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Power Transmission Channel Characteristics and Performance 5.1. Introduction This chapter mainly discusses the characteristics and performance of transmission lines. A very important problem in a power system is the load flow above the transmission line so that the voltage at various nodes is maintained within the specified limit. While these generally interconnected system issues will be discussed in Chapter 6, current attention is focused on the performance of a single transmission line, giving the reader a clear understanding of the principles involved. Transmission lines are usually operated with a balanced three-phase load; analysis can therefore be continued on a per phase basis. Transmission lines on a per-phase basis can be considered a two-port network, where the sending voltage Vs and current are related to the end-receiver voltage Vr and the current Ir through the ABCD constant "as

The following identities also apply to ABCD constants: AD - BC = 1 These constants can be easily determined for short and medium length lines with the appropriate approach to clump the line impedance and shunt entry. For the long line the right analysis must be done by considering the resistance distribution, the inductance and capacitance parameters and the ABCD constant of the line are determined from it. The equations for power flow in the lines and pie charts of recipients and shippers will also be developed in this chapter so that various types of tip conditions can be handled. The following nomenclature has been adopted in this chapter: z = series impedance / unit length / phase y = shunt entry / unit length / phase to neutral r = resistance / unit length / phase C = cepacitance / unit length / phase to neutral / = transmission line length Z = zl = total series / phase impedance Y = yl = total incoming shunt / phase to neutral

Note: Subscript S stands for end-sender quantity and subscript R stands for end-recipient quantity 5.2. Short Transmission Line For short lines with a length of 100 km or less, a total of 50 Hz shunt admittance (jwCl) is small enough to be negligible so as to produce a simple equivalent circuit of Figure 5.1

Figure 5.1. The circuit is equivalent to a short line This becomes a simple series, the connection between the voltage senders and the receiver current can be directly written as:

The phasor diagram for short lines is shown in Figure. 5.2 for the backing of the lagging current. From this picture we can write:

For overhead transmission lines, shunt admittance is mainly a capacitive susceptance (jwcl) because channel conductance (also called leakance) can always be ignored. The last term is usually a neglected order:

Extending binomially and maintaining the first order terms, we get:

The above equation is quite accurate for the normal load range.

Figure 5.2 Short-line phasor diagram for lagging currents

Voltage Settings The voltage regulation of the transmission line is defined as the increase in voltage on the receiving side, expressed as a percentage of full load voltage, when the full load on the specified power factor is thrown, i.e.

Regulation percent = where (VR0) = the endless amount of voltage receiver (VRL) = the amount of the receiver's final voltage (at the specified power factor

For Short lane =

Regulation percent =

In the above derivation, 0R has been considered positive for lagging loads. it will be negative for the leading charge.

Regulation percent = The voltage setting becomes negative (eg load voltage is more than without load voltage), when in Eq. (5.8)

This also follows from Eq. (5.8) for zero voltage regulation

Where 0 is the impedance angle of the transmission line. However, this is an approximate condition. The right conditions for zero regulation are determined as follows:

Figure 5.3 Phasor diagram in unregulated conditions

Figure 5.3 shows the phasor diagram under conditions of zero voltage regulation, namely:

Or This follows from the angle geometry at A, that for the regulation of zero voltage

From the discussion above it can be seen that the 6f line voltage setting is very dependent on the load power factor. voltage regulation increases (decreases) because the power factor of the lagging load increases and becomes zero on the main power factor given by Eq. (5.10). Example 5.1 The 50 Hz single-phase generator supplies 5,000 kW inductive loads at a power factor of 0'707 which is left through a 20 km overhead transmission line. The channel resistance and inductance are 0.0195 ohms and 0.63 mH per km 'The voltage at the receiving end must be kept constant at 10 kv. Find (a) voltage and voltage settings at the sender end; (B) the value of the capacitor to be placed in the load volume check so that the rule is reduced to 50vo from that obtained in part (a); and (c) comparing the transmission efficiency in parrs (a) and (b). Solution The line constant is R = 0.0195 x 20 = 0.39 ohm X = 3I4 x 0.63 x 10-3 x 20 = 3.96 ohm (a) This is a case of a short line with I = Ir = Is given by

From Eq. (5.5)

Voltage regulation =

(B) voltage regulation desire =

or new value lVs | = 11.09 kV Figure 5.4 shows the equivalent circuit of a channel with capacitive reactance which is placed parallel to the load

Assuming cos QR is now a load power factor and capacitive reactance put together, we can write (11.09 - 10) x 103 = lIR| (R cos QR + X sin QR) Because capacitance doesn't produce real power, we have

Solve Equations. (i) and (ii), we get cos QR = 0,911 lagging and lIRl= 549 A now Ic = IR – I = 549(0.911 – j0.412) – 707(0.707 – j0.707) = 0.29 + j273,7

Note that the actual part of 0.29 appears because the estimate in (i) ignores it, we have

Efisiensi transmisi

It should be noted that by placing the capacitor in parallel with the load, the receiver-end power factor improves (from 0.707 iug to 0.911 lag), the channel current decreases (from 707 A to 549 A), the sound channel regulation decreases (half of the previous value) and efficiency transmission increases (from 96.2 to 97.7%) 'Adding capacitors in parallel with the load is a powerful method for improving transmission system performance and will be discussed further towards the end of this chapter. Example 5.2 A substation as shown in Figure. 5.5 receives 5 MVA at 6 kv, the lagging factor of 0.g5 on the low voltage side of the transforner from the power plant via cable which has resistance per phase and reactance of 8 and 2.5 ohms respectively. The identical 6.6 / 33 kV transfer receivers are installed at each end of the channel. The sides are 6'6 kV from transfonner connected delta while the 33 kV side is connected to the star. The resistance and reactance of star-connected windings are 0.5 and 3'75 ohms, respectively and for connected windings delta arJ0.06 and 0.36 ohms. what is the voltage on the bus at the end of the power plant.

The solution here is comfortable using the unit method. Let's choose Basis MVA = 5 Basis Kv = 6.6 di low voltages = 33 di high voltages Capable impedance = (8 + j2.5) ohm/fasa = (8 + j2.5)/(33)2 = (0.037 + j.0.0115) pu The star impedance is equivalent to the transformer windings of 6.6 kv = 1/3 (0.06 + j0.36) = (0.02 + j0.12) ohm/fasa Impedansi transormator per unit Zt = (0.02 + j0.12)x5/(6.6)2 + (0.5 + j3.75)x5/(33)2 = (0.0046 + j0.030) pu Load Voltage = 6 / 6.6 = 0.91 pu Source voltage = 1 / 0.91 = 1.1 pu Using Equations. (5.5), we get lVs| = 0.91 + 1.1(0.046 x 0.85 + 0.072 x 0.527 = 0.995 pu = 0.995 x 6.6 = 6.57 Kv (line to line)

Example 5.3 The input to a single phase short channel shown in Figure 5.6 is 2,000 kw at a power factor of 0.8 lagging. This path has a series impedance (0.4 + j0.a) ohms. If the load voltage is 3 kV, find the load and power factor of the receiving end. Also measure the supply voltage.

Solution This is a problem with mixed-voltage conditions and the input power is determined. The right solution is outlined below: Sending-end active / reactive power = active active / reactive + active receiver power loses active line

Squaring (i) and (ii), adding and simplifying, we get

Note: This is, in fact, the same as Eq. (5.4) if I2 is canceled in full. For numeric values given.

5.3 MEDIUM TRANSMISSION PATH For channels with a length of more than 100 km, the charging current due to shunt entry cannot be ignored. For lines in the range of 100 km to 250 km. accurate enough to agglomerate all entry points on the receiving side to produce the same diagram as shown in Figure 5.7. Starting from the fundamental circuit equation, it is quite easy to write the equation of the transmission line in the constant form ABCD given below

T-Nominal Representation If all the shunt capacitance is equated in the middle of the line, it leads to the Z-nominal circuit shown in Figure 5.8.

Reordering the results, we get the following equation:

5.4 OUTSTANDING SOLUTIONS OF OLD TRANSMISSIONS For lines more than 250 km, the fact that line parameters are not unified but uniformly distributed along their length must be considered.

Figure 5.10 shows one phase and a neutral return (from zero impedance) of the transmission line. Let dx be part of the line element at the distance x from the receiving-receiver having the zdx series impedance and ydx shunt reception. Voltage increase * becomes neutral above the part of the element in the direction of increasing x is dV. We can write the following differential relations in all parts of the element:

It can be noted that the connection type (eg T or phi) assumed for the element part, does not affect this first-order differential relationship. Distinguishing Equations. (5.14) with respect to x, we obtain

Here Z, called line character impedance and 7 is called a propagation constant

Knowing VR, IR and line parameters, using Eq. (5.21) the complex number of the rms value of V, and I, at any distance x along the line can be easily found A more convenient form of voltage and current expression is obtained by introducing hyperbolic functions. Reorder Equations. (5.21), we get:

ABCD Constant Evaluation ABCD constants of long lines can be evaluated from the results given in Equation (5.24). It must be noted that y = yz root is generally a complex number and can be expressed as:

The hyperbolic function of complex numbers involved in evaluating ABCD constants can be calculated by one of the three methods given a punch. 5.5 LONG EQUATION INTERPRETATION As already stated in Eq. (5.26), y is a komprex number that can be expressed as

The real part a is called the attenuation constant and the imaginary part B is called the phase constant. Now v, from Eq. (5.2r) can be written as

Every time t, v.rl is distributed sinusoidally along the distance from the receiving end with amplitude increasing exponentially with distance, as shown in Figure 5.11 (a> 0 for channels that have resistance)

After time At distribution is forward in phase distance by (wAtlB). Thus this wave moves towards the receiving end and is a coming wave. Line losses cause the limit to decrease exponentially from sending to recipient. Now :

After time At voltage distribution slows the distance phase by (wAtlB). This is a reflected wave that moves from the receiving end to the sender's end with an amplitude decreasing exponentially from the receiving end to the sender's end, as shown in Figure.5.12. At any point along the line, voltage is the number of events and the reflected voltage wave is present at the point of Eq. (5.32). The same applies to the current wave. The expressions for reflected events and waves can now be written in the same way by continuing from Eq. (5.21). If Z "is pure resistance, current waves can be easily obtained from voltage waves by dividing it by Zr.

If the load impedance Zl = VR / IR = ZC, ie the line is terminated in its characteristic impedance, the reflected voltage wave is zero (VR - ZCIR = 0). A line that ends with its characteristic impedance is called an infinite line. An incident wave in this condition cannot distinguish between an infinite termination and continuation of a line. Power system engineers usually caII ZC impedance surges. It has a value of around 400 ohms for airways and phase angles typically varying from 0 "to - 15o. For Zc underground cables. Approximately one tenth of the value for air ducts. The term impedance surge is, however, uses a connection with a surge ( because of lightning or switching) or transmission lines, where loss of channels can be ignored so that:

Slurge Impedance Loading (SIL) of the transmission line is determined as the power supplied by the channel to a pure resistive load whose value is equal to the impedance of the channel surge. So for channels that have a wave impedance of 400 ohms,

Now for a typical power transmission line\

Now the time for the change in the 2phi phase is 1 / fs, where f = w / 2phi is the frequency in the cycle / s. During this time, waves travel the same distance as h i.e. one wavelength.

The actual speed of the wave propagation along the line will be slightly less than the speed of light. The 50 Hz power transmission wavelength is given by

The transmission line is practically shorter than this (usually several hundred kilometers). It should be pointed out, that the interesting waves in Figures.5.11 and 5.12 are for illustration only and are not related to real power transmission lines. 5.6 FERRANTI EFFECTS As illustrated in Exarnple 5.5, the channel capacitance effect is to cause the receiving end voltage to be no more than the sender's end voltage. The effect becomes clearer when the length of the line increases. This phenomenon is known as Ferranti. effect. A general description of this effect is explained below:

The above equation shows that at I = 0, the coming voltage wave (Eio) and reflected (Ero) are equal to Vr / 2. With reference to Figure 5.13, when I increase, the voltage wave comes increasing exponentially. It is clear from the geometry of this image that the phasor voltage generated is Vs (OF) such that lVrl> lVsl. A simple explanation of the Ferranti effect based on estimates can be stated by lumping in the inductance and channel capacitance parameters. As shown in Figure 5.14 the capacitance is equated to the end-receiving line.

Because C is small compared to L, wLI can be ignored in comparison with 1 / wCI. Thus 5.7 LINE POWER TUNED Equations (5-23) characterize long line performance. For the shunt conductance of the overhead channel, G can always be ignored and accurate enough to ignore R channel resistance too. with this estimate The transmission distance is too far from the point of view of cost and efficiency (note that channel resistance is ignored in the analysis above). For certain channels, setting length and freouency can be obtained by increasing L or C, that is by adding series inductance or shunt capacitance in several places along the line. This method is not practical and economical for power frequency channels and is adopted for tclephony where higher frequencies are used. The power line adjustment method currently being tested uses series capacitors to cancel the channel inductance effect and the shunt inductor to neutralize the channel capacitance. The long line is divided into several individually tuned parts. However, so far the practical method for increasing channel settings and power transfer capacity is to add series capacitors to reduce channel inductance; shunt capacitors under heavy load conditions; and inductor shunts in light or no-load conditions.

5.8 EOUIVALENT FRAMES FROM LINE LINE As far as the end conditions are concerned, the exact equivalent circuit of the transmission line can be formed in the form of a T-or zr network. Equivalent network parameters are easily obtained by comparing the network-z performance equation and transmission line in terms of the final number. Lv,

Thus is a factor where the thc series impedance of the nominal phi must be multiplied to obtain the parameter Z 'from equivalent -phi. Change Z 'from Eq. (5.50) in Eq. (5.51), we get:

So the (tanh y / 2 / y / 2) factor where the shunt arm enters from nominal-phi must be multiplied to get the shunt parameter (Y '/ 2) from the equivalent

Equivalent-T network parameters from the transmission line are obtained on the same channel. The T-equivalent network is shown in Figure 5.16. As we will see in Chapter 6, equivalent-phi (or nominal-r) networks are easily adopted to contain stream studies and, therefore, are used universally Comment We find from the example above that the results obtained by the nominal-zr method and estimates (5.28b) are practically the same and very close to those obtained with the right calculation (section (c)). On the other hand the results obtained with the short-line approach are quite a big mistake. Therefore, if the line is this long (around 300 knr), it is accurate enough to use nominal r (or estimate (5.28b)) which results in considerable savings in computing efforts. 5.9 POWER FLOW THROUGH LINE TRANSMISSION So far the transmission channel performance equation is presented in the form of a voltage and current relationship between the ends of the receiver and receiver. Because loads are more often expressed in terms of real power (wattslkW) and reactive (VARs / kVAR), it is easier to deal with the transmission line equation in the form of the power and voltage of the sending and receiving complexes. While power flow problems in public networks are discussed in the next chapter, the principles involved are illustrated here through a single transmission channel (2nodel2-bus system) as shown in Figure 5.17.

Let us take the receiver end voltage as the areference phasor (Vr = lVRl 0 ") and let the sendend voltage guide it to angle 6 (Vs = lVsl 16). Angle 6 is known as the torsion angle whose significance has been described in Chapter 4 and then will discussed in Chapter 12 when discussing stability issues. Complex forces that leave the receiving end and enter the sending end of the transmission line can be expressed as (based on per phase)

In the above equation Sr and Ss are complex voltamperes per phase, while Vr and Vs are expressed as volts per phase. If Vr and Vs are expressed in kv lines, then the three-phase por end-receiver complex is given by

This is indeed the same as Eq. (5.59). The same results apply to Ss. So we see Eq. (5.58) and (5.59) give three-phase MVA if Vs ina Vr is expressed in the kV line. If Equation (5'58) is expressed in real and imaginary parts, we can write real and reactive forces at the receiving end

The short line equation above will also apply to a long line when the line is replaced by the equivalent (or nominal-r) and the shunt receipt is combined with the receiver and the sender's generation. In fact, this technique is always used in load flow problems to be treated in the next chapter.

some important conclusions that are easy to follow from Eq. (5.71) to (5.74) mentioned below: 1.) For R = 0 (which is a valid estimate for the transmission line) the real power transferred to receivers is proportional to sin 6 (= 6 for small values d), while reactive power is proportional to the magnitude of the voltage drop across the line.

2.) The actual received strength is maximum for 6 = 90 and has an IVsIIVrI / X value. Of course, 6 is limited to values well below 90 from the stability considerations discussed in Chapter t 12. 3.) The maximum real power transferred for a particular channel (fixed X) can be increased by increasing the voltage level. It is from this consideration that the voltage level is progressively pushed upwards to send larger chunks of power at a greater distance guaranteed by the largesize generating station. For very long channels, the voltage level cannot be increased beyond the limit placed by the current high-voltage technology. To increase the power transmitted in such cases, the only choice is to reduce channel reactance. This is done by adding series capacitors in the row. This idea will be pursued further in chapter 12. The series of capacitors will of course increase the severity of the line more than the voltage under switching conditions 4.) As stated in 1 above, VARS (reactive power lagging) which is changed by a channel is proportional to the decrease in channel voltage and is not dependent on d. Therefore, in a transmission system if VARS demands a large load, the voltage profile at the Point tends sagged rather sharply. To maintain the desired voltage profile, VAR demand from the load must be fulfilled locally using a positive VAR generator (condenser). This will be discussed at length in Sec. 5.10. Results that are somewhat more accurate but close reveal a decrease in channel voltage in terms of active and reactive power can be written directly from Eq. (5.5), i.e.

Small 6 is needed from consideration of system stability which will be discussed at length in Chapter 12. 5.10 VOLTAGE CONTROL METHODS Practically every equipment used in a power system is rated for cprain voltage with the allowable voltage variation band. Voltage on various buses must be controlled in the specified regulatory numbers. This article will discuss two methods by which the voltage on the bus can be controlled.

Consider the two bus system shown in Figure 5.26 (akeady exemplified in Section 5.9). For the sake of simplicity, let the lines be characterized by a series of reactances (i.e. having negligible resistance). Furthermore, because the angle of torque d is small under practical conditions, the real and reactive power supplied by the channel for the final voltage lVr and the cell-receiver voltage specified are l {| can be written as below from Eq. (5.71) and (5.73).

Because the real power requested by the local must be given by a line Pr = Pd Varying the real power demand Pd is fulfilled by consequent changes in the torque angle 6. However, it should be noted that the reactive power received from the channel must remain in Qrs as given by Eq. (5.83) for fixed IVsI and certain types. Therefore, the channel will operate with the end-receiver voltage specified for only one Qd value given by Qd = Qsr Practical loads are generally lagging and such that VAR Qd demand can exceed Qsr. Easily follow from Eq. (5.83) that for Qd greater Qsr, the receiving end voltage must change from the specified IVI value, some IV values to meet the requested VAR. Thus Indeed, under light load conditions, channel charging capacitance can cause VAR demand to be negative resonance at end-receiver voltage that exceeds delivery-end voltage (this is the Ferrary effect described in Section 5.6). To regulate the channel voltage under various VAR demands, the two methods discussed below are used.

The negative sign in the quadratic solution is rejected because otherwise the solution will not match the specified end-voltage of the receiver which is only slightly less than the end-shipping vortage (the difference is less than 12 percent). Reactive Power Injection This follows from the discussion above that to maintain the receiver voltage at the specified IVI value, a fixed amount of VAR (QSR) must be taken from the channel. To achieve this in these conditions under varying QD VAR demand conditions. Local VAR generators (controlled reactive resources / compensation equipment) must be used as shown in Figure 5.27. The VAR balance equation at the receiver side now, Fluctuations in Qd are absorbed by the local VAR generator Qc so that the VAR pulled from the channel remains in the QSR. The final receiver voltage will remain at IVSRI (this of course assumes IVSI constant send_end voltage). Local VAR compensation can actually be made automatically by using a signal from the VAR meter installed at the receiving end.

Two types of VAR generators are used in static-type and rotating type. This is discussed below. Control by Transformer The VAR injection method discussed above does not have the flexibility and voltage control savings by changing the transformer tap. Change tap transformer is clearly limited to a narrow voltage control range. If the required vortex correction exceeds this range, tap changes are used together with the VAR injection method. The receiving-end voltage tends to sag because the vAR demanded by the load can be increased by simultaneously changing the sending faucet and the receiving-end transformer. Tap changes must be made, on_road, and can be done manually or automatically, the transformer is called the Tap Changing Under Load (TCUL) transformer.

Transmission Line Compensation The performance of the old AC EHV transmission system can be improved by reactive compensation from the type of series or shunt (parallel). Series capacitors and shunt reactors are used to artificially reduce series reactance and channel shunt vulnerability and thus they act as channel compensators. Path compensation results in increased system stability (Chapter 12) and convoluted stresses, in improving power transmission efficiency, facilitating path energy and reducing transient and transient stresses. The level of compensation effect depends on the number, location and arrangement of series capacitor and shunt reactor stations. While planning long-distance routes, in addition to the required average compensation level, it is necessary to know the most appropriate location of the reactor and capacitor bank, the optimal connection scheme and the number of intermediate stations. To find the operating conditions along the line, the ABCD constant of the line parts on each side of the capacitor bank, and the ABCD bank constant can be found first and then the equivalent constant of the line-capacitor-line combination can then be contacted using the given formula in Appendix B. It can be noted that for the same voltage increase, the reactive capacities of shunt capacitors are greater than series capacitors. Shunt capacitors increase Pf load while series capacitors have almost no impact on PF. Series capacitors are more effective for long lines to improve system stability. Thus, we see that in both the series and shunt compensation of the long transmission line it is possible to transmit large amounts of power efficiently with a flat voltage profile. The right type of compensation must be provided in the right amount in the right place to achieve the desired stress control. The reader is tried to read the details about Static Var Systems (SVS) in References 7, 8 and 16. For complete care about 'compensation', the reader can refer to Chapter 15.

PROBLEMS 5.1. A three-phase voltage of 11 kV is applied to a line having R = 10 ohm and X = 12 ohm per conductor. At the end of the line is a balanced load of PkW at a leading power factor. At what value of P is the voltage regulatior zero when the power factor of the load is (a) 0.707, (b) 0.85? 5.2. A long line with A = D = 0.9 1,5" and B= 150 65" ohm has at the loar end a transformer having a series impedance Zr = 100 67" ohm . The loar voltage and current are VL and IL. Obtain expressions for Vs and Is form of 5.3. A three-phase overhead line 200 km long has resistance = 0.16 ohm/km an conductor diameter of 2 cm with spacing 4 m, 5 m and 6 m transposed ( Find: (a) the ABCD constants using Eq. (5.28b), (b) the Vs, Is, pfs when the line is delivering full load of 50 MW at 132 kV and 0.8 lagging pf. (c) efficiency of transmission, and (d) the receiving-end voltage regulation 5.4. A short 230 kV transmission line with a reactance of 18 O/phase supplir a load at 0.85 lagging power factor. For a line current of 1,000 A thus receiving- and sending-end voltages are to be maintained at 230 kV Calculate (a) rating of synchronous capacitor required, (b) the loa current, (c) the load MVA. Power drawn by the synchronous capacitt may be neglected 5.5. A 40 MVA generating station is connected to a 'three-phase line havin Z = 300 75" ohm Y = 0.0025 90" ohm. The power at the generating station is 40 MVA at unity power factor a a voltage of 120 kV. There is a load of 10 MW at unity power factor a the mid point of the line. Calculate the voltage and load at the distant end of the line. Use nominal-T circuit for the line. 5.6. The generalized circuit constants of a transmission line are A = 0.93 + j0.016 B = 20 + j140 The load at the receiving-end is 60 MVA, 50 Hz 0.8 power factor lagging. The voltage at the supply end is 22O kV. Calculate the load voltage 5.7. Find the incident and reflected currents for the line of Problem 5.3 at the receiving-end and 200 km from the receiving-end 5.8. If the line of Problem 5.6 is 200 km long and delivers 50 MW at 22O kV and 0.8 power tactor lagging, determine the sending-end voltage, current, power factor and power. Compute the efficiency of transmission, characteristic impedance, wavelength, and velocity of propagation. 5.9. For Example 5.7 find the parameters of the equivalent-n circuit for the line 5.10. An interconnector cable having a reactance of 6 ohm. links generating stations 1 and 2 as shown in Fig. 5.18a. The desired voltage profile is IV1I = lV2l = 22 kV. The loads at the two-bus bars are 40 MW at 0.8 lagging power factor and 20 MW at 0.6 lagging power factor,

respectively. The station loads are equalized by the flow of power in the cable. Estimate the torque angle and the station power factors 5.11. A 50 Hz, three-phase,275 kV, 400 km transmission line has following parameters (per phase). Resistance = 0.035 ohm/km Inductance = 1 mH/km Capacitance = 0.01 uF/km If the line is supplied at 275 kV, determine the MVA rating of a shunt reactor having negligible losses that would be required to maintain 275 kV at the receiving-end, when the line is delivering no-load. Use nominalzr method. 5.12. A three-phase feeder having a resistance of 3 ohm and areactance of 10 ohm supplies a load of 2.0 MW at 0.85 lagging power factor. The receiving end voltage is maintained at 11 kV by means of a static condenser drawing 2.1 MVAR from the line. Calculate the sending-end voltage and power factor. What is the voltage regulation and efficiency of the feeder 5.13. A three-phase overhead line has resistance and reactance of 5 and 20 ohm, respectively. The load at the receiving-end is 30 MW, 0.85 power factor lagging at 33 kV. Find the voltage at the sending-end. What will be the kVAR rating of the compensating equipment inserted at the receiving-end so as to maintain a voltage of 33 kV at each end? Find also the maximum load that can be transmitted. 5.14. Construct a receiving-end power circle diagram for the line of Example 5.7. Locate the point coresponding to the load of 50 MW at 220 kV with 0.8 lagging power factor. Draw the circle passing through the load point. Measure the radius and determine therefrom lVrl. Also draw the sendingend circle and determine therefrorn the sending-end power and power factor 5.15. A three-phase overhead line has resistance and reactance per phase of 5 and 25 ohm , respectively. The load at the receiving-end is 15 MW, 33 kV, 0.8 power factor lagging. Find the capacity of the compensation equipment needed to deliver this load with a sending-end voltage of 33 kv. Calculate the extra load of 0.8 lagging power factor which can be delivered with the compensating equipment (of capacity as calculated above) installed, if the receiving-end voltage is permitted to drop to 28 kV.