A DC motor (k = 1.2 V·s/rad, Ra = 0.25 Ω) runs from 240 V. At full load it draws 30 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 240 Vk = 1.2 V·s/radRa = 0.25 ΩIFL = 30 A
SR = %
4.DC motorsHard
A DC chopper feeds a motor (k = 1.5 V·s/rad, Ra = 0.2 Ω) from a 240 V DC bus. The average armature voltage is D·240 V. Find the duty cycle D that runs the motor at 900 rpm with a load of 60 N·m.
A single-phase motor takes 40 kW at a power factor of 0.65 lagging from a 240 V supply. Find the reactive power Q it draws.
P = 40 kWpf = 0.65V = 240 V
Q = kvar
6.AC power & power factorHard
A 10 V (rms), 60 Hz source has an internal impedance Zs = 4 + j3 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 = 10 V (rms)f = 60 HzZs = 4 + j3 Ω
C = µF
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 30 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.92 lagging. Find the total real power P drawn by the load in kW.
VL = 690 VIL = 30 Acos φ = 0.92
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 60 kW at a power factor of 0.8 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 60 kWpf1 = 0.8pf2 = 0.92VL = 690 Vf = 50 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.224 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 − 12/40 = 0.7
Inductor voltage during the on-time
While S is on, the inductor sits directly across Vin.
VL = 12 V
Ripple current
ΔIL = VL·D/(L·f) = 12·0.7/(150·10⁻⁶·250·10³) = 0.224 A
Answer
ΔIL = 0.224 A
2.DC-DC converters
Answer: C = 109.1 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 = 6/15 = 0.4
Inductor ripple current
ΔIL = (15 − 6)·0.4/(330·10⁻⁶·25·10³) = 0.4364 A
Capacitance
C = 0.4364/(8·25·10³·20·10⁻³) = 109.1 µF
Answer
C = 109.1 µF
3.DC motors
Answer: SR = 3.226 %
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 − 30·0.25)/(2π·1.2) = 1850 rpm
Speed regulation
SR = (1910 − 1850)/1850·100 = 3.226 %
Answer
SR = 3.226 %
4.DC motors
Answer: D = 62.24 %
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.
Required armature voltage
Va = 1.5·94.25 + 40·0.2 = 149.4 V
Duty cycle
D = 149.4/240 = 62.24 %
Answer
D = 62.24 %
5.AC power & power factor
Answer: Q = 46.77 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(0.65) = 49.46°, tan φ = 1.169
Reactive power
Q = P·tan φ = 40·1.169 = 46.77 kvar
Answer
Q = 46.77 kvar
6.AC power & power factor
Answer: C = 884.2 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* = 4 − j3 Ω
Capacitance
XC = 3 Ω → C = 1/(2π·60·3) = 884.2 µF
Answer
C = 884.2 µF
7.Three-phase power
Answer: P = 32.99 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 = √3·690·30 = 35.85 kVA
Real power
P = S·cos φ = 35.85·0.92 = 32.99 kW
Answer
P = 32.99 kW
8.Three-phase power
Answer: QC = 19.44 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.