MODULE TITLE : ELECTRICAL SUPPLY AND DISTRIBUTION SYSTEMS

MODULE TITLE : ELECTRICAL SUPPLY AND DISTRIBUTION SYSTEMS TOPIC TITLE : THE OPERATION OF THREE-PHASE TRANSFORMERS IN PARALLEL AND GENERATORS ON INFINITE BUSBARS LESSON 2 : OPERATING CHARACTERISTICS OF THREE PHASE GENERATORS ON INFINITE BUSBARS ESD - 2-2

Published by School of Science & Engineering Teesside University Tees Valley, UK TS1 3BA +44 (0)1642 342740 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the Copyright owner. This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, re-sold, hired out or otherwise circulated without the publisher's prior consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser.

1 INTRODUCTION In this lesson we are introduced to another essential item of equipment required for an electrical supply system, the generator (or alternator as a.c. generators are often referred to). The lesson covers aspects of commercial generators and the electrical parameters that affect the efficient output of generated electrical energy. The lesson shows the importance of the 'capability' chart which is utilised to predict limits of the possible actual performance of the industrial generator against the theoretical ideal. YOUR AIMS At the end of this lesson you should be able to: correctly construct the capability chart for a synchronous generator indicate the performance limitations of a generator using the above chart. STUDY ADVICE You will need to have graph paper, a compass and a protractor available to complete this lesson.

2 GENERATOR CAPABILITY CHARTS This section deals with the concepts behind, and accurate scaled production of, a special chart which power station engineers and operators use to get the best out of the machinery without incurring electrical or mechanical damage. Although we only use graph paper to produce a scaled drawing, modern power stations display the chart on large monitor screens or in the form of a wallmounted, manually operated mimic diagram. MINDING YOUR Ps AND Qs A main power system generator is primarily a device for converting rotary mechanical power to electrical power. However, most industrial and commercial loads also require reactive power (VAr). This means that unless a consumer generates his own leading or lagging VArs they must be provided by the supply authority so that the output from the national generators will be a combination of real power (P) and reactive power (Q). Thus, MVA = phasor sum of MW + MVAr. Ignoring machine internal losses, the electric power output equals the mechanical power input from the prime mover. Consider FIGURE 1 opposite, which shows a mechanical prime mover driving a synchronous generator which is supplying both real and reactive power (P + Q).

3 Engine prime mover Synchronous generator 3φ power out (P + Q) Fuel in Mechanical power DC excitation (E) FIG. 1 Power Generation System You will see from the diagram that the power station operator has control over two parameters the engine's fuel which varies the mechanical power the direct current into the rotor which varies the excitation. Think about FIGURE 1 and the two variable parameters and decide which of the electrical powers, P or Q, is varied by altering the mechanical power and which is varied by altering the d.c. excitation. Variation of... = Variation of P Variation of... = Variation of Q

4 Since mechanical prime movers are only capable of providing real power (P), any reactive power (Q) requirement will have to be met by variation of the excitation of the synchronous machine. Therefore Variation of Mechanical Power = Variation of P Variation of Excitation = Variation of Q For example, if a machine is generating real power at unity power factor (u.p.f.), increasing the excitation will produce lagging reactive power. In the same way, decreasing the excitation from the u.p.f. position will produce leading reactive power. (It is the opposite when the machine is being used as a synchronous motor.) INFINITE CAPACITY Remember that the UK power system consists of about 175 synchronous generators operating in parallel and feeding the national grid system. The approximate total grid system capacity is: day 85 10 3 MW night 55 10 3 MW A national power system such as the national grid is thought of as having infinite capacity since the connection of any one load or loss of any one generator, does not markedly affect voltage and frequency. Therefore, the connection points or nodes of the national grid are known as infinite busbars.

5 AN INTRODUCTION TO GENERATOR CAPABILITY CHARTS When two or more synchronous machines are operating as generators on infinite busbars, what is known as the operating or capability chart for any particular machine is of considerable importance. A capability chart is an envelope or system of boundaries within which a particular machine may satisfactorily work without fear of damage due to overload or risk of instability. The limits of operation imposed to prevent damage to the generating set are the prime mover limit the rotor heating limit the stator heating limit the theoretical stability limit the practical stability limit. Consider FIGURE 2 which shows the circuit of a synchronous machine. X S R I E IX S IR V IZ S FIG. 2 Synchronous Machine Equivalent Circuit

6 Since the voltages around the circuit must equate, the phasor E must equal the phasor sum of the other volt drops. i.e. E = V + IXS + IR Since the phasor sum of IX S and IR = IZ S, E = V + IZS Take the equation E = V + IZ S and solve it for I. I = E Z S V Look at the equation for I. Since the solution contains a common denominator we should be in a position to conclude that the phasor I (the supply current) is made up of two other phasors. Decide what these two phasors are, which when added together, give the supply current of the machine and rearrange the equation to show this. Our answer is in the text that follows.

7 I = E V Z S so I = E Z S V Z S Now remember that I is the sum of the two phasors, so I E V = + Z Z S S Now providing the busbar voltage remains constant (which it does to within a few percent) and since Z S is constant V is constant in magnitude and direction and does not change with Z S variation in excitation or power input E alters in magnitude with excitation and in direction with power input. Z S FIGURE 3 shows the various currents and busbar voltage with the machine under-excited, i.e. leading power factor.

8 V Z S δ δ is known as the load angle ψ φ E Z S I ψ X ψ = tan ( S ) 1 R + V Z S This is the correct position for V since Z S is inductive Z S FIG. 3 Phasor Diagram showing leading power factor Note how the addition of the two phasors current I. V Z S and E Z S equals the machine When you multiply each current phasor by the busbar voltage (V), the phasor I becomes an indicator of machine voltamps (VI) so that the phasor diagram can be redrawn as FIGURE 4 showing this. Note that the phasor + V Z S has been dropped since its only function was to establish the position and direction of V. Z S

9 V 2 Z S Q δ ψ φ VE Z S VI E since V Z S is constant P FIG. 4 Phasor diagram showing the phasors in FIGURE 3 multiplied by the busbar voltage So what was originally a current diagram (FIGURE 3) becomes a power diagram (FIGURE 4) by multiplying the current phasors by the common busbar voltage. VE This means that V 2 Z is a voltamps phasor as is and of course the S phasor VI. Z S This is why the voltage reference phasor in FIGURE 3 has been replaced with a real power axis (P) and a corresponding reactive power axis (Q). The projection of VI onto the horizontal axis gives the real power in watts. Projection onto the vertical axis gives the reactive power in VArs. FIGURE 5 shows you how vertical locus lines (representing constant real power) and horizontal loci (representing constant reactive power) can also be drawn.

10 V 2 Z S Q E ψ I φ Constant VAr line P Constant power line Point P 1 FIG. 5 Phasor diagram with locus lines (It should be noted that the generated voltage E is proportional to the d.c. excitation current fed to the rotor of the generator to establish a magnetic field which when rotated induces the generated voltage E in the stator.) Look at FIGURE 5. It should be apparent that the amount of real power being generated at P can be read off. Clearly one needs to be careful to draw a diagram like this to a suitable scale. Think about each phasor, angle and axis and attempt to list up to six parameters in addition to real power (P) which could also be read off, if scales are known.

11 The list could include: the reactive power (Q) the supply current (I) the output power factor (φ) the output voltamps (VI) the load angle (δ) the percentage excitation (E) It is also very important to note that you can obtain unknown quantities from the basic diagram providing one or two parameters are drawn in. DRAWING A LOAD DIAGRAM Exercise The following exercise will take you step by step through the process of drawing and using a load diagram. You'll need some graph paper to do it on. Each step you have to take is indicated by a bullet. On a piece of graph paper long way up (portrait format), draw a vertical axis 40 mm in from the left-hand edge. Label this axis Reactive Power Lead at the top, and Reactive Power Lag at the bottom. Draw a horizontal axis across the centre and label it Real Power. Scale these axes off at 20 mm = 10 units. Do this for 50 units in each case. This represents maximums of 50 MW on the real power axis and 50 MVAr lead and lag on the reactive power axis.

12 FIGURE 6 has been included as the solution to this exercise and if you are really stuck (there's no benefit in just looking at it straight away) have a look at the diagram. Try not to look at it all just the area you have difficulty with. Start at the zero point and draw in the phasor V 2 using the values Z S length = 85 mm ψ = 80 Draw the current phasor I 1 using the values length = 60 mm φ = (power factor) = 30 lead Join the ends of V 2 and I 1 to give the excitation phasor E 1. Z S (The length of E 1 should be close to 85 mm if you've followed the instructions correctly.) We can now imagine that our machine is working under these conditions and if we read off the real and reactive power we should be delivering 26 MW and 15 MVAr lead. Check this for yourself on your diagram. What does your diagram show the load angle (δ ) to be?.......... (It should be forty-one degrees.) Now suppose National Control requests us to supply the same power but with 10 MVAr lag (i.e. 26 MW, 10 MVAr lag).

13 We don't need to touch the real power input but we must alter...? (the excitation E 1 ). To force the output current to move vertically downwards to 10 MVAr lag we'll have to increase excitation. Do this now and read off E 2 and I 2. Write the values you get in the space provided. E = I = 2 2 Now calculate these values as a percentage of their original values. You should get E 2 = 123 mm and and I 2 = 55 mm. Expressed as a % of their original values: 123 85 55 = 144.% 7 for E and = 91.% 7 for I 60 2 2 Measure the new load angle (δ ) and note it below. It should be reduced to about 22.5. FIGURE 6 shows you what you should have drawn for this exercise. Study it carefully to see if there are any differences you can't account for between FIGURE 6 and your own effort. It is important for you to understand the diagram fully at this stage.

14 Reactive power lead MVAr 50 V 2 Z S 40 δ 30 20 E 1 10 I 1 ψ 0 φ 10 20 30 40 50 E 2 Real power MW 10 I 2 20 30 40 Reactive power lag MVAr 50 FIG. 6 Solution to Exercise

15 SELF-ASSESSMENT QUESTION 1 Imagine you are working at your last position, i.e. E 2, I 2 on FIGURE 6. National Control now requests that your output should rise to 45 MW, 30 MVAr lag. What control adjustments (expressed as a percentage) are necessary to achieve this? Remember that you can only make two external adjustments, real power and excitation. Check your solution with that on page 44.

16 You should now be able to appreciate that you can move to any point on the chart, that is, to any specified output, by fiddling about with these two parameters. In this way the combined output of all the national generators can be juggled to meet the demand at any time. However, there are snags! What do you think would be the value of the excitation E and the load angle δ if all the real and reactive power were reduced to zero, i.e. I = 0? Write your answer and explanation in the space below and compare your answer in the text that follows. You should have concluded that as the powers, and thus the current, are reduced, E swings closer and closer to V Z S 2. Eventually, when the current is zero, it will lie coincident with V Z S 2 so that δ is also zero. This is the excitation required on no load and is termed 1.0 p.u. or 100% excitation.

17 GENERAL LOAD DIAGRAMS: STABILITY LIMITS Using the concepts described in the previous section, what is known as a general load diagram can be drawn. Study FIGURE 7 on the next page and concentrate on the excitation line. Note that since the machine resistance R is usually very small compared to the synchronous reactance X S, R is ignored. So X S becomes the machine impedance Z S and ψ is 90. This assumption will be made from now on since it greatly simplifies matters. We discovered previously that when the machine current is zero, excitation is vertical and lies along the 'Q' line as. V V or Z X 2 2 S S Remember that this special indeed unique situation is referred to as 100% excitation. If this value of excitation is being applied to the machine then, as the real power input is increased, the machine will generate and produce real power (active current) at its terminals. Since power is being produced and since one can think of the excitation and current phasors as being 'fastened together' at their tips, they both follow the arc100% excitation as the power increases. Find the line labelled Stability Limit on FIGURE 7.

18 Q lead V 2 Z S 50% Excitation 100% Excitation Excitation Theoretical stability limit MVA lead ψ P MVA lag Q lag FIG. 7 General Load Diagram for Synchronous Generator

19 At some particular value of power, excitation will lie along this line. Note how the load angle (δ) would increase as the power increases. Now suppose the mechanical power input has been further increased; but remember that the length of the excitation phasor has not altered Study FIGURE 7 and see if you can work out whether excitation would tend to swing beyond the steady-state stability limit. We should see that excitation would tend to swing above the stability limit if power input is increased. Now attempt Self-Assessment Question 2.

20 SELF-ASSESSMENT QUESTION 2 Consider the following: the horizontal axis in FIGURE 7 represents electrical power delivered to the busbars the load angle increases as the mechanical power input increases. What do you think would be the consequences of continuing to increase the mechanical power input? Make a sketch on a separate piece of paper to help you draw conclusions. Check your answers against the solution given on page 44.

21 But what happens to the excess power? It's a bit like opening the throttle in a powerful motor car. The rear wheels cannot convert all the power fed to them as forward motion and so they slip on the road surface and synchronism is lost between the power conversion mechanism (tyres) and the road. The same thing happens with the generator in that the excess mechanical power results in the rotor being torn from synchronism with the stator magnetic field (which is connected to the national grid) so that the rotor accelerates away. This results in mechanical overspeed as well as large electrical currents, so that one or more of the protective systems operate to shut down the generator set. It is important therefore that sets are not operated on, or even near, the theoretical stability limit. OPERATIONAL LIMITS ON GENERATOR SETS Bearing in mind what we've just learned about stability limits in the previous section, it's time to look in more detail at the five operational limits imposed on generator sets. (i) the turbine (or prime mover) limit This is also often referred to as the MW limit and is determined by the power output capabilities of the prime mover (engine). This limit gives the maximum active power output line on a capability chart.

22 (ii) the rotor heating limit This is the maximum direct current which can be drawn by the rotor for purposes of exciting the machine without leading to overheating. It is often referred to as the excitation limit and determines the maximum excitation circle on the chart and often the lagging reactive limit. (iii) the stator heating limit This is sometimes referred to as the MVA limit. It determines the maximum stator current the set can supply without leading to overheating. The stator heating limit produces the maximum MVA circle on the chart. (Note also that for constant current I and MVA, IX S is constant with its locus a circle with its centre being the origin.) (iv) the theoretical stability limit For any given value of excitation (rotor current) the ability of the machine to remain stable (i.e. in synchronism with the system) becomes less as the power output increases. If there is a sudden increase in power input, a point is reached when the restoring torque is zero. When this happens the machine quickly accelerates out of synchronism.

23 (v) the practical stability limit As you know it is undesirable for a machine to be operated to the limit of its theoretical stability and operators must know to what reasonably safe limit the machine may be taken. The practical stability limit is a safety factor which allows the machine to respond to a sudden increase in power demand with no corresponding increase in excitation. This limit is obtained by subtracting an assumed stability margin (usually 0.1 p.u. power) from each point on the theoretical stability limit. These five limits can now be drawn to suitable scales so that they form an envelope or system of boundaries (chart) within which a generating set may safely work without risk of damage or instability. (This chart is sometimes referred to as a shark fin chart.) Look at the example of an operating chart, FIGURE 8, and note the position of the various operating limits.

24 Q lead Theoretical stability limit Practical stability limit Turbine limit P Excitation limit Stator heat limit Q lag FIG. 8

25 CONSTRUCTING AN OPERATING CHART It is now time to look at the data required to construct an operating chart and how it's done. You'll need to have graph paper, a protractor and a pair of compasses to hand. Assumptions and Working Bases It is an accepted practice when producing and working with operating charts to neglect the machine resistance and consider only the synchronous reactance X S. It is usual to work in per-unit terms taking the following base values the machine rating as the power base S = 1.0 p.u. power the machine terminal voltage per phase as the voltage base V ph = 10. p.u. volts The operating chart, FIGURE 8, is constructed around the machine short circuit ratio (SCR). This is a comparison of the excitation required to produce normal terminal voltage on no-load with the excitation required to overcome the stator demagnetising m.m.f. (i.e. 90 lag load). i.e. SCR = excitation for rated volts on O/C excitation for rated current on S/C

26 We have seen the importance of the phasor 1 the short circuit ratio is that SCR =. XS V 2 X S. The reason for introducing If we know the SCR then you can find the length of V X S 2 as well. CONSTRUCTING THE MORE STRAIGHTFORWARD LIMITS The turbine (prime mover) limit, rotor heating limit, stator heating limit and theoretical stability limit are all relatively simple to draw in. The practical stability limit is more involved. All these are dealt with in the following exercise. Exercise 2 A 3-phase synchronous generator rated at 600 MVA is connected to infinite busbars. Using the data for the machine listed below we will methodically construct the machine's operating chart. Short circuit ratio = 0.5 p.u. Excitation limit = 2.8 p.u. Maximum MVA limit = 600 MVA Turbine Power limit = 500 MW Practical Stability margin = 10% full load power. You can now draw the more straightforward lines on the chart. SCR = 1 X S 1 XS = = SCR 20. p.u.

27 Busbar voltage per phase = 1.0 p.u. volts V 2 10. 10. p.u. So = Z 20. = 05. p.u. i.e. 0.5 p.u. lead. S ( ) The MVA limit is 1.0 p.u. power so 500 MW relates to a real power limit of 500 600 = 0. 833 p.u. Take a sheet of graph paper long way up and draw a horizontal reference across the middle (P axis) and a vertical reference 20 mm in from the edge (Q axis). Mark these off so that 10 mm = 0.1 p.u. Now draw in the theoretical stability limit at 0.5 p.u. lead, i.e. V Z2 S Put the point of the compass at zero. Now open it up to 1.0 p.u. power and swing it in the arc from 1.0 p.u. reactive lag to the real power axis. This gives the maximum voltamps the machine can deliver. This value is also known as the stator heat limit since exceeding this value draws more current than the machine is designed for, leading to overheating.

28 Next, draw in a vertical line at 0.833 p.u. real power. This indicates the turbine or prime mover limit such that operation to the right of this line will overload and damage the turbine. V 2 Next put the compass point at Z on the reactive lead scale, extend it S down the reactive scale for 140 mm and swing in the excitation limit arc. Why 140 mm? Well, at zero current the excitation phasor will be coincident with 2 V, i.e. 50 mm long. This is known as 100% or 1.0 p.u. ZS excitation. We are told that the excitation limit is to be 2.8 p.u. (or 280%) so that 2.8 50 mm = 140 mm. So the maximum length of the excitation phasor is 140 mm pivoting 2 from V. Z S FIGURE 9 shows the operating chart completed to this point.

29 Reactive lead p.u. 1.0 V 2 Z S 0.8 0.6 0.4 Turbine (or prime mover) limit Theoretical stability limit 0.2 0 0.2 0.4 0.6 0.8 1.0 Real power p.u. 0.2 0.4 0.6 Excitation limit (Rotor heating limit) 0.8 1.0 Reactive lag p.u. Stator heating limit (MVA limit) FIG. 9 Partial Operating Chart: Exercise 2

30 We are now ready to put in the practical stability limit. This has been left till last since it is the most complicated to construct. Constructing the practical stability limit Project the turbine limit on your operating chart up until it crosses the theoretical stability limit. Now measure the length of the portion of the theoretical stability limit line between the reactive power axis and the turbine limit and divide it by 10 (remember 10%, i.e. 1/10 practical stability margin in the data). This should give you 83.3 mm/10 = 8.33 mm Divide up the line into 10 equal spaces 8.33 mm long. (This is only to aid construction so faint pencil will do.) V 2 Now draw the 9 lines the full height of as in FIGURE 10. Z S

31 Q lead V 2 Z S Turbine limit Stability limit 8.33 mm P Q lag FIG. 10 Next, put the point of the compasses on V Z extending it along the stability limit line to the turbine limit line. 2 S Keeping this radius, swing an arc down until it cuts the next vertical line 8.33 mm to the left of the turbine limit. Put a small cross at the intersection of the vertical line and the arc. Now put the compass pencil on the line 8.33 mm to the left of the turbine limit along the stability limit line. Swing this arc down until it cuts the next vertical line to the left then mark the intersection with a cross.

32 Repeat the procedure for all the remaining vertical lines as in FIGURE 11. Compass point Compass pencil Turbine limit V 2 Z S Stability limit 8.33 mm SCR FIG. 11 We are now ready to join up the crosses except that there is one missing on the turbine limit line. It is therefore necessary to go beyond this line by 8.33 mm and swing an arc back as shown in FIGURE 12. Do this now. Joining the crosses should give you a nice smooth curve. We have now put in a practical stability limit based on 10% full load power. An old fashioned term (which sums it up better) is 10% power in hand.

33 V 2 Z S 8.3 mm Practical stability limit FIG. 12

34 The 10% Practical Stability Limit: The Theory Suppose a machine has a certain excitation on and is working along the practical stability limit line as shown below in FIGURE 13. V 2 Z S Theoretical stability limit E Practical stability limit I Turbine limit FIG. 13 Practical Stability Limit We know that the practical stability limit line has been constructed using 10% power in hand system. So, if the machine is working on the line, a sudden increase in real power input of 10% will move the excitation phasor to the next 10% power increment to the right. Since its length remains unaltered, it will lie coincident to the theoretical stability limit line. This is shown in FIGURE 14.

35 V 2 Z S E I Theoretical stability limit Practical stability limit 10% power increase Turbine limit FIG. 14 Real Power Increase on Practical Stability Line Can you now see what will happen when a machine working on a 10% of full load power stability limit line experiences a sudden increase in 10% of full load power? The increase will cause the machine to be working on its limit of stability. Providing a machine is working on, or within, the practical stability limit, it will always have a 10% power-in-hand factor before becoming unstable. For normal operation this is quite adequate. Look at the Capability Chart of FIGURE 8. You will see that a system of boundaries has been formed between the zero power axis, part of the excitation limit, part of the stator MVA limit, part of the turbine limit and the practical stability limit.

36 It is often helpful to lightly shade in along the inside of the envelope of boundaries as shown in FIGURE 15. (It will be beneficial if you photocopy two or three copies of this figure for future use in the lesson.)

37 Reactive power Q lead p.u. 0.8 Turbine Limit 0.6 Theoretical Stability Limit 0.4 Practical Stability Limit 0.2 0.83 Real power P p.u. 0 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 Excitation Limit 0.8 1.0 Stator Heating Limit Q lag p.u. FIG. 15

38 Providing the machine is operated within the boundaries, no risk of damage or instability should ensue. In other words, the machine is capable of working within the boundaries. Hence the name Capability Chart. Before we go any further, review what you have done in this section to make sure you can both read the capability chart and understand the steps needed to construct it. Then test yourself by doing Self-Assessment Question 3.

39 SELF-ASSESSMENT QUESTION 3 You are given the following data about a 600 MVA, 11 kv, 50 Hz, 3 ph, star connected, synchronous generator connected to an infinite busbar. SCR = 0.5 maximum power limit = 500 MW excitation limit is that which gives the rated MVA at p.f. = 0.707 lag Construct the operating chart of the generator assuming a stability limit of 10% of maximum power output. Compare your chart with that shown on page 46.

40 GENERATOR PERFORMANCE LIMITS This section takes the operating or capability chart you produced in the previous section and uses it to predict the performance limitations of any particular machine with regard to its real and reactive power output. Look at FIGURE 15 again the capability chart. If the machine is working at any point on the envelope of boundaries, the real and reactive powers can be read off the respective axes. Suppose the machine is operating at the intersection of the stator MVA limit and the excitation limit. You can now determine a certain amount of information about how the machine is working. From FIGURE 15 work out and note in the spaces below and compare with our answers on the next page. the real power output the reactive power output the p.u. excitation the operating power factor the apparent power output.

41 the real power output From the intersection of the two limit lines project up onto the p.u. real power axis and read the value. It should be in the order of 0.71 per unit. Since the basic power value of the machine was given as 600 MW the real power figure is found by multiplying the per-unit value by the base value, so that real power = 0.71 600 = 426 MW. the reactive power output In a similar manner project across from the intersection of the two limits and read off the p.u. reactive power. It should be in the order of 0.71 per unit. It is just a coincidence that this is the same as your previous reading. The actual reactive power = 0.71 600 = 426 MVAr lag. the p.u. excitation Since the machine is working on the excitation limit, the p.u. excitation must be 2.8 (from original data used to construct the chart). the operating power factor This can be obtained by using the real and reactive power in the form (MW jmvar) i.e. ( ) ( 426 j426) S = P jq = So S = 602 45 Therefore, the operating power factor = cos 45 = 0.707 p.f. lag. An added bonus is that we can find the apparent power output at the same time, although this will be checked using another method later.

42 An alternative method for finding the power factor is to draw in the current phasor from the zero power axis to the stator limit intersection and measure the power factor angle. (See FIGURE 16) Q lead φ P Excitation limit Stator limit Q lag FIG. 16 An alternative method of finding the power factor φ should be in the order of 44 so that cos 44 = 0.719 p.f. lag. Don't worry that the two methods of finding the power factor don't tie up exactly remember they have been obtained from a scale drawing. the apparent power output Since the machine is working on the stator MVA limit, the apparent power output must therefore be 600 MVA. Now test how well you can interpret the Capability Chart by doing Self- Assessment Question 4.

43 SELF-ASSESSMENT QUESTION 4 Look at Diagram 15 again, but suppose now that the machine is working on the practical stability limit and producing 0.6 p.u. real power. Find the apparent power the reactive power the p.u. excitation the operating power factor the control adjustments necessary to enable the machine to produce 0.4 p.u. real power but still work on the practical stability limit. Our answers start on page 47.

44 ANSWERS TO SELF-ASSESSMENT QUESTIONS 1. Real power increase = 45 26 26 = 73. 07% increase excitation increase See FIGURE 17. = = 178 123 123 44.% 7 increase 2. For a fixed value of excitation the load angle (δ) increases as the power input increases. This means that if the real power input continues to be increased when excitation lies along the stability limit, the tip of the phasor would swing back towards the zero power axis instead of moving away from it and the electrical power fed to the busbars would tend to decrease. This is a serious situation since there is now a mismatch between the mechanical power input and the electrical power output, i.e. the mechanical power input exceeds the electrical power output. 3. See FIGURE 18 following.

45 Reactive power Q lead MVAr 50 V 2 Z S 40 δ 30 20 E 1 10 I 1 ψ 0 φ 10 20 30 40 50 Real power MW E 2 10 I 2 20 30 40 Reactive power Q lag MVAr 50 FIG. 17 Solution to Self-Assessment Question 1

46 Reactive power Q lead p.u. 0.8 Turbine Limit 0.6 Theoretical Stability Limit 0.4 Practical Stability Limit 0.2 0.833 Real Power p.u. 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 Excitation Limit 0.8 Excitation limit pf = 0.707 lag 1.0 Stator Heating Limit Q lag p.u. FIG. 18

47 4. FIGURE 19 shows the construction the apparent power Put the compass point on the zero power axis and the pencil on the intersection of 0.6 p.u. real power and the practical stability limit. Swing the pencil down to cut the real power axis (at about 0.625 p.u.). The apparent power will be 0.625 600 = 375 MVA the reactive power Read this off in p.u. form. It should be about 0.17 p.u. lead. The new excitation length is 69 mm. So excitation must be reduced from 91 mm to 69 mm = 22 mm = = 22 % 91 24.% 2 reduction. So the reactive power is 0.17 600 = 102 MVAr lead.

48 Reactive power Q lead p.u. 0.8 Turbine Limit 0.6 Theoretical Stability Limit 0.4 Practical Stability Limit 0.2 φ 0.2 0.4 0.6 0.8 0.83 1.0 Real Power p.u. 0.2 0.4 0.6 Excitation Limit 0.8 1.0 Stator Heating Limit 0.833 p.u. Q lag p.u. FIG. 19

49 the p.u. excitation Put the compass point on operating point. V X S 2 and the pencil point on the Swing the pencil down to cut the reactive power axis. The length should be 69 mm from compass point to pencil point. Since 1.0 p.u. excitation = 50 mm p.u. excitation = 69 50 = 138p.u.. the operating power factor Find this as before. φ should be about 15.5 so that the power factor = cos φ = 0.96 lead. control adjustments to move to new position. Only two adjustments can be made, i.e. power input and excitation. To reduce power from 0.6 p.u. to 0.4 p.u. means a decrease in real power input of 0.2 p.u. = 0.2 600 = 120 MW decrease.

50 SUMMARY In this lesson we have seen that it is important to restrict the operation of power system generators to within certain limits in order to ensure system stability and avoid damage. We have developed the operating chart for a large generator, showing the four main limits which constitute operational boundaries. Finally, we have used the chart to determine the output performance limitations of a large generator which must remain within stated operational boundaries.