High voltage transmission line with ABCD complex equations

bizuputyi

Joined Aug 3, 2014
21
Figure shows a 50 Hz, high-voltage, transmission line. The relationships between the sending and receiving end voltages and currents are given by the complex ABCD equations:

$$V_S=V_R(A_1+jA_2)+I_R(B_1+jB_2)$$

$$I_S=V_R(C_1+jC_2)+I_R(D_1+jD_2)$$

where 'S' stands for sending-end and 'R' stands for receiving-end

(a) Given the parameter values in TABLE C and an open-circuit received voltage measured as 88.9 kV, calculate the values of $$V_S$$ and $$I_S$$ and hence the power $$P_{SO}$$ absorbed from the supply by the transmission line on open circuit.

(b) If the line is modelled by the T-circuit of FIGURE 3(b), see if you can estimate the primary line coefficients R, L, G and C. The line is 50 km long.

What I'm thinking is $$I_R=0$$ as this is an open-circuit and given $$V_R$$; $$V_S$$ and $$I_S$$ can be calculated with matrix.

$$V_S=77325+j3149 KV$$
$$I_S=j119.9A$$

But I don't think that is correct because $$V_S$$ should not be lower than $$V_R$$, also 50Hz frequency is given for a reason, I'm sure it has to be used somewhere.

And as for question (b) I have only got ideas. I can find coefficient of propagation γ and $$Zo$$ from R,L,G,C but I don't know how to produce them vice versa (from ABCD), I don't see where I could go from there either.

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shteii01

Joined Feb 19, 2010
4,644
Regarding (a).

77325+3149j kV is greater than 88.9 kV.
Or did you mean 77325+3149j V? That would be 77.325+3.149j kV, and yes, this is less than 88.9 kV.

bizuputyi

Joined Aug 3, 2014
21
I'm sorry, that's a typo. It's 77325+j3149 V. That's where it gets suspicious, it shouldn't be less than 88.9kV.
I got this result by using matrix equations shown in attachment with $$I_R=0$$.

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bizuputyi

Joined Aug 3, 2014
21
Wow, that's a new aspect. From what I read about Ferranti effect I reckon I've got the voltages and currents all right. And the power becomes $$P=Re[VI^*]=377565.1W$$
What I'm thinking if the impedance in the stem is $$\frac{V_R}{I_S}$$ then I get j741.45Ω from which

$$X_c=\frac{1}{\omega lC}$$ (where l is length in metres)

C=85.86pF that is far too low for such a long line.
Without putting 'length in metres' in that equations capacitance is 4.29μF.

Resistance can be neglected, so can conductance, from http://www.electrical4u.com/ferranti-effect-in-power-system/ source.

I don't feel I'm on the right track. I only found methods to calculate R,L,G,C with ∏-model not T like this one.

The Electrician

Joined Oct 9, 2007
2,952
Consider the T network in Fig. 3(b).

Label the two series resistors R/2, the two series inductors L/2, the shunt conductance G and the shunt capacitance C.

Now form the ABCD matrix for that T network. Have a look here:

http://en.wikipedia.org/wiki/Two-port_network

about halfway down under the heading "ABCD-parameters". There you will find info (under "Table of transmission parameters") about how to form an ABCD matrix from lumped components.

Having formed the symbolic ABCD matrix from R, L, C, G you can then equate each symbolic element in your ABCD matrix for the T network to the corresponding numeric values you have for the line. That will give you 4 equations in 4 unknowns. Solve them and you will get values for the line coefficients.

bizuputyi

Joined Aug 3, 2014
21
Please see attachments, I think that's the same approach. This is fantastic, I'm getting so close. However, I need time to figure those 4 equations.

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• 42.4 KB Views: 228

The Electrician

Joined Oct 9, 2007
2,952
You're on the right track.

However, because the T network is symmetric, the A and D elements are the same, so you will only have 3 independent equations. That means that one of R,L,G,C will have to be chosen arbitrarily. A good starting point would be to set G=0.

If you try a solution with 4 equations, you'll probably get an error. You might be able to do a least squares solution with 4 equations.

I notice that the numeric values for A,B,C,D form a matrix whose determinant should be exactly 1, but it isn't, so you haven't been given an artificial problem with an exact mathematical solution; the numbers perhaps come from a real transmission line.

bizuputyi

Joined Aug 3, 2014
21
$$A=D=0.8698+j0.03542=1+\frac{1}{2}(R+j\omega L)(G+j\omega C)l^2$$
$$B=47.94+j180.8=(R+j\omega L)+\frac{1}{4}(R+j\omega L)^2(G+j\omega C)l^3$$
$$C=j0.001349=(G+j\omega C)l$$

I'm sorry for my blindness but I don't see where to go from here and also where the least squares solution comes in. I'm so close just need a bit more advise.

The Electrician

Joined Oct 9, 2007
2,952
After giving a try to solving those equations, I think an easier approach would be this:

Consider a general ABCD matrix:

[ A B ]
[ C D ]

Now if you have a T network that has that matrix, the elements of the T network will be:

Left series element = (A-1)/C

Shunt element = 1/C

Right series element = (D-1)/C

You can work this out by considering how the ABCD matrix is formed from a T network.

Note that the numerical value your problem gave you for C is 0 + .001349 j

This means that G is zero, which is what I suggested you assume when solving those equations above.

But, now if the shunt element of the T network is 1/C , we just need to solve (because G=0):

1/C = 1/(.001349 j) = 1/(j ω Cs), where Cs (I designate it Cs instead of just C to avoid confusion with the C of the matrix) is the shunt capacitance of the T network.

R/2 will be the real part of (.8698 + .03542 j - 1)/(.001349 j)

and L/2 will be the imaginary part of (.8698 + .03542 j - 1)/(.001349 j) divided by ω.

These values of R, L and Cs are for the whole line; divide by the line length for the final result.

Last edited:

bizuputyi

Joined Aug 3, 2014
21
Is it really that simple? That's great, thank you very much indeed.

Do these appear to be correct?

R=0.0005252Ω
L=6.14$$\mu$$F

1/C = 1/(.001349 j) = 1/(j ω Cs)

The Electrician

Joined Oct 9, 2007
2,952
Here's what I get for the T network components before dividing by the line length. The R and L values are for each series arm of the T network. They need to be doubled for the actual R and L for the line:

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bizuputyi

Joined Aug 3, 2014
21
Of course. What's wrong with me?! I wanted this so bad I made silly mistakes.
Thank you for your effort to help!

CNC682

Joined Jan 23, 2015
27
Left series element = (A-1)/C

Shunt element = 1/C

Right series element = (D-1)/C

How do you derive these equations?

The Electrician

Joined Oct 9, 2007
2,952
I said in post #10:

"You can work this out by considering how the ABCD matrix is formed from a T network."

Look halfway down the page here: http://en.wikipedia.org/wiki/Two-port_network

Under the heading "Table of transmission parameters" you can see how to form the [a] matrix for a series impedance. Follow this with the [a] matrix for a shunt impedance, and finally the [a] matrix for another series impedance. Multiply the three matrices and get the overall ABCD matrix for a T network.

Then, you can work backwards to see what the T network elements would be if you're starting from an ABCD matrix.

viky1104

Joined Aug 9, 2015
1
Why we transmit power at high voltage ? If there is reduction in power loss i.e. I^2*R losses , then power is also depends on voltage as power P=(V^2)/R . Hence with high voltage power loss should be more with respect to voltage and power loss should be less with respect to current. Which one is correct ? and why ?

koupitan

Joined Dec 6, 2016
2
I said in post #10:

"You can work this out by considering how the ABCD matrix is formed from a T network."

Look halfway down the page here: http://en.wikipedia.org/wiki/Two-port_network

Under the heading "Table of transmission parameters" you can see how to form the [a] matrix for a series impedance. Follow this with the [a] matrix for a shunt impedance, and finally the [a] matrix for another series impedance. Multiply the three matrices and get the overall ABCD matrix for a T network.

Then, you can work backwards to see what the T network elements would be if you're starting from an ABCD matrix.
Hi All,

I am still struggling to understand how the left series element = (A-1)/C, Shunt element = 1/C , Right series element = (D-1)/C are obtained.
I found in my notes that ABCD matrix formed from the T-network looks like this:

What should be done next? Should i Plug for A in the matrix 0.8698+j0.03542 and so on and then equate A with?

The Electrician

Joined Oct 9, 2007
2,952

Papabravo

Joined Feb 24, 2006
20,583
Why we transmit power at high voltage ? If there is reduction in power loss i.e. I^2*R losses , then power is also depends on voltage as power P=(V^2)/R . Hence with high voltage power loss should be more with respect to voltage and power loss should be less with respect to current. Which one is correct ? and why ?
Ohms law says they are the same. The reason we use high voltage transmission is that transformers allow us to easily increase voltage while reducing current and vice versa.

https://en.wikipedia.org/wiki/Electric_power_transmission

Scroll down to the section on Losses

koupitan

Joined Dec 6, 2016
2
Hi The Electrician,
Sorry for not giving enough details.
I am referring to part b) of the original question which is:
" If the line is modelled by the T-circuit of FIGURE 3(b), see if you can estimate the primary line coefficients R, L, G and C. The line is 50 km long". There is also a table with values :

What I understand so far is to I have to obtain an ABCD matrix for the T network that looks like the one i posted in my firs post.
Reading your earlier replies, I understand you recommended doing so using [a] matrices for series impedance and shunt impedance from wikipedia Two port network article. Assuming we don't have to consider right hand impedance as this is an open circut , is it done by multiplying only these two matrices?

Is using the ABCD matrix obtained from the above matrices equivalent to the one in posted in my previous post?

I home the explanation of my problem is clear enough

Cheers