Transmission line calculation

Discussion in 'Homework Help' started by ham3388, Jul 25, 2012.

Jul 3, 2012
97
3
Hi dear friends...

I did my best but still im not sure the answer is correct

Please go through it and help me if possible.

My regards

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2. mlog Member

Feb 11, 2012
276
36
Look at your phase angle of the receiving current. PF = 0.9, but it is lagging, which means current should have a negative angle. Current lags voltage.

3. mlog Member

Feb 11, 2012
276
36
Part (c) is not difficult. Your text book should have the breakdown of how A, B, C, and D are derived in terms of Y and Z, or it isn't difficult to derive them yourself. I would suggest you try to derive them to help you better understand.

For example, A=Vs/Vr with the condition that Ir=0, B=Vs/Ir when Vr=0, and so on. You should be able to do the same with C and D. B is very simple. C and D are a little more complicated but not bad.

After you find ABCD in terms of Y and Z, then you decompose Y and Z in terms of their consituent components. I'll get you started. Z = R + jX. You were given the numerical values in a table with the problem statement, but with X you'll need to do a little more work to find the value of L. Follow the same Process with Y, the admittance, Y = G + jB.

The problem asked you to find R, L, G, and C. How do these relate to Y and Z?

Good luck and let us know how you finish.

ham3388 likes this.

Jul 3, 2012
97
3
Thank you very much for the reply.....

The ' a ' and ' b ' I did based on -25.84' the followings are the new figures:

Vsp=98.697 x 10^3 /__ 14.32'

Is=137.63 /_29.5'

Psend=39.33 x 10^6 mw

Plos=3.32 mw

I tried but still I coulnt do it

Consider the cable lenth is 50 km long and the Ï design of the line , where are you going to include them?

Please you do the first one as an example then I will retry again.

By the way there are some formulae but about the Ï model and R-G ladder only, Im including it as well.

My regards

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5. mlog Member

Feb 11, 2012
276
36
You have attached the equations for a long transmission line. This line is of a length that can be considered a medium length line. The long line is much more complicated because it requires differential equations and hyperbolic functions. The medium length line is simpler because it uses lumped parameters in either a nominal pi or or a nominal tee form.

Go back and look at the original problem you posted, which showed a tee form with a shunt admittance "Y" (sometimes shown as Y/2) that appears on either side of the pi. (Ref. Fig. 3(b).) The input value of Y is the same as the output value of Y. The 3rd element of the pi, which is between the shunt Y's is a "Z" for the impedance.

You should have lumped parameter equations that relate Y (or Y/2) and Z in terms of A, B, C, and D. Well, I'll work the easy one for you. B=Z, which will be of the form R + jX. Of course the "X" here is the inductive reactance at 50 Hz. The R is the line resistance and the X is the inductive reactance from the lines. You will use the values in Table A. For example, Z = B = 47.94 + j180.8 ohms. You know R, and you can find L given that the reactance is 180.8 ohms and the frequency is 50 Hz.

Now see if you can find the medium-length line equations for A, C, and D as a function of Y and Z. I think you will find it very easy now, especially since you know the value of Z.

Your line is only 50 km in length, so you shouldn't need to use the long-line equations. The 60 Hz break point for medium to long lines is 150 miles or 240 km, and I would expect it to be slightly longer for a 50 Hz line. The fact that the problem handed you the ABCD parameters should be a hint that you are expected to use the medium line equations.

ham3388 likes this.