A DC motor (k = 0.6 V·s/rad, Ra = 0.25 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 = 0.6 V·s/radRa = 0.25 ΩIFL = 40 A
SR = %
4.DC motorsHard
A 120 V DC motor has an armature resistance of 0.5 Ω. Find the series starting resistance Rst that limits the starting current to 20 A.
V = 120 VRa = 0.5 ΩImax = 20 A
Rst = Ω
5.AC power & power factorMedium
Two loads are connected in parallel to a 400 V single-phase supply: a 14 kW load at a power factor of 0.85 lagging and a 17 kW heater (unity power factor). Find the total current I drawn from the supply.
A single-phase load takes 45 kW at a power factor of 0.8 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 = 0.8pf2 = 0.9V = 400 Vf = 60 Hz
C = µF
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 120 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.88 lagging. Find the total real power P drawn by the load in kW.
VL = 208 VIL = 120 Acos φ = 0.88
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 150 kW at a power factor of 0.72 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 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 150 kWpf1 = 0.72pf2 = 0.92VL = 480 Vf = 60 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: IL = 33.33 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 − 24/40 = 0.4
Load current
Io = 40/2 = 20 A
Inductor current
IL = Io/(1 − D) = 20/0.6 = 33.33 A
Answer
IL = 33.33 A
2.DC-DC converters
Answer: C = 120 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 = 24/48 = 0.5
Inductor ripple current
ΔIL = (48 − 24)·0.5/(10·10⁻⁶·500·10³) = 2.4 A
Capacitance
C = 2.4/(8·500·10³·5·10⁻³) = 120 µF
Answer
C = 120 µF
3.DC motors
Answer: SR = 4.762 %
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·0.6) = 3501 rpm
nFL = 60·(220 − 40·0.25)/(2π·0.6) = 3342 rpm
Speed regulation
SR = (3501 − 3342)/3342·100 = 4.762 %
Answer
SR = 4.762 %
4.DC motors
Answer: Rst = 5.5 Ω
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.
Total resistance
Ra + Rst = 120/20 = 6 Ω
Starting resistor
Rst = 6 − 0.5 = 5.5 Ω
Answer
Rst = 5.5 Ω
The resistor is cut out step by step as the motor speeds up and the back-EMF grows.
5.AC power & power factor
Answer: I = 80.48 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.
Reactive power of each load
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Q1 = 14·tan(arccos 0.85) = 8.676 kvar
Q2 = 0 = 0 kvar
Totals
P = 14 + 17 = 31 kW
Q = 8.676 + 0 = 8.676 kvar
S = √(31² + 8.676²) = 32.19 kVA
Total current
I = S/V = 32191/400 = 80.48 A
Answer
I = 80.48 A
6.AC power & power factor
Answer: C = 198.2 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.
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(36.87°) = 33.75 kvar
Q2 = P·tan φ2 = 45·tan(25.84°) = 21.79 kvar
QC = Q1 − Q2 = 11.96 kvar
Capacitance
A capacitor across the supply delivers Q = V²·ωC.
C = QC/(2πf·V²) = 11956/(2π·60·400²) = 198.2 µF
Answer
C = 198.2 µF
7.Three-phase power
Answer: P = 38.04 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 = √3·208·120 = 43.23 kVA
Real power
P = S·cos φ = 43.23·0.88 = 38.04 kW
Answer
P = 38.04 kW
8.Three-phase power
Answer: QC = 80.68 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.