Q 1. A 3 −

EXACT SOLUTION OF A LONG LINE:

IS

VS

V +dV

I +dI ^{r.ds x.ds}

g.ds c.ds

V

I Ir

Vr

l

ds s

The difference in voltages between he ends of the assumed section of length ds is dV. The difference is caused by the voltage drop in the series impedance of the section:

𝑑𝑉 = 𝐼. 𝑧𝑑𝑠 𝑑𝑉

𝑑𝑠 = 𝐼. 𝑧 … … … … . (𝑖)

The difference dI of the currents between the two ends of the section due to leakage current through the shunt admittance y.ds of the section is given by

𝑑𝐼 = 𝑉. 𝑦𝑑𝑠 𝑑𝐼

𝑑𝑠 = 𝑦. 𝑉 … … … (𝑖𝑖)

Equations (i)and (ii) are solved to find V and I as functions of s. Differentiation of eq(i) w.r.to s gives 𝑑²𝑉

𝑑𝑠² = 𝑧.𝑑𝐼

𝑑𝑠 … … … . (𝑖𝑖𝑖) The value of ^𝑑𝐼

𝑑𝑠 is substituted from eq.(ii) in eq.(iii) to give 𝑑²𝑉

𝑑𝑠² = 𝑧𝑦 . 𝑉 … … … (𝑖𝑣) Now eq.(ii) is differentiated w.r.to s

𝑑²𝐼

𝑑𝑠²= 𝑧.𝑑𝑉

𝑑𝑠 … … … . . (𝑣) Substituting the value of ^𝑑𝑉

𝑑𝑠 in eq.(v) from eq.(i) gives 𝑑²𝐼

𝑑𝑠²= 𝑧𝑦. 𝐼 … … … . . (𝑣𝑖)

Equations (iv) and (vi) are similar in form and therefore their general solutions will also be the same. Putting zy

= 𝛾², we obtain

𝑑²𝑉

𝑑𝑠² = 𝛾². 𝑉 … … … (𝑣𝑖𝑖)

Equation (vii) is a linear differential equation of second order with constant coefficients. The general solution of this equation is of the form

𝑉 = 𝐶₁ 𝑒^𝛾𝑠 + 𝐶₂ 𝑒^−𝛾𝑠 … … … . . (𝑣𝑖𝑖𝑖)

Where 𝐶₁ and 𝐶₂ are arbitrary constants whose values depend upon the boundary conditions, i.e 𝐶₁ and 𝐶₂ are found from known values of V and I at some point of the line.

For finding the value of I, we differentiate eq.(vii) w.r. to s to get 𝑑𝑉

𝑑𝑠 = 𝐶₁ 𝛾 𝑒^𝛾𝑠 − 𝐶₂ 𝛾 𝑒^−𝛾𝑠 … … … … (𝑖𝑥) Putting the value of ^𝑑𝑉

𝑑𝑠 from eq(i) in eq(viii), we get

𝑧. 𝐼 = 𝐶₁ 𝛾 𝑒^𝛾𝑠 − 𝐶₂ 𝛾 𝑒^−𝛾𝑠

𝑧. 𝐼 = 𝛾(𝐶₁ 𝑒^𝛾𝑠 − 𝐶₂ 𝑒^−𝛾𝑠 ) 𝑧. 𝐼 = √𝑧𝑦 (𝐶1 𝑒^𝛾𝑠 − 𝐶₂ 𝑒^−𝛾𝑠 )

𝐼 = √𝑧𝑦

𝑧 (𝐶₁ 𝑒^𝛾𝑠 − 𝐶₂ 𝑒^−𝛾𝑠 ) 𝐼 = √𝑦

𝑧 (𝐶₁ 𝑒^𝛾𝑠 − 𝐶₂ 𝑒^−𝛾𝑠 ) … … … . (x)

The values of V and I at the receiving end where s=0 are 𝑉𝑟 and 𝐼𝑟 respectively. Substituting these values in wqs(viii) and (x), we obtain

𝑉_𝑟 = 𝐶₁+ 𝐶₂ … . (𝑥𝑖) 𝐼_𝑟 = √𝑦

𝑧 (𝐶₁+ 𝐶₂) 1

√𝑦 𝑧

𝐼_𝑟 = (𝐶₁+ 𝐶₂)

√𝑧

𝑦 . 𝐼_𝑟 = (𝐶₁− 𝐶₂) Putting 𝑍_𝑐 ≜ √^𝑧_𝑦 , we get

𝑍_𝑐 . 𝐼_𝑟 = (𝐶₁− 𝐶₂) … … … . . (𝑥𝑖𝑖) Where 𝑍𝑐 is called characteristic impedance.

After solving eqns. (xi) and (xii), we get, 𝐶₁= ¹

2 [𝑉_𝑟 + 𝑍_𝑐𝐼_𝑟] and 𝐶₂= ¹

2 [𝑉_𝑟− 𝑍_𝑐𝐼_𝑟]

Putting the values of 𝐶₁ and 𝐶₂ in eqns(viii) and (x), we get 𝑉 = 1

2 (𝑉_𝑟+ 𝑍_𝑐𝐼_𝑟) 𝑒^𝛾𝑠 +1

2 (𝑉_𝑟− 𝑍_𝑐𝐼_𝑟) 𝑒^−𝛾𝑠 𝑉 = 1

2 𝑉_𝑟 𝑒^𝛾𝑠 + 1

2 𝑍_𝑐𝐼_𝑟 𝑒^𝛾𝑠 +1

2 𝑉_𝑟 𝑒^−𝛾𝑠 − 1

2 𝑍_𝑐𝐼_𝑟 𝑒^−𝛾𝑠 𝑉 = 𝑉_𝑟 (𝑒^𝛾𝑠 + 𝑒^−𝛾𝑠

2 ) + 𝑍_𝑐𝐼_𝑟 (𝑒^𝛾𝑠 − 𝑒^−𝛾𝑠 2 ) 𝑉 = 𝑉_𝑟cosh 𝛾𝑠 + 𝑍_𝑐𝐼_𝑟 sinh 𝛾𝑠 … … … . . (𝑥𝑖𝑖𝑖) 𝐼 = √𝑦

𝑧 [1

2{𝑉_𝑟 + 𝑍_𝑐𝐼_𝑟 }𝑒^𝛾𝑠 −1

2{𝑉_𝑟− 𝑍_𝑐𝐼_𝑟}𝑒^−𝛾𝑠 ] Putting √^𝑦

𝑧 = ¹

𝑍_𝑐 ,we get

𝐼 = 1 𝑍_𝑐 [1

2{𝑉_𝑟 + 𝑍_𝑐𝐼_𝑟 }𝑒^𝛾𝑠 −1

2{𝑉_𝑟− 𝑍_𝑐𝐼_𝑟}𝑒^−𝛾𝑠 ] 𝐼 = 𝐼_𝑟 (𝑒^𝛾𝑠 + 𝑒^−𝛾𝑠

2 ) + 𝑉_𝑟

𝑍_𝑐 (𝑒^𝛾𝑠 − 𝑒^−𝛾𝑠

2 )

𝐼 = 𝐼_𝑟 cosh 𝛾𝑠 + 𝑉_𝑟

𝑍_𝑐 sinh 𝛾𝑠 … … … . . (𝑥𝑖𝑣)

The values of voltage and current at the sending can be found by substituting s=S in eqns.(xiii) and (xiv), we get 𝑉_𝑠= 𝑉_𝑟cosh 𝛾𝑙 + 𝑍_𝑐𝐼_𝑟 sinh 𝛾𝑙 … … … . . (𝑥𝑣)

𝐼_𝑠 = 𝑉_𝑟

𝑍_𝑐 sinh 𝛾𝑙 + 𝐼_𝑟 cosh 𝛾𝑙 … … … . (𝑥𝑣) Comparing with the standard equations, we obtain

𝐴 ≜ cosh 𝛾𝑙 𝐵 ≜ 𝑍_𝑐 sinh 𝛾𝑙 𝐶 ≜ 1

𝑍_𝑐 sinh 𝛾𝑙 𝐷 ≜ cosh 𝛾𝑙

: Numericals for practice

Q 1. A 3 − 𝑝ℎ𝑎𝑠𝑒, 50 𝐻𝑧 transmission line 300 km long has a total series impedance of (40 + 𝑗124)Ω and a total shunt admittance of 𝑗10

𝑚ℎ𝑜. The receiving end load is 50 𝑀𝑊 𝑎𝑡 220 𝐾𝑉 𝑤𝑖𝑡ℎ 0.8 𝑙𝑎𝑔𝑔𝑖𝑛𝑔 power factor. Find the sending end voltage, current, power , power factor, efficiency and voltage regulation using exact transmission line equation.

Soln.:

𝑧. 𝑙 = 𝑍 = (40 + 𝑗 125) Ω = 131.2 ∠73.2^𝑜 Ω 𝑦. 𝑙 = 𝑌 = 𝑗 10⁻³= 10⁻³∠90^𝑜 𝑚ℎ𝑜

|𝐼_𝑅𝑃| = 𝑃_3−𝜙

√3 𝑉𝐿cos 𝜙= 50 × 10⁶

√3 × 220 × 10³× 0.8= 164 𝐴 𝐼_𝑅𝑃= 164∠−36.86^𝑜 A

𝑉_𝑅𝑃=^220×103

√3 = 127.017 × 10³∠0^𝑜 𝑣𝑜𝑙𝑡𝑠 𝑉_𝑆𝑃 = 𝐴𝑉_𝑅𝑃+ 𝐵𝐼_𝑅𝑃 volts

𝐼_𝑆𝑃 = 𝐶. 𝑉_𝑅𝑃+ 𝐷. 𝐼_𝑅𝑃 Amp.

𝐴 = D = cosh 𝛾𝑙

𝛾𝑙 = 𝛼𝑙 + 𝑗 𝛽𝑙 = √𝑍𝑌 = √131.2∠73.2 × 10⁻³∠90 = √131.2 × 10⁻³∠163.2 𝛾𝑙 = 0.362∠81.2 = 0.055 + 𝑗0.3577

𝛾𝑙 = (𝛼𝑙 + 𝑗 𝛽𝑙)

𝛼𝑙 = 0.055 𝑁𝑒𝑝𝑒𝑟, 𝛽𝑙 = 0.3577 𝑟𝑎𝑑. = 20.5 𝑑𝑒𝑔𝑟𝑒𝑒

cosh 𝛾𝑙 = cosh(𝛼𝑙 + 𝑗 𝛽𝑙) = cosh 𝛼𝑙 cos 𝛽𝑙 + 𝑗 sinh 𝛼𝑙 sin 𝛽𝑙 = cosh(0.055) cos(20.5) + 𝑗 sinh(0.055) sin(20.5) cosh 𝛾𝑙 = 1.0 × 0.9366 + 𝑗 0.055 × 0.35 = 0.9366 + 𝑗 0.01925 = 0.9367∠1.177^𝑜

sinh 𝛾𝑙 = sinh(𝛼𝑙 + 𝑗 𝛽𝑙) = sinh 𝛼𝑙 cos 𝛽𝑙 + 𝑗 cosh 𝛼𝑙 sin 𝛽𝑙 = sinh(0.055) cos(20.5) + 𝑗 cosh(0.055) sin(20.5) sinh 𝛾𝑙 = 0.055 × 0.9366 + 𝑗 1.0 × 0.35 = 0.0515 + 𝑗 0.35 = 0.3537∠81.63^𝑜

𝑍_𝐶 = √^𝑧.𝑙

𝑦.𝑙= √^𝑍

𝑌= √^131.2∠73.2

10⁻³∠90 = √131200∠ − 16.8 = 362.21∠ − 8.4 Ω 𝐴 = D = cosh 𝛾𝑙 = 0.9367∠1.177^𝑜

𝐵 = 𝑍_𝐶sinh 𝛾𝑙 = 362.21∠ − 8.4 × 0.3537∠81.63 = 128.2∠72.65

𝐶 = 1

𝑍_𝐶 sinh 𝛾𝑙 = 1

362.21∠ − 8.4× 0.3537∠81.63 = 9.77 × 10⁻⁴∠90.4

𝑉_𝑆𝑃 = 𝐴𝑉_𝑅𝑃+ 𝐵𝐼_𝑅𝑃= 0.9367∠1.177 × 127.017 × 10³∠0 + 128.2∠72.65 × 164∠−36.86^𝑜= 136.97∠6.2^𝑜 𝐾𝑉 𝑉_𝑆𝐿= √3 × 𝑉_𝑆𝑃 = √3 × 136.97∠6.2^𝑜 = 237.23∠6.2^𝑜 𝐾𝑉

𝐼_𝑆𝑃 = 𝐶. 𝑉_𝑅𝑃+ 𝐷. 𝐼_𝑅𝑃= 9.77 × 10⁻⁴∠90.4 × 127.017 × 10³∠0^𝑜+0.9367∠1.177^𝑜× 164∠−36.86^𝑜 = 128.6∠15.3 𝐴 𝑠𝑒𝑛𝑑𝑖𝑛𝑔 𝑒𝑛𝑑 𝑝𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 = cos 𝜙_𝑆= cos(15.3 − 6.2) = cos 9.1 = 0.9874 leading

𝑠𝑒𝑛𝑑𝑖𝑛𝑔 𝑒𝑛𝑑 𝑝𝑜𝑤𝑒𝑟, 𝑃_𝑆= √3 × 𝑉_𝑆𝐿× 𝐼_𝑆𝐿× cos 𝜙_𝑆= √3 × 237.23 × 128.6 × 0.9874 = 52.15 𝑀𝑊

% 𝜂 = 𝑃_𝑅

𝑃_𝑆 × 100 = 50

52.15× 100 = 95.87%

% 𝑉. 𝑅 = 𝑉_𝑆𝑃

𝐴 − 𝑉^𝑅𝑃

𝑉_𝑅𝑃 × 100 =

136.97

0.9367 − 127.017

127.017 × 100 = 146.22 − 127.017

127.017 100 = 19.21

127.017100 = 15.12%

Receiving End Power: We know that

𝑆

= 𝑉

𝐼

= 𝑃

+ 𝑗𝑄

………….. (1)

𝐴 = |𝐴|∠𝛼 , 𝐵 = |𝐵|∠𝛽, 𝐶 = |𝐶|∠𝜃, 𝐷 = |𝐷|∠𝛼, 𝑉

= |𝑉

|∠𝛿, 𝑉

= |𝑉

|∠0, 𝐼

= |𝐼

|∠ − 𝜙, 𝐼

= |𝐼

|∠ + 𝜙,

𝑉

= 𝐴 𝑉

+ 𝐵 𝐼

𝐼

= 𝑉

𝐵 − 𝐴 𝑉

𝐵 = |𝑉

|

|𝐵| ∠𝛿 − 𝛽 − |𝐴||𝑉

|

|𝐵| ∠𝛼 − 𝛽

𝐼

=

|𝐵|

∠𝛽 − 𝛿 −

|𝐵|

∠𝛽 − 𝛼 ……… (2)

𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 𝑒𝑞𝑛 (2) 𝑏𝑦 |𝑉

|, 𝑤𝑒 𝑔𝑒𝑡

𝑉

𝐼

= |𝑉

||𝑉

|

|𝐵| ∠𝛽 − 𝛿 − |𝐴||𝑉

|

|𝐵| ∠𝛽 − 𝛼

𝑆

= |𝑉

||𝑉

|

|𝐵| ∠𝛽 − 𝛿 − |𝐴||𝑉

|

|𝐵| ∠𝛽 − 𝛼

𝑃

+ 𝑗𝑄

= |𝑉

||𝑉

|

|𝐵| ∠𝛽 − 𝛿 − |𝐴||𝑉

|

|𝐵| ∠𝛽 − 𝛼

𝑃

=

cos(𝛽 − 𝛿) −

cos(𝛽 − 𝛼) ……… (3)

𝑄

=

sin(𝛽 − 𝛿) −

sin(𝛽 − 𝛼) ………….. (4)

Soln.: This is an example of of Short transmission line.

𝑅 = 0.0195 × 20 = 0.39Ω, 𝐿 = 0.60 × 10

× 20 = 0.012 𝐻, 𝑋

= 2𝜋𝑓𝐿 𝑋

= 2𝜋 × 60 × 0.012 = 4.5216 Ω

𝑍 = 𝑅 + 𝑗 𝑋

= 0.39 + 𝑗4.5216 = 4.5238∠85.07 Ω

𝐴 = 𝐷 = 1, 𝐵 = 𝑍 = 4.5238∠85.07 Ω , 𝐶 = 0 𝐼

= 4500 × 10

10.2 × 10

× 0.8 = 551.47 ∠ − 36.86 𝐴

(𝑎) 𝑉_𝑆𝑃 = 𝐴𝑉_𝑅𝑃+ 𝐵𝐼_𝑅𝑃 = 1 × 10.2 × 10³∠0 +

4.5238∠85.07 × 551.47 ∠ − 36.86

=

𝑃_𝑆𝑃 = 𝑉_𝑆𝑃× 𝐼_𝑆𝑃× cos 𝜙_𝑆

𝐼_𝑆𝑃 = 𝐼_𝑅𝑃= 551.47∠ − 36.86, 𝜙_𝑆= 𝛿 + 𝜙_𝑅 = 8.9 + 36.86 = 45.76 cos 𝜙_𝑆= cos 45.76 = 0.6976 (𝑙𝑎𝑔𝑔𝑖𝑛𝑔)

𝑃_𝑆𝑃 = 12 × 10³× 551.47 × cos 45.76 = 4.6168 × 10⁶ 𝑊

%𝜂 = 𝑃𝑅𝑃

𝑃_𝑆𝑃 × 100 = 4.5 × 10⁶

4.6168 × 10⁶× 100 = 97.47

% 𝑉. 𝑅 = 𝑉𝑆𝑃

𝐴 − 𝑉^𝑅𝑃

𝑉_𝑅𝑃 × 100 = 𝑉_𝑆𝑃− 𝑉_𝑅𝑃

𝑉_𝑅𝑃 × 100 =(12.0 − 10.2) × 10³

10.2 × 10³ × 100 = 17.64%

(b) %𝑉. 𝑅^′ = 0.6 × 0.1764 × 100 = 10.58%

𝑉. 𝑅^′ = ^{𝑉𝑆𝑃′−𝑉𝑅𝑃}

𝑉_𝑅𝑃 = ^{𝑉𝑆𝑃′−10.2×103}

10.2×10³

0.1058 × 10.2 × 10³ = 𝑉_𝑆𝑃^′− 10.2 × 10³ 1.080 × 10³+ 10.2 × 10³= 𝑉𝑆𝑃′

𝑉_𝑆𝑃^′ = 11.28 × 10³∠𝛿^′ volts 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛 (3)

𝑃

= |𝑉

|

|𝑉

|

|𝐵| cos(𝛽 − 𝛿

) − |𝐴||𝑉

|

|𝐵| cos(𝛽 − 𝛼) 𝑃

= 11.28 × 10

× 10.2 × 10

4.5238 cos(85.07 − 𝛿

) − 1 × (10.2 × 10

)

4.5238 cos(85.07 − 0) 𝑃

= 25.4334 × 10

cos(85.07 − 𝛿

) − 23 × 10

× 0.086

𝑃

= 25.4334 × 10

cos(85.07 − 𝛿

) − 1.978 × 10

……… (5)

𝑑𝑢𝑟𝑖𝑛𝑔 𝑐𝑜𝑚𝑝𝑒𝑛𝑠𝑎𝑡𝑖𝑜𝑛, 𝑎𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑠 𝑠𝑎𝑚𝑒 𝑖. 𝑒. 𝑃

= 𝑃

= 4500 × 10

𝑊 4500 × 10

= 25.4334 × 10

cos(85.07 − 𝛿

) − 1.978 × 10

4500 × 10

+ 1.978 × 10

= 25.367 × 10

cos(85.07 − 𝛿

) 25.367 × 10

cos(85.07 − 𝛿

) = 6.478 × 10

cos(85.07 − 𝛿

) = 6.478 × 10

25.4334 × 10

= 0.2547 85.07 − 𝛿

= cos

0.2547 = 75.24

𝛿

= 85.07 − 75.24 = 9.826

𝑓𝑟𝑜𝑚 𝑒𝑞𝑛 (4)

𝑄

= |𝑉

||𝑉

|

|𝐵| sin(𝛽 − 𝛿) − |𝐴||𝑉

|

|𝐵| sin(𝛽 − 𝛼)

𝑄

= 25.4334 × 10

sin(85.07 − 9.85) − 23 × 10

sin(85.07 − 0)

𝑄

= 25.4334 × 10

sin(75.22) − 23 × 10

× sin(85.07)

𝑄

= 25.4334 × 10

× 0.967 − 23 × 10

× 0.9963 = 24.59 × 10

− 22.915 × 10

𝑄

= 1.675 × 10

𝑉𝐴𝑟

𝑄

= 𝑉

𝜔𝐶

𝐶 = 𝑄

𝜔𝑉

= 1.675 × 10

2𝜋 × 60 × (10.2 × 10

)

= 1.675 × 10

3.9222 × 10

= 42.70 𝜇𝐹

𝑃_𝑆𝑃^′ = 𝑉_𝑆𝑃^′× 𝐼_𝑆𝑃^′× cos 𝜙_𝑠 𝐼_𝑆𝑃^′ = 𝐼_𝐶+ 𝐼 ….. (6)

𝐼 = 551.47∠ − 36.86, 𝐼_𝐶 = 𝑗^𝑉𝑅𝑃

𝑋_𝐶 = 𝑗𝑉_𝑅𝑃𝜔𝐶 = 𝑗10.2 × 10³× 2𝜋 × 60 × 42.70 × 10⁻⁶ = 164.2 ∠90 A 𝐼𝑆𝑃′ = 𝐼𝑅𝑃= 𝐼𝐶+ 𝐼 = 164.2 ∠90 + 551.47∠ − 36.86 = 471.64∠ − 20.68 𝐴

𝜙_𝑆^′ =

𝛿

+ 20.68 = 9.826 + 20.68 = 30.5

𝑃𝑆𝑃′ = 11.28 × 10³× 471.64 × 0.8615 = 4.5832 × 10⁶ 𝑊

%𝜂^′= 𝑃_𝑅

𝑃_𝑆𝑃^′× 100 = 4.5 × 10³

4.5832 × 10⁶× 100 = 98.18 %

(c) By shunt compensation, power factor improves from 0.8 to 0.8615 but the power improves negligibly.