A DC motor (k = 1.2 V·s/rad, Ra = 1.2 Ω) runs from 240 V. At full load it draws 25 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 240 Vk = 1.2 V·s/radRa = 1.2 ΩIFL = 25 A
SR = %
4.DC motorsHard
A separately excited DC motor (k = 1 V·s/rad at full field, Ra = 1.5 Ω, V = 110 V) drives a constant load torque of 6 N·m. The field flux is reduced to 75 % of its full value. Find the new speed n.
V = 110 Vk = 1 V·s/radRa = 1.5 ΩT = 6 N·mΦ = 75 %
n = rpm
5.AC power & power factorMedium
Two loads are connected in parallel to a 230 V single-phase supply: a 21 kW load at a power factor of 0.65 lagging and a 4 kW load at a power factor of 0.95 leading. Find the overall power factor cos φ seen by the supply.
V = 230 VP1 = 21 kW, pf1 = 0.65 laggingP2 = 4 kW, pf2 = 0.95 leading
cos φ =
6.AC power & power factorHard
A 120 V (rms), 60 Hz source has an internal impedance Zs = 8 + j15 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.
Vs = 120 V (rms)f = 60 HzZs = 8 + j15 Ω
C = µF
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 75 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.9 lagging. Find the total real power P drawn by the load in kW.
VL = 415 VIL = 75 Acos φ = 0.9
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 200 kW at a power factor of 0.75 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 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 200 kWpf1 = 0.75pf2 = 0.95VL = 480 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: ΔIL = 1.6 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 − 48/64 = 0.25
Inductor voltage during the on-time
While S is on, the inductor sits directly across Vin.
VL = 48 V
Ripple current
ΔIL = VL·D/(L·f) = 48·0.25/(150·10⁻⁶·50·10³) = 1.6 A
Answer
ΔIL = 1.6 A
2.DC-DC converters
Answer: C = 3.064 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 = 7.2/12 = 0.6
Inductor ripple current
ΔIL = (12 − 7.2)·0.6/(47·10⁻⁶·500·10³) = 0.1226 A
Capacitance
C = 0.1226/(8·500·10³·10·10⁻³) = 3.064 µF
Answer
C = 3.064 µF
3.DC motors
Answer: SR = 14.29 %
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 = 60·240/(2π·1.2) = 1910 rpm
nFL = 60·(240 − 25·1.2)/(2π·1.2) = 1671 rpm
Speed regulation
SR = (1910 − 1671)/1671·100 = 14.29 %
Answer
SR = 14.29 %
4.DC motors
Answer: n = 1248 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.
Field weakening is how a DC drive runs above its base speed.
5.AC power & power factor
Answer: cos φ = 0.7325
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 each load
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Q1 = 21·tan(arccos 0.65) = 24.55 kvar
Q2 = −4·tan(arccos 0.95) = −1.315 kvar
Totals
P = 21 + 4 = 25 kW
Q = 24.55 + (−1.315) = 23.24 kvar
S = √(25² + 23.24²) = 34.13 kVA
Overall power factor
cos φ = P/S = 25/34.13 = 0.7325 lagging
Answer
cos φ = 0.7325
6.AC power & power factor
Answer: C = 176.8 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.
Conjugate match
The load reactance cancels the source reactance, and the load resistance equals the source resistance.
ZL = Zs* = 8 − j15 Ω
Capacitance
XC = 15 Ω → C = 1/(2π·60·15) = 176.8 µF
Answer
C = 176.8 µF
7.Three-phase power
Answer: P = 48.52 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 = √3·415·75 = 53.91 kVA
Real power
P = S·cos φ = 53.91·0.9 = 48.52 kW
Answer
P = 48.52 kW
8.Three-phase power
Answer: Cph = 424.6 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 0.75) = 200·0.8819 = 176.4 kvar
Q2 = 200·tan(acos 0.95) = 200·0.3287 = 65.74 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 110.6 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.