Calculation on a 3 Phases Unbalanced Delta

What you could do, Mildred, is assume 3300W between all three phases, work it out, and the switch one of the loads off. Show your working. Show you're working.

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Hi Guys,

Thanks for answers. I believe there is a misunderstanding (may be because I used the word school): I am not a student trying to cheat. I am a 59 years old electrical supervisor working on a construction site and we have a switchboard which is not yet energized. This switchboard is equipped with a 3x20A breaker (C curve) and as mentioned above we have 3300W of space heaters (then pf=1) load between L1-L3 and the same between L2 and L3. Nothing between L1-L2. Note that a student would have used A,B,C for the phases. L1,2,3 is industrial. So this is an open delta. I have search on the internet but cannot find a clear answer. I find examples for unbalanced wye (which all focus on current in neutral) but nothing for unbalanced delta. I don't remember how we did it in school (so long time ago). I'm quite sure that doing the vector sum of I2 and I1 is the way but what is the formula? I can assume that the I1 and the I2 line current will be P/U=I so 3300/220=15A. but how to calculate I3? Of course when the board will be energise it will be easy to measure it, but this will happen in several months. We want to confirm if the breaker is correctly sized. I would say yes but demonstrate it by calculation is better.

Thanks to help if you know the answer or can point to web site.

By the way English isn't my native language and I didn't catch the joke, if there is any, with Milred…

Thanks

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Jmp0

Hi,

or is it just Ix1.732 = 25.98A

Thanks

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Jmp0

Under British Standard 7671, breaker sizing depends upon cable size, length and method of installation. It has nothing to do with the load!

If the voltages are symmetric [V12=V23=V31 and the angle will be 0,-120,+120] then[ in complex]

I12=0+0j; I23=-7.5-13j; I31=-7.5+13j and then:

I1=I12-I31; I1=7.5+13J and in module I1=SQRT(7.5^2+13^2)=15 A

I2=I23-I12; I2=-7.5-13j;I2=SQRT((-7.5)^2+(-13)^2)=15 A

I3=I31-I23; I3=-7.5+13J+7.5+13J=26j; I3=26 A.

Sorry, I forgot to mention this [typical]:

I31[complex]=I31*cos(angle)+I31*sin(angle)j

I31=P31/V31=3300/220=15A

angle=120

I31[complex]=15*cos(120)+15*sin(120)j=15*(-0.5)+15*0.866j=-7.5+12.99j

For I23[complex] I23=3300/220=15 angle=-120.

Hello,

Thanks everybody, so all agree the answer is 26A. I found in several place on the web that line current is phase current xSQRT(3). So that is 15x1.732=26A. That's the easyest way.

However I would like to undestand the math. Unfortunately I'm too bad in math (or this is too far in my memory...).

@spades : thanks for your detailed answer, but I would like to understand how to get the value from the formulas. Let's take for exemple L2

L2 =√(ab²+bc²)+(ab x bc)

we now that the result is 15. but since a=0 this would give

L2 =√(0²+15x15²)+(0) = this doesn't equal 15 but 58 so I have missed something.

by approximation I found: SQRT(a²+b²+ab); SQRT(b²+c²+bc); SQRT(c²+a²+ac) will give the correct value of 15, 26, 15. Is this you were meaning? Please can you detail your formulas: in L1 =√(ab²+ac²)+(ab x ac) is it L1 =√((ab²+ac²)+(ab x ac)) or L1 =√((ab)²+(ac)²)+(ab x ac). I've tried different combination but exept the one mentioned above cannot get the value. Thanks anyway I appreciate you help.

@7anoter4: I also appreciate your answer and the time you take to do it. But sorry I don't undestand as well. What is j? I have also tried different way to get the result thru the formulas but couldn't. If you can detail more I would highly appreciate. Thanks anyway.

@PWSlack: Yes in international standard as well (as in any standard I believe) the breaker is defined according the size of the cable. Unfortunately this doesn't prevent users to overload and breakers to do the job: tripping. And this is what is going to happen for me in this situation we 26A on 1 phase of 3x20A(C) breaker. I have to fix this.

Again thanks all

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Jmp0

t=sec Voltage V a=pi Vmax

Sorry, JumpZero, but working with complex [or imaginary] numbers is an elementary electrical education [I don't intend to insult you ].If I will have time I'll try to explain you.

For now I'll try to explain that:

An A.C. voltage[V] measured all the time[t] presents a sinusoid:

V=Vmax*sin(k*t)

Instead of t[sec] you may use angle a=k.t [a=0 sin(a)=0 a=90 sin(a)=1 a=pi sin(a)=0 ]

constant k depends on frequency k=2*pi*f.

If the time 0 the voltage could be different [not 0] then a supplementary angle is required so the actual formula it is V=Vmax.sin(k.t+fi) and at t=0 V=Vmax.sin(fi).

If we shall take fi=0 for phase 12 then for phase 23 will be fi=-120 degrees and for phase 31 fi=-240 [or +120 degrees].

You can represent voltage or current in complex numbers using an imaginary number j[or i]=sqrt(-1) and then V=Vmax/sqrt(2).[cos(fi)+sin(fi).j].As you see this way helps us to solve simpler most of electrical problems.

It is absolute nonsense to uprate the breaker without uprating the cable. I'd fire you if you worked for me.

Crabtree,

The OP never said anything about the conductor sizes but I'm sure that this excercise not only told him that an existing breaker isn't up to the job and therefore not a stroke of luck but something that also needs to be added to the shopping list, it also has given him the figures to be able to select and purchase the correct sized conductors in seven months time.

Hi everybody,

Thanks again for your very quick answer. This forum is very responsive this is really a pleasure.

@spades: Yes! Now it's limpid, 99% clear. So to be 100% clear I would say

L3 =√(ac²+bc²)+(ac x bc)

we should read

L3 =√((ac²+bc²)+(ac x bc))

Because the first one is for me SQRT(225+225)+225 = 246.21

and the second one is SQRT(225+225+225) = 25.98 approx=26

@7anoter4: No you don't insult me, and yes I learnt the complex at school in 1974-75 and have never used them since then. But thanks for your explanation, again I sincerely appreciate the time you guys take to answer and even include graph. That's great.

@Crabtree: Oh! I see now why on the picture you wear a policeman uniform, you want to find somebody guilty and punish him. Ok this is humor huumm, a joke, I don't want to offense anybody here specially after having receive so high quality answers. Where have you seen I want to increase the breaker size? I am just telling in this case the load exceed the breaker rated current and this must be solved. There are several solutions. Actually if you want to know the details: We have a HV switchboard of 33 cubicles, each cubicle is equipped with an anti-condensation heater of 100W. The vendor has designed this switchboard in this way: from the site he receives auxiliary LV heater power from a 3 phases, 3 wires network 220Vac. The main breaker is 3x20A(C) then the 33 cubicles heaters are connected in between L1-L3. Each heater has an individual breaker in each cubicle. Then we have several HV motors, each with anti-condensation heater the total is 3348W. To make it simple in my initial post I said 3300W to make it same as cubicles heaters. All of these motor heaters are in between L2-L3. Each one also has its own breaker in the HV motor contactor cubicle. Nothing is connected in between L1-L2 (except when some equipment are in test position, but most of the time it wont be the case, so we can say nothing).So I am now with a switchboard that has a design mistake: as demonstrated L3 will pull 26A on a breaker of 20A. Even if a C curve can accept something like 120% of the rated current for ever it's not acceptable. So my intention is not to change the breaker but report the problem to the vendor and ask him to find a solution. Of course if the breaker is to be changed the panel internal wiring (and no cable) must be changed (or confirmed it is still ok). But there is a smarter solution that I would use if it was my own problem: balance the circuit, move some heaters to L1-L2. Do you still fire me boss?

BTW: Mildred?...

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Jmp0

AS you were describing that in more detail I was thinking why L!L2 couldn't be used in other cabinets just as you suggested at the end.

Seems sensible to me.

"@spades: Yes! Now it's limpid, 99% clear. So to be 100% clear I would say

L3 =√(ac²+bc²)+(ac x bc)

we should read

L3 =√((ac²+bc²)+(ac x bc))

Because the first one is for me SQRT(225+225)+225 = 246.21

and the second one is SQRT(225+225+225) = 25.98 approx=26"

The root sign covers the entire equation although I can't show that adequately here, so both of those formulae that you show will yield the same result of 25.98 unless you do it on MS Excel or via some of the more mundane online solvers. Because it covers the entire equation, you do the sums in each set of brackets first, add them together, and then find the root of that number.

If you try to solve this on an Excel spreadsheet using the SQRT function, or on some online solvers, then you would have to add the extra brackets as they require them to tell the program that the root applies to the entire equation, otherwise they will see the SQRT function as applying only to the first set of brackets, but the extra set make no difference if solving manually as you know that the entire equation is under the root.

If you want to test what I have said - here is a good online equation solver that works correctly, takes a bit of learning, but it's a good one. Use the function menu at the top. Below are screen shots of both equations solved, bit hard to read but visible if you enlarge, and you can see that both solutions are identical at 25.98.

As I already said JumpZero , these are interesting formulas but the final relation has to be:

L3 =√(ac²+bc²)-2x(ac x bc)xcos(angle)

and for "angle" between ac and bc vectors[let's say "phasors"] of 120o it turns in:

L3 =√(ac²+bc²)+(ac x bc) [as you said]

As you can see for any other angle an error occurs.

In my opinion Mildred she is a protagonist of British comedy George and Mildred:

http://en.wikipedia.org/wiki/George_and_Mildred_%28film%29

You may also see other comedies connected with as for instance:

http://en.wikipedia.org/wiki/Caf%C3%A9_Ren%C3%A9

http://en.wikipedia.org/wiki/Officer_Crabtree

Is there any possibility of rearranging perhaps the motor heaters to balance the loads?

2,200 W on each phase would be ideal!

Hi Guys!

New to this Forum

Great discussion on 3phase.

Jumpzero, I would like to ask you if the 220V you mention is the inbound 3Phase Line to neutral voltage or is it the line to line voltage?

You should confirm Line to line voltage by measurement before making any decisions on beaker rating changes

If 220v is line to line I agree with the results here, however if 220v is Line to neutral the line to line voltage will be 380V (220Vxsqrt(3)) which gives I23 and I13=8.7Amps and L12=17.5Amps.. approx.

Hello,

@spades: Yes you're right. I was mislead by the symbol √. Actually what is misleading is the upper line which is missing in this symbol. When you write it by hand it's clear because the line will cover the entire expression you want the square root of.

@Geoffrey36 yes I believe it will be the final choice (of the vendor) to reorganize the distribution to get a better balanced circuit.

@LINKS53 no neutral, 3 phases 3 wires only as mentionned in initial post.

Thanks to everybody for the support, the quality and the speed of the answers.

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Jmp0

Nice to hear from a satisfied contributor.....seems you have the information you need now to proceed with secure confidence.

One thing though,

L1L2 220V and same for other phases..... what's your location?

Lots of possibilities.

Look them up here

Not that many actually.

Wondering which one the OP is in.

Somewhere wet and cold given the panel and motor heater requirements.

Hello,

the project is a FPSO:

ok what is a FPSO; look here here

that will operate in the USA waters in gulf of Mexico.

HV=6.6kV

LV=440V for motors etc.. power equipment

220V for lighting and small power

everything 60Hz

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Jmp0

A nominal 440 v 3ø supply has a phase-neutral voltage of 254 v

A nominal 220 v 3ø supply has a phase-neutral voltage of 124 v

Please confirm that the arrangement is like this (the loads lumped together.)

Very interesting. Not my environment. Nice to learn...

So, no neutrals. All isolated.(right?)

Hi,

@Wal: correct no neutral, everything is 3 phases, 3 wires, IT system, whatever it's 6.6kV 440V or 220V.

@Geoffrey36: there are step down transformers 440/220V. Then 220V is IT system, no neutral. The arrangement is as per your drawing. We can consider this topic as solved, as I have received the answer above.

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Jmp0

Thanks, appreciate the reply.

Work safe, hope to hear from you again.