An ideal inverting buck-boost converter turns 36 V into −12 V for a 10 Ω load at 40 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 36 VVo = −12 Vf = 40 kHzR = 10 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor (k = 1 N·m/A, Ra = 0.4 Ω) runs from 240 V and drives a load torque of 40 N·m. Find its speed n in rpm.
V = 240 Vk = 1 N·m/ARa = 0.4 ΩT = 40 N·m
n = rpm
4.DC motorsHard
A separately excited DC motor on 120 V (Ra = 0.4 Ω) runs at 1800 rpm with an armature current of 15 A. The load changes and the current becomes 30 A, with the field unchanged. Find the new speed n2.
V = 120 VRa = 0.4 Ωn1 = 1800 rpmIa1 = 15 AIa2 = 30 A
n2 = rpm
5.AC power & power factorMedium
A single-phase motor takes 35 kW at a power factor of 0.6 lagging from a 240 V supply. Find the reactive power Q it draws.
P = 35 kWpf = 0.6V = 240 V
Q = kvar
6.AC power & power factorHard
A single-phase load takes 45 kW at a power factor of 0.7 lagging from a 400 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 = 45 kWpf1 = 0.7pf2 = 0.92V = 400 Vf = 60 Hz
QC = kvar
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 100 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.
VL = 208 VIL = 100 Acos φ = 0.85
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 75 kW at a power factor of 0.72 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 75 kWpf1 = 0.72pf2 = 0.95VL = 415 Vf = 60 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: Imax = 1.12 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 = 2.7/9 = 0.3
Average and ripple current
Io = 2.7/2.5 = 1.08 A
ΔIL = VL·D/(L·f) = 6.3·0.3/(47·10⁻⁶·500·10³) = 0.08043 A
Peak current
Imax = 1.08 + 0.08043/2 = 1.12 A
Answer
Imax = 1.12 A
2.DC-DC converters
Answer: Lcrit = 70.31 µH
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 = 12/(36 + 12) = 0.25
Boundary condition
Set the average inductor current equal to half the ripple current and solve for L.
Lcrit = (1 − D)²·R/(2f)
Critical inductance
Lcrit = (1 − 0.25)²·10/(240·10³) = 70.31 µH
Answer
Lcrit = 70.31 µH
3.DC motors
Answer: n = 2139 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 = 40/1 = 40 A
Back-EMF and speed
E = 240 − 40·0.4 = 224 V
ω = E/k = 224 rad/s → n = 2139 rpm
Answer
n = 2139 rpm
4.DC motors
Answer: n2 = 1705 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 = 120 − 15·0.4 = 114 V
E2 = 120 − 30·0.4 = 108 V
New speed
n2 = 1800·108/114 = 1705 rpm
Answer
n2 = 1705 rpm
5.AC power & power factor
Answer: Q = 46.67 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 = 35·1.333 = 46.67 kvar
Answer
Q = 46.67 kvar
6.AC power & power factor
Answer: QC = 26.74 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 = 45·tan(45.57°) = 45.91 kvar
Q2 = P·tan φ2 = 45·tan(23.07°) = 19.17 kvar
QC = Q1 − Q2 = 26.74 kvar
Answer
QC = 26.74 kvar
7.Three-phase power
Answer: P = 30.62 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 = √3·208·100 = 36.03 kVA
Real power
P = S·cos φ = 36.03·0.85 = 30.62 kW
Answer
P = 30.62 kW
8.Three-phase power
Answer: QC = 47.64 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.