A DC motor (k = 0.5 V·s/rad, Ra = 0.8 Ω) runs from 48 V. At full load it draws 2 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 48 Vk = 0.5 V·s/radRa = 0.8 ΩIFL = 2 A
SR = %
4.DC motorsHard
A separately excited DC motor (k = 0.25 V·s/rad at full field, Ra = 0.5 Ω, V = 48 V) drives a constant load torque of 2 N·m. The field flux is reduced to 90 % of its full value. Find the new speed n.
A single-phase motor takes 4 kW at a power factor of 0.9 lagging from a 230 V supply. Find the reactive power Q it draws.
P = 4 kWpf = 0.9V = 230 V
Q = kvar
6.AC power & power factorHard
A 200 kVA single-phase transformer supplies a load of 55 kW at a power factor of 0.85 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 = 200 kVAP1 = 55 kWpf1 = 0.85
ΔP = kW
7.Three-phase powerMedium
A balanced Δ-connected load with an impedance of 20 + j15 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 = 20 + j15 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 200 kW at a power factor of 0.75 lagging from a 415 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 = 200 kWpf1 = 0.75pf2 = 0.95VL = 415 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: ΔIL = 1.92 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 − 48/120 = 0.6
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) = 48·0.6/(150·10⁻⁶·100·10³) = 1.92 A
Answer
ΔIL = 1.92 A
2.DC-DC converters
Answer: C = 18.18 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 = 6/12 = 0.5
Inductor ripple current
ΔIL = (12 − 6)·0.5/(330·10⁻⁶·25·10³) = 0.3636 A
Capacitance
C = 0.3636/(8·25·10³·100·10⁻³) = 18.18 µF
Answer
C = 18.18 µF
3.DC motors
Answer: SR = 3.448 %
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 and full-load speed
n₀ = 60·48/(2π·0.5) = 916.7 rpm
nFL = 60·(48 − 2·0.8)/(2π·0.5) = 886.2 rpm
Speed regulation
SR = (916.7 − 886.2)/886.2·100 = 3.448 %
Answer
SR = 3.448 %
4.DC motors
Answer: n = 1849 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.
Field weakening is how a DC drive runs above its base speed.
5.AC power & power factor
Answer: Q = 1.937 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 = 4·0.4843 = 1.937 kvar
Answer
Q = 1.937 kvar
6.AC power & power factor
Answer: ΔP = 142.1 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 = 55·tan(arccos 0.85) = 34.09 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(200² − 34.09²) = 197.1 kW
Extra real power
ΔP = 197.1 − 55 = 142.1 kW
Answer
ΔP = 142.1 kW
7.Three-phase power
Answer: P = 4.153 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| = √(20² + 15²) = 25 Ω
Iph = 208/25 = 8.32 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·8.32²·20 = 4.153 kW
Answer
P = 4.153 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 568.1 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 = 200·tan(acos 0.75) = 200·0.8819 = 176.4 kvar
Q2 = 200·tan(acos 0.95) = 200·0.3287 = 65.74 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 110.6 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.