What Is the Sending-End Voltage in a Star-Connected Load?

AI Thread Summary
The discussion centers on calculating the sending-end voltage for a star-connected load of 3300 kVA at 18 kV with a power factor of 0.8 lagging. Participants express confusion about the concept of "sending-end voltage" and whether to use line or phase voltage for calculations. The impedance of the transmission line is noted, with a resistance of 4Ω and inductive reactance of 7Ω per phase, leading to questions about how to determine the voltage drop. There is a debate on whether to use rated power or true power for efficiency calculations. Overall, the thread highlights the complexities of power calculations in three-phase systems and the need for clarity on voltage definitions.
Michael V
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1. Statement

A star connected load of 3300 kVA at 18 kV and a power factor of 0,8 lagging is supplied by a three-phase transmission line having a resistance of 4Ω per phase and an inductive reactance of 7Ω per phase.

Determine:
a. the sending-end voltage
b. the percentage regulation
c. the efficiency of the line

2. The attempt at a solution

I've attached my attempt at this.
i'm confused on what ''sending-end voltage'' is. also do i need to workout the impedance for this.
 

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Michael V said:
i'm confused on what ''sending-end voltage'' is. also do i need to workout the impedance for this.

There is a voltage drop on the transmission line due to its impedance. That means the 'sending voltage' (the voltage source attached at the other end of the transmission line) is different than the voltage at the load.
 
So then, because it's star-connected, do I use the line voltage or the phase voltage to start with? Using VL = \sqrt{3} × Vph, but I can only do that when IL = Iph?
 
For (a), first i took the resistance and reactance to get impedance: Zphase = 4 + j7 = 8,062 |60,26° Ω

then i use the rating: S = I × V (not sure if i should use VLine or Vphase)

Iphase = \frac{3300 kVA}{18 kV / \sqrt{3}} = 317,543 A or ILine = \frac{3300 kVA}{18 kV} = 183,333 A

then to find voltage drop: E = I × Z (which current do i use?)
 
Last edited:
Heres my full attempt attached, just struggling on c. whether to use rated power or the true power.
 

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Michael V said:
then to find voltage drop: E = I × Z (which current do i use?)

I'd use phase voltages and phase currents to find the voltage drop in a single supply line, then use that to find the phase voltage at the supply end.

It has been a very long time since I have done anything with power so it would be better if someone else came along to check your work. I'd need to do a review first.
 

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