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​ = 12 Acos φ = 0.75lagging

A balanced three-phase load draws a line current of 12 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.75 lagging. Find the total real power P drawn by the load in kW.

VL = 400 VIL = 12 Acos φ = 0.75
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 0.65 lagging from a 415 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 75 kWpf1 = 0.65pf2 = 0.9VL = 415 Vf = 60 Hz
QC = kvar
3.TransformersMedium
RL​ = 4 Ω511200 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 5:1 has 1200 V applied to its primary winding. A 4 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 5:1V1 = 1200 VRL = 4 Ω
I1 = A
4.TransformersHard
pf 0.9250 kVA, η = 98.5%

A single-phase 250 kVA transformer has a core loss of 2000 W. At rated load and a power factor of 0.9 lagging its efficiency is 98.5%. Find the full-load copper loss Pcu,fl in watts.

S = 250 kVAPcore = 2000 Wpf = 0.9η = 98.5%
Pcu,fl = W
5.AC power & power factorMedium
P = 33 kWQ = ?Sφ

A single-phase motor takes 33 kW at a power factor of 0.9 lagging from a 400 V supply. Find the reactive power Q it draws.

P = 33 kWpf = 0.9V = 400 V
Q = kvar
6.AC power & power factorHard
P = 35 kWQ1​S1​φ1​QC​

A single-phase load takes 35 kW at a power factor of 0.7 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the reactive power QC the capacitor must supply.

P = 35 kWpf1 = 0.7pf2 = 0.92V = 240 Vf = 60 Hz
QC = kvar
7.AC impedance & phasorsMedium
50 V60 HzR = 40 ΩL = 200 mHI

A resistor and an inductor are connected in parallel across a 50 V (rms), 60 Hz supply. Find the rms current I.

V = 50 V (rms)f = 60 HzR = 40 ΩL = 200 mH
I = A
8.AC impedance & phasorsHard
12 V60 HzR = 50 ΩL = 100 mHC = 47 µF

A series RLC circuit is connected to a 12 V (rms), 60 Hz supply. Find the resonant frequency f0 of the circuit.

V = 12 V (rms)f = 60 HzR = 50 ΩL = 100 mHC = 47 µF
f0 = Hz
9.DC motorsMedium
T (N·m)n (rpm)9 N·mn0​nFL​

A DC motor (k = 0.6 V·s/rad, Ra = 0.25 Ω) runs from 110 V. At full load it draws 15 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 110 Vk = 0.6 V·s/radRa = 0.25 ΩIFL = 15 A
SR = %
10.DC motorsHard
+−240 VRa​ = 0.4 ΩIa​: 30 → 12 AM+−En = ?T

A separately excited DC motor on 240 V (Ra = 0.4 Ω) runs at 1500 rpm with an armature current of 30 A. The load changes and the current becomes 12 A, with the field unchanged. Find the new speed n2.

V = 240 VRa = 0.4 Ωn1 = 1500 rpmIa1 = 30 AIa2 = 12 A
n2 = rpm

3Solutions

1.Three-phase power

Answer: P = 6.235 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 = √3·400·12 = 8.314 kVA

Real power

P = S·cos φ = 8.314·0.75 = 6.235 kW

Answer

P = 6.235 kW
2.Three-phase power

Answer: QC = 51.36 kvar

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 0.65) = 75·1.169 = 87.68 kvar
Q2 = 75·tan(acos 0.9) = 75·0.4843 = 36.32 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 51.36 kvar

Answer

QC = 51.36 kvar
3.Transformers

Answer: I1 = 12 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 = 1200/5 = 240 V
I2 = 240/4 = 60 A

Primary current

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

I1 = I2/5 = 12 A

Answer

I1 = 12 A

Check: the load seen from the primary is a²·RL = 100 Ω, and 1200/100 gives the same current.

4.Transformers

Answer: Pcu,fl = 1426 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 = 250·0.9 = 225 kW
Pin = 225000/0.985 = 228426 W

Total and copper losses

Ploss = 228426 − 225000 = 3426 W
Pcu,fl = 3426 − 2000 = 1426 W

Answer

Pcu,fl = 1426 W
5.AC power & power factor

Answer: Q = 15.98 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(0.9) = 25.84°, tan φ = 0.4843

Reactive power

Q = P·tan φ = 33·0.4843 = 15.98 kvar

Answer

Q = 15.98 kvar
6.AC power & power factor

Answer: QC = 20.8 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.

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 = 35·tan(45.57°) = 35.71 kvar
Q2 = P·tan φ2 = 35·tan(23.07°) = 14.91 kvar
QC = Q1 − Q2 = 20.8 kvar

Answer

QC = 20.8 kvar
7.AC impedance & phasors

Answer: I = 1.415 A

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

XL = ωL = 2π·60·0.2 = 75.4 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

IR = V/R = 50/40 = 1.25 A
IX = V/XL = 50/75.4 = 0.6631 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(IR² + IX²) = √(1.25² + 0.6631²) = 1.415 A

Answer

I = 1.415 A
8.AC impedance & phasors

Answer: f0 = 73.41 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|.

Resonance

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.1·47·10⁻⁶)) = 73.41 Hz

Answer

f0 = 73.41 Hz
9.DC motors

Answer: SR = 3.529 %

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 = 60·110/(2π·0.6) = 1751 rpm
nFL = 60·(110 − 15·0.25)/(2π·0.6) = 1691 rpm

Speed regulation

SR = (1751 − 1691)/1691·100 = 3.529 %

Answer

SR = 3.529 %
10.DC motors

Answer: n2 = 1547 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 = 240 − 30·0.4 = 228 V
E2 = 240 − 12·0.4 = 235.2 V

New speed

n2 = 1500·235.2/228 = 1547 rpm

Answer

n2 = 1547 rpm

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