The LCOE Illusion: What 100% Wind and Solar Electricity Really Costs

The levelised cost of electricity, or LCOE, is one of the most frequently quoted numbers in energy debates.

Wind can supposedly produce electricity for a few cents per kilowatt-hour. Solar can supposedly produce it for even less. Those figures are then compared with the retail price of electricity or with the cost of nuclear, gas, coal, or other firm power.

That comparison is usually misleading.

LCOE measures the cost of generating electricity from a particular type of power plant. It does not necessarily measure the cost of providing electricity to customers at the time they need it.

A city does not consume “average annual electricity.” It consumes electricity at 7:00 a.m., at 3:00 p.m., and at 11:00 p.m. It consumes electricity during heat waves, cold snaps, cloudy winter weeks, and periods when the wind is weak across an entire region.

A wind turbine producing cheap electricity during a period of low demand does not solve the problem of supplying electricity during a windless evening. A solar farm producing power at noon does not by itself provide electricity after sunset.

If the electricity is unavailable when customers need it, then it is not equivalent to firm electricity.

The relevant question is therefore not:

What does it cost to generate one megawatt-hour from wind or solar?

The relevant question is:

What does it cost to provide reliable electricity to customers every hour, using wind and solar as the primary energy sources?

That requires counting the entire system.

What LCOE includes

A conventional LCOE calculation generally includes:

• Construction of the generating facility
• Financing costs
• Fuel, where applicable
• Routine operation and maintenance
• Expected electricity production
• Sometimes decommissioning costs

A simplified version is:

$\text{LCOE}=\frac{\text{Lifetime costs}}{\text{Lifetime electricity generated}}$

For a wind or solar project, the denominator is usually the electricity generated by the project over its lifetime.

That can be a useful number when comparing similar generation technologies or deciding whether to build one additional plant in an already reliable grid.

But it does not answer the full-system question.

What a standalone renewable system must also provide

A city relying entirely on wind and solar must pay for much more than the turbines and panels:

• Generation capacity far above average demand
• Batteries for daily balancing
• Long-duration storage for multi-day shortages
• Potentially seasonal storage
• Power-conversion equipment
• Transmission and substations
• Grid-forming inverters and controls
• Reserve capacity
• Curtailment of surplus electricity
• Storage charging losses
• Storage discharge losses
• Replacement of batteries, inverters, and other equipment
• Protection against periods when both wind and solar output are low
• Enough capacity to meet peak demand, not merely average demand

These costs are not optional additions. They are part of the system required to turn intermittent electricity into reliable electricity.

The 1-million-person city

Consider a city of 1 million people.

For this example, assume the city uses electricity for ordinary electrical demand: homes, businesses, services, lighting, appliances, data centers, and industry. This is not yet an estimate of all primary energy used for transportation, industrial heat, aviation, or other fuels.

Assume:

• Annual electricity consumption: 6 TWh
• Annual consumption in kilowatt-hours: 6 billion kWh
• Average demand: approximately 685 MW
• Peak demand: 1,000 MW
• No fossil-fuel, nuclear, hydroelectric, or external-grid backup
• Wind and solar are the primary energy sources
• Stored renewable energy must cover nights, weak-wind periods, and extended shortages

The average load is:

$\frac{\mathrm{6,000}\text{ GWh}}{\mathrm{8,760}\text{ hours}}=685\text{ MW}$

The city may average only 685 MW, but the system must be capable of serving a peak approaching 1,000 MW.

Designing only for average demand would leave the city unable to meet its largest loads.

Wind and solar capacity

Assume the system uses:

• 4 GW of wind
• 3 GW of solar

For wind, assume a 40% capacity factor:

$4\text{ GW}\times \mathrm{8,760}\text{ hours}\times 40\mathrm{\%}=14.016\text{ TWh/year}$

For solar, assume a 25% capacity factor:

$3\text{ GW}\times \mathrm{8,760}\text{ hours}\times 25\mathrm{\%}=6.570\text{ TWh/year}$

Total annual generation is therefore:

$14.016+6.570=20.586\text{ TWh/year}$

The city only consumes 6 TWh per year, so the system generates approximately:

$\frac{20.586}{6}=3.43$

times as much annual electricity as the city consumes.

That does not mean 70% of the electricity is automatically wasted. Some of the surplus is used to charge storage, and some is lost during conversion and discharge. But significant curtailment is unavoidable because the generation profile will not match the city’s demand profile.

The system needs this level of overbuilding because generating 6 TWh annually is not enough. The electricity must also be available at the right hour.

Storage requirement

A daily battery system can shift electricity from sunny afternoons to evenings and from windy periods to nearby periods of weak generation.

Assume:

• Battery power capacity: 2 GW
• Battery energy capacity: 48 GWh

At a 1-GW load, 48 GWh would represent approximately 48 hours of energy. In practice, not all battery capacity is usable, and the system must retain reserves, so the practical coverage period would be shorter.

Assume an installed battery cost of:

$\mathrm{\$}200\mathrm{/}\text{kWh}$

The battery cost is:

$48\text{ GWh}=\mathrm{48,000,000}\text{ kWh}$

$\mathrm{48,000,000}\text{ kWh}\times \mathrm{\$}200\mathrm{/}\text{kWh}=\mathrm{\$}9.6\text{ billion}$

This battery does not solve the full reliability problem. It mainly handles daily balancing and some short-term weather variation.

A regional wind drought can last several days. In some locations, low wind and low solar output can persist much longer. Batteries sized for only a few hours or one evening are not equivalent to firm generation.

Long-duration storage

The system also needs longer-duration storage.

For this example, assume a hydrogen-based system or another comparable technology with:

• Approximately 1 GW of power-conversion capacity
• Around 100–500 GWh of stored energy
• Equipment to convert surplus electricity into stored energy
• Storage tanks or underground storage
• Turbines, fuel cells, or other equipment to convert the energy back into electricity

The exact storage volume depends on:

• Local weather
• Geographic diversity
• Whether outside electricity can be imported
• The required reliability standard
• Whether the system must survive a one-week, two-week, or longer renewable drought
• Whether seasonal storage is required

A 100-GWh storage reserve would provide only 100 hours at a continuous 1-GW output before conversion losses. That is roughly four days. It is not a guarantee against a two-week regional shortage.

For the financial example, use a midrange installed cost of $4 billion for the long-duration storage system. A more conservative design using much larger seasonal reserves could cost substantially more.

Capital cost

The illustrative system is:

Component	Size	Assumed cost
Wind generation	4 GW	$6.0 billion
Solar generation	3 GW	$2.7 billion
Batteries	2 GW / 48 GWh	$9.6 billion
Long-duration storage	1 GW plus storage inventory	$4.0 billion
Transmission, substations, controls, and interconnection	—	$3.0 billion
Total initial capital	—	$25.3 billion

These are not universal prices. Actual costs vary by country, terrain, labour costs, financing terms, commodity prices, permitting, transmission distance, and technology.

The purpose of the example is to include the system components that a simple wind or solar LCOE normally does not include.

Annualised capital cost

To convert capital costs into annual costs, assume:

• 7% cost of capital
• 30-year economic life for wind, solar, and grid infrastructure
• 15-year economic life for batteries and storage equipment

The capital-recovery factor for 7% financing over 30 years is approximately:

${\text{CRF}}_{30}=\frac{0.07(1.07{)}^{30}}{(1.07{)}^{30}-1}\approx 8.06\mathrm{\%}$

The capital-recovery factor for 7% financing over 15 years is approximately:

${\text{CRF}}_{15}\approx 10.98\mathrm{\%}$

Wind

$\mathrm{\$}6.0\text{ billion}\times 8.06\mathrm{\%}=\mathrm{\$}483.6\text{ million/year}$

Solar

$\mathrm{\$}2.7\text{ billion}\times 8.06\mathrm{\%}=\mathrm{\$}217.6\text{ million/year}$

Transmission and grid infrastructure

$\mathrm{\$}3.0\text{ billion}\times 8.06\mathrm{\%}=\mathrm{\$}241.8\text{ million/year}$

Batteries

$\mathrm{\$}9.6\text{ billion}\times 10.98\mathrm{\%}=\mathrm{\$}1.054\text{ billion/year}$

Long-duration storage

$\mathrm{\$}4.0\text{ billion}\times 10.98\mathrm{\%}=\mathrm{\$}439.2\text{ million/year}$

Total annual capital recovery is:

$483.6+217.6+241.8+\mathrm{1,054}+439.2=\mathrm{\$}\mathrm{2,436.2}\text{ million/year}$

Therefore:

$\boxed{{\displaystyle \text{Annual capital recovery}\approx \mathrm{\$}2.44\text{ billion}}}$​

Operation and maintenance costs

Now add annual operation and maintenance costs.

Assume:

• Wind fixed O&M: 2.5% of initial capital
• Solar fixed O&M: 1.5%
• Battery O&M: 2.0%
• Long-duration storage O&M: 3.0%
• Grid and transmission O&M: 2.0%

Wind O&M

$\mathrm{\$}6.0\text{ billion}\times 2.5\mathrm{\%}=\mathrm{\$}150\text{ million/year}$

Solar O&M

$\mathrm{\$}2.7\text{ billion}\times 1.5\mathrm{\%}=\mathrm{\$}40.5\text{ million/year}$

Battery O&M

$\mathrm{\$}9.6\text{ billion}\times 2.0\mathrm{\%}=\mathrm{\$}192\text{ million/year}$

Long-duration storage O&M

$\mathrm{\$}4.0\text{ billion}\times 3.0\mathrm{\%}=\mathrm{\$}120\text{ million/year}$

Grid O&M

$\mathrm{\$}3.0\text{ billion}\times 2.0\mathrm{\%}=\mathrm{\$}60\text{ million/year}$

Total routine O&M is:

$150+40.5+192+120+60=\mathrm{\$}562.5\text{ million/year}$

Round this to:

$\boxed{{\displaystyle \mathrm{\$}563\text{ million/year}}}$

Replacement reserves

Wind turbines, solar inverters, battery cells, power electronics, control systems, and storage equipment will not all last for the full 30-year project life.

A serious calculation needs money set aside for:

• Battery replacement
• Inverter replacement
• Turbine component replacement
• Solar inverter replacement
• Control-system upgrades
• Storage-system refurbishment
• Unexpected major repairs
• Decommissioning and repowering

Assume a replacement and contingency reserve of:

$\boxed{{\displaystyle \mathrm{\$}150\text{ million/year}}}$​

This may be too low for a system with a very large battery fleet, but it provides a reasonable allowance for this illustration.

Total annual system cost

Add the annual costs:

Cost category	Annual cost
Wind capital recovery	$483.6 million
Solar capital recovery	$217.6 million
Battery capital recovery	$1,054.0 million
Long-duration storage capital recovery	$439.2 million
Grid capital recovery	$241.8 million
Routine O&M	$562.5 million
Replacement and contingency reserve	$150.0 million
Total	$3,148.7 million/year

Thus, the complete system costs approximately:

$\boxed{{\displaystyle \mathrm{\$}3.15\text{ billion per year}}}$

Cost per delivered kilowatt-hour

The city consumes:

$6\text{ TWh/year}=\mathrm{6,000,000,000}\text{ kWh/year}$

Therefore:

$\frac{\mathrm{\$}3.1487\text{ billion}}{6\text{ billion kWh}}=\mathrm{\$}0.5248\mathrm{/}\text{kWh}$

Rounded:

$\boxed{{\displaystyle \mathrm{\$}0.52\mathrm{/}\text{kWh}}}$

That is the approximate all-in system cost in this illustrative case.

It is not the cost of raw electricity produced at a wind turbine or solar panel. It is the cost of constructing and operating enough infrastructure to provide the city with electricity throughout the year while relying on wind and solar energy and stored renewable electricity.

Why the result is so different from the advertised LCOE

Suppose a wind farm reports an LCOE of 5¢/kWh.

That figure might be calculated using the wind farm’s own annual production:

$\text{Wind-farm LCOE}=\frac{\text{Wind-farm lifetime costs}}{\text{Wind-farm lifetime generation}}$

But the city does not need electricity only when the wind farm produces it.

The city needs:

$\text{electricity at every hour}$

The system therefore requires:

$\text{wind}+\text{solar}+\text{overbuilding}+\text{batteries}+\text{long-duration storage}+\text{transmission}+\text{controls}+\text{replacement}$

The relevant denominator is not all electricity generated by the wind and solar farms. It is the electricity actually delivered to customers:

$\text{delivered electricity}=6\text{ billion kWh/year}$

That is why the denominator matters so much.

If the system generates 20.6 TWh but customers consume 6 TWh, dividing total system costs by 20.6 TWh produces a much lower-looking number. But customers cannot use electricity that was curtailed, lost during storage, or generated at the wrong time.

The proper calculation divides total system cost by useful delivered electricity.

The hidden cost of curtailment

Curtailment occurs when wind and solar are producing more electricity than the city can use and more than the storage system can absorb.

For example, imagine a sunny and windy afternoon when:

• City demand is 700 MW
• Solar output is 2,000 MW
• Wind output is 1,500 MW
• Batteries are already full
• Long-duration storage is full or charging at its maximum rate

The system has 3,500 MW of renewable output but only 700 MW of immediate demand. Much of the remaining output must be curtailed.

That electricity has a production cost even though it produces no useful electricity for the customer. The turbines and solar panels still had to be built, financed, maintained, and eventually replaced.

Curtailment is not free. It is the cost of building generation that cannot always be used.

The hidden cost of storage losses

Storage is not 100% efficient.

A battery might have an approximately 85–95% round-trip efficiency depending on the system and operating conditions. Hydrogen systems can have much lower electricity-to-electricity efficiency.

A simplified hydrogen pathway might be:

$\text{electricity}\to \text{electrolysis}\to \text{compression/storage}\to \text{turbine or fuel cell}\to \text{electricity}$

If the full round-trip efficiency is 35%, then:

$1\text{ MWh stored}\to 0.35\text{ MWh delivered}$

To deliver 1 MWh to customers through that pathway, the system must generate:

$\frac{1}{0.35}=2.86\text{ MWh}$

before accounting for transmission losses and other losses.

This is why long-duration storage can be valuable for reliability while still being expensive. It does not merely require storage tanks. It requires extra generation to compensate for the energy that disappears during conversion.

The hidden cost of reliability

A system designed for ordinary weather is not the same as a system designed for difficult weather.

Consider two designs:

Design A: Handles normal nights and short wind lulls.

Design B: Handles a rare, region-wide event lasting one or two weeks, with poor solar production and weak wind.

Design B requires much more:

• Generation capacity
• Stored energy
• Power-conversion capacity
• Transmission
• Reserve margin
• Equipment capable of sitting unused for long periods

That equipment may operate only a few times per year, or perhaps only a few times per decade. Nevertheless, it must be paid for all year.

This is the same basic reason that emergency departments, fire stations, and spare aircraft cost money even when they are not used constantly. Reliability requires capacity that is idle much of the time.

“Zero outages” needs a defined standard

No electrical system can realistically promise that a customer will never experience an outage under every imaginable circumstance.

Transmission lines fail. Transformers fail. Fires, floods, cyberattacks, equipment defects, and extreme weather occur.

A practical analysis must define a reliability target, such as:

• 99.9% availability
• 99.99% availability
• A specified loss-of-load expectation
• A specified maximum number of outage hours per year
• Survival of a defined renewable drought scenario

The calculation above treats reliability as a design requirement rather than assuming that the city can simply import electricity whenever wind and solar production are inadequate.

If outside grid electricity is allowed, the system becomes cheaper. But then the city is depending on other regions, which may themselves be experiencing the same shortage.

If gas turbines are retained for rare emergency use, the system becomes cheaper still. But then the city is not receiving all of its electricity from wind and solar.

Those are legitimate design choices, but they must be stated openly.

A range of possible outcomes

The 52¢/kWh estimate is not a universal price. It is an example based on a city that must provide its own reliability.

The outcome could be lower if:

• The city is connected to a very large geographic grid
• Wind and solar resources are excellent
• Neighboring regions have complementary weather
• Existing transmission is available
• The required reliability standard is modest
• Storage prices fall substantially
• Demand can be shifted voluntarily
• Hydro, nuclear, geothermal, or fossil backup is allowed

It could be higher if:

• The city must operate as an electrical island
• It must survive multi-week renewable droughts
• Seasonal storage is required
• The city is located in a poor wind or solar region
• Transmission must be built over long distances
• Financing costs are high
• Battery replacement costs rise
• Electrification significantly increases peak demand

Approximate all-in ranges might look like this:

System design	Approximate delivered cost
Wind and solar added to a large interconnected grid	15–30¢/kWh
City with substantial batteries and regional transmission	25–45¢/kWh
Mostly self-sufficient city with long-duration storage	40–70¢/kWh
Strictly local system designed for severe multi-week shortages	70¢–$1.00+/kWh

These figures are system costs, not necessarily the final retail bill. Taxes, utility administration, customer service, distribution charges, and profit could increase the amount paid by customers.

What this means for Thorium, Fission and other firm-energy proposals

The point is not that wind and solar are useless. They are valuable energy sources, and their fuel is free. But their economics suck.

The point is that their low generation cost does not automatically equal a low cost of reliable electricity.

A firm-energy technology—whether nuclear, hydroelectric, geothermal, fossil fuel with carbon capture, or another technology—reduces the amount of storage and overbuilding required. Its comparison should therefore be made against the cost of firm renewable electricity, not against the raw LCOE of wind or solar.

The fair comparison is:

$\text{firm nuclear cost}\text{versus}\text{firm wind-plus-solar system cost}$

not:

$\text{nuclear LCOE}\text{versus}\text{wind LCOE}$

and not:

$\text{nuclear LCOE}\text{versus}\text{solar LCOE}$

Those are different products.

A wind turbine and a solar panel produce variable electricity. A firm generator produces dispatchable electricity. Comparing their raw LCOEs without valuing timing, reliability, storage, transmission, and capacity is like comparing the price of an inexpensive ingredient with the price of a finished meal.

The central conclusion

LCOE is not mathematically fraudulent, but it’s manner of presentation is. It is simply being asked to answer a question it was not designed to answer.

It can estimate the cost of electricity generated by one project in one moment of time. Effectively useless measure.

It does not, by itself, estimate the cost of supplying a city with dependable electricity every hour.

For the illustrative 1-million-person city:

• Annual electricity demand: 6 TWh
• Average demand: 685 MW
• Peak demand: 1 GW
• Wind capacity: 4 GW
• Solar capacity: 3 GW
• Battery storage: 48 GWh
• Long-duration storage: approximately 100–500 GWh
• Total system capital: approximately $25 billion
• Annual system cost: approximately $3.15 billion
• Delivered electricity: 6 billion kWh
• All-in system cost:

$\boxed{{\displaystyle \frac{\mathrm{\$}3.15\text{ billion}}{6\text{ billion kWh}}\approx \mathrm{\$}0.52\mathrm{/}\text{kWh}}}$​

The exact result will vary with geography, financing, technology, reliability standards, and storage requirements. But the structure of the calculation does not change:

$\boxed{{\displaystyle \text{real cost of renewable electricity}=\text{generation}+\text{overbuild}+\text{storage}+\text{losses}+\text{curtailment}+\text{transmission}+\text{reserves}+\text{replacements}}}$

If those costs are omitted, the result is not the cost of reliable electricity. It is only the cost of producing electricity when the weather happens to cooperate.