An ideal boost converter turns 9 V into 36 V for a 4 Ω load at 250 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 9 VVo = 36 Vf = 250 kHzR = 4 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor (k = 0.3 N·m/A, Ra = 0.2 Ω) runs from 48 V and drives a load torque of 7.5 N·m. Find its speed n in rpm.
V = 48 Vk = 0.3 N·m/ARa = 0.2 ΩT = 7.5 N·m
n = rpm
4.DC motorsHard
A DC motor with k = 0.12 V·s/rad and Ra = 2 Ω must drive a load of 0.24 N·m at 1200 rpm. Find the armature voltage V it needs.
k = 0.12 V·s/radRa = 2 ΩT = 0.24 N·mn = 1200 rpm
V = V
5.AC power & power factorMedium
A single-phase motor takes 28 kW at a power factor of 0.9 lagging from a 400 V supply. Find the line current I.
P = 28 kWpf = 0.9V = 400 V
I = A
6.AC power & power factorHard
A 200 kVA single-phase transformer supplies a load of 128 kW at a power factor of 0.85 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
Smax = 200 kVAP1 = 128 kWpf1 = 0.85
ΔP = kW
7.Three-phase powerMedium
A three-phase induction motor delivers 110 kW of mechanical output power at full load. At this load its efficiency is 90% and its power factor is 0.88 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.
Pout = 110 kWη = 90%cos φ = 0.88VL = 480 V
IL = A
8.Three-phase powerHard
A balanced three-phase load takes 250 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 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 250 kWpf1 = 0.65pf2 = 0.9VL = 480 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 0.4301 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 = 7.5/15 = 0.5
Average and ripple current
Io = 7.5/20 = 0.375 A
ΔIL = VL·D/(L·f) = 7.5·0.5/(68·10⁻⁶·500·10³) = 0.1103 A
Peak current
Imax = 0.375 + 0.1103/2 = 0.4301 A
Answer
Imax = 0.4301 A
2.DC-DC converters
Answer: Lcrit = 0.375 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 = 1 − 9/36 = 0.75
Boundary condition
Set the average inductor current equal to half the ripple current and solve for L.
Lcrit = D(1 − D)²·R/(2f)
Critical inductance
Lcrit = 0.75(1 − 0.75)²·4/(2250·10³) = 0.375 µH
Answer
Lcrit = 0.375 µH
3.DC motors
Answer: n = 1369 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 = 7.5/0.3 = 25 A
Back-EMF and speed
E = 48 − 25·0.2 = 43 V
ω = E/k = 143.3 rad/s → n = 1369 rpm
Answer
n = 1369 rpm
4.DC motors
Answer: V = 19.08 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 = 0.12·125.7 = 15.08 V
Armature current
Ia = 0.24/0.12 = 2 A
Armature voltage
V = 15.08 + 2·2 = 19.08 V
Answer
V = 19.08 V
5.AC power & power factor
Answer: I = 77.78 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.
Line current
Only the real part of the power does work, so divide by the power factor.
I = P/(V·cos φ) = 28000/(400·0.9) = 77.78 A
Answer
I = 77.78 A
6.AC power & power factor
Answer: ΔP = 55.6 kW
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 of the existing load
Q1 = 128·tan(arccos 0.85) = 79.33 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(200² − 79.33²) = 183.6 kW
Extra real power
ΔP = 183.6 − 128 = 55.6 kW
Answer
ΔP = 55.6 kW
7.Three-phase power
Answer: IL = 167.1 A
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.
Electrical input power
Pin = Pout/η = 110/0.9 = 122.2 kW
Line current
IL = Pin/(√3·VL·cos φ) = 122222/(√3·480·0.88) = 167.1 A
Answer
IL = 167.1 A
8.Three-phase power
Answer: Cph = 1971 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 = 250·tan(acos 0.65) = 250·1.169 = 292.3 kvar
Q2 = 250·tan(acos 0.9) = 250·0.4843 = 121.1 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 171.2 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.