Load Characteristics and Utilisation

The energy is the area under the curve. Since the curve is piecewise constant, the integral is a sum of rectangles:

The hours must sum to 24 — the first arithmetic check, and one that catches most transcription errors immediately.

The three headline numbers:

The average demand is the height of a rectangle of the same area — the flat load that would deliver the same energy over the day. It is not the arithmetic mean of the seven demand levels, which would be 67.14 MW by coincidence here but is a meaningless quantity in general because the periods have different lengths.

Reading the shape. Four features are visible and every one of them has a cause:

The evening peak is the defining feature of a mixed domestic-commercial system: it occurs because domestic demand rises just as commercial demand has not yet fallen. A purely industrial area has no evening peak, and its load curve is nearly flat.

The ratio that governs everything else:

The system must have plant capable of 110 MW, but for six hours a day it needs only 30 MW. That mismatch — capacity idle most of the time, yet indispensable for four hours — is the whole economic problem of electricity supply, and every quantity in this chapter is a way of measuring one aspect of it.

Annualising, as a first approximation. Taking this as a typical day:

In practice weekends run 20–30% below weekdays and the seasonal variation is larger still, so a real annual figure requires eight or twelve representative day-types weighted by their frequency. The single-day approximation typically overstates annual energy by 10–15%.

Why the chronological curve matters as well as the reordered one of Problem 3:

The ramp between 14:00 and 18:00 is 25 MW in four hours, and between 22:00 and midnight the load sheds 65 MW in two. Reordering the curve by magnitude destroys exactly that information, which is why both forms are kept.

The definition:

The last form is the useful one: it is the energy actually supplied divided by the energy that would have been supplied had the maximum demand persisted for the whole period. It is a dimensionless number between 0 and 1.

For this system:

A daily load factor of 61%, which is typical for a mixed urban supply. Note the period must be stated: the same system's annual load factor is lower, because the annual peak exceeds any single day's peak while the annual average does not rise correspondingly.

Why it matters is a cost argument, not an engineering one. The cost of supply splits into two parts:

The fixed component per unit of energy is inversely proportional to load factor. Doubling the load factor halves the capital cost carried by every kilowatt-hour sold, without changing the fuel cost at all.

Quantified for this system. With annual fixed cost \(F = \) Rs 8000/kW and running cost \(R = \) Rs 2.50/kWh:

Using \(F/(8760\times\text{LF})\). Raising the load factor from 0.30 to 0.61 cuts the cost of supply by 28%, and every rupee of that saving comes from spreading the same capital over more kilowatt-hours.

Typical values, which are worth knowing because they differ by a factor of four:

The individual domestic figure is the striking one: a household with a 4 kW maximum demand consuming 300 kWh a month has a load factor of 0.10. It is only aggregation that raises the system figure to 0.61 — the subject of Problem 5.

How a utility raises load factor, since it cannot change consumers' habits directly:

Every one of these is an attempt to buy energy in the trough and sell it at the peak, and each is worth up to the difference between peak and off-peak marginal cost — computed for this system in Problem 11.

The construction. For each demand level, count the total time at or above it:

The result is a monotonically decreasing staircase from (0 h, 110 MW) to (24 h, 30 MW). Time is no longer clock time — it is duration, and the horizontal axis has lost all information about when anything happened.

Verify the area. Integrating horizontally, in strips of constant demand:

Each term is a horizontal strip: its height is the step in demand between consecutive levels and its width is the duration at the upper level. The total matches Problem 1 exactly, which is the structural check that the reordering was done correctly.

What is preserved and what is lost:

The last row deserves emphasis. The duration curve says the load exceeds 85 MW for 8 hours a day; it does not say whether that is one continuous 8-hour block or two separate 4-hour blocks. For plant that takes 6 hours to start, those are completely different problems.

Load factor read off the duration curve. It is the ratio of the area to the enclosing rectangle:

Which gives the load factor an immediate geometric meaning: how well the load duration curve fills its bounding rectangle. A perfectly flat load fills it completely; a single brief spike fills almost none of it.

Why the duration curve is the right tool for plant mix. A unit of capacity placed at height \(P\) on the curve:

So the duration curve reads directly as a schedule of how hard each successive megawatt of capacity will work — 24 hours a day for the first 30 MW, 4 hours a day for the last 25 MW. That is exactly the information the capital-versus-fuel trade-off needs, and it is why Problems 9–11 are built on this curve rather than the chronological one.

The annual load duration curve is the form actually used in planning:

An annual curve built from 8760 hourly readings is smooth rather than a staircase, and its characteristic shape — a steep drop over the first few hundred hours, then a long flat tail — is why a small amount of very expensive peaking capacity is economic while a large amount is not.

Connected load is the arithmetic sum of the nameplate ratings of every device that could be switched on:

It is a purely notional quantity — the demand that would occur if literally everything were switched on simultaneously, which never happens and in most systems could not physically be supplied.

The demand factor:

Less than half the connected load is ever drawn at once, and this is a high figure for a system with a large domestic component. Demand factor is always less than 1 — a system in which it approached 1 would be one where every consumer switched everything on together, which is precisely the condition that never occurs.

Why it is always less than one. Two independent reasons, easily confused:

The overall figure of 0.447 contains both effects. Separating them is the subject of Problem 5, and doing so matters because the two are influenced by completely different things.

Typical group demand factors:

The trend is systematic: the more purposefully the connected equipment was chosen, the closer the demand factor is to 1. Street lighting is the limiting case — nothing is installed that will not be used.

Applying the group demand factors to get each group's maximum demand:

166.8 MW against a system maximum of 110 MW. The 56.8 MW difference is entirely diversity — the groups do not peak together — and it is quantified in Problem 5.

What demand factor is used for in practice:

It is a design tool. A distribution engineer sizing the transformer for a new housing estate knows the connected load from the drawings and has no load curve at all; a demand factor of 0.55 turns one into an estimate of the other. Published tables of demand factors by consumer class exist for exactly this purpose, and the estimate is only as good as the table's match to the local mix.

The definition:

Note the direction. Unlike load factor and demand factor, diversity factor is greater than one — a source of endless confusion, and the reason some texts use its reciprocal under the name coincidence factor.

Between the four groups:

The four groups' peaks, added arithmetically, exceed the system peak by 52%. That 56.8 MW is capacity that never has to be built, and it is a direct consequence of the groups peaking at different times of day.

Why they miss each other is visible in the load curve of Problem 1:

The system peak occurs at the moment when the sum happens to be largest, which is generally not any individual group's own peak. Diversity is therefore not a random cancellation but a systematic consequence of daily rhythms.

Within the domestic group:

Smaller than the between-group figure of 1.5164, and for a good reason: 60 000 households do broadly the same thing at the same time each evening, so their peaks coincide far more than a domestic peak coincides with an industrial one.

The hierarchy of diversity, which is the practical structure of the whole idea:

Each level multiplies. The cumulative effect from a single appliance to the system is a factor of five to ten, and it is the single largest economy in the whole industry — the reason a household with 8 kW of appliances contributes only about 1 kW to the system peak.

The design consequence. A distribution transformer is sized from:

For 200 houses at 2.0 kW connected with DF 0.55 and a distributor diversity of 1.3: \((200\times2.0\times0.55)/1.3 = 169\) kW, so a 200 kVA transformer. Ignoring diversity would give 220 kW and a 250 kVA unit — 25% more iron for no benefit. This one calculation, repeated a million times, is where diversity earns its money.

Where diversity is falling, which is a live concern:

The first three are the reason distribution networks designed on 1970s diversity assumptions are being reinforced. A weather-driven load has almost no diversity at all: when it is cold, every heat pump runs, and the factor approaches 1.0.

Plant capacity factor — how hard the installed plant works:

The plant produces 51.6% of what it could produce running flat out. Note that "installed capacity" means all of it, including the units standing as reserve.

Plant use factor — how hard the plant works while it is running:

If the whole 130 MW is synchronised for all 24 hours, PUF = CF = 0.5160. But that is not how a system is run: at 03:00 only 30 MW is needed, so most of the plant is shut down. If, say, an average of 100 MW is synchronised across the day, PUF = \(1610/(100\times24) = \) 0.6708 — noticeably higher.

Utilisation factor — how much of the plant the peak actually calls for:

85% of the plant is needed at the peak; the remaining 15% is reserve. Utilisation factor is the one plant factor that says nothing about energy — it is purely about capacity.

The identity that ties the plant to the load:

Exactly. Two immediate corollaries: capacity factor can never exceed load factor, and it equals load factor only when there is no reserve at all.

The four factors compared, since keeping them apart is most of the difficulty:

The first two are properties of what the consumers do; the last two of what the utility built. A question about consumer behaviour is never answered by a plant factor, and vice versa.

Typical annual capacity factors, which show the plant-mix logic directly:

The first four are chosen capacity factors — the plant could run more but is not despatched. The last three are imposed: the resource is not there. Comparing a wind farm's 0.30 with a gas turbine's 0.10 and concluding the wind farm works harder is a category error, because one figure is a constraint and the other a decision.

What capacity factor is used for, and one thing it is not:

It is the standard way of converting a capacity into an energy, and therefore into a revenue. It is not a measure of efficiency, availability or reliability: a plant with 100% availability and a capacity factor of 0.05 is a perfectly healthy peaking unit doing exactly the job it was built for.

The three quantities, which are routinely confused:

Installed reserve is a planning quantity; spinning reserve an operating quantity; firm capacity a contractual and statistical one. Only the first is a simple subtraction.

The adequacy test most commonly applied is the single-largest-unit criterion:

The reserve is inadequate. Losing the 40 MW unit at the evening peak leaves 90 MW of plant against 110 MW of demand — a 20 MW shortfall, requiring immediate load shedding of 18% of the system.

What is required:

So 20 MW more must be built, or the largest unit must be split into smaller ones. Note the second option: two 20 MW units in place of one 40 MW unit would make the existing reserve adequate at once, at the cost of a worse heat rate and higher capital cost per MW. That trade-off — unit size against reserve requirement — is a standing question in system planning.

Spinning reserve through the day, assuming units are committed to give at least 15% headroom:

Spinning reserve must be synchronised and loadable within seconds — a shut-down unit is not spinning reserve however quickly it could start. At the peak the whole installed capacity is committed and the spinning reserve equals the installed reserve, which is why the peak is the binding hour for the adequacy test above.

Firm capacity is what remains after allowing for forced outages. With a forced outage rate of 8% per unit:

Above the 110 MW peak on average, but "on average" is not a planning criterion — the whole point is what happens on the bad days. The proper treatment is probabilistic, using the loss-of-load-probability method, and it typically demands a reserve margin of 15–25% rather than the single-largest-unit rule of thumb.

Reserve margin as usually quoted:

18.2% would be adequate for a large system, and is not adequate here — precisely because this system is small enough that one unit is 36% of its peak. Reserve requirements are set by the ratio of unit size to system size, not by the margin percentage alone.

The interconnection alternative. Rather than build 20 MW, the utility could contract for it:

Interconnection is almost always cheaper, because the reserve is shared: two systems each carrying their own largest unit as reserve need less total reserve when interconnected than apart. This sharing of reserve, rather than energy trading, was the original economic justification for building interconnected grids, and it remains the largest single benefit.

The rule. With base capacity \(C\), at each instant:

The base unit runs at whatever the demand is until it saturates at \(C\); the peaking plant supplies the excess. This is exactly a horizontal cut across the load duration curve at height \(C\) — everything below the cut is base energy, everything above is peak energy.

Worked at \(C = 70\) MW, block by block:

1350 + 260 = 1610 MWh ✓. The base unit supplies 84% of the energy from 64% of the peak capacity — the essential asymmetry that makes the plant-mix decision worth making.

The full relationship:

The base unit's own load factor is its energy divided by \(C\times24\). Note the trade-off in the last two columns: more base capacity captures more energy but works each megawatt less hard.

The shape is concave, and that is the whole point:

The derivative of base energy with respect to base capacity is simply the duration read off the load duration curve. Since that duration decreases as \(C\) rises, the curve is concave — each successive megawatt of base capacity earns less energy than the one before.

Checking the derivative numerically between two rows:

And the load duration curve says the demand is at or above 80 MW for exactly 12 hours a day ✓. This identity is what makes the economic optimisation of Problem 10 a one-line calculation instead of a search.

The 30 MW row is special. Below the minimum demand, base capacity runs at 100% load factor:

The first 30 MW of capacity is unambiguously worth building as base plant, whatever the cost figures — it runs continuously and displaces the most expensive energy on the system. This is why the minimum demand, not the average, sets the floor for base-load capacity.

The origin of the division is the load duration curve. The first 30 MW of capacity runs 24 hours a day; the last 25 MW runs 4 hours a day:

A megawatt that runs 24 hours and one that runs 4 hours are being asked to do completely different jobs, so it would be surprising if the same machine were best for both.

The two cost structures:

The first two rows are the essential ones and the rest follow from them: a plant with high capital cost must run continuously to recover it, so it is designed for efficiency rather than flexibility, which in turn makes it slow to start and unsuited to cycling.

The annual cost of each type as a function of running hours:

Two straight lines per MW of capacity, with \(F_b > F_p\) and \(R_b < R_p\). They must cross, and where they cross is the whole answer.

For this system, with representative figures:

The base plant costs six times as much to build and 42% as much to run. Both figures are per megawatt of installed capacity per year.

The break-even running time, obtained by equating the two lines:

Any megawatt that will run more than 7.83 hours a day should be base plant; any that will run less should be peaking plant. The load duration curve says which is which.

Applying it to the curve:

The switch occurs at exactly 85 MW, where the duration drops from 8 hours (just above \(h^*\)) to 4 hours (well below). Problem 10 confirms this by direct cost comparison.

The middle category that this two-way split omits:

Real systems have three or four bands rather than two, and the analysis extends directly: with \(n\) plant types there are \(n-1\) break-even durations, each defining a boundary on the load duration curve. Challenge 1 works a three-type case.

The cost function. Total annual cost with base capacity \(C\) and peaking capacity \(110-C\):

with \(E_b(C)\) and \(E_p(C)\) the daily base and peak energies from Problem 8, in MWh, converted to kWh for the running-cost terms.

Evaluated across the range, in crore rupees (10⁷ Rs) per year:

The minimum is at \(C = \mathbf{85\ MW}\), confirming the break-even argument of Problem 9. Note how shallow the minimum is between 80 and 90 MW — a feature of the problem, discussed below.

Why the optimum is exactly 85 MW, and not somewhere between the LDC steps:

This is negative (build more base) when \(t(C) > 7.83\) h/day and positive (build less) when \(t(C) < 7.83\). Since the duration is a staircase, the derivative changes sign at the step where the duration crosses 7.83 — between 8 h (at 85 MW) and 4 h (above it). The optimum sits exactly on the step.

The savings achieved:

The asymmetry is striking. Building all base plant costs 4.6% too much; building all peaking plant costs 40.5% too much. Erring towards base plant is far less costly than erring towards peaking plant, which is exactly what one would expect from a system with a load factor of 0.61 — most of the energy is base energy however you build it.

The shallowness of the optimum is worth quantifying, because it changes how much effort the calculation deserves:

A 12% error in the base capacity costs under 1% in total cost. Since the cost data themselves are uncertain by 10–20%, refining the optimum beyond the nearest available unit size is pointless — and, more importantly, other considerations (unit sizes actually on the market, fuel security, emissions, ramping capability) can be allowed to decide without much economic penalty.

The resulting average cost of supply:

Which becomes the basis of the tariff in Problems 14–16. Note it exceeds the base plant's running cost of Rs 2.50/kWh by 82%, the difference being the capital charge and the peaking plant's expensive energy.

What the model omits, and it omits a good deal:

Every one of these pushes towards more base capacity than the simple calculation gives, and the fifth is usually the largest: a base unit built for today's load duration curve will see a higher duration every year of its 40-year life.

The independence. The break-even condition equates two annual costs per megawatt:

Only cost data appear. The load curve does not enter at all — it decides how many megawatts fall on each side of \(h^*\), but not where \(h^*\) is. That separation is what makes the screening-curve method possible: the plant comparison can be done once and applied to any system.

A realistic set of plant types, with annual fixed costs in million rupees per MW and running costs in rupees per kWh:

The two columns run in opposite directions, as they must for the comparison to be interesting. Note that efficiency does not correlate simply with running cost — the CCGT is the most efficient plant on the list but burns the most expensive fuel.

The three adjacent break-even points:

Three thresholds dividing the year into four bands. That they come out in increasing order is not automatic — it is the condition for all four plant types to be economic somewhere, and a plant type for which the ordering fails is dominated and should never be built.

The screening chart is the resulting merit table:

Read the annual load duration curve at each of the three thresholds, and the megawatt levels found there are the capacity boundaries of the four plant types. The entire capacity-planning problem reduces to reading three points off one curve.

Checking the dominance condition. A plant type is worth building only if it is cheapest somewhere:

Suppose a fifth plant type had \(F = 8.0\) and \(R = 5.00\). Against OCGT its break-even is \((8-2)/(6.00-5.00) = 6000\) h; against CCGT it is \((8-6)/(5.00-4.00) = 2000\) h. Since 6000 > 2000, there is no band in which it wins — CCGT is cheaper wherever it would be, and the plant is dominated. Testing for dominance before doing any further work saves a great deal of effort.

Sensitivity to fuel price, which is what actually moves these boundaries:

The third and fourth rows give a break-even above 8760 hours, which means coal is never cheapest at any utilisation. A 25% swing in one fuel price, or a moderate carbon price, is enough to eliminate an entire plant category from the mix — which is why fuel-price and carbon-price forecasts dominate capacity planning far more than the engineering does.

What the screening method omits:

The last is the modern complication and it changes the answer qualitatively. With substantial wind and solar, the curve to screen against is the residual load duration curve — demand minus renewable output — which is far steeper and shorter-tailed. It shifts the whole mix towards flexible peaking plant and away from base load, which is the technical statement of a change now visible in every developed system's plant mix.

Scale the daily curve to an annual one. Taking the same shape with a peak of 120 MW and the same durations expressed as fractions of the year:

Scaling by 120/110 = 1.0909. A real annual curve would be smoother and would have a longer tail near the peak, but the exercise is identical.

Read the three thresholds off the curve. For each break-even duration, find the demand level:

Interpolating within the staircase steps, since a real curve is continuous. The three levels partition the 120 MW peak into four capacity bands.

The recommended mix:

Plus reserve. From Problem 7 the reserve must cover the largest unit, so if the nuclear plant is a single 40 MW unit the installed capacity must be 160 MW — and the extra 40 MW would sensibly be gas turbines, being the cheapest capacity to leave idle.

The immediate practical objection is unit size. A 40 MW nuclear station does not exist commercially:

A 120 MW system cannot use the plant types the screening chart recommends, because they do not come in small enough pieces. This is the central difficulty of small isolated systems, and the reason such systems are dominated by diesel engines and gas turbines with capacity factors that the screening chart says are uneconomic.

The realistic recommendation for a 120 MW isolated system:

Note what has changed: unit sizes are small so that the reserve requirement is manageable, and the mid-merit band is filled with engines rather than a CCGT. The economic penalty relative to the ideal mix is perhaps 5–10%, and it is unavoidable.

If the system is interconnected, the answer changes completely:

A 120 MW area inside a 20 000 MW grid does not build its own plant mix at all — it buys from the pool, and the screening chart is applied at the level of the whole grid where 800 MW units are a small fraction of the peak. The economies of scale in generation are large enough that this is almost always the better answer, and it is why isolated systems are rare outside islands and remote mines.

The growth law:

The rule of 72 gives \(72/6 = 12\) years, correct to 1%. Six per cent is a high growth rate typical of a developing system; mature systems run at 1–2% and some are now flat or declining.

When the existing 130 MW becomes inadequate. Requiring reserve to cover the 40 MW largest unit:

Which is already violated at \(n = 0\) — the system is short of reserve today, as Problem 7 found. Taking the weaker criterion that installed capacity merely exceed the peak:

The capacity timetable:

The requirement grows by about 10 MW every year and a half. With a coal station taking five years to build and a gas turbine eighteen months, the commitment decisions must be taken well ahead of the shortfall — which is why a capacity plan is drawn for 10–15 years and revised annually.

How growth changes the plant-mix answer. A unit built today will see a rising duration every year of its life:

A megawatt of capacity at the 85 MW level starts at 8 hours a day, comfortably above \(h^* = 7.83\), and rises to 20 hours within fifteen years. Averaged over a 30-year plant life its utilisation is far above the break-even, so the case for base plant is much stronger than the static calculation suggests.

The general principle:

The correct comparison is between the present values of the two cost streams over the plant life, not between one year's costs. Doing that here would move the optimum from 85 MW towards 95–100 MW of base capacity — but since Problem 10 showed the cost surface is flat to within 1% over that range, it barely matters in cost terms. What it does change is the direction of the error one should prefer to make.

The countervailing risk is that growth may not materialise:

The asymmetry runs the other way from the cost calculation. Peaking plant is cheap to build and cheap to abandon; base plant is neither. In a system where the growth forecast is uncertain, the option value of building short-lead-time peaking plant and deferring the base-load decision can outweigh its higher running cost — which is a large part of why gas turbines proliferated in liberalised markets during the 1990s and 2000s.

Forecasting the growth rate is the weakest link:

The last is now standard practice, precisely because the others produce a single number with false confidence. A capacity plan robust across several scenarios is worth more than one optimal for the most likely.

The three categories, distinguished by what they depend on:

The middle category is the interesting one and the one that gives the subject its shape. Semi-fixed costs are capital costs, so they do not vary with output — but their size was determined by the maximum demand the system had to meet.

The cost equation that follows:

Three terms, and this three-term structure is the origin of the three-part tariff. Everything else in tariff design is a simplification or elaboration of it.

Apportionment for this system, using the Problem 10 result of Rs 266.69 crore a year:

The split is roughly 40:60 between capacity and energy, which is typical. In a hydro-dominated system it would be 80:20; in a system running mostly gas turbines, 20:80.

Why an energy-only tariff is unjust. Consider two consumers taking the same annual energy:

Both buy the same energy, so an energy-only tariff charges them the same. But B requires four times as much plant, four times the transformer, four times the cable. Under an energy-only tariff, A subsidises B — and the subsidy is a direct incentive for B not to improve.

The cost each actually causes, using Rs 973 per kW of MD per year and Rs 2.72/kWh:

B costs 14.6% more to supply for identical energy. A tariff that recovers that difference is not a penalty on B — it is an accurate bill, and it gives B a financial reason to raise his load factor, which benefits everyone.

The tariff forms in use, each a way of recovering the three terms:

The block-rate tariff is worth a note: by charging less for later blocks it makes the average rate fall with consumption, which mimics a two-part tariff without requiring a maximum-demand meter. It was the standard domestic tariff for most of the twentieth century for exactly that reason, and smart metering is now displacing it with time-of-day pricing.

The tariff:

The demand charge is on kVA, not kW — the Hopkinson form. That choice matters and is the subject of Problems 17–19; for now take unity power factor so kVA = kW.

Express the energy in terms of the demand and the load factor:

Divide by the energy to get the average rate:

The maximum demand has cancelled entirely. The average rate depends only on the load factor, not on the size of the consumer — a large and a small consumer with the same load factor pay the same rate per unit. That is the defining property of a two-part tariff, and it is exactly the right one: it charges for the pattern of consumption rather than penalising size.

Numerically:

Setting \(\bar c = 3.00\) gives \(\text{LF} = 0.20548/0.80 = \mathbf{0.2568}\). A consumer at 26% load factor pays Rs 3.00/kWh; one at 80% pays Rs 2.46 — 18% less for identical energy.

The shape is a rectangular hyperbola asymptotic to \(B\):

The steepness at low load factor is the tariff's whole point. A consumer at 0.10 pays 77% more than one at 1.00, and the marginal benefit of improving is greatest exactly where the load factor is worst — which is where the incentive is most needed.

Setting \(A\) and \(B\). The utility must recover its costs, so from Problem 14:

The tariff quoted (\(A = 1800\), \(B = 2.20\)) recovers more through the demand charge and less through energy than a strict cost apportionment would. That is a deliberate choice: loading the demand charge sharpens the load-factor incentive, at the price of a bill that varies less with consumption and is therefore less popular with small consumers.

Checking that the tariff recovers the revenue. With a system load factor of 0.6098 and a peak of 110 MW at unity power factor:

Against costs of Rs 274.69 crore. The tariff under-recovers by 46% — because the system's diversity means the sum of consumers' individual maximum demands (166.8 MW, Problem 5) is far more than the system peak, and the demand charge is levied on each consumer's own maximum, not on his share of the system peak. Recomputing with 166.8 MW of billed demand gives Rs 159.3 crore, still short. The tariff as stated is simply too low, and the exercise illustrates that setting a tariff requires the billed-demand total, not the system peak.

Convert each consumer's data into billable quantities:

A and B are identical in every respect except when they run. That is the comparison the tariff is designed to expose.

The bills:

A and B pay exactly the same demand charge — Rs 1.059 crore — because their maximum demands are identical. A buys 2.3 times as much energy and pays 1.98 times as much in total, so his average rate is 13% lower.

Verify against the load-factor formula of Problem 15, adjusted for power factor:

The formula reproduces both, and shows explicitly that load factor and power factor enter identically — a consumer can improve his average rate by raising either.

The commercial consumer sits between them, which repays a moment's thought:

C has a worse load factor than A but a better power factor than B, and the product \(\text{LF}\times\cos\phi\) — 0.405 against A's 0.680 and B's 0.298 — is what actually determines the rate. C is ten times smaller than A and B, and that size difference has no effect whatever on the rate.

What B can do about it. Three options, with their effect on his average rate:

The second row is a real saving of Rs 14.04 lakh a year for a capacitor bank costing about Rs 10 lakh — Problem 19 works it through. The third row is not a saving in the same sense: B would buy far more energy, so his bill rises even as his rate falls; the benefit is that he produces more output for it. The fourth row is the important negative: a demand charge based on the consumer's own maximum gives no incentive to avoid the system peak, which is a genuine defect of the two-part tariff and the reason time-of-day pricing exists.

Whose demand actually causes cost. The defect above is worth stating precisely:

A consumer whose maximum demand falls in the overnight trough imposes no capacity cost at all, yet pays the same demand charge as one peaking at 19:00. Correcting this requires measuring demand only during system peak hours — which is what a maximum-demand charge levied on a restricted time window, or a time-of-day tariff, does.

The revenue perspective. Which consumer is most valuable to the utility?

Using the Problem 14 cost basis of Rs 973/kW of MD and Rs 2.72/kWh. Every consumer is loss-making, because the tariff of Problem 15 was shown there to under-recover by 46%. What the comparison does show is where the shortfall is concentrated: the high-load-factor consumer buys the most energy at a rate below cost, so he generates the largest absolute loss. Setting \(B\) below the true running cost is the specific error, and it is a common one in practice — usually a deliberate cross-subsidy from somewhere else.

Every item of plant is rated in kVA, not kW. That is the whole argument:

Only the prime mover is rated in kW. Everything electrical is rated by current, and current is set by kVA. So a consumer taking 5000 kW at 0.85 power factor requires the same plant as one taking 5882 kW at unity — and should pay accordingly.

The size of the effect:

The last column is the one that surprises: at 0.70 power factor the \(I^2R\) losses in every conductor between the generator and the consumer are doubled, because loss goes as the square of current and current goes as \(1/\cos\phi\).

The reactive power itself:

At 0.70 power factor the reactive demand exceeds the real demand. That reactive power has to be generated and transmitted like any other, and it does no useful work at the consumer's premises.

Where the reactive power comes from and why it is the consumer's responsibility:

The second row is the practical heart of the matter: an oversized motor running lightly loaded has a dreadful power factor, and oversizing motors is extremely common. Correcting the tariff to charge for kVA is what makes the consumer notice.

The three ways a tariff can signal power factor, all in use:

The three are not equivalent. A kVA demand charge signals the peak reactive requirement; a kVArh charge signals the reactive energy. Since capacity cost is driven by the peak, the kVA form is the more accurate — but the kVArh form is easier to meter with older equipment, which is why it persists.

Quantifying the system-wide cost. If the whole 110 MW system ran at 0.85 rather than 0.98 power factor:

Plus the increased losses: \((0.98/0.85)^2 = 1.329\), so a 33% rise in all \(I^2R\) losses. At 6% system losses that is an extra 2% of all energy generated — about Rs 3.2 crore a year on this system's running cost. The total is around Rs 5 crore a year, or 1.8% of the cost of supply, for a power factor difference that a few crore of capacitors would eliminate.

The principle. A capacitor supplies leading reactive power locally, so the reactive power drawn from the supply falls while the real power is unchanged:

The real power is unchanged because an ideal capacitor consumes none. The kW demand charge, if there were one, would not move at all — only the kVA falls.

The sizing formula:

One line, and it is the only formula needed. The angles come from \(\phi = \arccos(\text{pf})\), and the result is in the same units as \(P\).

Correction to 0.95:

And the apparent power falls from \(5000/0.85 = 5882\) kVA to \(5000/0.95 = 5263\) kVA — a reduction of 619 kVA.

Correction to 0.98:

Note the diminishing return. Going from 0.85 to 0.95 needs 1455 kVAr and saves 619 kVA; the further step to 0.98 needs an extra 628 kVAr and saves only a further 161 kVA. The marginal ratio has fallen from 0.43 to 0.26 kVA saved per kVAr installed.

The diminishing return, tabulated:

Correcting all the way to unity requires more than twice the capacitors needed for 0.95 and delivers only 42% more saving. This is why correction targets in practice cluster at 0.95–0.98 and never at 1.00 — and why the economic optimum, computed in Problem 19, falls in exactly that range.

Why over-correction is actively harmful:

The second row is the serious one in modern installations. A capacitor bank and the supply transformer's leakage inductance form a parallel resonant circuit, typically at the 5th to 13th harmonic; if a nearby drive injects current at that order, the resulting voltage distortion can be severe. Detuned capacitor banks — with a small series reactor moving the resonance below the lowest harmonic present — are the standard remedy.

Where to install, which changes what the correction achieves:

A bank at the incoming supply reduces the bill but does nothing for the consumer's own cables and transformer. Individual motor correction relieves the whole chain and is technically ideal, but it needs many small units, each of which must be sized below the motor's magnetising kVAr to avoid self-excitation on disconnection.

Set up the annual saving and cost as functions of the target power factor \(\cos\phi_2\):

with \(A = 1800\) Rs/kVA/yr, \(c = 500\) Rs/kVAr and \(k = 0.12\) the annual charge rate. Both are proportional to \(P\), so the consumer's size cancels — the economic power factor is the same for a 5 MW works and a 50 kW workshop.

The net annual benefit:

Differentiate with respect to \(\phi_2\) and set to zero. Using \(d(\sec\phi)/d\phi = \sec\phi\tan\phi\) and \(d(\tan\phi)/d\phi = \sec^2\phi\):

The optimality condition:

A strikingly clean result. The economic power factor depends only on the ratio of the capacitor's annual cost per kVAr to the tariff's demand charge per kVA — not on the consumer's size, load factor, or initial power factor.

Numerically:

Essentially unity. With a demand charge as high as Rs 1800/kVA and capacitors as cheap as Rs 500/kVAr, it pays to correct almost completely — the tariff is signalling very strongly indeed.

The sensitivity is what makes the result useful:

The economic power factor is only weakly sensitive to the demand charge over the practical range: a fourteen-fold reduction in \(A\) moves it from 0.999 to 0.954. Only when the demand charge falls below about Rs 150/kVA does correction stop being worthwhile — and that is far below any real industrial tariff.

Why real installations stop at 0.95–0.98 when the arithmetic says 0.999. Four reasons the model omits:

The first is decisive. A works whose load falls to 20% overnight would go strongly leading with a bank sized for full-load unity correction, raising both voltage and kVA. Automatic power factor correction with switched stages solves this, and its cost per kVAr is two to three times that of a fixed bank — which is what really sets the practical limit at 0.95–0.98.

The practical answer for Industry B, correcting to 0.98 with a switched bank at Rs 900/kVAr:

A sixteen-month payback on a device with a twenty-year life. Power factor correction is routinely the highest-return investment available to an industrial consumer, which is why it is nearly universal wherever the tariff charges for kVA.

The benefit the consumer does not see but the system does:

A quarter of the \(I^2R\) loss in every conductor upstream of the capacitor disappears. If the bank is at the incoming supply, that benefit accrues entirely to the utility; if at the motors, the consumer gets it too, in his own cables and transformer. That is the technical argument for distributing the correction rather than concentrating it at the meter.

Step 1 — the load, characterised completely:

Eight numbers, from which every subsequent decision follows.

Step 2 — the plant, from the screening analysis:

Note the last row: with the reserve corrected to 150 MW, the capacity factor falls to 0.4472 — which coincidentally equals the demand factor, a numerical accident of this example rather than any identity.

Step 3 — the cost, with the corrected installed capacity:

The reserve capacity has added Rs 8 crore a year, raising the cost of supply by 2.9% over the Problem 10 figure. Reserve is not free, and this is what it costs.

Step 4 — a tariff that recovers it. Splitting the cost into demand-related and energy-related:

The demand charge must be recovered from the total billed demand, which by Problem 5 is 166.8 MW, not the system peak of 110 MW:

The tariff:

taking an average system power factor of 0.90. The demand charge is far higher than the Rs 1800/kVA assumed in Problem 15, which is why that tariff under-recovered so badly. Check: \(6637\times185\,333+2.717\times587.65\times10^6 = 123.0+159.7 = \) Rs 282.7 crore ✓.

Step 5 — what consumers pay under the corrected tariff:

using \(\bar c = A/(8760\,\text{LF}\cos\phi)+B\). The system-average row uses the load factor computed on billed demand — \(587.65\times10^6/(166\,800\times8760) = 0.4022\) — not the system load factor of 0.6098, because the demand charge is levied on each consumer's own maximum. The spread is large — the domestic consumer pays 2.2 times the three-shift industrial rate for the same energy. This is cost-reflective and correct, but a domestic tariff of Rs 8/kWh would be politically impossible almost anywhere, which is why domestic supply is cross-subsidised in nearly every system in the world.

Step 6 — the consistency checks. Six identities, each computed by an independent route:

Every one uses a different route from the quantity it checks, so agreement is evidence and not tautology.

Step 7 — the leverage available. What would most reduce the cost of supply?

Load factor is the largest lever the utility controls, and it is the one every tariff device in this chapter — the demand charge, the kVA basis, time-of-day pricing — exists to pull. Fuel price is larger still but is not within the utility's control at all.