Homework #3
Randall Robert
10.19.4
9771
Problem 1.
Distribution voltage RMS value measured at unloaded bus is E = 2.4kV, its short circuit
power ia 1MVA and the reactance to resistance ratio ξ = 2.
Calculate the load
voltage
RMS value V, power loss in the supply and draw diagrams of the CRMS values when the
bus supplies a load that at the voltage RMS value V = 2.4kV has the active power P =
50kW and the reactive powers:
Parameters:
Source
E = 2.4 kV
S
sc
= 1MVA
ξ
s
= 2
Load
V
RMS
= 2.4kV
P = 50kW
Reactive Powers
1.
Q = 50kVAr
2.
Q = 0
3.
Q = 50kVAr
1. Q = 50kVAr
S = P+jQ
S = VI*,
I* = S/V,
I* = (50k+j50kVAr)/2.4kV
I* = 70.7k∟45°A
I* = 29.5∟45°A
V/I = Z
L
= 2.4kV/(29.5∟45°A)
Z
L
= 81.4∟45°Ω = 57.6 + j57.6
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View Full DocumentZ
s
= E²/Ssc = (2.4kV²)/(1MVA)
Z
s
= 5.76Ω
R
s
= Z
s
/(√1+ξ²) = 5.76Ω/(√5)
R
s
= 2.58Ω
X
s
= ξR
x
= 5.15Ω
Therefore Z
x
= R
s
+ jX
s
= 5.76∟63.4°Ω
I = E/(ZL + ZS) = 2.4kV/(63.4 + j60.18) = 2.4kV/(87.4∟43.5°)
I = 27.5(∟43.5°)A
V = E – Izs = 24kV – (27.5(∟43.5°)A)( 5.76(∟63.4°)Ω)
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 Fall '08
 Staff
 Volt, 2.5k, 2.25K, voltage RMS value, 1.37°, 2.35k

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