An ideal buck-boost converter turns 24 V into −72 V for a 10 Ω load at 100 kHz. The output capacitor is 10 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.
A DC motor with k = 0.1 V·s/rad runs from 24 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
V = 24 Vk = 0.1 V·s/rad
n0 = rpm
4.DC motorsHard
A DC motor with k = 1 V·s/rad and Ra = 0.25 Ω must drive a load of 30 N·m at 900 rpm. Find the armature voltage V it needs.
k = 1 V·s/radRa = 0.25 ΩT = 30 N·mn = 900 rpm
V = V
5.AC power & power factorMedium
A single-phase motor takes 4 kW at a power factor of 0.9 lagging from a 240 V supply. Find the reactive power Q it draws.
P = 4 kWpf = 0.9V = 240 V
Q = kvar
6.AC power & power factorHard
A single-phase load takes 10 kW at a power factor of 0.8 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the new line current Inew.
P = 10 kWpf1 = 0.8pf2 = 0.95V = 400 Vf = 50 Hz
Inew = A
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 80 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.
VL = 690 VIL = 80 Acos φ = 0.82
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 200 kW at a power factor of 0.7 lagging from a 690 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 = 200 kWpf1 = 0.7pf2 = 0.98VL = 690 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 4.954 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 = 19.2/48 = 0.4
Average and ripple current
Io = 19.2/4 = 4.8 A
ΔIL = VL·D/(L·f) = 28.8·0.4/(150·10⁻⁶·250·10³) = 0.3072 A
Peak current
Imax = 4.8 + 0.3072/2 = 4.954 A
Answer
Imax = 4.954 A
2.DC-DC converters
Answer: ΔVo = 5400 mV
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 = 72/(24 + 72) = 0.75
Load current
Io = 72/10 = 7.2 A
Output ripple
ΔVo = 7.2·0.75/(100·10³·10·10⁻⁶) = 5400 mV
Answer
ΔVo = 5400 mV
3.DC motors
Answer: n0 = 2292 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.
No-load speed
ω₀ = V/k = 24/0.1 = 240 rad/s
n0 = 60·240/(2π) = 2292 rpm
Answer
n0 = 2292 rpm
4.DC motors
Answer: V = 101.7 V
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
E = 1·94.25 = 94.25 V
Armature current
Ia = 30/1 = 30 A
Armature voltage
V = 94.25 + 30·0.25 = 101.7 V
Answer
V = 101.7 V
5.AC power & power factor
Answer: Q = 1.937 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 = 4·0.4843 = 1.937 kvar
Answer
Q = 1.937 kvar
6.AC power & power factor
Answer: Inew = 26.32 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 = P·tan φ1 = 10·tan(36.87°) = 7.5 kvar
Q2 = P·tan φ2 = 10·tan(18.19°) = 3.287 kvar
QC = Q1 − Q2 = 4.213 kvar
New line current
Inew = P/(V·cos φ2) = 10000/(400·0.95) = 26.32 A
Answer
Inew = 26.32 A
7.Three-phase power
Answer: P = 78.4 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 = √3·690·80 = 95.61 kVA
Real power
P = S·cos φ = 95.61·0.82 = 78.4 kW
Answer
P = 78.4 kW
8.Three-phase power
Answer: Cph = 910.5 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 0.7) = 200·1.02 = 204 kvar
Q2 = 200·tan(acos 0.98) = 200·0.2031 = 40.61 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 163.4 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.