ECE5500 Winter 2013

POWER SYSTEM ANALYSIS

Homework Assignment # 1 Due Date: January 15th

, 2013

1) A single phase load operates at 20 kW, 0.8 pf lagging. The load voltage is 220 0 V rms at 60Hz. The impedance of the line is 0.09+j0.3 . Determine the voltage and power factor at the input (source) to the line (, ).

Circuit for Problem 1)

Solution:

Apparent power S consumed by the load is:

= cos =20,0000.8 = 25,000

Therefore, at the load, complex power is:

= 25,000 = 25,000 cos = 25,00036.87 = 20,000 + #15,000

Since % = &' , the load current is:

' = )25,00036.87

2200 *= 113.6436.87

The complex power losses in the line are:

-./ = 012-./ = 3113.644130.09 + #0.34 = 1162.26 + #3874.21

The complex power at source (generator) is:

= + -./ = 21,162.26 + #18,874.21 = 28,356.2541.73

Hence, the generator voltage is:

= ||0 =28,356.25113.64 = 249.53789

The generator power factor is:

cos341.734 = 0.75:;

'? = &KL2MNLO + 2P

= 12001 + #0.5 + 4.49 + #2.09 = 19.7725.26 A rms

The phase voltage at the load is then Q = |AD| , where: &AD = '?2E

= 319.7725.26434.9524.954 = 97.860.31 V rms

Therefore, the line currents and load phase voltages are:

'?A = 19.7725.26789&AD = 97.860.31789 '@B = 19.77145.26789&BD = 97.86120.31789 'C = 19.77+94.74789&CD = 97.86+119.69789

In order to determine the phase currents in the delta-connected load, the original circuit should be

revisited. Although the wye-connected load, with which we replaced the delta-connected load, is

equivalent at the external terminals, we must use the original circuit to determine internal voltages or

currents. Since:

&AD = 97.860.31789 Then:

&AB = 3 97.860.31 + 30 = 169.529.69 V rms

Hence,

'AB = &AB2 =169.529.69

24 + #9 = 6.619.13 A rms

The remaining phase currents in the delta are:

'BC = &ST2 = 6.61110.87 A rms 'CA = &TU2 = 6.61+129.13 A rms

3) A balanced three-phase source serves three loads as follows:

Load 1: 24 kW at 0.6 pf lagging

Load 2: 10 kW at unity power factor

Load 3: 12 kVA at 0.8 pf leading

The line voltage at the loads is 208 V rms at 60 Hz. The line impedance is 2-./ = 0.05 +#0.02.Determine: a) The line current (0) b) The combined factor of the loads (-?I) c) The line voltage and power factor at the source ( , 4.

Solution:

The total load is:

= 24,000 + #32,000 1 = 10.000 + #0 F = 12,000 36.9 = 9,600 #7,200

Therefore: = + 1 + F = 43,600 + #24,800 = 50,16029.63

0 = ||3 =50,1603208 = 139.23

The combined power factor for all three loads is:

-?I = cos29.63 = 0.869:;

= |XYZ[\]|F^_ = 220.87 Vrms And the power factor at the source is:

= cos29.17` = 0.873:;