Shunt admittance - from farads

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
Hey guys

Just muling over a problem here. Studying for a test Monday. The lecturer has given a problem - in place of a siemens value, for a shunt capacitor, on a power line model; he has instead given a value in Farads. Nanofarads to be precise... and to be even more precise 9.95nF/km on a 130km line at 50Hz... Yes that's right Dorothy you ain't in Kansas no more :D

Anywho do I just calculate the impedance, then just give the inverse of that value?
Like is it = 3.13*10^-6 S/Km?
then just multiply that by the length?
 

WBahn

Joined Mar 31, 2012
30,286
Without saying whether you are right or wrong, let's look at it from a few different viewpoints.

First viewpoint: Does the answer make sense?

If the impedance goes up, what do we expect to happen to the admittance?

If you double the length of the line, does the impedance go up or down?

If you do what you suggest and double the length of the line, does the impedance go up or down?

Second viewpoint: Do the units work out?

Where did your S/km come from. Did those units arise naturally by tracking them, or did they come from wishful thinking (i.e., tacked on at the end because that's what you hope they are)?

Third viewpoint: What does the math say?

X = 1/(ω·C)

where C is given by

C = C'·D

where C' is the capacitance per unit length and D is the length (using D instead of L to avoid confusion with inductance).

So

X = 1/(ω·C'·D)

Since the admittance, Y, is the reciprocal of impedance

Y = 1/X = ω·C'·D
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
Hmmm see I thought that was right but now I'm not so sure. If I arrange the formula as you have suggested 1/C'D. Then double the length then (C_2) < (C_1).

But I was thinking about:-

C= (1/ ω·C')D.

I didn't arrive at the idea purely by wishful thinking. There's a lot of information out there about the Siemens unit. Probably for different applications. But some do suggest that it's basically the reciprocal of the Ohm.

So I thought ok well, just do the normal calculation for the inductance of the capacitor = 1/ω·C. Then once you've got the value. Then it should be the reciprocal of that answer...
No???

Please tell me Wbahn... I don't have much time.
 

WBahn

Joined Mar 31, 2012
30,286
Hmmm see I thought that was right but now I'm not so sure. If I arrange the formula as you have suggested 1/C'D. Then double the length then (C_2) < (C_1).
I never suggested 1/C'D. I very carefully suggested 1/(C'D).

Say D = 100 km. Then C'D is 995 nF. If I double the length, it becomes 1990 nF.

But what about the impedance?

If I increase the capacitance, the impedance goes down. So if I double the length, the impedance should go down. That means that D needs to be in the denominator of the formula for impedance.

But since admittance is the reciprocal of impedance, if the impedance goes down the admittance goes up. That means that the D needs to be in the numerator of the formula for admittance.

But I was thinking about:-

C= (1/ ω·C')D.
Nope. Check the units.

C' = 9.95 nF/km

So (ω·C') is going to have units of 1/(Ω·km) and 1/(ω·C') is going to have units of Ω·km. So

C= (1/(ω·C'))D is going to have units of Ω·km².

I didn't arrive at the idea purely by wishful thinking. There's a lot of information out there about the Siemens unit. Probably for different applications. But some do suggest that it's basically the reciprocal of the Ohm.
That exactly what it is. Just as with ohms, there is a difference between pure resistance and reactive components, the same is true with admittance. But the unit S is A/V which is the same as 1/Ω.

For what it is worth, the reciprocal of resistance is called conductance, the reciprocal of reactance is called susceptance, and the reciprocal of impedance is called admittance. The different terms just make it a bit easier for us to easily distinguish when we are talking about the real, the imaginary, or the full complex values involved.

So I thought ok well, just do the normal calculation for the inductance of the capacitor = 1/ω·C. Then once you've got the value. Then it should be the reciprocal of that answer...
No???
Other than I think you meant "impedance" or "reactance" and not "inductance" you are fine.

Well, other than you are still being sloppy since 1/ω·C = C/ω and NOT 1/(ω·C).

Please tell me Wbahn... I don't have much time.
What more is there to tell you than what I did in the prior post? Look at the last line of that post and I explicitly give you the formula.
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
Now I'm really confused how is 1/ω·C not equal to 1/(ω·C)???

And how' it equal to C/ω?

Using the value 9.95 for convenience

152680735840953237339.jpg
 

WBahn

Joined Mar 31, 2012
30,286
Now I'm really confused how is 1/ω·C not equal to 1/(ω·C)???
Simple. Standard -- pretty much universal -- order of operations.

Multiplication and division have equal precedence and are left associative (i.e., given the choice, the left one is performed before the right one). So adding explicit parens it becomes

1/ω·C = (1/ω)·C = C/ω

You REALLY need to stop seeing what you want to see (i.e., interpreting the '/' as a long horizontal bar with an arbitrary expression underneath it instead of a single operator with two operands, one immediately to the left and one immediately to the right) as it will get you in all kinds of trouble in the age of computers.

Don't believe me?

Open a spreadsheet and type =100/2*5 and see what you get. According to you, it should be 10. But according to virtually an program and most calculators the answer is 250.

This bites people writing code or setting up spreadsheets or using calculators all the time.

Another one is people writing

Req = R1*R2/R1+R2

They see what they want to see, but what they wrote (and what they will get if they put it into a program or a spreadsheet) is

Req = ((R1*R2)/R1)+R2

Which, sans roundoff error, is the same as

Req = 2*R2

I've seen numerous people make exactly this mistake and spend hours trying to understand why they are getting the wrong answer and have even had several of them tell me with certainty that the compiler or the spreadsheet or whatever has a bug.

And I'm not even going to look at the image since the briefest glance shows that you are once again ignoring units.
 
Last edited:

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
I said I was using a value for convenience... It's my value, therefore I get to express the units as a unitless number, so :p

I put in my rads/s

Haha it was a trick all along... Even though tbh we were in fact talking about capacitance...so fair enough... I should be beaten :D

Didn't know that about the "/" sign I'll remember it for the future. However I believe it is an accepted norm especially on forums, that this / represents a division with everything after it expressed as a combined product.

So just to be 100% here. I can work out the impedance, then just do the reciprocal?
1/(Z) ?
 

WBahn

Joined Mar 31, 2012
30,286
I said I was using a value for convenience... It's my value, therefore I get to express the units as a unitless number, so :p

I put in my rads/s

Haha it was a trick all along... Even though tbh we were in fact talking about capacitance...so fair enough... I should be beaten :D
Glad we agree on that one, at least. :D

Didn't know that about the "/" sign I'll remember it for the future. However I believe it is an accepted norm especially on forums, that this / represents a division with everything after it expressed as a combined product.
While I'm willing to concede that it is commonly seen, I don't know that I would agree that it is an "accepted norm".

If nothing else, it introduces needless ambiguity into something, namely mathematical expressions, that have no business being ambiguous.

How would you express something written as

\(\frac{a}{b}+c\)

in line? Most people would simply write a/b+c or perhaps a/b + c.

Yet according to your "accepted norm", that should be interpreted as a/(b+c) in either case. In reality, some people will interpret them one way and some the other way. Not the best way to achieve good communication with others.

So just to be 100% here. I can work out the impedance, then just do the reciprocal?
1/(Z) ?
Yes.
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
284
Haha cheers, you are the best !

See what you mean about the (a/b)+c...

Kk back to the grind here. 1 more day and she's all done, off for a whole year on placement. Whoop whoop!
 
Top