In a short transmission line the shunt conductance and shunt capacitance are neglected and so only the series resistance and inductive reactance are to be considered. The equivalent circuit of a short transmission line is shown in Fig. 5.1 (a) where R and X represent the total resistance and inductive reactance of both the conductors (i.e., go and return) of a single phase transmission line.
Since the effect of shunt capacitance is neglected, the series resistance and reactance are taken as lumped. Thus the equivalent circuit of a short transmission line is a simple ac series circuit.
The equivalent circuit shown in Fig. 5.1 (a), will represent a 3-phase short transmission line if R and X represent the resistance and inductive reactance to neutral, I is the current in one conductor, VR is the receiving-end voltage to neutral, VS is the sending-end voltage to neutral, and the load is that in one phase; the voltage drop found in this way is the drop between a phase and neutral.
ADVERTISEMENTS:
From the equivalent circuit shown in Fig. 5.1 (a) it is clear that,
Receiving-end voltage, VR = VS – I (R + j X) = VS – IZ … (5.4)
The quantity I Z is the voltage drop along the line.
Phasor diagram, taking phasor I as the reference phasor, for an inductive load (lagging power factor load) is shown in Fig. 5.1 (b) where phasor OI, OA AB, BC, AC and OC represent the load current I, receiving-end voltage VR, resistive drop I R in line, reactive drop I X in line, line impedance drop I Z and sending-end voltage VS respectively.
ADVERTISEMENTS:
From phasor diagram shown in Fig. 5.1 (b),
Hence, knowing the conditions existing at one end of the line, those at the other end can easily be determined by means of above relations. Problems may also be solved graphically by drawing the phasor diagram to scale, but owing to very small size of impedance triangle ABC compared with rest of the diagram, accurate results cannot be obtained. So it is advisable to calculate all the quantities trigonometrically, using the graphical method as a rough check,
For determination of an approximate value of sending-end voltage, VS may be taken equal to its component along VR.
From the above expression it is obvious that voltage regulation of the line depends upon the resistance and reactance of the line for supplying a given load at a given voltage—larger the value of line resistance and line reactance, greater the percentage regulation.
where vr and vx are the per unit values of resistance and reactance of the line.
ADVERTISEMENTS:
Power delivered to load = VRI cos ɸR
Line losses = I2 R
Power sent from supply end = VR I cos ɸR + I2 R
Note:
The approximate formula for sending-end voltage VS [Eq. (5.9)] gives fairly accurate results for lagging power factors (i.e., inductive loads) but an appreciable error is caused for leading power factors (i.e. capacitive loads).
Effect of Load Power Factor on Transmission Efficiency and Voltage Regulation of Line:
To deliver given power at a given voltage at consumer end, line current is inversely proportional to the power factor of the load. The current, hence, the power loss in the line goes on increasing as the power factor of the load goes on decreasing. Thus efficiency of the transmission line decreases with the fall in power factor and vice-versa.
Since regulation of the line is given by:
where Z is line impedance and is equal to 
and θ is the line impedance angle and is equal to tan-1 X/R
The regulation is zero when (ɸR – θ) = -π/2 and then cos ɸR = sin θ = X/Z. If (R + j X) is the impedance of the line, generator or transformer, θ is positive since X is positive. It is then necessary that ɸR be negative for zero regulation. If it is the impedance of a synchronous condenser, θ may be negative, so that ɸR may be positive for zero regulation.
The formula given above for zero regulation viz. ɸR = θ – (π/2) is approximate. The exact relation is determined as follows:
The regulation is negative i.e., the receiving-end voltage is greater than the sending-end voltage, if ɸR is negative and numerically greater than the value given in this equation.
From the expression for regulation viz. I R cos ɸR + IX sin ɸR the following conclusions can be drawn:
(i) When the load power factor, cos ɸR is lagging or unity or such leading that I R cos ɸR exceeds I X sin ɸR, the voltage regulation is positive (receiving-end voltage is lesser than the sending-end voltage) and increases with the decrease in power factor for a given load current to be delivered at a given voltage at receiving-end.
(ii) When the load power factor, cos ɸR is leading to the extent that IX sin ɸR > I R cos ɸR the voltage regulation is negative (the receiving-end voltage, is more than sending- end voltage) and decreases with the decrease in power factor for a given load current to be delivered at a given voltage at receiving-end.
Note:
For leading power factor loads the Eq. (5.10) for voltage regulation of a transmission line is slightly modified. This modification is due to the fact that phase angle ɸR is considered positive for lagging power factor loads and negative for leading power factor loads and sin (- ɸR) = – sin ɸR.
The modified expression for voltage regulation for leading power factor loads is given below:
Mixed Sending-and Receiving-End Conditions:
Sometimes the mixed conditions arise for example when the sending-end voltage and the receiving-end power and power factor are given and the sending-end power and power factor and the receiving-end voltage are to be determined. In such circumstances the solution becomes more complicated.
Such problems can be dealt with as follows:
From the above expression we will get two values for VR depending on whether the + ve or – ve sign is used. The solution giving the higher value of VR and consequently higher transmission efficiency is the one required in practice, and therefore the + ve sign is usually taken.
After determining the value of VR, other quantities can be evaluated.
Graphical Solution:
The value of receiving-end voltage VR can also be determined graphically as explained below:
OA is drawn representing the voltage to neutral at sending end of the line and AO’ is drawn making an angle π/2 + ɸR – θ with AO where θ = tan-1 X/R (Fig. 5.2).
BO’ is drawn perpendicular to OA and an arc of circle, OCA is drawn with O’ as centre and O’A as radius.
Now with B, mid-point of OA, as centre and BC as radius an arc cutting the arc OCA at C is drawn where, BC = 
Then OC represents the unknown voltage at the receiving end, VR.
Maximum Power with a Given Regulation:
The maximum power that can be transmitted with a given regulation can be determined as follows:
For a given line, i.e., given values of R and X, and for given receiving-and sending-end voltages, the above Eq. (5.19) gives relation between P and Q. Active power P delivered, varies with the variation in reactive power Q delivered i.e., the above equation gives P as an implicit function of Q. For determining the condition for a maximum value of power P, let us differentiate the above Eq. (5.19) w.r.t Q and put dP/dQ = 0
i.e., for maximum value of P, the value of Q is a constant negative value.
The maximum value of P is determined by substituting this value of Q in Eq. (5.19) which then becomes, taking + ve sign of the square root.
If the reactance of the line can be varied, Pmax will vary, and will have a final maximum when dPmax/dX = 0.
