In this article we will discuss about:- 1. Introduction to Long Transmission Lines 2. Analysis of Long Transmission Lines (Rigorous Method of Solution of Long Transmission Lines) 3. Surge Impedance 4. Long Transmission Line with Series Impedance 5. Network of a Long Transmission Line 6. Constant Voltage Transmission in Long Transmission Lines and Other Details.
Contents:
- Introduction to Long Transmission Lines
- Analysis of Long Transmission Lines (Rigorous Method of Solution of Long Transmission Lines)
- Surge Impedance of Long Transmission Lines
- Long Transmission Line with Series Impedance
- Network of a Long Transmission Line
- Constant Voltage Transmission in Long Transmission Lines
- Synchronous Phase Modifiers or Compensators in Long Transmission Lines
- Tuned Power Lines
1. Introduction to Long Transmission Lines:
It is well known that the line constants (resistance, inductance, capacitance and conductance) of a transmission line are uniformly distributed over the entire length of the line. So far it has been assumed that the line has lumped constants and the line calculations made with such assumptions gave results with reasonable accuracy. If such an assumption of lumped line constants is applied to long transmission lines (lines having length more than 200 km), serious errors are introduced in the line performance calculations.
ADVERTISEMENTS:
Therefore, the performance calculations of long transmission lines are made with line constants uniformly distributed over the entire length of the line so that the results with fair degree of accuracy are obtained. Rigorous mathematical treatment is required for the solution of such transmission lines. The equivalent circuit of a 3-phase long transmission line is represented schematically by Fig. 6.1.
The whole line is divided into n-sections, each section having line constants 1/nth of those for the whole line.
The following points are worth noting:
ADVERTISEMENTS:
(i) The line constants (resistance, inductive reactance, capacitive susceptance and conductance) are uniformly distributed over the entire length of the line as is actually the case.
(ii) The resistance (R) and inductive reactance (X) are the series elements.
(iii) The capacitive susceptance (B) and leakage conductance (G) are shunt elements. The leakage susceptance is due to the fact the capacitance exists between line and neutral. The leakage conductance (G) takes into account the energy losses occurring through leakage over the insulators, or due to corona effect between the conductors.
Thus shunt admittance,
(iv) The leakage current through the shunt admittance is maximum at the sending end of the transmission line and decreases continuously as we move towards the receiving end of the line and becomes zero at the receiving end.
2. Analysis of Long Transmission Lines (Rigorous Method of Solution of Long Transmission Lines):
The exact equivalent circuit of a transmission line is shown in Fig. 6.2.
Consider an infinitely small length dx of the line at a distance x from the receiving end.
Let r = Resistance per unit length of line
ADVERTISEMENTS:
x = Reactance per unit length of line
b = Susceptance per unit length of line
g = Conductance per unit length of line
V = Voltage per phase at the end of element towards receiving end
V + d V = Voltage per phase at the end of element towards sending end
VR = Voltage per phase at receiving end
VS = Voltage per phase at sending end
IR = Current per phase at receiving end
IS = Current per phase at sending end
I + d I = Current entering the element dx
and I = Current leaving the element dx
Now the series impedance of element dx of the line – z dx
The shunt admittance of element dx of the line = y dx
Obviously the rise in voltage over the element length in the direction of increasing x
dV = I z dx, voltage drop in the element dx
Similarly, the difference of current entering the element and that of leaving the element,
d I = V y d x, the current drawn by this element
Differentiating expression (6.1) w.r.t. x we get,
The solution of above differential equation is,
where A1 and A2 are unknown constants. Differentiating Eq. (6.4) w.r.t. x we get,
and from Eq. (6.1) we have,
Equations (6.4) and (6.5) thus give the expressions for V and I in the form of unknown constants A1 and A2.
The values of A1 and A2 can be determined by applying receiving-end conditions as under:
At receiving end x = 0, V = VR and I = IR
Substituting these values in Eqs. (6.4) and (6.6) we get
Thus the expressions for V and I become,
Expanding and re-arranging Eqs. (6.9) and (6.10) we get,
The sending-end voltage, VS and sending-end current, IS can be obtained by substituting x = l in the above Eqs. (6.11) and (6.12),
where Z is the total impedance of the line and Y is the total admittance of the line.
The expressions for sending-end voltage and sending-end currents are:
Comparing above equations with the general voltage and current equations of the line we have,
Equations (6.13) and (6.14) can be written in matrix form as:
Thus sending-end values can be obtained if VR, IR and line parameters are known. In case VS, IS are known, VR and IR can be easily found by inverting the above Eq. (6.17),
Evaluation of Constants ABCD:
ABCD constants can be evaluated by three methods given below:
1. Convergent Series (Complex Angle) Method:
In this method the hyperbolic sine and cosine are expressed in terms of their power series.
The expansions are:
It is never worthwhile going beyond the seventh power.
We then get,
2. Convergent Series (Real Angle) Method:
In this method the hyperbolic sines and cosines are expanded by the trigonometrical formulae given below, and tables of the trigonometric functions and hyperbolic functions of real numbers are used.
Thus we express √YZ in the form:
3. By Use of Hyperbolic Charts:
Woodruff has prepared a convenient set of charts for evaluating the hyperbolic functions of complex variables, usually required in transmission line calculations. But such charts are usually not available.
In general, method 1 is less laborious to use than method 2; moreover the tables of the sinh and cosh of complex numbers are not easy to use, and are not given for small differences.
3. Surge Impedance of Long Transmission Lines:
The square root of the ratio of line impedance (Z) and shunt admittance (Y) is called the ‘surge impedance’ (Z0) of the line,
where Z = R + jX and Y = G + jB
Characteristic impedance (ZC) has also been defined in the same way. Actually characteristic impedance is a more correct nomenclature. Surge impedance is the characteristic impedance of a loss-free line.
For a heavy copper conductor and well insulated line the resistance (R) and leakage conductance (G) can be taken as zero and therefore, surge impedance, Z0 = √L/C which is a
pure resistance. Its value varies between 400 Ω and 600 Ω in case of overhead transmission lines and between 40 Ω and 60 Ω incase of underground cables.
The surge impedance can also be evaluated by measuring the line impedance at the sending end when:
(i) The line at the receiving end is open-circuited.
(ii) The line at the receiving end is short-circuited.
When the line is open-circuited (IR) is zero so, the sending-end voltage (VS) and sending -end current (IS) are given as:
Similarly, when the line is short-circuited at the receiving end, the receiving-end voltage (VR) reduces to zero and the sending-end voltage (VS) and the sending-end current (IS) are given as:
Multiplying Eqs. (6.22) and (6.23) we have,
Surge Impedance Loading:
This is defined as the load (of unity power factor) that can be delivered by the line of negligible resistance.
where VRL is the receiving-end voltage in kV and Z0 is the surge impedance in ohms.
PR is called the ‘surge impedance loading’ or ‘natural power’ of the line.
The above expression gives a limit of the maximum power that can be delivered by a line and is useful in the design of transmission lines. This can be used for the comparison of loads that can be carried on the transmission lines at different voltages.
The power transmitted through a long transmission line can be increased either by increasing the value of receiving-end line voltage (VRL) or by reducing surge impedance (Z0).
Nowadays the trend is of employing higher and higher voltages for transmission; therefore, this is the most commonly adopted method for increasing the power limit of heavily loaded long transmission lines. But there is a limit beyond which it is neither economical nor practicable to increase the value of receiving-end line voltage.
Since the spacing between the conductors, which depends upon the line voltage employed, cannot be reduced much, so the value of surge impedance (Z0) cannot be varied as such. However some artificial means, such as series capacitors (capacitors in series with the transmission lines) or shunt capacitors (capacitors in parallel with the transmission lines), can be used to reduce the value of surge impedance (Z0).
For a loss free transmission line surge impedance, Z0 = √L/C and propagation constant, g = jω √LC – jβ, where β is the phase shift, which determines the torque angle δ between VS and VR and hence the system stability.
By use of series capacitors surge impedance (Z0) and the phase shift (β) get reduced due to decrease of line inductance (L). This improves the system stability limit also. But it causes difficulties under short-circuit conditions of the system as no satisfactory protection of capacitors has yet been devised. These capacitors are also helpful in reducing line drops and so voltage variations.
By use of shunt capacitors though the surge impedance (Z0) is reduced but the phase shift β is increased due to increase in the value of C. Hence stability conditions become worse specially when synchronous machines are used as loads. So this method is not used in case of long transmission lines specially where stability limits are present.
4. Long Transmission Line with Series Impedance:
i. Long Transmission Line with Series Impedance at the Receiving End:
Line auxiliary constants, as determined before, are changed due to insertion of series impedance at the receiving end of a transmission line. Normally impedance is supposed to consist of resistance and inductance and the current flowing through it is the same as the receiving-end current, IR. The circuit is shown in Fig. 6.3.
Let the auxiliary constants of the transmission line be A, B, C and D and Zse by the value of series impedance of the line at the receiving- end. If VR is the phase voltage at the load end and IR is the load current, then the phase voltage at the end of transmission line will be,
V’R = VR + IR Zse … (6.27)
and VS = A V’R + B IR = A (VR + IR Zse) + B IR = A VR + (A Zse + B) IR … (6.28)
and IS = C V’R + D IR
Substituting the value of V’R from Eq. (6.27) we have,
IS = C (VR + IR Zse) + D IR = CVR + (C Zse + D) IR … (2.29)
Comparing with the general voltage and current expressions of the transmission line we have,
A0 = A; B0 = A ZSe + B; C0 = C and D0 = C Zse + D … (6.30)
ii. Long Transmission Line with Series Impedance at the Sending End:
Figure 6.4. represents a transmission line with a series impedance Zse at the sending end. In this case the current passing through the series impedance Zse is the same as the sending end current IS
If A, B, C and D are the auxiliary constants of the transmission line, VR and VS are the phase voltages at the load and supply ends respectively, IR is the load current and V’s is the phase voltage at the starting end of the transmission line.
then V’S = AVR + B IR … (6.31)
and IS = C VR + D IR … (6.32)
Also VS = V’S + ISZse … (6.33)
Substituting the values of V’S and IS from Eqs. (6.31) and (6.32) in Eqs. (6.33) we have,
VS = A VR + BIR + (C VR + D IR) Zse = (A + C Zse) VR + (B + D Zse) IR … (6.34)
Comparing Eqs. (6.34) and (6.32) with the general expressions for voltage and current we have,
A0 = A + C Zse; B0 = B + D Zse; C0 = C and D0 = D = A … (6.35)
iii. Long Transmission Line with Series Impedances at Both Ends:
In this case there are two series impedances, one at the receiving end and the other at the sending end, as shown in Fig. 6.5.
Now we have,
V’R = VR + IRZseR … (6.36)
and V’S = A V’R + BIR = A (VR + IR ZseR) + BIR = AVR + (B + A ZseR) IR … (6.37)
IS = CV’R + DIR = C (VR + IR ZseR) + DIR = CVR + (D + C ZseR) IR … (6.38)
and VS = V’S + IS ZseS
= AVR + (B + A ZseR) IR + {C VR + (D + C ZseR) IR} ZseS
= (A + C ZseS) VR + {B + A ZseR + (D + C ZseR) ZseS} IR … (6.39)
Comparing Eqs. (6.39) and (6.38) with the general expressions for voltage and current of the transmission line we have,
A0 = A + C ZseS
B0 = B + A ZseR + (D + C ZseR) ZseS … [6.40 (a)]
= B + A ZseR + (A + C ZseR) ZseS ∵ A = D
= B + A (ZseR + ZseR) + C ZseR ZseS … [6.40 (b)]
C0 = C … [6.40 (c)]
D0 = D + C ZseR … [6.40 (d)]
iv. Long Transmission Line with Transformers at Both Ends:
The transmission line may have a step-up transformer at the sending end and step-down transformer at the receiving end. The transformer has series impedance ZT and shunt admittance YT. The shunt admittance of transformer is usually so high that in most of the cases it may be taken as infinite without much loss of accuracy in which case transformers are just equivalent to series impedances and the equivalent circuit is, as shown in Fig. 6.5.
In case the shunt admittance of transformers is to be taken into account, the equivalent circuit will be as shown in Fig. 6.6.
In this case the sending-end voltage and current for the transmission line are V’S and I’S respectively.
Also the receiving-end voltage and current for the transmission line are V’R and I’R respectively.
ITS and ITR are the magnetising currents taken by the transformers at the sending end and receiving end of the transmission line respectively.
Now applying the transmission line treatment to the line we have,
V’S = A V’R + BI’R … (6.41)
and ITR = YTR VR
I’R = IR + ITR = IR + YTR VR … (6.42)
∴ V’R = VR + I’R ZTR = VR + (IR + ITR) ZTR = VR + (IR + YTR VR) ZTR … (6.43)
Substituting the value of V’R from Eq. (6.43) and value of I’R from Eq. (6.42) in Eq. (6.41) we get,
V’S = A {VR + (IR + YTR VR) ZTR} + B (IR + YTR VR)
Considering the sending-end current for the line,
I’S = C V’R + D I’R = C {VR + (IR + YTR VR) ZTR} + D (IR + YTR VR)
Also VS = V’S + I’S ZTS
= A {VR + (IR + YTR VR) ZTR} + B (IR + YTR VR)
+ C ZTS {VR + (IR + YTR VR) ZTR} + D ZTS (IR + YTR VR)
= [A (1 + YTR ZTR + ZTS YTR) + B YTR + C ZTS (1 + YTR ZTR)] VR
+ [A ZTR + B + C ZTS ZTR + A ZTS] IR … (6.44)
∵ A = D
and IS = I’S + ITS
= C {VR + (IR + YTR VR) ZTR} + D (IR + YTR VR) + YTS VS
= C {VR + (IR + YTR VR) ZTR} + A (IR + YTR VR)
+ [{A (1 + YTR ZTR + YTR ZTS) + B YTR + C ZTS (1 + YTR ZTR)} VR
+ {A ZTR + B + C ZTS ZTR + A ZTS} IR] YTS
= [C + C YTR ZTR + A YTR + A YTS + A YTR ZTR YTS + A YTR ZTS YTS
+ B YTR YTS + C ZTS YTS + C ZTS YTR ZTR YTS] VR
+ IR [C ZTR + A + A ZTR YTS + B YTS + C ZTS ZTR YTS + A ZTS YTS]
= [C (1 + YTR ZTR + ZTS YTS + ZTS ZTR YTR YTS) + B YTR YTS
+ A (YTR + YTS + YTR YTS ZTR + YTR YTS ZTS] VR
+ [A (1 + YTS ZTR + ZTS YTS) + B YTS + C ZTR (1 + ZTS YTS)] IR … (6.45)
Comparing Eqs. (6.44) and (6.45) with the general expressions for voltage and current of the transmission line we have,
For identical transformers impedances ZTR and ZTS will be equal and also shunt admittances YTR and YTS will be equal.
Let ZTR = ZTS = ZT (say) and YTR = YTS = YT (say)
Substituting the above values of ZTR ZTS, YTR and YTS in Eq. (6.46) we get,
5. Network of a Long Transmission Line:
i. Equivalent T-Network of a Long Transmission Line:
The equivalent T-network of a long transmission line is determined to represent the line accurately by assuming suitable values of lumped constants.
Let the transmission line having auxiliary constants A, B, C and D be represented by a T-network shown in Fig. 6.7.
Now we have,
Comparing Eqs. (6.49) and (6.48) with the general expressions for voltage and current of the transmission line we have,
A = D = 1 + ½ YZ … (6.50)
B = Z + ¼ YZ2 … (6.51)
C = Y … (6.52)
From Eqs. (6.50) and (6.52) we have,
Y = C … (6.53)
and Z = 2(A – 1)/C … (6.54)
ii. Equivalent π–Network of a Long Transmission Line:
Sometimes the equivalent π-network of a long transmission line is required to be determined to represent the line accurately by assuming suitable values of lumped constants.
Let the transmission line having auxiliary constants A, B, C and D be represented by a -network shown in Fig. 6.8.
The sending-end voltage and current are given as:
Comparing Eqs. (6.55) and (6.56) with the general expressions for voltage and current of the transmissions line we have,
A = D = 1 + ½ YZ … (6.57)
B = Z … (6.58)
C = Y + ¼ Y2Z … (6.59)
From Eqs. (6.50) and (6.52) we have,
Z = B … (6.60)
and Y = 2(A – 1)/B … (6.61)
6. Constant Voltage Transmission in Long Transmission Lines:
For constant voltage transmission, specially designed synchronous motors, called the synchronous phase modifiers, are installed at the receiving end, which maintain the voltage drop along the line constant. With the change in load, power factor of the system is changed by the synchronous motors and thus voltage drop along the line remains constant.
The merits and demerits of the system are given below:
Merits:
(i) Possibility of carrying increased power for a given conductor size in case of long distance heavy power transmission.
(ii) Improvement of power factor at times of moderate and heavy loads.
(iii) Possibility of better protection for the line due to possible use of higher terminal reactances.
(iv) Availability of steady voltage at all loads at the line terminals.
(v) Improvement in system stability due to inertia effect of synchronous phase modifier and reduction in effect of sudden changes in load.
Demerits:
(i) Increase of short-circuit current of the system and, therefore, increase in the circuit breaker ratings.
(ii) Increase risk of interruption of supply due to falling of synchronous motors, called the synchronous phase modifiers, out of synchronism.
(iii) Lower reserve of lines in case of line trouble.
7. Synchronous Phase Modifiers or Compensators in Long Transmission Lines:
It is well known that a synchronous motor can be made to take either a lagging or leading current from the line by altering its excitation. Idle-running synchronous motors were first employed in connection with power plants to correct for low power factor of the load, and thus reduce the current and power losses in the feeders and generators. The synchronous machines used for power factor improvement are usually referred to as ‘synchronous condensers’ as they are always required to take a leading current.
However, an idle-running synchronous motor can be used for voltage regulating purposes by connecting it in parallel with the load at the receiving end of a line. Since the machines used for voltage regulation are required to run part of the time with leading current and part of the time with lagging current depending upon the load conditions, therefore, it is more appropriate to refer to these machines as ‘synchronous phase modifiers’. They are usually of salient pole design with 6 or 8 poles with ratings up to 60 MVA, 11 kV and connected to high voltage system through transformers.
Synchronous phase modifiers differ from the ordinary synchronous motors in as much as they are built for the highest economical speeds and provided with smaller shafts and bearings and special attention is paid for securing a high overall efficiency.
Standard machines for this purpose are designed to give their full-load output at leading power factor, and can carry about 50% of their rated capacity at lagging power factor. Machines can, however, be designed to operate at full rating on both leading and lagging power factors but they are larger in size, have poor efficiency and are more expensive than standard machines.
A transmission line needs lagging VARs (reactive volt-amperes) at the receiving end during peak load conditions to avoid excessive voltage drop and these lagging VARs are supplied by the synchronous phase modifier when it is over-excited as it absorbs leading VARs when over-excited. During off-peak hours the transmission line needs leading VARs at the receiving end to prevent any voltage rise.
These leading VARs are supplied by the synchronous phase modifier when it is under-excited as it absorbs lagging VARs when under- excited. Thus synchronous phase modifier or compensator is a very flexible source of supply of reactive VARs (leading or lagging as per need) and is well suited for voltage regulation (or compensation).
8. Tuned Power Lines:
The behaviour of a transmission line is described by the Eqs. (6.15) and (6.16), as given below:
For an overhead line shunt conductance G is always negligible and it is sufficiently accurate to neglect line resistance R as well. So with this approximation,
Y = j ω C l and Z = j ω L l
where L and C are the inductance and capacitance per unit length and I the length. Thus
So the characteristic equations become,
If there is no load, the receiving-end current IR is zero.
The charging current becomes:
Thus sending-end voltage VS and sending-end current IS are in quadrature. If the relation,
In both the cases the receiving-end voltage and current are numerically equal to the sending-end values, so that there is no voltage drop on load. Such a line is called a tuned line.
For 50 Hz, the length of line for tuning is,
It is too long a distance of transmission from the point of view of cost and efficiency. For a given line (length of line and frequency being fixed), the line may be tuned by increasing L or C i.e. by adding series inductances or shunt capacitances at intervals along the line length, as illustrated in Figs. 6.10 (a) and 6.10 (b). This method is impractical and uneconomical for power transmission and is used in telephony where higher frequencies are employed.
Another method of tuning power lines, which is being presently experimented with, is shown in Fig. 6.11. The capacitors in series with the line neutralize the voltage drop due to the line inductance, and the inductors across the line neutralize the charging current due to line capacitance. This method is known as the method of tuning by the use of compensating sections.
The values of L’ and C’ may be found with fair accuracy by replacing the line by its nominal- equivalent. The two capacitances C’ are equivalent to a single capacitance ½ C’ which will neutralize the inductance L of the line impedance Z if ω2 L (1/2 C’) = 1. The inductance L’ will neutralize the charging current provided ω2 L’ (1/2 C) = 1. The exact values required in long lines, can be obtained from the -equivalent shown in Fig. 6.8. Thus 1/2 C’ must neutralize Z, which is given as,
