Question

AC Circuit Analysis

3. Three balanced three-phase loads are connected in parallel. Load 1 is Y- connected with an impedance of 400 + j300N/0 load 2 is A-connected with an impedance of 2400 – 18001/¢, and load 3 is 270 +39000kVA. The loads are fed from a distribution line with an impedance of 1 + j4N/0. The magnitude of the line-to-neutral voltage at the load end of the line is 1673 kv. a) Calculate the total complex power at the sending end of the line (20 pts) b) What percentage of the average power at the sending end of the line is delivered to the loads? (10 pts).

Answer 1

ANS:

First calculate active a reactive power of all the loads.

(1) Y-connected load

Load impedance per phase

Zy 400+ 7300

  

300 = 4002 + 3002 Ztan 400

  

500236,862

Line to neutral voltage at the load end  

V = 16V3kV

Current in the phase of Y-connected load,

P Іү Zy

  

16V3 x 103 500

55.42A

Active power in the load per phase

= 1} Ry

  

= 55.422 x 400

= 1228.55k

Total active power of load   

Py = 3 x 1228.55 = 3685.65kW

Reactive power in the load per phase

= 1} Xy

  

= 55.424 х 300

921.41KV Ar

Total reactive power in the load

Qy = 3 x 921.41 = 2764.33kV Ar

Complex power of the Y connected load ,

Sy Py +jQy

Sy = 3685.65 + 12764.33kVA

(2)

Delta connected load

Delta connected load per phase

- 2400 - 1800Ω

  

= 24002 + 180022 – tan 1800 2400

  

სი98'9$ -7ბბბ¢ =

Line to neutral voltage at the load end  

V = 16V3kV

Load is balanced, therefore line voltage

VL = 3V P

= V3 x 16V3

= 48kV

Magnitude of phase current in delta connected load

IA VL ZA

  

48 x 103 3000

= 16A

Active power in the load per phase

1² RA

  

= 162 х 2400

= 61ᏎᏎᎬᏉ

Total active power of load   

PA = 3 X 614.4= 1843.2kW

Reactive power in the load per phase

= 1 X4

  

= 162 x (-1800

-460.8KV Ar

Total reactive power in the load

QA = 3 X (-460.8) = -1382.4kV Ar

Complex power of the delta connected load ,

SA PA+QA

SA = 1843.2 - 31382.4kV A

(3)

Load 3

S3 = 270 +79000KV A

Pa = 270kW
  
Q3 9000KV Ar

Total active power of all load

Pload Py + PA+ P3

- 3685.65 + 1843.24 270

  

5798.85kW

Total reactive power of all loads

Qload Qy +QA+Q3

2764.33-1382.4 + 9000

= 10381.93KV Ar

Complex powe of load

Sload Pload + iQload

= 5798.85 +j10381,Ꮽ3ᏂᏙᎪ

​​​​​​​

Reactive Power (Q) Apparant Power (S) Active Power (P)

Powefactor angle of combined load ,

o = tan -1 Qload Pload

o = tan 10381.93 5798.85

  

Recall the equation of power for three phase balanced load

$P=\sqrt3V_LI_Lcos(\phi)$

(o)soɔTAE IL = P

Put $P=P_{load}=5798.85kW$   $V_L=48kV$  $\phi=60.81^0$

$I_L=\frac{P}{\sqrt3V_Lcos(\phi)}$

$=\frac{5798.85\times10^3}{\sqrt3\times48\times10^3cos(60.81^0)}$

= 143.01A

Now find complex power of line .

Line impedance per phase$Z_L=1+j4\Omega$

$S_{line}=3I_L^2Z_L$​​​​​​​

  $=3\times 143.01^2 (1+j4)$

  $=61.355+j245.422kVA$

(a)

Total complex power at the source,

Ss = SOUT ce Sline + Sload

  $=61.355+j245.422+5798.85+j10381.93$

  

5860.205+310627352EVA

(b)

Average power refers to active power.

Active power at source

Ps = 5860.205kW

Active power at load $P_{load}=5798.85kW$

%Average power delivered to load =  $\frac{P_{load}}{P_S}\times100$

$=\frac{5798.85}{5860.205}\times 100$

=98.95%