# 3-phase, 4-wire, 208V/120V distribution system

Discussion in 'Homework Help' started by Radiotechman, Sep 22, 2011.

May 3, 2011
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Hi Everyone,

As a first time poster, I'd first like to say that this website has been invaluable throughout my university studies.

Now onto my question.

For the system in the included picture you are to calculate the currents in the secondary windings.

Motor M1: 50kVA, cosθ=0.5 lagging
Motor M2: 160kVA, cosθ=0.8 lagging

Now the answer given is 777A and the reason i'm skeptic that it's right, is i know it came from a companion textbook we are using that has had a plethora of mistakes in it.

So, here is my attempt at this solution, hopefully someone can let me know if i'm on the right track or not.

To me the easiest way is to investigate a single phase and since the three phase loads are balanced this should be fine to do so. Starting with the single phase load

I = P/E = 30e3/120 = 250A, so each single phase load will have a current of 250A

Now my understanding regarding the three phase loads is that the actual power i.e with power factor must be determined and the divided by three, the number of phases. So the values become.

M1=(50e3*0.5)/3 = 25e3/3 = 8.33e3
M2=(160e3*0.8)/3 = 128e3/3 = 42.67e3

then using the equation S =√3 E I, I can be found. But since the Δ connected I is the line current divided by √3, the found ( I ) will need to be multiplied by √3, or as i have done, cancle √ 3 from the equation.

So I = S/E, where E is the line voltage. Applying this results in.

M1: I = 8.33e3/120 = 69.42 A
M2: i = 42.67e3/120 = 355.58 A

The total line current per phase is then the sum of all of these, which is

I = 250+69.42+355.58 = 675 A.

This answer is off by 102A and I personally know three phase is a real weak point for me. I'm a radio tech by trade and I simply don't have much to do with it. Unfortunately it's not even worth contacting my lecturer( i'm a correspondance student) as he is always un-available(still waiting for a reply on a question i asked 1 month ago.

I have tried asking local electricians about it, but they are practical sorts of people and have no idea about the math ( which is sort of scary when you think about it).

Hopefully someone out there can help me out with this, as i really need to know what's what before i attend my final exam next friday.

Dylan Whittaker

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
You've only calculated the in-phase [real power] current values. You've forgotten to add the reactive power current values.

I have

I_line=675-j386.95 Amps

which is a magnitude of 778 Amps with 0.868 pf lagging.

May 3, 2011
5
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t_n_k

Please don't feel i'm being ignorant, but I'm not sure I understand what you mean. During the course we haven't really covered motors and have never had any situations like this to assess.

I have tried re-doing the calculations using the 120 phase angle that would exist between the phases, but this has not provided the solution. However i'm fairly certain that this is not what you mean, as it is my understanding that within the phase there shouldn't be a phase shift when moving from the Y source to the delta load.

I must say i haven't heard of reactive power current before and have yet to find any reliable reference to it in my textbooks or on the web.
Is the reactive power current a value that is calculated seperately and then added to the real power, or have i made an error in my calculations/understanding.

Some direction regarding this last part will help me to focus my effort and hopefully prevent me from going up the wrong path. Based on what you posted, i at least feel that i'm somewhere in the ball-park.

Thankyou for your help so far.

Dylan Whittaker

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
If you've not heard of reactive power then this will be a problem beyond the scope of your learning thus far.

Are you sure you've not run across the concept of Real, Reactive and Apparent Powers to date? The power triangle perhaps?

May 3, 2011
5
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t_n_k

Yes i have heard of Real, reactive, and apparent power, unless a load is purely resistive and is in a unity power factor circuit( or a compensated circuit) it will have a reactive component. In normal circuits the apparent power is determined by S = √P^2 +Q^2. I'm aware of many different formula's for calculating the apparent three phase power, however they are all generally in the form of S = √3 EI and other variations.

Perhaps i was a bit unclear, i haven'tn heard of reactive power "current" and had been under the impression the current as a whole was infulenced by the total circuit conditions.

What i am not totally clear about is whether the reactive component is calculated based on the total apparent power i.e. 50kVA and 160kVA, or do i need to reduce this to a per phase apparent power, or should it include the power factors as well.

It is obvious that i still have alot to learn regarding 3 phase, however we have only covered two weeks of three phase over the whole term and do not have the benefit of a contactable lecturer and for the better part have had to rely on teaching ourselves.

Based on your response, i feel, i should direct my searches to 3 phase reactive power to obtain some information that may be useful.

Many thanks

Dylan Whittaker

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
OK - then you will probably be making sense of all this pretty soon.

I'm could post a solution if you are not sure of how to proceed. Let me know and I'd be happy to oblige - if someone else doesn't do it for you anyway.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Here is the soln ....

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May 3, 2011
5
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t_n_k

I see now, the magnitude of the 3-phase loads is solved using the basic formula S = √3 E I, where E is the line voltage and for a Y circuit is the phase to neutral voltage x √3, or the line to line voltage.

The phase angle is then determined by the power factor -sin^-1(0.5 and 0.8), which combines with the magnitude to provide the reactive component

It is quite apparent now that the way i was approaching it was not a very efficient way of calculating the line current.

Whilst this has solved my immediate problem, i will no doubt benefit from further readings in this area.

Many thanks t_n_k, you have been immensely helpful to me.

Kind Regards Dylan Whittaker

May 3, 2011
5
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opps,

Must have been really tired when I posted this

"The phase angle is then determined by the power factor -sin^-1(0.5 and 0.8), which combines with the magnitude to provide the reactive component".

It should in fact be -Cos^-1(p.f), incase someone had been following and didn't know what i was on about.

Glad i wasn't operating heavy machinery yesterday!!

Anyway, thanks for all the help on this.

Cheers dylan