Question

3) Consider the below 3 phase network. There are distributed loads at only ACB and ADB branches. The ACB branch has )=50W/m and ADB branch has J-30 W/m. Allowable maximum voltage reduction is 6%. Rated voltage of the system is 380V. Please find the cable cross-section for both ACB and ADB branches. (Cables are copper and underground. = 56 m/2mm) Please note that there is not any voltage reduction at TA and TB lines which means you will not consider these lines 200m 600m C 400m 15kW 15kw B 500m D 300m 40kW

Answer 1

ANSWER :
consider the network. 200m 6oom 400m B 15kW 15kW 500m 300m 40kw The value of I for ACB branch is 50 W/m. The length of the ACB branch is 200+600 + 400 = 1200m power of the ACB branch is PACB 50x1200 = 60,000 W dissipated in The voltage across the branch is soov write the formula for the power the resistore P. R (380)2 60,000 - RACB RACE = 2.412 write the formula for the & siesistance

a l R- ll A SA G=56 m Lamm - 56 x 10°172-m. 1200 2-41= 568106 AACB A ACB 8.89x106m2 2 8.89 mm (AACB The The value of j for ADB branch is 30 wlm The length of the ABB branch pg 500+800 = 800m power of the ADB branch is. 30X800 = 24,000 W across the branch Pg 380v write the formula for the power dissipated in the resistor PACB The voltage P= 2 R (380)? 24,000 = RADBI 6.022 RADB write the formula for the resistance

al R= pl A SA 3= 56 m 12 mm-56X 1061/2-m 800 6.02= 56X106 APPS AADB 2:37x106m2 AADB = 2.37 mm²