Notes 10: Conductor Sizing & an Example s1

Transmission Matrix Comments on Example 1, and Lumped Circuit Equivalent

1.0 Introduction

In these notes, I want to write the equations from last time in a matrix form. Then I will say a few words about example 4.1. Then I am going to skip over examples 4.2, 4.3, section 4.2, then jump to Sections 4.3 and 4.4.

2.0 Transmission matrix (Section 4.3) We jump to Section 4.3 because it is easy to do so. Recall the equations developed from last time relating the sending end voltage and current (V1, I1) to receiving end voltage and current (V2, I2) of a transmission line.

V1  V2 coshl  ZC I 2 sinhl (1)

V2 I1  I2 coshl  sinhl (2) ZC

1 These equations can be written in matrix form:  coshl Z sinhl V  C V  1   1  2   sinhl coshl   (3) I1   I2   ZC  Reference Module D1 of your EE 303 notes, Fig. D1.16, repeated here for convenience:

I1 I2 V1   1Y2Z Z V2        V1 Y1 Y2 V2 I1  Y1 Y2  ZY1Y2 1Y1Z I2 

Fig. D1.16: ABCD parameters for “” circuit [1] So eq. (3) is the relationship that we call the “ABCD” parameters for a two-port network:

V1  A BV2        (4) I1  C DI2  Clearly, in this case, we see that: 1 A  coshl C  sinhl ZC (5)

B  ZC sinhl D  coshl

2 The ABCD parameters are also referred to as the transmission parameters, where the following matrix is defined as the transmission matrix: A B T    (6) C D With this notation, eq. (4) becomes:

V1  V2     T   (7) I1  I2  Two comments about this: a. It is easy to get the receiving end (V2, I2) as a function of the sending end quantities (V1, I1):

V2  1 V1     T   (8) I2  I1  The reason why this is easy is because of the hyperbolic identity: cosh 2 x  sinh 2 x 1 and therefore the determinate of T is |T|=AD-BC=1 and therefore

3 1  D  B T    (9)  C A  b. It is easy to express sending end or receiving end quantities as a function of the other when there are multiple two- ports chained together. Figure 1 illustrates the case of two 2-ports chained together.

I1a I2a I1b I2b

a b V1a V2b

Fig. 1 If we know Ta and Tb, where

Aa Ba  Ab Bb  Ta    Tb    Ca Da  Cb Db  then it is easy to show that

V1  V2  V2  1 V1     T   ,    T   I1  I 2  I2  I1  where 1 1 1 T  TaTb , T  Ta Tb

4 3.0 Comments on Example 4.1

The text provides three nice examples, Examples 4.1-4.3 on pp 93-96. I will not go through them (you can read for yourself in the text) but rather I will just focus (for now) on the first one and draw out what are some conceptual implications of which you should be aware. Please consider (look for, think about, connect with) these concepts as you review these examples on your own.

Example 4.1: The main idea of this example is to give you some practice in using the equations derived last time and repeated here for convenience:

V1  V2 coshl  ZC I 2 sinhl (1)

V2 I1  I 2 coshl  sinhl (2) ZC The example gives you information to compute everything on the right hand side:  The line is l=225 miles long.

5  V2 and I2 (the voltage and current phasors on receiving end of the transmission line)  y and z and therefore γ and ZC. And from this, we compute V1 and I1 (the voltage and current phasors on sending end of the transmission line) using eqs. (1), (2).

Note it requires evaluation of the hyperbolic functions. We illustrate details of this. With γl=0.4688∟83.95° =0.0494+j0.466 ZC=387.3∟-6.05° Ω and recalling that el  e l el  e l coshl  ; sinhl  2 2  2coshl  el  e l ; 2sinhl  el  e l

6 Evaluation of 2coshγl: 2coshl  el  el e0.0494 j0.466  e0.0494 j0.466 e0.0494e j0.466  e0.0494e j0.466 e0.0494 (cos0.466  j sin 0.466)  e0.0494 (cos0.466  j sin 0.466) 1.0506(0.8934  j0.4493)  0.9518(0.8934 - j0.4493) 0.9386 + j0.4721 0.8503- j0.4276 1.7889 + j0.0445 1.78951.425 Aside: In evaluating the sin and cos terms above, the arguments (which come from Im(γl)) are radians, so do not plug them into your calculator as if they are in degrees unless you first convert them to degrees. Comment: Another useful relation [2, pg. 205] is that the hyperbolic sines and cosines of the complex argument x=(+j)x may be separated into real and imaginary parts by use of the following:

7 cosh x  cosh(  j )x  coshx cos x  j sinhxsin x sinh l  sinh(  j )x  sinhx cos x  j coshxsin x Therefore: 1.78951.425 coshl   0.89481.425 2 which is about what the text gives at the bottom of page 93.

Two other notes: 1. Eqs. (1) and (2) were derived for a per- phase model and therefore we must use line-neutral voltages. 2. The statement on page 94: “It is convenient to pick ∟V2=0°,…” Remember, we can always pick any one quantity in a circuit to be the reference.

Applying eqs. (1) and (2), the text gets the following result: V1 =89,280∟19.39° I1 =162.42∟14.76°

8 I want to work this problem using a lumped parameter π-equivalent model, where Z =z*l=(0.807∟77.9°)Ω/mile*225miles =181.6∟77.9° Y =y*l=(j5.38*10-6)Mhos/mile*225miles =j.00121 We model half of Y, i.e., Y/2=j0.000605, on the receiving end and on the sending end. So our circuit appears as in Fig. 2.

IZ I1 I2=184.1∟-18.195° Z=181.6∟77.9° IY1 IY2

Y/2= Y/2= Load V1 j0.000605 j0.000605 V2=76,200∟0°

Fig. 2 Using this lumped parameter π-model, we can make the following calculations: IY2 =V2Y/2=76,200(j.000605)=j46.101 IZ =IY2+I2 =j46.101+184(cos18.195-jsin18.195) =j46.101+174.8 - j57.454 =174.8-j11.353=175.1683∟-3.7161°

9 The impact of line charging on current into the load is to make IZ almost 1.0 pf. V1 =V2+IZZ =76,200+175.1683∟-3.7161°(181.6∟77.9°) =76,200+31,811∟74.1839° =76,200+8670.1+j30,607=90,220∟19.83°

IY1=V1Y/2=90,220∟19.83°(j.000605) =54.5831∟109.83° I1 =IY1+IZ =54.5831∟109.83°+175.1683∟-3.7161° =161.3198∟14.35°

Compare results: V1 I1 Distributed 89,280∟19.39° 162.42∟14.76° parameter model Lumped 90,220∟19.83° 161.3198∟14.35° parameter π-model Lumped parameter π-model is not bad!

10 Would it be better or worse if the line was only 50 miles long?

We are interested to see if we can devise the relation between the two models to see if we can better understand the difference between them. 4.0 Lumped circuit model Our question is the following. Can we create a lumped parameter π-model that gives exactly what the distributed parameter model gives? Our work here relates to that of Section 4.4 in the text.

In other words, we want to find parameters Y’ and Z’ that make the relation between sending and receiving end quantities in Fig. 3 be exactly the same as the distributed parameter model gives.

11 IZ I1 I2 Z’ IY1 IY2

Y’/2 Y’/2 V2 V1

Fig. 3 Note that Y’, Z’ are NOT the same as Y, Z as used in the example of Fig. 2 above.  Y and Z are the lumped parameters obtained from zl and yl, respectively. Our example shows that these values do NOT give the same results as the distributed parameter model gives.  Y’ and Z’ are the lumped parameters that give exactly the same results as the distributed parameter model gives.

We know that the distributed parameter model relates sending and receiving lines according to eqs. (4) and (5), repeated here for convenience:

12 V1  A BV2        (4) I1  C DI2  1 A  coshl C  sinhl ZC (5)

B  ZC sinhl D  coshl So our plan of attack materializes at this point. We must 1. Derive the ABCD-parameters for the network of Fig. 3 in terms of Y’ and Z’. 2. Equate them to the appropriate expressions in eq. (5). 3. Solve for Y’ and Z’. We have actually done the equivalent of Step 1 in the calculations for the Exercise 1 lumped parameter model done above (except we used Z there and here we use Z’). We can express this calculation here:

13 Voltage relation: V1 =V2+drop across Z’. =V2+IZZ’ =V2+(I2+IY2)Z’ =V2+(I2+V2Y’/2)Z’ =(1+Y’Z’/2)V2+Z’I2 Current relation: I1 =IY1+IZ =IY1+IY2+I2 =V1Y’/2+V2Y’/2+I2 Use voltage relation to eliminate V1, to get: I1 =[(1+Y’Z’/2)V2+Z’I2]Y’/2+V2Y’/2+I2 2 =(Y’/2+Y’ Z’/4+Y’/2)V2+(1+Z’Y’/2)I2 =Y’(1+Y’Z’/4)V2+(1+Z’Y’/2)I2

From the voltage and current relations, we infer the following expressions for A,B,C,D: A=(1+Y’Z’/2) C=Y’(1+Y’Z’/4) B=Z’ D=1+Z’Y’/2

Equating these expressions with eqs. (5) results in:

14 A  coshl  1  Y' Z'/ 2

B  ZC sinhl  Z' 1 (10) C  sinhl  Y'(1  Y' Z'/ 4) ZC D  coshl  1  Z'Y'/ 2 Solving for Z’ is very easy (second equation). Solving for Y’ is more difficult. The text does both (pg 100), and also provides a bit of manipulation that I will avoid. The result is: sinhl Z' Z sinhl  Z C l 2 l tanh(l / 2) Y' tanh  Y ZC 2 l / 2

Note that the right-hand-side of expressing Z’ and Y’ relate it to Z=zl and Y=yl. In

15 other words, the multipliers of Z and Y in those expressions are “correction factors” that need to be applied to Z and Y in order for the lumped parameter model, when based on zl and yl, to give results equal to the distributed parameter model.

A useful exercise to prove the middle terms equal the right-hand-side terms in the above is to show that Z Z Z Z ZY Z ZY ZY Y Y Z Z Z         C l zyl zlyl ZY ZY ZY Y Y Y Y Y 2 Y  ... ZC l / 2

16 1[] ISU EE 303 Class Notes: “Introduction to Distribution Systems.” 2[] O. Elgerd, “Electric Energy Systems Theory: An Introduction,” 2nd edition, McGraw-Hill, 1982.