As there has not been any activity in here lately...

[D Skelton]

I think i see where i went wrong..... 39.2 microfarads!

No, you're not far off, too far however for it to be a rounding issue.

Go through all of your calculations

 

[Simonslimline]

Ok 42.9.

 

[D Skelton]

I got 42.78.

Take me through your calculations. Explain where you think you went wrong on each occasion and I'll see if I can help make things clearer.

 

[Simonslimline]

Without rounding off any calculations I get a power factor of 0.7981392393 and a capacitor value of 43.43079103.

 

[telectrix]

50 years ago, i'd have had a punt on this, but nowadays it's as relevant to my work as camelsh!t.

 

[Simonslimline]

RT = 26.47ohms XL=20.74ohms XC=50.75ohms

IR=63A Ixl=80.38A Ixc=32.85A

IT=78.92A power factor =0.79827 Rounded=(0.80 inductive)

Reactive I = 47.53A.

IR/0.95=66.31 √(66.31²-63²)=20.69. √(63²+20.69²)=66.31.

63/66.31=0.95

47.53−20.69=26.84

1667/26.84=62.10

2x3.142x60x62.10=23414.184 10 to the six /23414.184=42.70

My rounded off my calcs slightly differently this time, still not the exact value you got though.

I got 42.78.

Take me through your calculations. Explain where you think you went wrong on each occasion and I'll see if I can help make things clearer.

 

[Simonslimline]

Im not quite sure what you mean! i have worked that 1 out now and moved on to another.

So i suppose there is a quicker way of working these out?

Have you done complex numbers before, it's the easiest way to the solution.

 

[Silly Sausage]

Simon,

The maths is much simpler in complex form. Sounds like a contradiction!

If I pull my finger out I'll post up my solution some time.
I've still got to work on my question I mentioned earlier.

Keep watching this thread though, it's an oasis of sanity in a sea of madness!!! :smilielol5:

 

[Simonslimline]

Ok thanks, it would be interesting to learn how.

Simon,

The maths is much simpler in complex form. Sounds like a contradiction!

If I pull my finger out I'll post up my solution some time.
I've still got to work on my question I mentioned earlier.

Keep watching this thread though, it's an oasis of sanity in a sea of madness!!! :smilielol5:

 

[D Skelton]

RT = 26.47ohms XL=20.74ohms XC=50.75ohms

IR=63A Ixl=80.38A Ixc=32.85A

IT=78.92A power factor =0.79827 Rounded=(0.80 inductive)

Reactive I = 47.53A.

IR/0.95=66.31 √(66.31²-63²)=20.69. √(63²+20.69²)=66.31.

63/66.31=0.95

47.53−20.69=26.84

1667/26.84=62.10

2x3.142x60x62.10=23414.184 10 to the six /23414.184=42.70

My rounded off my calcs slightly differently this time, still not the exact value you got though.

Bang on the nose mate!

Slightly different rounding up/down that's all

 

[Simonslimline]

Got there in the end with some help. Thanks for posting these up, its helped me a lot

Bang on the nose mate!

Slightly different rounding up/down that's all

 

[D Skelton]

No worries mate, happy to help

It's what the training section is for afterall lol.

C'mon Archy, pull yer weight

 

[Simonslimline]

What does j represent?

Or the easy way...

Z = R + j.2.pi.f.L

 

[Silly Sausage]

C'mon Archy, pull yer weight

Geez, hassle, hassle, hassle!

OK......

New install, 7 circuits, wired out in 4000m of 2.5mil. T&E.
All IRs out at >2G @ 500V
No equipment connected, i.e. all cable ends are clean cut.

Perfect supply subject to no external influences etc, so zero current flowing in any of its conductors when energised. TNS 230V 50Hz.

So, when the installation is powered up, how many Amps will be flowing in the supply's Earth conductor, if any?

For simplicity, assume the Line to Neutral capacitance, of the installation cables, is zero.



You may need to do some research to find the answer, the clues are in the question!

Answers must include any info found, source info and workings!

 

[Simonslimline]

Interesting question, suppose i better do some research and try and answer it.