A DC motor with k = 1.5 V·s/rad runs from 110 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
V = 110 Vk = 1.5 V·s/rad
n0 = rpm
4.DC motorsHard
A DC motor (k = 0.3 V·s/rad, Ra = 0.8 Ω) on a fixed 48 V supply drives a load of 1.8 N·m. Find the series resistance Rs that slows it down to 1000 rpm.
An inductive load takes 16 A and 1440 W from a 120 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.
V = 120 VI = 16 AP = 1440 W
R = Ω
6.AC power & power factorHard
A 100 kVA single-phase transformer supplies a load of 31 kW at a power factor of 0.65 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
Smax = 100 kVAP1 = 31 kWpf1 = 0.65
ΔP = kW
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 20 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.9 lagging. Find the total real power P drawn by the load in kW.
VL = 690 VIL = 20 Acos φ = 0.9
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 75 kW at a power factor of 0.72 lagging from a 690 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 Y. Find the capacitance Cph of each capacitor in µF.
P = 75 kWpf1 = 0.72pf2 = 0.95VL = 690 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: ΔIL = 2.909 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 − 48/80 = 0.4
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) = 48·0.4/(330·10⁻⁶·20·10³) = 2.909 A
Answer
ΔIL = 2.909 A
2.DC-DC converters
Answer: C = 360 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/(100·10⁻⁶·25·10³) = 1.44 A
Capacitance
C = 1.44/(8·25·10³·20·10⁻³) = 360 µF
Answer
C = 360 µF
3.DC motors
Answer: n0 = 700.3 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.
No-load speed
ω₀ = V/k = 110/1.5 = 73.33 rad/s
n0 = 60·73.33/(2π) = 700.3 rpm
Answer
n0 = 700.3 rpm
4.DC motors
Answer: Rs = 1.964 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.
Current and back-EMF
Ia = 1.8/0.3 = 6 A
E = 0.3·104.7 = 31.42 V
Series resistance
Ra + Rs = (48 − 31.42)/6 = 2.764 Ω
Rs = 1.964 Ω
Answer
Rs = 1.964 Ω
The resistor wastes 70.7 W; a chopper controls the speed with far less loss.
5.AC power & power factor
Answer: R = 5.625 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 = 1440/16² = 5.625 Ω
Answer
R = 5.625 Ω
6.AC power & power factor
Answer: ΔP = 62.2 kW
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 the existing load
Q1 = 31·tan(arccos 0.65) = 36.24 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(100² − 36.24²) = 93.2 kW
Extra real power
ΔP = 93.2 − 31 = 62.2 kW
Answer
ΔP = 62.2 kW
7.Three-phase power
Answer: P = 21.51 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 = √3·690·20 = 23.9 kVA
Real power
P = S·cos φ = 23.9·0.9 = 21.51 kW
Answer
P = 21.51 kW
8.Three-phase power
Answer: Cph = 265.4 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 = 75·tan(acos 0.72) = 75·0.9639 = 72.29 kvar
Q2 = 75·tan(acos 0.95) = 75·0.3287 = 24.65 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 47.64 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.