A DC motor (k = 0.8 V·s/rad, Ra = 0.5 Ω) runs from 220 V. At full load it draws 40 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 220 Vk = 0.8 V·s/radRa = 0.5 ΩIFL = 40 A
SR = %
4.DC motorsHard
A DC motor with k = 0.6 V·s/rad and Ra = 0.4 Ω must drive a load of 15 N·m at 1500 rpm. Find the armature voltage V it needs.
k = 0.6 V·s/radRa = 0.4 ΩT = 15 N·mn = 1500 rpm
V = V
5.AC power & power factorMedium
A load with impedance Z = 8 + j6 Ω is connected to a 230 V (rms) single-phase supply. Find the reactive power Q it absorbs.
V = 230 VZ = 8 + j6 Ω
Q = var
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 0.9. Find the capacitance C needed.
P = 45 kWpf1 = 0.7pf2 = 0.9V = 400 Vf = 60 Hz
C = µF
7.Three-phase powerMedium
A three-phase induction motor delivers 90 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.
Pout = 90 kWη = 91%cos φ = 0.82VL = 690 V
IL = A
8.Three-phase powerHard
A balanced three-phase load takes 150 kW at a power factor of 0.8 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 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 150 kWpf1 = 0.8pf2 = 0.98VL = 415 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.1364 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 = 18/24 = 0.75
Inductor voltage during the on-time
While S is on, the inductor sits between Vin and Vo.
VL = 24 − 18 = 6 V
Ripple current
ΔIL = VL·D/(L·f) = 6·0.75/(330·10⁻⁶·100·10³) = 0.1364 A
Answer
ΔIL = 0.1364 A
2.DC-DC converters
Answer: C = 211.8 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 = 9.6/24 = 0.4
Inductor ripple current
ΔIL = (24 − 9.6)·0.4/(68·10⁻⁶·100·10³) = 0.8471 A
Capacitance
C = 0.8471/(8·100·10³·5·10⁻³) = 211.8 µF
Answer
C = 211.8 µF
3.DC motors
Answer: SR = 10 %
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 and full-load speed
n₀ = 60·220/(2π·0.8) = 2626 rpm
nFL = 60·(220 − 40·0.5)/(2π·0.8) = 2387 rpm
Speed regulation
SR = (2626 − 2387)/2387·100 = 10 %
Answer
SR = 10 %
4.DC motors
Answer: V = 104.2 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 = 0.6·157.1 = 94.25 V
Armature current
Ia = 15/0.6 = 25 A
Armature voltage
V = 94.25 + 25·0.4 = 104.2 V
Answer
V = 104.2 V
5.AC power & power factor
Answer: Q = 3174 var
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| = √(8² + 6²) = 10 Ω
I = V/|Z| = 230/10 = 23 A
Reactive power
Q = I²·X = 23²·6 = 3174 var
Answer
Q = 3174 var
6.AC power & power factor
Answer: C = 399.8 µF
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(25.84°) = 21.79 kvar
QC = Q1 − Q2 = 24.11 kvar
Capacitance
A capacitor across the supply delivers Q = V²·ωC.
C = QC/(2πf·V²) = 24115/(2π·60·400²) = 399.8 µF
Answer
C = 399.8 µF
7.Three-phase power
Answer: IL = 100.9 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 = 90/0.91 = 98.9 kW
Line current
IL = Pin/(√3·VL·cos φ) = 98901/(√3·690·0.82) = 100.9 A
Answer
IL = 100.9 A
8.Three-phase power
Answer: Cph = 421.2 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 = 150·tan(acos 0.8) = 150·0.75 = 112.5 kvar
Q2 = 150·tan(acos 0.98) = 150·0.2031 = 30.46 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 82.04 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.