Two loads are connected in parallel to a 400 V single-phase supply: a 14 kW load at a power factor of 0.65 lagging and a 7 kW load at a power factor of 0.9 leading. Find the total current I drawn from the supply.
V = 400 VP1 = 14 kW, pf1 = 0.65 laggingP2 = 7 kW, pf2 = 0.9 leading
I = A
6.AC power & power factorHard
A single-phase load takes 20 kW at a power factor of 0.6 lagging from a 230 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the new line current Inew.
P = 20 kWpf1 = 0.6pf2 = 0.92V = 230 Vf = 50 Hz
Inew = A
7.Three-phase powerMedium
A three-phase induction motor delivers 37 kW of mechanical output power at full load. At this load its efficiency is 94% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.
Pout = 37 kWη = 94%cos φ = 0.82VL = 690 V
IL = A
8.Three-phase powerHard
A balanced three-phase load takes 300 kW at a power factor of 0.65 lagging from a 415 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 300 kWpf1 = 0.65pf2 = 0.9VL = 415 Vf = 50 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.1309 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 − 12/30 = 0.6
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) = 12·0.6/(220·10⁻⁶·250·10³) = 0.1309 A
Answer
ΔIL = 0.1309 A
2.DC-DC converters
Answer: Io,min = 0.03409 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 = 3/12 = 0.25
Ripple current
ΔIL = 3·0.75/(330·10⁻⁶·100·10³) = 0.06818 A
Boundary load current
Io,min = ΔIL/2 = 0.03409 A
Answer
Io,min = 0.03409 A
Below this current the converter enters discontinuous conduction and Vo rises above D·Vin.
3.DC motors
Answer: n = 2139 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 = 8/1 = 8 A
Back-EMF and speed
E = 240 − 8·2 = 224 V
ω = E/k = 224 rad/s → n = 2139 rpm
Answer
n = 2139 rpm
4.DC motors
Answer: Rs = 3.133 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 = 8/1 = 8 A
E = 1·78.54 = 78.54 V
Series resistance
Ra + Rs = (110 − 78.54)/8 = 3.933 Ω
Rs = 3.133 Ω
Answer
Rs = 3.133 Ω
The resistor wastes 200.5 W; a chopper controls the speed with far less loss.
5.AC power & power factor
Answer: I = 61.72 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 0.65) = 16.37 kvar
Q2 = −7·tan(arccos 0.9) = −3.39 kvar
Totals
P = 14 + 7 = 21 kW
Q = 16.37 + (−3.39) = 12.98 kvar
S = √(21² + 12.98²) = 24.69 kVA
Total current
I = S/V = 24686/400 = 61.72 A
Answer
I = 61.72 A
6.AC power & power factor
Answer: Inew = 94.52 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 = P·tan φ1 = 20·tan(53.13°) = 26.67 kvar
Q2 = P·tan φ2 = 20·tan(23.07°) = 8.52 kvar
QC = Q1 − Q2 = 18.15 kvar
New line current
Inew = P/(V·cos φ2) = 20000/(230·0.92) = 94.52 A
Answer
Inew = 94.52 A
7.Three-phase power
Answer: IL = 40.17 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 = 37/0.94 = 39.36 kW
Line current
IL = Pin/(√3·VL·cos φ) = 39362/(√3·690·0.82) = 40.17 A
Answer
IL = 40.17 A
8.Three-phase power
Answer: QC = 205.4 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.