Question

A balanced delta connected load of 15+?18 Ω per phase is connected at the end of a three-phase line as shown in Figure 1. The line impedance is 1+?2 Ω per phase. The line is supplied from a three-phase source with a line-to-line voltage of 207.85 V rms. Taking the phase voltage ??? as reference, determine:

1+ 32 12 ao |VL = 207.85 V b ba 15 + j18 12 Co- Figure 1 Circuit for problem 1

1. Total complex powers of the line and the load.

2.A delta connected capacitor bank is connected in parallel with the load to improve the load power factor to unity. Find the kvar and capacitance of the capacitor.(F = 60hz)

Answer 1

Soln Guien that 1 + 32 uz AML IVL1--207.85V 10602 wwwm 15+91812 mm 4 we are 4 to analysis the system analyse the Juien Syatem on Per-Phase basis . on her-Phase basis we need to Convert the A- Connected lood to Ogunalent Isten - Connected Ovad. 1) Determine total Comhlea Porces of the line and the load. 4 Converting A to y Zy = ZO .. ( for balance A-lood) i. Zy=15+318 = 5+86 . 3 Circuit om Per-Phase. basis IL 1tj2 um 1 Van- 207.85 13 15+j6 Ĉ

I = Van 1+j2+5+j6 >IL 120 6+j8 = 12 /-53-13 total Complex Porner of the line 4 To find total Complex porner of the line we need to find the Voltage acriss line umfadame. lime (ph) = Iph Zlimelen) Vine (pa) = 12 (-53.13 (1+j2) Vine (en) = 26-833/10-3° v Now Total Complex Power of line ( Sline) =3 Viime on I" + Slime = 3x 26-833/10-3° x 12 [53-13 3 Siime = 966163.4° VA | Slime ) = 966 VA

Total Complex Pruer of load Ven Cload) = IL Z Load = 12 -53-13 ~ (5+j6) 9 Ven (load) = 93.723 /-3° V Sload (38) = 3 Vpn It → Sload 13-0) = 3 x93.723 +3° x 12 (153-130 Sload = 3374 50-13 ° VA Sload = 3374 VA 2 ) we have, = 3374 150:13° =(2162.89+32 589.5) "va Heno PL (real Pruer) = 2162.89 W Sload (38) & P. f = ls 50.13 = 0.64 lagging Li Nowo Uill need to Cinneet a A - Connected Capacitor bank in parallel to load which can improve the hermes fator to cinity •
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Let the recetimo Pous Sydhly by the Caparilan bank is Qc - bulib Perguried ? : Qc + PL ( tan 50:13 - tam oo ) ľ Qc = 2162:89 (tam so'130) 7Qc = 2.58 KVAR Value of Capacitos... Qe on = Q3.6 2.59 - - 0.8633 KVAR 3 3 & Q con z Vph Xc lobo ven » Xe(Y-ph) = 2 Qcph 2 » 3 Xc (A) = Ven l'. Zy = Z2 / Qc ph

R π ·3 WC Vph Qcph а с 3xQc ph Vph x 20f 키 CE 2:59 (93.723)*x 201x60 C = 782.12 MF