A DC motor (k = 0.5 N·m/A, Ra = 0.8 Ω) runs from 120 V and drives a load torque of 10 N·m. Find its speed n in rpm.
V = 120 Vk = 0.5 N·m/ARa = 0.8 ΩT = 10 N·m
n = rpm
4.DC motorsHard
A DC chopper feeds a motor (k = 0.12 V·s/rad, Ra = 2 Ω) from a 48 V DC bus. The average armature voltage is D·48 V. Find the duty cycle D that runs the motor at 500 rpm with a load of 0.24 N·m.
A load with impedance Z = 12 + j9 Ω is connected to a 230 V (rms) single-phase supply. Find the real power P it absorbs.
V = 230 VZ = 12 + j9 Ω
P = W
6.AC power & power factorHard
A 12 V (rms), 50 Hz source has an internal impedance Zs = 5 + j40 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 = 50 HzZs = 5 + j40 Ω
Pmax = W
7.Three-phase powerMedium
A balanced Y-connected load with an impedance of 12 + j5 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.
VL = 480 VZ = 12 + j5 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 100 kW at a power factor of 0.65 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 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 100 kWpf1 = 0.65pf2 = 0.98VL = 480 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.3706 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 = 1 − 9/30 = 0.7
Inductor voltage during the on-time
While S is on, the inductor sits directly across Vin.
VL = 9 V
Ripple current
ΔIL = VL·D/(L·f) = 9·0.7/(68·10⁻⁶·250·10³) = 0.3706 A
Answer
ΔIL = 0.3706 A
2.DC-DC converters
Answer: C = 68.18 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 = 3/12 = 0.25
Inductor ripple current
ΔIL = (12 − 3)·0.25/(330·10⁻⁶·25·10³) = 0.2727 A
Capacitance
C = 0.2727/(8·25·10³·20·10⁻³) = 68.18 µF
Answer
C = 68.18 µF
3.DC motors
Answer: n = 1986 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 = 10/0.5 = 20 A
Back-EMF and speed
E = 120 − 20·0.8 = 104 V
ω = E/k = 208 rad/s → n = 1986 rpm
Answer
n = 1986 rpm
4.DC motors
Answer: D = 21.42 %
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 = 0.12·52.36 + 2·2 = 10.28 V
Duty cycle
D = 10.28/48 = 21.42 %
Answer
D = 21.42 %
5.AC power & power factor
Answer: P = 2821 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.
Current
|Z| = √(12² + 9²) = 15 Ω
I = V/|Z| = 230/15 = 15.33 A
Real power
P = I²·R = 15.33²·12 = 2821 W
Answer
P = 2821 W
6.AC power & power factor
Answer: Pmax = 7.2 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* = 5 − j40 Ω
Maximum power
I = Vs/(2Rs) = 12/10 = 1.2 A
Pmax = I²·RL = Vs²/(4Rs) = 12²/(4·5) = 7.2 W
Answer
Pmax = 7.2 W
7.Three-phase power
Answer: P = 16.36 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 = 480/√3 = 277.1 V
Phase current
|Z| = √(12² + 5²) = 13 Ω
Iph = 277.1/13 = 21.32 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·21.32²·12 = 16.36 kW
Answer
P = 16.36 kW
Check: cos φ = R/|Z| = 0.9231, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 1112 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 = 100·tan(acos 0.65) = 100·1.169 = 116.9 kvar
Q2 = 100·tan(acos 0.98) = 100·0.2031 = 20.31 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 96.61 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.