LESSON 3 Transmission Revised)

Lesson 3

Prepared by: Mr. John Mateo

Three characteristics, or parameters: its resistance, inductance, and capacitance.


The dc resistance R of a conductor of length l and cross-sectional area A is R = pl/A (in ohms) where p is the resistivity of the material of the conductor in ohm-meters The temperature dependence of resistance is quantified by the relation R2 = R1[1 + α(T2 – T1)] Prepared by: Mr. John Mateo

Resistivities and Temperature Coefficients of Resistance


The inductance per conductor of a two-wire, singlephase transmission line is given by (in henrys per meters) where µ0 = 4p × 10-7 H/m (the permeability of free space), D is the distance between the centers of the conductors, and r is the radius of the conductors. The total, or loop, inductance is then

where r' = re-1/4 is known as the geometric mean radius (GMR) of the conductor.


The per-phase (or line-to-neutral) inductance of a three-phase transmission line with equilaterally spaced conductors is where r is the conductor radius and D is the spacing between conductors

Transposition is the interchanging of the positions of the conductors at regular intervals along the line. Due to unequal inductances in the three phases, leading to unequal voltage dropst he cause by the usual nonsymmetrical spacing


The average per-phase inductance for a transposed line is still given by D is equivalent to De, obtained from De = (DabDbcDca) 1/3 where the distances Dab, Dbc, and Dca are shown in Figure.


The shunt capacitance per unit length of a single-phase, two-wire transmission line is given by (in farads per meter) is the permittivity of free space

For a three-phase line with equilaterally spaced conductors, the per-phase (or line-to-neutral) capacitance is



A transmission-line cable consists of 19 strands of identical copper conductors, each 1.5mm in diameter. The length of the cable is 2km but, because of the twist of the strands, the actual length of each conductor is increased by 5 percent. What is the resistance of the cable? Take the resistivity of copper to be 1.72 × 10-8 Ω m. Prepared by: Mr. John Mateo

Allowing for twist, we find that l = (1.05)(2000) = 2100m. The cross-sectional area of all 19 strands is 19(╥/4)(1.5 × 10-3 )2 = 33.576 × 10-6 m2 . R = rl/A = (1.72 × 10-8 × 2100)/(33.576 × 10-6) = 1.076 Ω


The per-phase line loss in a 40-km long transmission line is not to exceed 60k Ω while it is delivering 100A per phase. If the resistivity of the conductor material is 1.72 × 10-8 Ω.m, determine the required conductor diameter.


The line loss is to be, at most, I2 R = (100)2 R = 60 × 103 R = 6. substitute A = ╥ D2 /4 to R = pl/A yields ╥ D2 /4 = (1.72 × 10-8 )(40 × 103 )/6 D = 1.208cm Prepared by: Mr. John Mateo

A sample of copper wire has a resistance of 50 Ω at 10°C. What must be the maximum operating temperature of the wire if the resistance is to increase by at most 10 percent? Take the tenperature coefficient at 10°C to be α = 0.00409°C-1


Here we have R 1 = 50 Ω and R 2 = 50 + 0.1 × 50 = 55 Ω. Also, T 1 = 10°C, and we require T 2. 55 = 50[1 + 0.00409(T 2 – 10)] or T 2 = 34.45°C


Determine the resistance of a 10km long solid cylindrical aluminum conductor with a diameter of 250 mils, at (a) 20°C and (b) 120°C.


To find the cross-sectional area of the conductor, we note that 250 mils = 0.25 in = 0.35 cm so A = (π /4)(0.635) 2= 0.317cm2 Also, from Table 1, r = 2.83µ Ω.cm and a = 0.0039°C-1 at 20°C (a) At 20°C, ( R20 = rl/A = 2.83 × 10-8 × (10 × 103) / (0.317 × 10-4) = 8.93 Ω (a) At 120°, R120 = R20[1 + a(120 – 20)] = 8.93(1 + 0.0039 × 100) = 12.41 Ω


A single-circuit, three-phase, 60Hz transmission line consists of three conductors arranged as shown in the Figure. If the conductors are the same as that in Question no 5, find the inductive reactance of the line per kilometer per phase


De = (5 × 5 × 8) 1/3 = 5.848m r = ½ × 0.635 × 10-2m, so that De/r = (5.848 × 2 × 102)/0.635 = 1841.9 and In(De/r) = 7.52. Hence, we have, for each kilometer of length, L = 2(1/4 + 7.52) × 10-7 × 103= 1.554 mH/km The inductive reactance per kilometer is then XL = ω L = 377 × 1.554 × 10-3= 0.5858Ω


Discuss the effect of TL parameters on the operation and performance of transmission lines. Evaluate the losses, efficiency, and voltage regulation of transmission lines Determine the consequence of such performance characteristics on the operation of a power system.

Short line, - considered valid for lines up to 80km long - the shunt effects (conductance and capacitance) are neglected - represented by a lumped resistance in series with a lumped inductance

Medium line - generally range from 80 to 240km in length - the shunt capacitances are lumped at a few predetermined locations along the line; Long lines - longer than 240km - have uniformly distributed

The short transmission line is represented by the lumped parameters R and L is the inductance (per phase) of the entire line. The line is shown to have two ends: the sending end (designated by the subscript S) at the generator, and the receiving end (designated R) at the load

Quantities of significance here are the voltage regulation and efficiency of transmission. These quantities are defined as follows for lines of all lengths: Percent voltage regulations = = [|VR(no load)| - |VR(load)|]/|VR(load)| × 100 Efficiency of transmission = Power at receiving end / power at sending end = PR/PS where VR is the receiving-end voltage.

In a medium-length transmission line the shunt effect of the transmission line due to the line capacitance is not negligible Two representations for a medium-length transmission line a. the nominal-II circuit and b. the nominal-T circuit

The parameters of a long line are considered to be distributed over the entire length of the line. One phase (with return through neutral) of a long line, of length L, is shown.

The voltage V at any point along this line is given by

where γ = √yz, γ is known as the propagation constant y is the shunt. admittance per unit length of the line, and z is the series impedance per unit length,

The voltage V at any point along this line

The current I at any point along the line is given by

Zc = √ z/y is called the characteristic impedance of the line

A transmission line may be viewed as a four-terminal network.

the terminal voltages and currents are related by Vs = A VR + B IR Is = C VR + D IR

where the constants A, B, C, and D are called the generalized circuit constants or ABCD constants and are, in general, complex. By reciprocity, they are related to each other as follows: AD – BC = 1

On a long transmission line, the voltage V and current I everywhere along the line satisfy a relation called the wave equation. For a lossless transmission line, such that z and y are purely reactive, the wave equation may be written as

the superscripts + and – denotes, respectively, waves traveling in the +x and –x directions along the transmission line

A wave such as V+(t – x/u) that is traveling in the positive x direction is called a forwardtraveling wave,

and one that is moving in the negative x direction is a backward-traveling wave

for a lossy line Zc = √z/y,

For a lossless line Zc = √L/C = Rc V+ / I+ = √L/C and V- / I- = √L/C

A transmission line of total length L that terminates in a resistance RL and is driven by a pulse voltage source having an open-circuit voltage waveform Vs(t) as shown and an internal resistance RS. To determine the terminal voltages V(0,t) and V(L,t) and terminal currents I(0,t) and I(L,t) as functions of time we consider the portion of the line at the load). At x = L, we must have V(L,t) = RL I(L,t)

If RL = RC; in that case there is no backward-traveling wave and the line is said to be matched at the load. But the discontinuity in the produced by the load resistor then results in a wave being reflected in the form of a backward-traveling wave.

A voltage reflection coefficient at the load is the ratio of the amplitudes of the backward- and forwardtraveling voltage waves at x = L;

a current reflection coefficient at the load may be defined as

The current reflection coefficient is thus the negative of the voltage reflection coefficient

A 10km long, single-phase short transmission line has 0.5<60°Ω /km impedance. The line supplies a 316.8kW load at 0.8 power factor lagging. What is the voltage regulation if the receiving-end voltage is 3.3kV?

Essential Formula cos θ = power factor (p.f.) Ohm’s law : V = I x Z Percent Regulation = (Vno load –Vload )/ Vno load x 100

cos -1 0.8 = 36.87° and Z = (0.5∠60°)(10) = 5 ∠ 60°Ω I = [(316.8 × 103)/(3.3 × 103 × 0.8)]<-36.87° = 120 ∠ -36.87°A IZ = (5 ∠ 60°)(120 ∠ -36.87°) = (551.77 + j235.69)V VS = (3300 + j0) + (551.77 + j235.69) = (3851.77 + j235.69)V |VS| = 3858.97V

Percent voltage regulation = [(3858.97 – 3300)/3300] × 100 = 16.94 percent

The per-phase impedance of a short transmission line is (0.3 + j0.4) Ω. The sending-end line-to line voltage is 3300V, and the load at the receiving end is 300 kilowatts per phase at 0.8 power factor lagging. Calculate (a) the receiving-end voltage and (b) the line current

Essential Formulas per-phase voltage : Vs= V/ √ 3 Ohm’s law : V = I x R ; P = I x V Percent Regulation = (Vno load –Vload )/ Vno load x 100

Vs2 = (VRcosΦR + RI)2 + (VR sinΦR + XI)2

(a) On a per-phase basis, Vs= 3300/ √ 3 = 1905.25V and I = (300 × 103)/(0.8) VR = (3.75 × 105)/ VR A Vs2 = (VRcosΦR + RI)2 + (VR sinΦR + XI)2 1905.252 = [0.8 VR + (0.3 × 3.75 × 105)/ VR] 2 + [0.6 VR × (0.4 × 3.75 × 105)/ VR ] 2

VR = 1805V (b) we have I = (3.75 × 105)/1905 = 207.75A

What is the maximum power that can be transmitted over a three-phase short transmission line having a per-phase impedance of (0.3 + j0.4) Ω if the receivingend voltage is 6351 volts per phase and the voltage regulation is not to exceed 5 percent?

Essential Formulas: impedance magnitude: Z 2= R2 + X2 per-phase voltage : Vs= V/ √ 3 Ohm’s law : V = I x R ; P = I x V VS = VR + I(R + jX) I = IR + ICR

On a per-phase basis, VR = 6351V VS = (1 + 0.05)(6351) = 6668.6V Z = √[(0.3) 2 + (0.4) 2] = 0.5 Ω Pmax = (6351/0.5) 2(0.5 × 6668.6 / 6351 0.3) = 36.3MW/phase and the maximum total power that can be transmitted is 3× 36.3 = 108.9MW

The per-phase parameters for a 60Hz, 200km long transmission line are R = 2.07Ω, L = 310.8mH, and C = 1.4774µF. the line supplies a 100MW, wye-connected load at 215kV (line-t0-line) and 0.9 power factor lagging. Calculate the sending-end voltage, using the nominal-II circuit representation

Essential Formulas: per-phase voltage : Vs= V/√ 3 Ohm’s law : V = I x R ; P = I x V VS = VR + I(R + jX) I = IR + ICR

To use the nominal-II circuit, we first express VR and IR per phase as follows: VR = (215 × 103)/√3 = 124.13kV IR = (100 × 106)/(√3 × 215 × 103 × 0.9) = 298.37<-25.8° I CR = V R /X Rc/2 = (124.12 × 103<0°)/[1/(377 × 0.5 × 1.4774 × 10-6) <90°] = 34.57<90°A

I = I R + I CR = 298.37<-25.8° + 34.57<90° = 285 ∠ -29.5°A R + jXL = 2.07 + j377 × 0.3108 » 117.19 ∠ 88.98° 10-6 I(R + jXL) = 285<-19.5° × 117.19 ∠ 88.98° = 33.4 ∠ 69.48°kV VS = VR + I(R + jXL) = 124.13<0° + 33.4<69.48° = 139.39 ∠ 12.97°kV/phase

Determine the ABCD constants for the nominal-T circuit of a transmission line for which R = 10 Ω, X = 20 Ω, and Y = 400µS for each phase.

A = D = (1 + ½YZ) = 1 + j[(4 × 10 -4)/2](10 + j20) = 0.996<0.115° B = Z(1 + ¼YZ) = (10 + j20)[1 + ¼ (j4 × 10 -4)(10 + j20)] = 22.25<63.45°W C = Y = j4 × 10 -4 = 4 × 10 -4<90°S

A 138-kV three phase short transmission line has a per-phase impedance of (2 +j4) ohms. If the line supplies a 25 MW load at 0.8 power factor lagging, calculate (a) the efficiency of transmission and (b) the sending end voltage and power factor.

ANSWER: (a) 98.78 percent (b) 139.5 kV, 0.99

DIRECT & TRANSFER SURGE CALCULATION

Overhead transmission lines and cables under transient conditions can be represented by seriesconnected distributed inductances and shuntconnected distributed capacitances. They can be considered to have a surge impedance, which at any locations relates the surge voltage and surge current by the equation: Surge voltage = Surge current x Surge impedance

The surge impedance of an overhead transmission line is of the order of 500 OHMs, and that of a cable of the order of 50 OHMs. With power transformer, the surge impedance varies from 500 OHM to 800 OHM

BEWLEY LATTICE DIAGRAM

Trace the propagation of a travelling wave in overhead transmission lines or cables.

Distance between P1 and P2 (ft) = Travelling wave velocity (ft per microsecond) x Transit time (microsecond)

where the travelling wave is 550ft per microseconds or 167.6 metre per microsecond

The travelling wave with 1 p.u. magnitude on reaching position P2 is reflected back into the line with a reflection coefficient where and is refracted into further lines beyond position P2 with a refraction coefficient , where are the surge impedances of the line P1 P2 and that of the further line beyond P2.

Example:

A 132 kV cable is to cross a river, one mile wide through a 132 kV overhead line and is to connect to another 132 kV cable at the other side of the river. Draw the Bewley lattice diagram and trace the surge voltage at the other side of the river crossing.

LESSON 3 Transmission Revised)

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