Delta Source with Wye Load problem

Thread Starter

BenderIsGreat34

Joined Feb 21, 2015
3
Hello, I am looking for some help with this Delta-Wye source/load problem.
I attached a picture for better visualization.
Here's my thought process:
1) Convert the delta source to a single phase equivalent Wye circuit.
This means V_y would be 1/√3 * V_delta.
So V_y = 714/√(3)
It also means the source impedance in my delta source would be 1/3 the impedance in my equivalent Wye source.
So Z_src = (1/3)(4.5+j3) = 1.5 + j1

Now I construct my single phase equivalent Wye circuit, with:
V_y=714/√3
Z_src=1.5 + j1
Z_line=6.5 + j15
Z_load=1192 + j1584

So, for part 1, it asks for the line to line voltage at the terminals of the load.
First, I find the magnitude of the phase current going through my equivalent circuit.
I_a = V_y / Z_total = (714/√3) / (1200 + j1600) = .206A

Now, I find the magnitude of the line to neutral voltage, V_an, across my Z_load.
V_an = I_a * Z_load = .206(1192+j1584) = 408V

I know that for a Wye-load, the line to neutral voltage and line to line voltage of a single phase is related by this:
V_ab = √3 * V_an
Where V_ab is my line-line voltage.

So, my final answer for the line-to-line voltage, V_ab, should be:
V_ab = √3 * V_an = √3 * 408 = 708V

I tried this answer, and also several other answers that I got with different thinking but none of them were correct. Any help would be much appreciated.
 

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WBahn

Joined Mar 31, 2012
29,979
I only took a quick look at your approach and didn't spot anything wrong, but it appears that you might be suffering from cumulative round off error if you are taking intermediate results and using those in future work. If you are going to do that, always keep a couple of additional sig figs on the intermediate results.
 

Thread Starter

BenderIsGreat34

Joined Feb 21, 2015
3
I only took a quick look at your approach and didn't spot anything wrong, but it appears that you might be suffering from cumulative round off error if you are taking intermediate results and using those in future work. If you are going to do that, always keep a couple of additional sig figs on the intermediate results.
Thanks for the reply. However, this online hw program is usually pretty lenient when considering sig figs. I just wanted to make sure that my process of converting delta->wye sources was correct among other things.
 

Thread Starter

BenderIsGreat34

Joined Feb 21, 2015
3
Your line current looks Ok but I have Vab as 713.39 volts with minuscule phase shift.
Hi all, thanks for the replies. My teacher just contacted us and told us the answer the program was looking for was wrong.
So I think my thought process is correct.
Thanks.
 

WBahn

Joined Mar 31, 2012
29,979
Your line current looks Ok but I have Vab as 713.39 volts with minuscule phase shift.
I don't have the answer I got written down, but it pretty much agreed with the TS at something like 707.7V. I didn't calculate the phase shift, but it would have been very small.
 

t_n_k

Joined Mar 6, 2009
5,455
Ah yes I misread the question as requiring the supply side value of Vab rather than the load side. A value of 707.7 volts makes sense.
 
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