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jaydub

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Hi guys, hopefully, maybe, someone will be able to help me with this one.

Trying to get my head around true power calculations and am really struggling to understand it.

So the scenario goes;

Three coils of resistance 40 Ω and inductive reactance 30 Ω are connected in delta to a

400 V 50 Hz three phase supply. The total power will be:

Now, I know people get annoyed with people like me asking people to do their homework for them but I genuinely do not understand how to calculate this.

I am doing the C&G L3 course and aiming to take my 302 unit exam as soon as possible.

I have the learning material but it's just lots of formulas and there are no examples for this particular section. Which normally help me to understand how it all works.

So even if someone could create an example of a calculation for this that would really help.

I understand about resistance and inductive reactance but I really struggle with anything 3 phase.

Sorry for the waffle.

Thanks
 
50 years ago i'd have no problem with calculating that, but never having had to use the theory since, it's long forgotten. :(
 
I can't even remember how many years ago I forgot about it, so that's me out as well..

Surely there's an App for this nowadays..
 
Not 100% sure on this....

I'd assume the load is purely inductive as the question doesn't mention power factor (i.e. cos θ). So, you should be able to get a value for impedance (Z²=R²+X²).

Per-phase power P=Vphase²/Z.

Otherwise you could think in terms of phase current and voltage for delta (i.e. Vphase = Iphase x Z and then calculate power per phase).

The total power in delta is three times the per-phase power.
 
Hi guys, hopefully, maybe, someone will be able to help me with this one.

Trying to get my head around true power calculations and am really struggling to understand it.

So the scenario goes;

Three coils of resistance 40 Ω and inductive reactance 30 Ω are connected in delta to a

400 V 50 Hz three phase supply. The total power will be:

Now, I know people get annoyed with people like me asking people to do their homework for them but I genuinely do not understand how to calculate this.

I am doing the C&G L3 course and aiming to take my 302 unit exam as soon as possible.

I have the learning material but it's just lots of formulas and there are no examples for this particular section. Which normally help me to understand how it all works.

So even if someone could create an example of a calculation for this that would really help.

I understand about resistance and inductive reactance but I really struggle with anything 3 phase.

Sorry for the waffle.

Thanks
 
Kl

I'm currently just finishing my C&G Level 3 2365,
To work this out the formula would be P= V x I x pf x 1.73 (square route of 3)

So first I would work out your impedanance Z = square route of R2 + XL2 which is 50 ohms

Then get the power factor which is R\Z. 40/50 = 0.8

Next you need the current which will be Phase voltage over resistance which is 400/40 = 10 A

The line voltage = 10A x square route of 3 which is 17.32 A

Now true power = 400 x 17.32 x 0.8 x square route of 3 (1.73) = 9.59 KW
 
Thanks for your help on this everyone! JaySpark have you taken the 302 exam yet? If so, how was it? Any tips to prepare?

Also I have a list of four possible answers for my question that I started this thread with and both your answer and M-Ty's answer are possible options! Although Jay Sparks working looks more along the right lines.

What is a 3-4-5 triangle M-Ty?

Cheers!
 
A 3,4,5 triangle is a right-angled one, like its mate the 5,12,13 triangle, where the hypotenuse and the sides are whole numbers that are worth remembering (for over 40 years!) because they seem to crop up in questions more often than you'd expect by chance and save having to calculate the square root of the sum of the squares when you spot them.
 
So using the information that I've gathered from everyone this is my take on the equation, but it seems to differ from the other two haha

P = V x I x pf x 1.73
V = 400V
I = V/Z so: 8A
pf = R/Z so: 0.8
Z = square root of Rsquared + Xlsquared So: 50ohms

So: P = 400 x 8 x 0.8 x 1.73 = 4428.8W

So how come three of us have different answers? And is this the power of one phase? Do i need to multiply this answer by 3 in order to calculate the ENTIRE power of all three phases together?
 
I took the 400V to be the line voltage, whereas you and Jay Spark have taken it to be the phase voltage.
(3 answers now: reminds me of Call My Bluff!)
 
Haha, it's such a nightmare, I've been waiting for a call back from my tutor about this particular topic for a couple of days now so hopefully as soon as he calls me it will all be cleared up! So many small variations that can be made throughout the equation, I guess because of the way that the question can be interpreted...
 
Instead of P=I^^R, I should have written something more like I^2.R. I meant I squared times R.
You'll have noticed that this is the same as I times p.f., but I find it easier to calculate and not bother with p.f..
Anyway, a couple of valid approaches and some good questions to ask your tutor.
(Including "What did he mean by 400V?")
Best fun is when the question is ambiguous AND all the given answers are wrong, but you can work out which one has just a 1-digit printing error and the other 4 that are completely wrong whichever interpretation you give to the question!
I vaguely remember one class where we complained to the lecturer that he'd included a value for something that was irrelevant to the calculation and he said "Yes; real life's like that!" He was the only one to ever do that, and he was right...)
All the best,
Martin.
 
Last edited:
So using the information that I've gathered from everyone this is my take on the equation, but it seems to differ from the other two haha

P = V x I x pf x 1.73
V = 400V
I = V/Z so: 8A
pf = R/Z so: 0.8
Z = square root of Rsquared + Xlsquared So: 50ohms

So: P = 400 x 8 x 0.8 x 1.73 = 4428.8W

So how come three of us have different answers? And is this the power of one phase? Do i need to multiply this answer by 3 in order to calculate the ENTIRE power of all three phases together?


I'll mate yah just passed my 302 exam last month, it is quite hard to be fair, make sure you revise on motors, lighting and this kind of thing with power in A.C Circuits.

But yh Your answer is similar to mine only thing I've done different is is I divided 400 by the resistance rather than the impedance, I done this because to get the line current we must first get the phase current and 4o is the resistance over each phase on the delta supply. This gave me 10
A of phase current, then to get the line current I done 10 x 1.73 (square root of 3). Which gave me 17.32 A
Then from there I added this to the rest of the formula to get 9590 W
 

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