A DC motor with k = 2 N·m/A and Ra = 1.5 Ω is connected to 240 V. Find its stall torque Tstall, the torque at zero speed.
V = 240 Vk = 2 N·m/ARa = 1.5 Ω
Tstall = N·m
4.DC motorsHard
A DC motor with k = 0.6 V·s/rad and Ra = 0.5 Ω must drive a load of 4.8 N·m at 400 rpm. Find the armature voltage V it needs.
k = 0.6 V·s/radRa = 0.5 ΩT = 4.8 N·mn = 400 rpm
V = V
5.AC power & power factorMedium
A single-phase motor takes 19 kW at a power factor of 0.6 lagging from a 400 V supply. Find the line current I.
P = 19 kWpf = 0.6V = 400 V
I = A
6.AC power & power factorHard
A single-phase load takes 17 kW at a power factor of 0.65 lagging from a 400 V, 50 Hz supply. A 150 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 17 kWpf1 = 0.65V = 400 Vf = 50 HzC = 150 µF
I = A
7.Three-phase powerMedium
A balanced Δ-connected load with an impedance of 24 + j18 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 = 24 + j18 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 75 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. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 75 kWpf1 = 0.65pf2 = 0.9VL = 415 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: ΔIL = 1.43 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 − 24/80 = 0.7
Inductor voltage during the on-time
While S is on, the inductor sits directly across Vin.
VL = 24 V
Ripple current
ΔIL = VL·D/(L·f) = 24·0.7/(47·10⁻⁶·250·10³) = 1.43 A
Answer
ΔIL = 1.43 A
2.DC-DC converters
Answer: ΔVo = 5.984 mV
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 = 12/48 = 0.25
Inductor ripple current
ΔIL = (48 − 12)·0.25/(100·10⁻⁶·200·10³) = 0.45 A
Output ripple
ΔVo = 0.45/(8·200·10³·47·10⁻⁶) = 5.984 mV
Answer
ΔVo = 5.984 mV
3.DC motors
Answer: Tstall = 320 N·m
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.
Stall current
Ia = V/Ra = 240/1.5 = 160 A
Stall torque
Tstall = 2·160 = 320 N·m
Answer
Tstall = 320 N·m
4.DC motors
Answer: V = 29.13 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 = 0.6·41.89 = 25.13 V
Armature current
Ia = 4.8/0.6 = 8 A
Armature voltage
V = 25.13 + 8·0.5 = 29.13 V
Answer
V = 29.13 V
5.AC power & power factor
Answer: I = 79.17 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.
Line current
Only the real part of the power does work, so divide by the power factor.
I = P/(V·cos φ) = 19000/(400·0.6) = 79.17 A
Answer
I = 79.17 A
6.AC power & power factor
Answer: I = 52.51 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 = 17·tan(arccos 0.65) = 19.88 kvar
QC = V²·2πf·C = 400²·2π·50·150·10⁻⁶ = 7540 var = 7.54 kvar
Q2 = 19.88 − 7.54 = 12.34 kvar
New apparent power
S2 = √(17² + 12.34²) = 21 kVA
New line current
I = S2/V = 21004/400 = 52.51 A
Answer
I = 52.51 A
7.Three-phase power
Answer: P = 13.78 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 = 415 V
Phase current
|Z| = √(24² + 18²) = 30 Ω
Iph = 415/30 = 13.83 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·13.83²·24 = 13.78 kW
Answer
P = 13.78 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 316.4 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 0.65) = 75·1.169 = 87.68 kvar
Q2 = 75·tan(acos 0.9) = 75·0.4843 = 36.32 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 51.36 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.