A DC motor draws 2 A from a 24 V supply. Its armature resistance is 0.4 Ω and its rotational (friction, windage and iron) losses are 3 W. Find its efficiency η.
V = 24 VIa = 2 ARa = 0.4 ΩProt = 3 W
η = %
4.DC motorsHard
A DC motor with k = 0.25 V·s/rad and Ra = 0.25 Ω must drive a load of 1 N·m at 300 rpm. Find the armature voltage V it needs.
k = 0.25 V·s/radRa = 0.25 ΩT = 1 N·mn = 300 rpm
V = V
5.AC power & power factorMedium
An inductive load takes 11.5 A and 1380 W from a 240 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.
V = 240 VI = 11.5 AP = 1380 W
R = Ω
6.AC power & power factorHard
A 12 V (rms), 60 Hz source has an internal impedance Zs = 2 + j10 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
Vs = 12 V (rms)f = 60 HzZs = 2 + j10 Ω
Pmax = W
7.Three-phase powerMedium
A balanced Y-connected load with an impedance of 15 + j20 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 V. Find the total real power P absorbed by the load in kW.
VL = 415 VZ = 15 + j20 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 50 kW at a power factor of 0.8 lagging from a 400 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 = 50 kWpf1 = 0.8pf2 = 0.95VL = 400 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.6818 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 − 12/48 = 0.75
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) = 12·0.75/(330·10⁻⁶·40·10³) = 0.6818 A
Answer
ΔIL = 0.6818 A
2.DC-DC converters
Answer: C = 0.1309 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/(220·10⁻⁶·500·10³) = 0.05236 A
Capacitance
C = 0.05236/(8·500·10³·100·10⁻³) = 0.1309 µF
Answer
C = 0.1309 µF
3.DC motors
Answer: η = 90.42 %
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.
Powers
Pin = 24·2 = 48 W
Pdev = 23.2·2 = 46.4 W
Pout = 46.4 − 3 = 43.4 W
Efficiency
η = 43.4/48·100 = 90.42 %
Answer
η = 90.42 %
4.DC motors
Answer: V = 8.854 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 = 0.25·31.42 = 7.854 V
Armature current
Ia = 1/0.25 = 4 A
Armature voltage
V = 7.854 + 4·0.25 = 8.854 V
Answer
V = 8.854 V
5.AC power & power factor
Answer: R = 10.43 Ω
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.
Resistance from the real power
R = P/I² = 1380/11.5² = 10.43 Ω
Answer
R = 10.43 Ω
6.AC power & power factor
Answer: Pmax = 18 W
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* = 2 − j10 Ω
Maximum power
I = Vs/(2Rs) = 12/4 = 3 A
Pmax = I²·RL = Vs²/(4Rs) = 12²/(4·2) = 18 W
Answer
Pmax = 18 W
7.Three-phase power
Answer: P = 4.133 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.
Phase voltage
Vph = 415/√3 = 239.6 V
Phase current
|Z| = √(15² + 20²) = 25 Ω
Iph = 239.6/25 = 9.584 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·9.584²·15 = 4.133 kW
Answer
P = 4.133 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 116.4 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 = 50·tan(acos 0.8) = 50·0.75 = 37.5 kvar
Q2 = 50·tan(acos 0.95) = 50·0.3287 = 16.43 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 21.07 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.