of shagah Name 120 96] Proble m 1: A balanced Y. connected, 50-Hz abc source is

Question

of shagah Name 120 96] Proble m 1: A balanced Y. connected, 50-Hz abc source is connected to the following loads via 0.5+0.25 j 0/phase line. A Y-connected 900 kVA at 0.7 pf lagging load. · A ?-connected 500 kw at 0.6 pf lagging load. A Y-connected 350 kVAR at 0.7 pf lagging load. · 4-connected 600 kW at 0.6 pf lagging load. A Part 1: If the load voltage is always kept at 11 kV a) what is the total complex power consumed by the total load? Stot b) Find the three phase line current supplying the total load.I c) Find the three phase supply phase voltages. Van,Vbn, Vcn (4%) SOLUTION a) What is the total complex power consumed by the total load? S

Explanation / Answer

load1:

900*10^3=1.732*11*10^3*I1

I1=[900*10^3]/[1.732*11*10^3]=47.23<-inverse cosine of 0.7=47.23<-45.57 amps.

load 2:

500*10^3=1.732*11*10^3*I2*0.6

I2=[500*10^3]/[1.732*11*10^3*0.6]=43.73<-inverse cosine of 0.6=43.73<-53.13 amps.

load 3:

350*10^3= 1.732*11*10^3*I3*SINE(INVERSE COSINE OF 0.7)

I3=[350*10^3]/[1.732*11*10^3*0.714]=25.72<-INVERSE COSINE OF 0.7=25.72<-45.57 amps.

load 4:

600*10^3=1.732*11*10^3*I4*0.6

I4=[600*10^3]/[1.732*11*10^3*0.6]=52.48<-INVERSE COSINE OF 0.6=52.48<-53.13 amps.

b) total load current=I1+I2+I3+I4=168.79<-49.87 amps.

a) s=3*[11000/1.732]*168.79<-49.87=3215881.414<-49.87 VA

c)per phase load voltage=11000/1.732=6350.85 volts.

Van=6350.85<0+(0.5+j0.25)*168.79<-49.87=6437.616<-0.3322 volts

Vbn=6437.616<(-0.3322-120)=6437.616<-120.3322 volts.

Vcn=6437.616<(-0.3322+120)=6437.616<119.6678 volts.

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