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​ = 400 VIL​ = 20 Acos φ = 0.88lagging

A balanced three-phase load draws a line current of 20 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.88 lagging. Find the total real power P drawn by the load in kW.

VL = 400 VIL = 20 Acos φ = 0.88
P = kW
2.Three-phase powerHard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.7 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 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 75 kWpf1 = 0.7pf2 = 0.95VL = 480 Vf = 60 Hz
Cph = µF
3.TransformersMedium
RL​ = 15 Ω21480 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 2:1 has 480 V applied to its primary winding. A 15 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 2:1V1 = 480 VRL = 15 Ω
I1 = A
4.TransformersHard
pf 0.8550 kVA, 50% load

A single-phase 50 kVA transformer has a core loss of 300 W and a copper loss of 600 W at rated load. It supplies 50% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.

S = 50 kVAPcore = 300 WPcu,fl = 600 Wx = 0.5pf = 0.85
Ploss = W
5.AC power & power factorMedium
230 V110 kW, 0.8228 kW, 0.9I = ?

Two loads are connected in parallel to a 230 V single-phase supply: a 10 kW load at a power factor of 0.8 lagging and a 28 kW load at a power factor of 0.9 leading. Find the total current I drawn from the supply.

V = 230 VP1 = 10 kW, pf1 = 0.8 laggingP2 = 28 kW, pf2 = 0.9 leading
I = A
6.AC power & power factorHard
12 V60 HzRs​ = 8 ΩXs​ = j5 ΩZL​ = ?

A 12 V (rms), 60 Hz source has an internal impedance Zs = 8 + j5 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.

Vs = 12 V (rms)f = 60 HzZs = 8 + j5 Ω
Pmax = W
7.AC impedance & phasorsMedium
Vf0​ = 500 HzRL = 50 mHC = ?

A series RLC circuit with L = 50 mH must resonate at 500 Hz. Find the capacitance C needed.

f0 = 500 HzL = 50 mH
C = µF
8.AC impedance & phasorsHard
VfR = 5 ΩL = 10 mHC = 0.022 µF

A series RLC circuit has R = 5 Ω, L = 10 mH and C = 0.022 µF. Find its quality factor Q at resonance.

R = 5 ΩL = 10 mHC = 0.022 µF
Q =
9.DC motorsMedium
T (N·m)n (rpm)9 N·mn0​n = ?

A DC motor (k = 0.6 N·m/A, Ra = 2 Ω) runs from 240 V and drives a load torque of 9 N·m. Find its speed n in rpm.

V = 240 Vk = 0.6 N·m/ARa = 2 ΩT = 9 N·m
n = rpm
10.DC motorsHard
+−220 VRa​ = 0.8 ΩIa​: 12 → 6 AM+−En = ?T

A separately excited DC motor on 220 V (Ra = 0.8 Ω) runs at 1440 rpm with an armature current of 12 A. The load changes and the current becomes 6 A, with the field unchanged. Find the new speed n2.

V = 220 VRa = 0.8 Ωn1 = 1440 rpmIa1 = 12 AIa2 = 6 A
n2 = rpm

3Solutions

1.Three-phase power

Answer: P = 12.19 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 = √3·400·20 = 13.86 kVA

Real power

P = S·cos φ = 13.86·0.88 = 12.19 kW

Answer

P = 12.19 kW
2.Three-phase power

Answer: Cph = 199 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 = 75·tan(acos 0.7) = 75·1.02 = 76.52 kvar
Q2 = 75·tan(acos 0.95) = 75·0.3287 = 24.65 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 51.86 kvar

Capacitance per phase

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

V = 480 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 17288/(377·480²) = 199 µF

Answer

Cph = 199 µF
3.Transformers

Answer: I1 = 8 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 = 480/2 = 240 V
I2 = 240/15 = 16 A

Primary current

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

I1 = I2/2 = 8 A

Answer

I1 = 8 A

Check: the load seen from the primary is a²·RL = 60 Ω, and 480/60 gives the same current.

4.Transformers

Answer: Ploss = 450 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 = 0.25·600 = 150 W

Total losses

The core loss is the same at any load.

Ploss = 300 + 150 = 450 W

Answer

Ploss = 450 W

The efficiency at this load is η = 97.93%.

5.AC power & power factor

Answer: I = 167.3 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 = 10·tan(arccos 0.8) = 7.5 kvar
Q2 = −28·tan(arccos 0.9) = −13.56 kvar

Totals

P = 10 + 28 = 38 kW
Q = 7.5 + (−13.56) = −6.061 kvar
S = √(38² + (−6.061)²) = 38.48 kVA

Total current

I = S/V = 38480/230 = 167.3 A

Answer

I = 167.3 A
6.AC power & power factor

Answer: Pmax = 4.5 W

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.

Conjugate match

The load reactance cancels the source reactance, and the load resistance equals the source resistance.

ZL = Zs* = 8 − j5 Ω

Maximum power

I = Vs/(2Rs) = 12/16 = 0.75 A
Pmax = I²·RL = Vs²/(4Rs) = 12²/(4·8) = 4.5 W

Answer

Pmax = 4.5 W
7.AC impedance & phasors

Answer: C = 2.026 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·500 = 3142 rad/s
C = 1/(ω₀²·L) = 1/(3142²·0.05) = 2.026 µF

Answer

C = 2.026 µF
8.AC impedance & phasors

Answer: Q = 134.8

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·√(0.01·0.022·10⁻⁶)) = 10730 Hz

Quality factor

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

Q = (1/R)·√(L/C) = (1/5)·√(0.01/(0.022·10⁻⁶)) = 134.8

Answer

Q = 134.8
9.DC motors

Answer: n = 3342 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 = 9/0.6 = 15 A

Back-EMF and speed

E = 240 − 15·2 = 210 V
ω = E/k = 350 rad/s → n = 3342 rpm

Answer

n = 3342 rpm
10.DC motors

Answer: n2 = 1473 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.

Back-EMF before and after

E1 = 220 − 12·0.8 = 210.4 V
E2 = 220 − 6·0.8 = 215.2 V

New speed

n2 = 1440·215.2/210.4 = 1473 rpm

Answer

n2 = 1473 rpm

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