A DC motor (k = 1 N·m/A, Ra = 0.4 Ω) runs from 120 V and drives a load torque of 40 N·m. Find its speed n in rpm.
V = 120 Vk = 1 N·m/ARa = 0.4 ΩT = 40 N·m
n = rpm
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 25 A.
V = 120 VRa = 0.5 ΩImax = 25 A
Rst = Ω
5.AC power & power factorMedium
Two loads are connected in parallel to a 230 V single-phase supply: a 4 kW load at a power factor of 0.85 lagging and a 6 kW heater (unity power factor). Find the total current I drawn from the supply.
A single-phase load takes 16 kW at a power factor of 0.7 lagging from a 400 V, 60 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 16 kWpf1 = 0.7V = 400 Vf = 60 HzC = 100 µF
I = A
7.Three-phase powerMedium
A three-phase induction motor delivers 75 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.85 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.
Pout = 75 kWη = 91%cos φ = 0.85VL = 690 V
IL = A
8.Three-phase powerHard
A balanced three-phase load takes 80 kW at a power factor of 0.8 lagging from a 480 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 80 kWpf1 = 0.8pf2 = 0.98VL = 480 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.5455 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 = 16/(48 + 16) = 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/(220·10⁻⁶·100·10³) = 0.5455 A
Answer
ΔIL = 0.5455 A
2.DC-DC converters
Answer: C = 1047 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 = 19.2/48 = 0.4
Inductor ripple current
ΔIL = (48 − 19.2)·0.4/(220·10⁻⁶·25·10³) = 2.095 A
Capacitance
C = 2.095/(8·25·10³·10·10⁻³) = 1047 µF
Answer
C = 1047 µF
3.DC motors
Answer: n = 993.1 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 = 40/1 = 40 A
Back-EMF and speed
E = 120 − 40·0.4 = 104 V
ω = E/k = 104 rad/s → n = 993.1 rpm
Answer
n = 993.1 rpm
4.DC motors
Answer: Rst = 4.3 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 = 120/25 = 4.8 Ω
Starting resistor
Rst = 4.8 − 0.5 = 4.3 Ω
Answer
Rst = 4.3 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 = 44.79 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 = 4·tan(arccos 0.85) = 2.479 kvar
Q2 = 0 = 0 kvar
Totals
P = 4 + 6 = 10 kW
Q = 2.479 + 0 = 2.479 kvar
S = √(10² + 2.479²) = 10.3 kVA
Total current
I = S/V = 10303/230 = 44.79 A
Answer
I = 44.79 A
6.AC power & power factor
Answer: I = 47.56 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 before and after
The real power P does not change; only the reactive power drawn from the supply does.
Q1 = 16·tan(arccos 0.7) = 16.32 kvar
QC = V²·2πf·C = 400²·2π·60·100·10⁻⁶ = 6032 var = 6.032 kvar
Q2 = 16.32 − 6.032 = 10.29 kvar
New apparent power
S2 = √(16² + 10.29²) = 19.02 kVA
New line current
I = S2/V = 19024/400 = 47.56 A
Answer
I = 47.56 A
7.Three-phase power
Answer: IL = 81.13 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 = 75/0.91 = 82.42 kW
Line current
IL = Pin/(√3·VL·cos φ) = 82418/(√3·690·0.85) = 81.13 A
Answer
IL = 81.13 A
8.Three-phase power
Answer: Cph = 604.5 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 = 80·tan(acos 0.8) = 80·0.75 = 60 kvar
Q2 = 80·tan(acos 0.98) = 80·0.2031 = 16.24 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 43.76 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.