A DC motor draws 30 A from a 240 V supply. Its armature resistance is 0.8 Ω and its rotational (friction, windage and iron) losses are 259 W. Find its efficiency η.
V = 240 VIa = 30 ARa = 0.8 ΩProt = 259 W
η = %
4.DC motorsHard
A small DC motor has an armature resistance of 1.2 Ω and runs from 24 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 24 VRa = 1.2 Ω
Pmax = W
5.AC power & power factorMedium
A single-phase motor takes 20 kW at a power factor of 0.6 lagging from a 400 V supply. Find the reactive power Q it draws.
P = 20 kWpf = 0.6V = 400 V
Q = kvar
6.AC power & power factorHard
A single-phase load takes 10 kW at a power factor of 0.6 lagging from a 400 V, 50 Hz supply. A 68 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 10 kWpf1 = 0.6V = 400 Vf = 50 HzC = 68 µF
I = A
7.Three-phase powerMedium
A balanced Δ-connected load with an impedance of 16 + j12 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.
VL = 208 VZ = 16 + j12 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 75 kW at a power factor of 0.8 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. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 75 kWpf1 = 0.8pf2 = 0.95VL = 690 Vf = 50 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: ΔIL = 1.023 A
DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.
Duty cycle
D = 9/36 = 0.25
Inductor voltage during the on-time
While S is on, the inductor sits between Vin and Vo.
VL = 36 − 9 = 27 V
Ripple current
ΔIL = VL·D/(L·f) = 27·0.25/(330·10⁻⁶·20·10³) = 1.023 A
Answer
ΔIL = 1.023 A
2.DC-DC converters
Answer: C = 225 µF
DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.
Duty cycle
D = 4.5/9 = 0.5
Inductor ripple current
ΔIL = (9 − 4.5)·0.5/(100·10⁻⁶·50·10³) = 0.45 A
Capacitance
C = 0.45/(8·50·10³·5·10⁻³) = 225 µF
Answer
C = 225 µF
3.DC motors
Answer: η = 86.4 %
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.
Powers
Pin = 240·30 = 7200 W
Pdev = 216·30 = 6480 W
Pout = 6480 − 259 = 6221 W
Efficiency
η = 6221/7200·100 = 86.4 %
Answer
η = 86.4 %
4.DC motors
Answer: Pmax = 120 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 = 12 V
Maximum power
Pmax = (V/2)²/Ra = V²/(4Ra) = 24²/(4·1.2) = 120 W
Answer
Pmax = 120 W
At this point the efficiency is only 50 %, so real motors run well above half speed.
5.AC power & power factor
Answer: Q = 26.67 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(0.6) = 53.13°, tan φ = 1.333
Reactive power
Q = P·tan φ = 20·1.333 = 26.67 kvar
Answer
Q = 26.67 kvar
6.AC power & power factor
Answer: I = 35.21 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 = 10·tan(arccos 0.6) = 13.33 kvar
QC = V²·2πf·C = 400²·2π·50·68·10⁻⁶ = 3418 var = 3.418 kvar
Q2 = 13.33 − 3.418 = 9.915 kvar
New apparent power
S2 = √(10² + 9.915²) = 14.08 kVA
New line current
I = S2/V = 14082/400 = 35.21 A
Answer
I = 35.21 A
7.Three-phase power
Answer: P = 5.192 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.
Phase voltage
Vph = VL = 208 V
Phase current
|Z| = √(16² + 12²) = 20 Ω
Iph = 208/20 = 10.4 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·10.4²·16 = 5.192 kW
Answer
P = 5.192 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: QC = 31.6 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.