Hi guys
I'm struggling with this question below:

A three phase, three wire, 240 V, ABC system has a Δ‐connected load with ZAB=10<0 Ω, ZBC=10<30 Ω and ZCA=15<-30 Ω.
a) Find line currents of the system IA, IB and IC
b) Plot the phasor diagram of the voltage and current
c) Find total power of the load
Answers:
IA=28.84<-33.69A
IB=46.36<195A
IC=34.87<53.42A
PL=14.07 kW
______________________________________________________

I've been doing all the balanced systems without much of a problem. Though tbh I don't usually calculate on delta connections, preferring to divide by 3 and have a nice simple Y connection to calculate instead.

I've been trying to back calculate using the answers and find some type of relationship, using omhs law. But there doesn't seem to be one.

Can anyone give me a push in the right direction here. Just a first step would be good.

 

It's just a circuit that happens to have three sources in it. Analyze the circuit just like you would if you had never heard the term "three phase".

As usual... SHOW YOUR WORK!

 

Kk so I've done this badboy ^... The answers though aren't the same as the ones given...

Can you settle something please Wbahn? According to most of the vids and lecturers ive seen a positive abc sequence goes
if
Vab = angle 0
then
Vbc = 120
or is it -120...

I have it as -120 in this pic...
I thought inverting it would give the same answer as my lecturer
but its actually something completely different as it results in an angle of 150 degrees...
I think perhaps there's been a error somewhere in the notes, easy enough mistake to make I suppose...

Whats your opinion?

 

All I see are a bunch of disembodied equations. It would sure help if you provided a properly annotated schematic.

A positive phase sequence is one in which the voltage peaks, relative to ground, occur in the order 'a' then 'b' then 'c'.

This means that

Van = Vo ∠ 0°
Vbn = Vo ∠ -120°
Vcn = Vo ∠ +120°

That should be enough to let you figure out Vab, Vbc, and Vca.

 

Lol I've missed you...
Well if you ask me i've chosen to go with Vbc as 120. But I've heard about a million different answers now. Only the Wbahn knows

 

 

Lol I've missed you...
Well if you ask me i've chosen to go with Vbc as 120. But I've heard about a million different answers now. Only the Wbahn knows

Just giving an angle for one doesn't tell you much since it is the phase relationships between them that matters.

I've given you the phase relationship for the phase voltages in a positive sequence system. Now you can translate them into the relationships for the line-to-line voltages by using the definition of what the line to line voltages are.

Vab = Va - Vb = Va - Vb + (Vn - Vn) = (Va - Vn) - (Vb - Vn) = Van - Vbn

Vab = (Vo ∠ 0°) - (Vo ∠ -120°)

Take it from here.

 

0-(-120) = 120

 

In the original problem, it doesn't say whether the 240 V values are the line-line voltages or the phase voltages. Since it states that it is a three-wire system, I suspect that they are line-line voltages.

 

0-(-120) = 120

Do you just subtract phase angles when you subtract one complex value from another?

 

Well I did think so... k if I convert to rectangular mode and subtract I get 240<60

 

Well I did think so... k if I convert to rectangular mode and subtract I get 240<60

Show your work!

I can't tell where you are going wrong if you don't show your work.

What is Van in rectangular?

What is Vbn in rectangular?

What is Vab in rectangular?

What is Van in polar?

 

 

Much better, although you are dropping your units again.

Notice that 415.69 V is √3·240 V

Since you are given that the line-line voltage magnitude is 240 V (I believe), then you have 240 V at an angle of 30°.

So that gives you Vab.

Now what are Vbc and Vca?

Once you get all three, then you can adjust them so that Vab is at 0° (if that is the reference you are supposed to use) by simply subtracting 30° from all three.

Since you are given the correct answers (supposedly), you can use them to figure out what the line-line voltages are that the author used.

Plus, without finding the phase currents you should be able to determine the real power delivered to the load fairly readily and confirm that it is, indeed, the claimed 14.07 kW. Assuming 240 Vrms line-line, I got 14.76 kW. So something is amiss. Either the answer is wrong or I am making a wrong assumption about the information given.

 

Interesting, and pretty cool, so what are you doing?
Sqrt3 (Vm)(Im) cos (phi)
for all three phases, then adding them together?
Thanks for the help as always Wbahn.

 

Assuming that the 240 V is the line-line voltage, each of the three legs of the delta load has a fixed 240 V voltage source directly across it. So each leg has absolutely no impact on either of the other two legs, each is simply an impedance driven by a voltage source, which means you can find the power for each leg independently. The apparent power is just V²/|Z| and the real power is then just this multiplied by the cosine of that leg's impedance angle.

 

K sorry I was just looking over a few things. Most people start with Van = then they derive this Vab = 30 degrees
-90
-210
from the relationships.

However this is a three line system. Does that n stand for neutral ?
And if a three line is without a neutral, is that simply a -120 relationship?

 

K sorry I was just looking over a few things. Most people start with Van = then they derive this Vab = 30 degrees
-90
-210
from the relationships.

Most people start with Van =.... what?

Does it really matter what most people get? More to the point, shouldn't YOU be able to get those same results? You've already gotten the 30° for Vab. What do YOU get for Vbc and Vac?

However this is a three line system. Does that n stand for neutral ?

Yes. The neutral is generally Earth ground that everything is referred to.

And if a three line is without a neutral, is that simply a -120 relationship?

I have no idea what "that" refers to?

 

K sorry just a thought. I'm wondering where my lecturer went wrong, he's not usually one for mistakes... but anyways.

Ok so the closest calculation I can find, for what I expect to get for Vbc, is:
(240 ∠ -120°) - (240 ∠ 120°)
=(415.69 ∠ 90°).

 

Assuming 240 Vrms line-line, I got 14.76 kW. So something is amiss. Either the answer is wrong or I am making a wrong assumption about the information given.

Lul I actually used your method and got his answer
Apparent power (q) Vab = 5760Var
Power = 5760W

Vbc
q= 5760
Power = 4988.31

Vca
q = 3840
p = 3325.54

ptot = 14073.85w