Load Characteristics, Tariffs and Power Factor Improvement

Parts 2 to 6 of this book treated the load as a number. A line was analysed for a receiving-end power, a load flow was solved for fixed injections, a stability study assumed a constant \(P_m\). Part 7 removes that convenience. The load is not a number, it is a time series, it is set by millions of people acting independently, and the system operator has no control over it whatsoever.

That would be tolerable if electricity could be stockpiled. It cannot — not in the quantities a grid deals in. A power station is not a factory that builds to inventory; it is a machine that must produce, at every instant, exactly what is being consumed at that instant, because the alternative is a frequency excursion of the kind Chapter 33 studies. Generation therefore follows demand continuously, and demand varies by a factor of two or three over a single day.

Three consequences follow, and between them they account for most of what Part 7 is about.

Plant must be built for the peak. The installed capacity is fixed by the highest demand of the year, plus a reserve. Every megawatt of that capacity carries interest, depreciation, insurance and staffing whether it runs for eight thousand hours or for eight. Since the average demand is well below the peak, a large part of the investment is idle a large part of the time.

Efficiency suffers. An alternator and its turbine reach best efficiency near rated output. A single machine sized for the peak would spend the night at a fraction of its rating, at poor efficiency and high heat rate. The remedy is several smaller units, brought on and off as demand moves — which improves running cost but raises capital cost per kW and floor area, and creates the scheduling problem of Chapters 31 and 32.

Auxiliary plant must track the load. Raising output means raising the flow of fuel, air and feedwater in step. That calls for controls and margin in every auxiliary, and it is why a station's response rate is finite — the ramping constraint that reappears in unit commitment.

A load curve is the plot of load against time. Readings are taken half-hourly or hourly and plotted in sequence, giving a daily load curve; averaging the daily curves of a month at each hour gives the monthly load curve, used for setting rates; assembling the monthly curves gives the annual load curve, from which the annual load factor is computed.

Because the ordinate is power in kW and the abscissa is time in hours, the area under the curve is energy in kWh. This is the single most useful property of the diagram, and every quantity in Section 30-3 is read from it.

The load curve answers "what was the load at 3 p.m.?" A planner more often asks "for how many hours in the year was the load above 40 MW?", and reading that off a chronological plot means measuring dozens of separate intervals. Rearranging the same data in descending order of magnitude answers it at a glance. The result is the load duration curve: the highest load at the left, falling monotonically to the lowest at the right, with the abscissa now reading "hours for which the load equalled or exceeded this value".

No information is lost or gained in the rearrangement — only the chronology, which the planner does not need. In particular the area under the two curves is identical, because the same rectangles of energy have merely been sorted. Extending the abscissa from \(0\) to \(8760\) hours summarises an entire year of demand in one monotone curve, and that annual load duration curve is the working document for deciding how much of each kind of plant to build.

Four quantities describe a single consumer or a single station, and each is defined so that it can be read from the load curve without further data.

The connected load is the arithmetic sum of the continuous ratings of every piece of equipment on the installation. It is a nameplate figure and nothing more; nobody switches everything on at once.

The maximum demand is the greatest load actually drawn during a stated period, taken as an average over a defined interval — usually thirty minutes, because it is the thermal time constant of transformers and cables that matters, not an instantaneous spike. The averaging interval must always be quoted with the figure, since a shorter interval yields a larger maximum demand from the same load.

The ratio of the two defines the demand factor, which is what a designer actually uses to size a supply.

The average load over any period is the energy consumed divided by the duration, which is the area under the load curve divided by its base:

Finally, the ratio of average load to maximum demand is the load factor, and it is the most informative single number about any load whatsoever. Multiplying numerator and denominator by the duration \(T\) turns it into a ratio of areas, which is why it can be read off the diagram directly as the fraction of the enclosing rectangle that the load curve fills.

A high load factor means the plant is being used near its maximum most of the time. Since capacity is bought to meet the maximum demand, a high load factor spreads that fixed investment over more units of energy and lowers the cost of each — a statement made quantitative in Section 30-7. A low load factor means expensive plant standing idle.

The previous section described one load. A station serves many, and their peaks do not coincide: the domestic peak is in the evening, the industrial peak in mid-morning, the irrigation peak at night. The station therefore never has to supply the arithmetic sum of its consumers' individual maximum demands, and the ratio of the two measures how much the mismatch is worth.

The two factors work in the same direction. A large demand factor and a small diversity factor both push the station's maximum demand up, requiring more plant and more capital for the same energy sold. Deliberately mixing consumer classes on one station — some domestic, some industrial, some agricultural — raises the diversity factor and is one of the cheapest ways to reduce the capacity required per unit of energy delivered.

Two further ratios compare what the plant produced with what it could have produced. The plant capacity factor uses the installed capacity as the reference:

Dividing this by the load factor makes the relationship between the two explicit, and shows exactly what the gap between them measures:

So the plant capacity factor equals the load factor exactly when the maximum demand equals the plant capacity — that is, when there is no reserve at all. In every real station the capacity exceeds the maximum demand, so the capacity factor is the smaller of the two, and the gap between them is a direct measure of the reserve carried.

The plant use factor asks a narrower question: while the plant was actually running, how hard was it worked?

A peaking station may show a capacity factor of \(8\%\) and a use factor of \(80\%\): it is idle most of the year, but when it does run it runs nearly flat out. Quoting the capacity factor alone would make it look like a badly utilised asset rather than what it is — a machine bought for a small number of expensive hours.

The load curve of a system is the sum of the curves of its consumer classes, and each class has a characteristic shape, load factor and weather sensitivity. Knowing which classes a station serves is what allows its curve to be predicted before it is measured.

Domestic load — lighting, fans, refrigeration, television, small water pumps. It is concentrated in the evening, with a smaller morning peak, and is almost dormant in the middle of the night. Its load factor is the worst of any class, typically \(10\%\) to \(12\%\), which makes it the most expensive class to serve per unit sold and explains why domestic tariffs are rarely the cheapest despite the small demands involved.

Commercial load — shop and office lighting, air conditioning, refrigeration in restaurants. It runs for more hours of the day than domestic load and so has a better load factor, but it is strongly seasonal, since air conditioning and space heating dominate the extremes of the year.

Industrial load — motors, furnaces, process plant. Magnitude varies from a few kilowatts in a small workshop to tens of megawatts in a steel or aluminium works; conventionally, small-scale industry is taken as up to about \(25\) kW, medium scale between \(25\) and \(100\) kW, and large scale above \(500\) kW. Industrial load is largely weather-independent and, where the plant runs in shifts, has by far the best load factor of any class. It is also the class with the worst natural power factor, for the reason set out in Section 30-8.

Municipal load — street lighting, water supply and drainage pumping. Street lighting is essentially constant through the hours of darkness. Water pumping is discretionary in timing, and utilities deliberately schedule it into the small hours, filling overhead tanks overnight and improving the system load factor at no cost to anyone.

Irrigation load — pump sets in agricultural areas, again supplied largely at night, typically for a block of about twelve hours, and again used as a filler for the overnight trough.

Traction load — trams, trolleybuses, suburban and main-line railways. It has the widest swing of any class, peaking sharply in the morning and again in the evening as people travel, and its rapid fluctuation makes it a difficult load for a generator to follow.

Draw a horizontal line across the load duration curve. Everything below the line is demand that persists for the whole period — the base load. Everything above it appears for a few hours and vanishes — the peak load. The two have completely different economics, and separating them is the central idea of station planning.

Base-load energy is large in quantity and predictable in timing, so it should be produced by whatever plant has the lowest running cost per unit, however expensive that plant is to build: large thermal sets, nuclear stations, run-of-river hydro. Peak-load energy is small in quantity and needed at short notice, so it should come from plant that is cheap to build and quick to start, however expensive its fuel: gas turbines, diesel sets, pumped storage.

Attempting to serve the whole curve with one kind of plant is wrong in both directions. Base-load plant asked to follow the peak is cycled, which wears it and worsens its heat rate; peaking plant asked to carry the base burns expensive fuel for eight thousand hours a year. The split is quantified in Example 2, where the base unit finishes with a load factor of \(84\%\) and the peaking unit with \(17\%\) — and both figures are correct for their purpose.

The oldest and best method of improving the picture further is to stop treating a station as an island. An interconnected grid — several stations operating in parallel, which is the arrangement Chapter 2 described physically and Chapters 16 to 20 analysed electrically — delivers a list of benefits that all trace back to diversity:

Every one of these is an economic argument, and every one of them is bought with the transmission network of Parts 2 and 3 and paid for with the stability constraints of Part 6. That trade — reliability and economy against synchronism — is the permanent tension in power system operation.

A tariff is not an arbitrary price list. It is an attempt to recover a cost, and the shape of the cost decides the shape of the tariff. The annual cost of supplying a consumer separates into three parts.

Fixed cost is independent of both the maximum demand and the energy: land, buildings, the cost of the distribution network's existence, management and metering.

Semi-fixed cost is proportional to the maximum demand, because that is what determines the size of the generator, transformer, switchgear and conductor that must be provided. It consists of interest and depreciation on that plant, plus a share of the salaries and maintenance that scale with plant size. It is incurred whether or not a single unit is drawn.

Running cost is proportional to the energy actually taken: fuel, lubricating oil, water, and the wear that scales with output.

Now divide by the units consumed to obtain the cost per unit, and eliminate the energy using the load-factor relation of Section 30-3, \(\text{kWh}=\text{MD}\times\text{L.F.}\times8760\). Ignoring the small constant \(a\):

That single expression is the justification for the whole family of tariffs. Any tariff that charges for energy alone forces the high-load-factor consumers to subsidise the low ones; any tariff that charges for demand alone gives away energy. The honest structure has two terms, matching the two terms of the cost.

A workable tariff also has to satisfy several practical requirements at once: it must recover the total cost with a fair return, be simple enough that a consumer can understand and predict the bill, be cheap to meter, apportion the cost in proportion to the burden each consumer places on the system, and encourage behaviour that improves the system — a flatter curve and a better power factor.

Chapter 3 defined the power triangle: for a three-phase load drawing line current \(I\) at line voltage \(V_L\), the apparent power is \(S=\sqrt3\,V_LI\), the active power is \(P=S\cos\phi\) and the reactive power is \(Q=S\sin\phi\). Only \(P\) does work. Yet every piece of equipment between the generator and the load is sized by \(S\) or by \(I\), never by \(P\), and that mismatch is what the power factor clause on a bill is about.

Invert the definition to see the damage. At fixed \(P\) and \(V_L\), the current a load draws is

A load at \(\cos\phi=0.7\) draws \(43\%\) more current than the same load at unity, and causes \(104\%\) more copper loss — the losses have roughly doubled for no extra useful output. Four separate costs follow from this one relation:

Larger conductors and plant. Cables, transformers, switchgear and alternators are rated in kVA, so serving \(P\) at a poor power factor ties up more capacity. An \(800\) kVA transformer delivers \(800\) kW at unity power factor and only \(560\) kW at \(0.7\).

Higher losses. The \(1/\cos^2\phi\) term above, paid for as fuel every hour of every year.

Worse regulation. Chapter 13 showed that the voltage drop along a line is approximately \(I(R\cos\phi + X\sin\phi)\); a larger current at a larger angle raises the drop on both counts, forcing larger conductors or more tap-changing.

Reduced system capability. The reactive current occupies the thermal rating of every element it passes through, so a system running at poor power factor reaches its limits while delivering less real power than it was built for.

The causes are structural rather than accidental. Most industrial load is induction motors, which draw magnetising current at close to \(90^\circ\) lagging; a motor's power factor is worst when it is lightly loaded, since the magnetising component is nearly constant while the load component falls. Arc and induction furnaces, welding sets, transformers on light load and discharge lighting all behave the same way. Left alone, an industrial installation settles at \(0.7\) to \(0.8\) lagging.

Since the utility's costs scale with kVA and its revenue under a simple tariff scales with kW, it recovers the difference by billing on the kVA maximum demand, or by adding a penalty below and a rebate above a stated power factor. Whichever form is used, the consumer's incentive is the same: raise \(\cos\phi\).

Correction means supplying the load's reactive demand locally instead of drawing it over the network. A capacitor connected in parallel with the load draws a leading current that cancels part of the lagging current the load takes; the active component is untouched, so \(P\) stays fixed and the triangle collapses inward.

Work in reactive power directly. Before correction the load draws \(Q_1 = P\tan\phi_1\); after correction the network must supply only \(Q_2 = P\tan\phi_2\). The difference is what the capacitor provides.

Three devices supply the leading kVAr, and the choice between them is a matter of size and duty. Static capacitors are cheap, have negligible losses, need almost no maintenance and can be switched in steps; against that, they are damaged by over-voltage, have a limited life, and — since \(Q\propto V^2\) — deliver least reactive power exactly when the voltage has sagged and it is most needed. Synchronous condensers are over-excited synchronous motors running on no load; they give continuously variable output of either sign and support the voltage smoothly, but they have rotating losses, need starting equipment and are only economic in large sizes. Phase advancers are fed into the rotor circuit of a wound-rotor induction motor to supply its magnetising ampere-turns at slip frequency, correcting the motor at source. Chapter 34 returns to all of these as instruments of voltage control rather than of billing.

The bank itself is sized from the kVAr. For a delta-connected three-phase bank across a line voltage \(V_L\), each of the three capacitors sees the full line voltage, so

The delta connection needs one third the capacitance for the same kVAr, at the price of full line-voltage insulation on each unit — the same trade that appears whenever a three-phase bank is connected either way.

How far should the correction be taken? Not to unity. Improving the power factor saves money on the demand charge but costs money in capacitors, and beyond a certain point the second effect wins. The optimum can be found exactly, and the answer is remarkably clean.

Let a consumer take \(P\) kW at \(\cos\phi_1\), let the annual maximum-demand charge be Rs \(A\) per kVA, and let the annual cost of the correction plant — interest and depreciation on the capacitors — be Rs \(B\) per kVAr. Assume the load is steady, so the maximum demand in kVA is \(P/\cos\phi_2\) after correction. The total annual cost as a function of the corrected angle is

Differentiate with respect to \(\phi_2\), remembering that \(\phi_1\) and \(P\) are constants:

The second derivative is positive at this point, so it is a minimum; and since \(B < A\) in any realistic case, \(\sin\phi_2 = B/A\) always has a solution. Unity power factor would require \(B=0\), which is to say free capacitors.

The minimum is a shallow one, and that matters practically. For the numbers of Example 6 the total annual cost varies by less than \(0.4\%\) between \(\cos\phi_2 = 0.90\) and \(0.95\), so a designer may choose a standard bank size anywhere in that range with no real penalty — but pushing on to \(0.99\) costs more than leaving the power factor uncorrected.

Take a year as \(8760\) hours throughout. Where a tariff is given in rupees the numbers are illustrative; the method is what is being tested.