 5 Ω
I = V/|Z| = 120/5 = 24 A

Apparent power

S = V·I = 120·24 = 2880 VA

Answer

S = 2880 VA
6.AC power & power factor

Answer: cos φ = 0.8303

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 = 23·tan(arccos 0.8) = 17.25 kvar
QC = V²·2πf·C = 240²·2π·50·100·10⁻⁶ = 1810 var = 1.81 kvar
Q2 = 17.25 − 1.81 = 15.44 kvar

New apparent power

S2 = √(23² + 15.44²) = 27.7 kVA

New power factor

cos φ = P/S2 = 23/27.7 = 0.8303

Answer

cos φ = 0.8303
7.AC impedance & phasors

Answer: φ = 83.51 °

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·500 = 3142 rad/s
XL = ωL = 3142·0.15 = 471.2 Ω
XC = 1/(ωC) = 1/(3142·10·10⁻⁶) = 31.83 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(50² + (439.4)²) = 442.2 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(439.4/50) = 83.51°

Answer

φ = 83.51°
8.AC impedance & phasors

Answer: |Z| = 76.14 Ω

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|.

Reactance

XC = 1/(2π·200·10·10⁻⁶) = 79.58 Ω

Parallel combination

Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.

Zp = R₂·(−jXC)/(R₂ − jXC) = (R₂·XC² − jR₂²·XC)/(R₂² + XC²)
Zp = 39.79 − j40 Ω

Total impedance

Z = 25 + 39.79 − j40 = 64.79 − j40 Ω
|Z| = √(64.79² + 40²) = 76.14 Ω

Answer

|Z| = 76.14 Ω
9.DC motors

Answer: n = 2578 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 = 0.16/0.08 = 2 A

Back-EMF and speed

E = 24 − 2·1.2 = 21.6 V
ω = E/k = 270 rad/s → n = 2578 rpm

Answer

n = 2578 rpm
10.DC motors

Answer: D = 8.355 %

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.

Required armature voltage

Va = 0.12·41.89 + 5·1 = 10.03 V

Duty cycle

D = 10.03/120 = 8.355 %

Answer

D = 8.355 %

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