Tutorial 2 Solution

Problem 1 If ν = 141.4 sin (ωt+30°) V and i = 11.31cos (ωt-30°) A, find for each; a) The maximum value, b) The rms value, c) The phasor expression in polar and rectangular form if voltage is the reference. Is the circuit inductive or capacitive? Solution: (a) Maximum values: Vmax=141.4 V Imax=11.31 A (b) rms values: V=141.42=100 V I=11.312=8 A (c) Phasor expression in polar and rectangular form: V=100∠0°=100+j0 V I=8∠-60°=4-j6.93 A Problem 2 If the circuit of Problem 1 consist of a purely resistive and a purely reactive element, find R and X; a)

if the elements are in series

b)

if the elements are in parallel

Solution: (a) Elements in series: Z= 100∠0°8∠-60°=12.5∠60° Ω=6.25+j10.83 Ω R=6.25 Ω XL=10.83 Ω (b) Elements in parallel: Y= 1Z=0.08∠-60°=0.04-j0.0693 R= 10.04=25 Ω XL= 10.0693=14.43 Ω

Problem 3 Three identical impedances of 10 ∠-15° Ω are Y-connected to balanced threephase line voltages of 208 V. Specify all the line and phase voltages and the 1

c

Vca

a

currents as phasors in polar form with Vca as reference for a phase sequence of abc. Solution:

Vcn Van

Van=120 ∠210° V Vab=208 ∠240° V

Vbc

Vbn

Vbn=120 ∠90° V Vbc=208 ∠120° V

Vab

b

Vcn=120 ∠-30° V Vca=208 ∠0° V

Ia= VanZ= 120 ∠210°10 ∠-15°=12∠225° A Ib= VbnZ= 120 ∠90°10 ∠-15°=12∠105° A Ic= VcnZ= 120 ∠-30°10 ∠-15°=12∠-15° A

Problem 4 A balanced-∆ load consisting of pure resistances of 15 Ω per phase is in parallel with a balanced-Y load having phase impedances of 8 + j6 Ω. Identical impedances of 2 + j5 Ω are in each of the three lines connecting the combined loads to a 110-V three phase supply. Find the current drawn from the supply and line voltage at the combined loads. Solutions: Convert ∆ to equivalent Y having 15/3 = 5 Ω/phase 58+j65+8+j6= 40+j3013+j6×13-j613j6=700+j50205=3.41+j0.732=3.49∠12.1° Ω Current drawn at supply: Z=2+j5+3.41+j0.73=5.41+j5.73=7.88∠46.65° Ω I=11037.88=8.06 A from supply Letting Vt equal voltage at the load, line-to-line voltage: Vt=8.06×3.49=28.13 V to neutral Line-to-line V2=3×28.13=48.72 V Problem 5 A three-phase load draws 250 kW at a power factor of 0.707 lagging from a 440V line. In parallel with this load is a three-phase capacitor bank which draws 60 kVA. Find the total current and resultant power factor.

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Solution: Letting S1 and S2 represent the load and capacitor bank, respectively, S1=250+j250 S2=0-j60 where S1+S2=250+j190=314∠37.23° kW I=314,0003×440=412.0 A p.f=cos37.23°=0.796 lag

Problem 6 A three-phase motor draws 20 kVA at 0.707 power factor lagging from a 220-V source. Determine the kilovolt ampere rating of capacitors to make the combined power factor 0.90 lagging, and determine the line current before and after the capacitor are added. 14.14 Solution; From the figure, θ=cos-10.9=25.84°

20

6.85

14.14tan25.84°=6.85 14.14-6.85=7.29 kvar Without capacitor:

θ=cos10.9=25.84°

I=20,0003×220=52.5 A With capacitor: I=14.14+j6.85×10003×220=41.2 A

Problem 7 Three loads are connected in parallel across a 1400-Vrms, 60-Hz single-phase supply as shown in Figure 1. Load 1: Inductive load, 125kVa at 0.28 power factor Load 2: Capacitive load, 10kW and 40kvar Load 3: Resistive load of 15kW 3

a) Find the total kW, kvar, kVA, and the supply power factor.

Figure 1 An inductive load has a lagging power factor, the capacitive load has a leading power factor, and the resistive load has a unity power factor. For load 1: θ1=cos-1(0.28)=73.74° lagging The load complex power is S=P+jQ=S1+S2+S3 =35+j120+10-j40+15+j0 =60 kW+j80 kvar=100∠53.13 kVA

The total current is, I= S*V* =100,000∠-53.13°1400∠0°=71.43∠-53.13°A The supply power factor is, PF=cos53.13=0.6 lagging

(b) A capasitor of negligible resistance is connected in parallel with the above loads to improve the power factor to 0.8 lagging. Determine the kvar rating of this capacitor and the capacitance in µF.

Total real power P = 60 kW at the new power factor of 0.8 lagging results in the new reactive power Q’. θ'=cos-10.8=36.87° 4

Q'=60tan36.87°=45 kvar Therefore, the required capacitor kvar is, Qc=80-45=35 kvar and Xc=V2Sc*=14002j35,000=-j56 Ω C=1062π6056=47.37 μF and the new current is, I'=S'*V*=60,000-j45,0001400∠0°=53.57∠-36.87° A Note the reduction in the supply current from 71.43 A to 53.57 A.

Problem 8 A three-phase line has an impedance of 0.4+j2.7Ω per phase. The line feeds two balanced three-phase loads that are connected parallel. The first load is absorbing 560.1kVA at 0.707 power factor lagging. The second load absorbs 132kW at unity power factor. The line-to-line voltage at the load end of the line is 3810.5V. Determine: a) The magnitude of the line voltage at the source end of the line. b) Total real and reactive power loss in the line. c) Real power and reactive power supplied at the sending end of the line.

Figure 2 a) The phase voltage at the load terminals is

The single-phase equivalent circuit is shown in figure 2. The total complex power is

With the phase voltage V2 as reference, the current in the line is

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The phase voltage at the sending end is

The magnitude of the line voltage at the sending end of the line is

b) The three-phase power loss in the line is

c) The three-phase sending power is

It is clear that the sum of load powers and the line losses is equal to the power delivered from the supply, i.e,

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