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Answers Please. Electrical Technology.
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2736 Posts
1. A load of 2.5kW takes a current of 12 Amp from a 230 Volt supply. Calculate the power factor.

2. Three single phase loads are to be connected to a three phase supply. All at unity power factor. The loads are,

i, 60 Amp.

ii, 50 Amp.

iii, 40 Amp.

By any method determine the neutral current.

Z.
26 Replies
AJJewsbury
2499 Posts
1. about 0.906
2. 17.32A

  - Andy.
ebee
1150 Posts
90.57971 by plastic brain - a.k.a a calculator..

15.5A by a rule and pen on the back of an envelope. (Guessed the angle by the "Rack of the eye" method)

Next calculation I might use a "Guessing Stick" Well that`s what our tutors used to call a slide rule as calculators (plastic Brain) were just starting to appear on the scene. 
I can`t remember a slang name for those adding machines that were like a mini cash register with levers, a handle and moveable carriage. We thought they were brill though, 4 in the whole school and was a treat to get to use one for a few minutes , I guess it was the second half of the 60s and I think they were about £40 each which was a lorra lorra money back then

Mind you, our school had a maths teacher (not my teacher, I had one of the others) that actually gave one girl 101% in a maths test - 101% ? yes 101%, she answered all the questions correctly plus her work was so neat and so well set out that he gave her an extra one per cent on top. 101% and that was at a Grammar type school !
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2736 Posts
ebee:
90.57971 by plastic brain - a.k.a a calculator..

15.5A by a rule and pen on the back of an envelope. (Guessed the angle by the "Rack of the eye" method)

Next calculation I might use a "Guessing Stick" Well that`s what our tutors used to call a slide rule as calculators (plastic Brain) were just starting to appear on the scene. 
I can`t remember a slang name for those adding machines that were like a mini cash register with levers, a handle and moveable carriage. We thought they were brill though, 4 in the whole school and was a treat to get to use one for a few minutes , I guess it was the second half of the 60s and I think they were about £40 each which was a lorra lorra money back then

Mind you, our school had a maths teacher (not my teacher, I had one of the others) that actually gave one girl 101% in a maths test - 101% ? yes 101%, she answered all the questions correctly plus her work was so neat and so well set out that he gave her an extra one per cent on top. 101% and that was at a Grammar type school !

Who threw that? Stop talking at the back. Put that gum in the bin Don't forget that the power factor has to be less than one. Put that away John.

Z.

ebee
1150 Posts
Sorry Sir, I was expressing the power factor as a percentage. I shoulda/menta said 0.905791 which agrees with AJJs answer
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2736 Posts
Very good so far.

Any more answers please?

Z.
In = +17.32A >/ 64deg
edit: whoops wrong angle and direction....didn't realise that editing can take place well after posting...
Legh
mapj1
3421 Posts
showing the (known to be erroneous - see comments below) workings.

12A 230V is 2760VA
PF is 2500/2760 = 250/276  = 0.90597 
So 25 degree phase angle between P and Q

2) Subtract 40A from each phase as that has no effect on neutral current.
Problem now  reduces to completing a parallelogram  with one side length 20 units, and the other side length 10 units and calculating the diagonal

resolve parallel and orthogonal to the phase carrying the 20A excess.
Orthogonal to the  dominant phase is 10A *sin120deg =  8.66A
in-phase  is 10A* cos 120 + the original 20A.  = (-5A + 20A) = 15A

use pythag to get the magnitude is sqrt[(15)2 + (8.6666)2] =  sqrt[(225) + (75)] = sqrt(525) = 22.9A

Edit That is wrong as Geoff points out below - should read ..
Magnitude  sqrt[(15)2 + (8.6666)2] =  sqrt[(225) + (75)] = sqrt(300) = 17.32A
Angle is arctan 15/8.6 ~ 30 degrees off the the phase carrying 60A, rotated  in the direction of the phase carrying 50.
Mike.

 
Blimey I got s stress headache now
Sorry Mike only 8/10.  225 + 75 = 300.  SQRT(300) = 17.32A.

Now for your homework - calculate the phase to neutral voltages and the voltage between your neutral terminal and the the star point of the supply transformer 😎

Regards
Geoff Blackwell
Correction - the above should say do the calculation assuming a broken supply neutral

Regards
Geoff Blackwell
As a foot note - Mike, you do have the correct angle,  although I prefer 330 degrees (anti-clockwise rotating phasors).
 How do you edit these posts!

Regards
Geoff Blackwell
ebee
1150 Posts
You wait ages for a Blackwell posting then three turn up at one go.
Welcome back Geoff lad.
Nice to see you
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2736 Posts
Zoomup:
1. A load of 2.5kW takes a current of 12 Amp from a 230 Volt supply. Calculate the power factor.

ANSWER: P.F. = 2500/230 X 12.

                             = 0.906

2. Three single phase loads are to be connected to a three phase supply. All at unity power factor. The loads are,

i, 60 Amp.

ii, 50 Amp.

iii, 40 Amp.

By any method determine the neutral current.

ANSWER = 16.5 Amps.

Answers & questions supplied from a City & Guilds and I.E.T. joint publication.

Z.

 

Well  City & Guilds and the I.E.T. joint publication have not produced an accurate answer then.  I don't know what method they have used but  a 'proper' calculation produces the result that I, and others,  gave (Mike's answer would have been the same but for a simple arithmetic error).

A decent phasor diagram would also come close to the correct answer.
BTW I got my angle wrong in my earlier post🙄
Regards
Geoff Blackwell
 
My method:

In = I1+ I2 + I3 where all terms are phasor quantities

In polar form

60 @ 0 degrees + 50 @ 120 degrees + 40 @ 240 degrees

In rectangular form (you can obtain this using the Polar to Rectangular function on a calculator or trigonometry)

by calculator function

I1 = 60 + j0
I2 = -25 + j43.3
I3 = -20 - j34.64

collecting terms
Re    Img
60    0
-25    43.3
-20    -34.64

15  + j8.66

in polar form (by calculator)

17.32A @ 30 degrees

Which by the way proves that I got my angle wrong in my earlier post.

Regards
Geoff Blackwell
mapj1
3421 Posts
let down by failure to use the book of tables , I knew the new fangled calculator was a mistake 😂.
Mind you it seems C and G have a similar problem.

To edit your own posts click on the 3 spots on the high right hand side in the post listing.  And then as courtesy add an 'edit' note if the change is more than cosmetic to avoid confusion to those who have already read it.

to do crossing out, like this in the editor change to source edit, and add strike and /strike inside <> signs around the offending section. And then copy and save all the text  somewhere , as swapping back from source view to normal view before posting sometimes crashes for reasons not clear to me.

I'll be back with the PEN lift voltage tonight, work is calling...
regards
Mike.
Thanks Mike

On the angle question I originally obtained my answer using a program I wrote some years ago that includes a facility to calculate the parameters of a broken neutral fault (as it stands, just for unity power factor).  This gave me my 330 º.  This is, in fact, correct for a positive phase sequence i.e: L1 @ 0 º, L2 @ 240 º, L3 @ 120 º - anticlockwise rotation which is the normal system in use.

30 º comes from a negative sequence.

Regards
Geoff Blackwell
 
mapj1
3421 Posts
Agree 330E  =30 W and all that - but in my defence,  the original question did not specify rotation.
Of course if it had, I'd have had a 50% chance of getting that wrong as well...
Looking at your workings, they are elegant and general, but I must admit while I fully understand complex no.s and use them for work I find to couch explanations as magnitude and angles them more visual in some way - I like to actually close my eyes and see the cogs rotating in the mathematical model, and imagine the electrons as rushing around, rather than just crunching the numbers (it also greatly increases my accuracy - the chance of spotting an error is increased by knowing from the visualisation roughly what sort of size and direction  the answer ought to  be.. Without that my chances of being off by  pi, or two or -1  or j increase rapidly with the no. of steps in the calculation).  (not good enough for spotting 17 instead of 20 something though)
It does get some funny looks when I do this at work, and refer to 'gear ratios' and 'missing teeth' when comparing harmonic frequencies and phases in complicated systems though.
M.
mapj1
3421 Posts
As promised, neutral offset voltage with open PEN.

3 loads  R1= 230/60 ohms  R2 =230/50 ohms and R3 = 230/40 ohms.

Let us ground the 40A phase (call it L3 )and consider L1 and L2 to be two phases of 400V relative to it, one at 60 degrees phase offset relative to the other.
We can transform our voltages back to a 'centre earth' at the end, and I prefer to work with only 2 moving parts in the model.

V 3 is the drop across the R3 resistor due to the sum of the two currents coming in through R1 and R2. They are not orthogonal and cannot simply be superposed, so we calc current in R3 from V1 with V2 grounded, then current in R3 from V2with V1 grounded, then
do the in phase and orthogonal thing to get the magnitude and angles.

see figure below



 
Thanks Mike
My software considers a broken neutral with all the phases intact.

I used Millman's theorem to calculate the various voltages and currents.  In this example it is not too dramatic as the loads are not too far out of balance (worst case voltage is Vbs ( voltage between blue phase and the star point of the transformer - blue phase whats that!) = 254V @ 123º.  Hardly destructive but this is usually a dynamic system so things could go pear shaped.

The voltage between the transformer star point and the neutral on the consumers side of the break Vsn =27v @ 330º.  So well below the 70 V maximum for EV systems.

Regards
Geoff Blackwell
Jam
82 Posts
Just wanted to say thanks all above for the enjoyable exercise; I've been playing with programming in Python lately and (once I'd realised I needed angles in radians not degrees...) it was satisfyingly tidy to use Millman's Theorem on the phasors directly. 👍
Tony Sung
1 Posts
How about drawing out the phasor diagram for the three currents on a piece of graph paper and use a ruler to measure the resultant neutral current to be around 17A at 30 degrees lagging L1? Will it be sufficient to answer the question?
The original post said   -  "By any method determine the neutral current."
So a diagram should be acceptible and, in fact, would be a quick way of obtaining the answer during an examination.
Regards
Geoff Blackwell

 
GeoffBlackwell:
The original post said   -  "By any method determine the neutral current."

I got 150A since my method was to connect all three loads to the same phase........

mapj1
3421 Posts
Err yeees but, no , but ...
I may have worked on a job where someone did something like that.  It might have been better if 2 of the 3 phases had been isolated prior to power being re-applied after the  fitting of the single phasing strap across the output terminals of the incoming switch.

The element of shadenfreude as it was clearly not me was unavoidable. Still the electrons did make it very clear which fuses should have been pulled to do the isolation, with a sort of noise emitting diode effect ....
M.

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