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Charecterization of a Long Lossless Line

A lossless transmission line, terminated at an impedance-matched load resistor (box on right). Red color indicates high voltage, and blue indicates low voltage. Black dots represent electrons.

For a lossless line, the line resistance is assumed to be zero. The characteristic impedance then becomes a pure real number and it is often referred to as the surge impedance. The propagation constant becomes a pure imaginary number.

By defining the propagation constant as γ = jβ and replacing l by x eq.(24) & eq.(25)can be written as:

$V=V_{R}cos\beta x+jZ_{C}I_{R}sin\beta x$ (27)

$I=jV_{R}{\frac {sin\beta x}{Z_{C}}}+I_{R}cos\beta x$ (28)

Voltage-current relationship in naturally loaded line.

The term surge impedance loading or SIL is often used to indicate the nominal capacity of the line. The surge impedance is the ratio of voltage and current at any point along an infinitely long line. The term SIL or natural power is a measure of power delivered by a transmission line when terminated by surge impedance and is given by :

$SIL=P_{n}={\frac {{V_{0}}^{2}}{Z_{C}}}$

Where V₀ is the rated voltage of the line.

At SIL, $Z_{C}={\frac {V_{R}}{I_{R}}}$ and hence from equation (27) & (28) we get :

$V=V_{R}e^{\gamma x}=V_{R}e^{-j\beta x}$ (29)

$V=I_{R}e^{\gamma x}=I_{R}e^{-j\beta x}$ (30)

Voltage and Current Charecteristics

Voltage and current profile over a loaded lossless transmission line

Voltage and current profile over the same lossless transmission line, now unloaded. Ferranti effect can be noticed here.

For the analysis, we have to assume that the magnitudes of the voltages at the two ends are the same. The sending and receiving voltages are given by :

$V_{S}=|V_{S}|\angle \delta \quad \&\quad V_{R}=|V_{R}|\angle 0^{o}$

where δ is angle between the sources and is usually called the load angle. As the total length of the line is l, we replace x by l to obtain the sending end voltage from eq.(22) as :

$V_{S}={\frac {|V_{R}|+Z_{C}I_{R}}{2}}e^{j\theta }+{\frac {|V_{R}|-Z_{C}I_{R}}{2}}e^{-j\theta }$

or,

$V_{S}=|V_{R}|cos\theta +jZ_{C}I_{R}sin\theta$ (31)

Solving the above equation we get,

$I_{R}={\frac {|V_{S}|\angle \delta -|V_{R}|cos\theta }{jZ_{C}sin\theta }}$ (32)

Substituting eq.(29) in eq.(24), the voltage equation at a point in the transmission line that is at a distance x from the receiving end is obtained as :

$V=|V_{R}|\cos \beta x+{\frac {|V_{S}|\angle \delta -|V_{R}|cos\theta }{jZ_{C}sin\theta }}(jZ_{C}sin\beta x)$

or,

$V={\frac {|V_{S}|\angle \delta sin\beta x+|V_{R}|sin(\theta -\beta x)}{sin\theta }}$ (33)

In a similar way, the current at that pint can be given by :

$I={\frac {-j}{Z_{C}}}\left[{\frac {|V_{S}|\angle \delta cos\beta x-|V_{R}|cos(\theta -\beta x)}{sin\theta }}\right]$ (34)

Mid point Voltage and Current

Derivation of Mid-point Voltage

The mid point voltage of a transmission line is of significance for the reactive compensation of transmission lines. To obtain an expression of the mid point voltage, it is assumeed that the line is loaded (i.e., the load angle δ is not equal to zero). At the mid point of the line, x = l/2 such that βx = θ/2. Denoting the midpoint voltage by V_M and the line is symmetric, i.e., |V_S| = |V_R| = V. Equation.(33) can be written as :

$V_{M}={\frac {(V\angle \delta +V)sin{\frac {\theta }{2}}}{sin\theta }}$ (34)

Again, $V\angle \delta +V=V(1+cos\delta =jsin\delta )=V{\sqrt {2+2cos\delta }}tan^{-1}\left({\frac {sin\delta }{1+cos\delta }}\right)$

or, $V\angle \delta +V=2Vcos{\frac {\delta }{2}}\angle \left({\frac {\delta }{2}}\right)$

So, from eq.(34), the following expression of the mid point voltage can be deduced:

$V_{M}={\frac {Vcos({\frac {\delta }{2}})}{cos({\frac {\theta }{2}})}}\angle \left({\frac {\delta }{2}}\right)$ (35)

The mid point current is similarly given by :

$I_{M}={\frac {Vsin({\frac {\delta }{2}})}{Z_{C}sin({\frac {\theta }{2}})}}\angle \left({\frac {\delta }{2}}\right)$ (36)

The phase angle of the mid point voltage is half the load angle always. Also the mid point voltage and current are in phase, i.e., the power factor at this point is unity. The variation in the magnitude of voltage with changes in load angle is maximum at the mid point. The voltage at this point decreases with the increase in δ. Also as the power through a lossless line is constant through out its length and the mid point power factor is unity, the mid point current increases with an increase in δ.

Thevenin equivalent of Transmission Line

It is of some interest to find the Thevenin equivalent of the transmission line looking from the mid point. It is needless to say that the Thevenin voltage will be the same as the mid point voltage. To determine the Thevenin impedance we first find the short circuit current at the mid point terminals. This is computed through superposition principle, as the short circuit current will flow from both the sources connected at the two ends. From (2.52) we compute the short circuit current due to source VS (= V∠δ) as :

$I_{SC1}={\frac {V\angle \delta }{jZ_{C}sin({\frac {\theta }{2}})}}$

Similarly the short circuit current due to the source V_R (= V) is :

$I_{SC2}={\frac {V}{jZ_{C}sin({\frac {\theta }{2}})}}$

Thus, we have :

$I_{SC}=I_{SC1}+I_{SC2}={\frac {V\angle \delta }{jZ_{C}sin({\frac {\theta }{2}})}}+{\frac {V}{jZ_{C}sin({\frac {\theta }{2}})}}$ (37)

The Thevenin Impedance is given by :

$Z_{TH}=jX_{TH}={\frac {V_{M}}{I_{SC}}}$

or,

$Z_{TH}=j{\frac {Z_{C}}{2}}tan({\frac {\theta }{2}})$ (38)

Power in Lossless Line

The power flow through a lossless line can be given by the mid point voltage and current equations given in eq.(36) and eq(37).

Since the power factor at this point is unity, real power over the line is given by

$P_{e}=V_{M}I_{M}^{*}={\frac {V^{2}}{Z_{C}sin\theta }}sin\delta$ (39)

Calculation for the Power Flow Equation

If V = V₀, the above expression can be written as :

$P_{e}={\frac {P_{n}}{sin\theta }}sin\delta$ (40)

Where, P_n is the natural power of the line and $P_{n}={\frac {V^{2}}{Z_{C}}}$

For a short transmission line :

$Z_{C}sin\theta \cong {\sqrt {\frac {l}{C}}}(\omega {\sqrt {lC}}\tau =\omega l\tau =X$

Where X is the total reactance of the line. Equation (39) then can be modified to obtain the well known power transfer relation for the short line approximation as :

$P_{e}={\frac {V^{2}}{X}}sin\delta$ (41)

In general it is not necessary for the magnitudes of the sending and receiving end voltages to be same. The power transfer relation given in (40) will not be valid in that case.

To derive a general expression for power transfer, it is assumed that :

$V_{S}=|V_{S}|(cos\delta +jsin\delta )=|V_{R}|cos\theta +jZ_{C}{\frac {P_{R}-jQ_{R}}{V_{R}|}}sin\theta$

Equating real and imaginary parts of the above equation :

$|V_{S}|cos\delta =|V_{R}|cos\theta +{\frac {Z_{C}Q_{R}}{|V_{R}|}}sin\theta$ (42)

and

$|V_{S}|sin\delta ={\frac {Z_{C}P_{R}}{|V_{R}|}}sin\theta$ (43)

Rearranging eq.(43), the power flow equation for a losslees line can be given as :

$P_{e}=P_{S}=P_{R}={\frac {|V_{S}||V_{R}|}{Z_{C}sin\theta }}sin\delta$ (44)

Similarly, rearranging eq.(42) can give us the expression of the reactive power delivered to the receiving end as :

$Q_{R}={\frac {|V_{S}||V_{R}|cos\theta -|V_{R}|^{2}cos\theta }{Z_{C}sin\theta }}$ (45)

Real Power and Apparent Power

Again from equation (34) it is obtained that,

$I_{S}={\frac {-j}{Z_{C}}}\left[{\frac {|V_{S}|cos\theta \angle \delta -|V_{R}|}{sin\theta }}\right]$ (46)

Real power flow and reactive power consumed by the same lossless transmission line

So,apparent power at the sending end is given by :

$P_{S}+jQ_{S}=V_{S}I_{S}^{*}=|V_{S}|\angle \delta {\frac {j}{Z_{C}}}\left[{\frac {|V_{S}|cos\theta \angle {-\delta }-|V_{R}|}{sin\theta }}\right]$

or,

$P_{S}+jQ_{S}=j{\frac {|V_{S}|^{2}cos\theta -|V_{S}||V_{R}|(cos\delta -jsin\delta )}{Z_{C}sin\theta }}$ (47)

Equating the imaginary parts of equation(47), the generation of reactive power at the source end can be written as :

$Q_{S}={\frac {|V_{S}|^{2}cos\theta -|V_{S}||V_{R}|cos\delta }{Z_{C}sin\theta }}$ (48)

SO, from eq. (45) and eq. (48), the reactive power that is absorbed by the line can be given by :

$Q_{L}=Q_{S}-Q_{R}$

Or,

$Q_{L}={\frac {(|V_{S}|^{2}+|V_{R}|^{2})cos\theta -2|V_{S}||V_{R}|cos\delta }{Z_{C}sin\theta }}$ (49)

Now, if the magnitude of the voltage at the two ends is equal, i.e. $|V_{S}|=|V_{R}|=V$. The reactive powers at the two ends will have the same magnitude but with opposite polarity (i.e. $Q_{S}=-Q_{R}$). Under this scenerio, the net reactive power absorbed by the line becomes two times the sending end reactive power (i.e. $Q_{L}=2Q_{S}$).

But, as cos θ ≈ 1 for small values of θ, cos θ becomes 1 for short transmission line.

So, the reactive powers for a short transmission line can be written as :

$Q_{S}={\frac {V^{2}cod\theta -V^{2}cos\delta }{Z_{C}sin\theta }}\equiv {\frac {V^{2}}{X}}(1-cos\delta )=-Q_{R}$ (50)

And the reactive power absorbed by a short transmission line :

$Q_{L}={\frac {2V^{2}}{X}}(1-cos\delta )$ (51)