A single-phase motor takes 30 kW at a power factor of 0.9 lagging from a 240 V supply. Find the reactive power Q it draws.
P = 30 kWpf = 0.9V = 240 V
Q = kvar
6.AC power & power factorHard
A single-phase load takes 30 kW at a power factor of 0.65 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the capacitance C needed.
P = 30 kWpf1 = 0.65pf2 = 0.92V = 400 Vf = 50 Hz
C = µF
7.Three-phase powerMedium
A balanced Δ-connected load with an impedance of 6 + j8 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.
VL = 208 VZ = 6 + j8 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 80 kW at a power factor of 0.75 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 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 80 kWpf1 = 0.75pf2 = 0.95VL = 690 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: IL = 4 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 − 12/24 = 0.5
Load current
Io = 24/12 = 2 A
Inductor current
IL = Io/(1 − D) = 2/0.5 = 4 A
Answer
IL = 4 A
2.DC-DC converters
Answer: Lcrit = 5.625 µH
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 = 56/(24 + 56) = 0.7
Boundary condition
Set the average inductor current equal to half the ripple current and solve for L.
Lcrit = (1 − D)²·R/(2f)
Critical inductance
Lcrit = (1 − 0.7)²·5/(240·10³) = 5.625 µH
Answer
Lcrit = 5.625 µH
3.DC motors
Answer: n = 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.
Armature current
Ia = T/k = 2.4/0.6 = 4 A
Back-EMF and speed
E = 48 − 4·1 = 44 V
ω = E/k = 73.33 rad/s → n = 700.3 rpm
Answer
n = 700.3 rpm
4.DC motors
Answer: Rs = 11.15 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 = 18/1.2 = 15 A
E = 1.2·31.42 = 37.7 V
Series resistance
Ra + Rs = (220 − 37.7)/15 = 12.15 Ω
Rs = 11.15 Ω
Answer
Rs = 11.15 Ω
The resistor wastes 2510 W; a chopper controls the speed with far less loss.
5.AC power & power factor
Answer: Q = 14.53 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(0.9) = 25.84°, tan φ = 0.4843
Reactive power
Q = P·tan φ = 30·0.4843 = 14.53 kvar
Answer
Q = 14.53 kvar
6.AC power & power factor
Answer: C = 443.5 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 = 30·tan(49.46°) = 35.07 kvar
Q2 = P·tan φ2 = 30·tan(23.07°) = 12.78 kvar
QC = Q1 − Q2 = 22.29 kvar
Capacitance
A capacitor across the supply delivers Q = V²·ωC.
C = QC/(2πf·V²) = 22294/(2π·50·400²) = 443.5 µF
Answer
C = 443.5 µF
7.Three-phase power
Answer: P = 7.788 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 = VL = 208 V
Phase current
|Z| = √(6² + 8²) = 10 Ω
Iph = 208/10 = 20.8 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·20.8²·6 = 7.788 kW
Answer
P = 7.788 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 98.63 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 0.75) = 80·0.8819 = 70.55 kvar
Q2 = 80·tan(acos 0.95) = 80·0.3287 = 26.29 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 44.26 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.