Roads are an obvious candidate to double as solar farms. They’re everywhere and there’s no good reason we couldn’t put a solar “roof” on most. But how much coverage would we need to satisfy all of our energy needs and is that cost effective?
Using roads as solar farms makes some practical sense. The seasonality of Michigan makes the project unattractively expensive, but not outlandish
For about $200 billion we could produce as much solar power as Michigan currently produces power from non-renewable resources
About 42% of that cost is for automatic snow removal. Using wind avoids this cost (there’s no seasonality to the presence of wind), but then the roads are less useful
Only 37% of the cost is the solar cells themselves and the last 21% of the cost is to store one week of power
If we had 100% efficient solar cells, the surface area of I-75 would produce enough power to replace all non-renewable energy produced in the state
With modern cost effective cells, we’d need about 6 times the surface area of I-75 to replace all non-renewable energy produced in the state, something we could easily achieve by covering more roads
Is it a good idea to use roads as solar farms?
That’s a hard set of questions to answer, so let’s break it down. Let’s just consider Michigan and let’s just consider one road. To make the best case possible, let’s pick the biggest road in the state. I-75. We’ll also consider the case of building the solar array on a platform above the road rather than on the road itself. That has the dual goals of minimizing disruption to traffic on a major thoroughfare and protecting the road by keeping snow and ice off of it.
A few reasons. Because it traverses the length of the state, it can send power everywhere. Being a major interstate, it has a lot of surface area to cover. Lastly and perhaps most importantly, it had readily accessible data.
I-75 is 395.92 miles long in Michigan. To find the full surface area we also need to know how wide it is. For most of its length, it’s a 4 lane highway. However, in some urban areas it’s as much as 10.
Let’s just call it an average of 5. This is an underestimate for urban areas, and a bit over for the vast rural stretches. However, we could reasonably assume that we’d want to extend across the entire roadway and part of the shoulder as well. If that gives us a half lane in each direction, that makes for 5 lanes even in the rural areas.
Assuming a lane is 12 feet across, this gives us 60 feet across, for a total surface area of
395.92 miles • 5,280 feet/mile • 60 feet = 1,505,114,265.6 feet^2
Luckily, lots of resources exist to estimate annual solar irradiance of a region. For Michigan, we’ll use an average of 3 kW • h / m2 / day. It is worth noting that is an annual average. The actual amount of energy would vary by season and day.
For our estimate, we won’t use any fancy solar panels. These are 17% efficient, are a little over 3 feet wide and 5 feet tall. Each one will produce a modest amount of power.
5.41 feet • 3.25 feet • 0.17 (Efficiency) • 3.0 kW • h / m2 / day • 365 day / year ÷ 10.76 feet^2 / m2 = 301.94 KW • h / panel / year
How many panels will it take to cover I-75?
1,505,114,265.6 feet^2 ÷ (5.41 feet • 3.25 feet) = 85,699,697.34 panels
How much power will they harvest?
A lot. Almost 26 billion kilowatt hours of power annually.
85,699,697.34 panel • 301.94 KW • h / panel / year = 25,875,968,080.70 KW • h / year
Is that really a lot?
To answer that question let’s compare against the current net energy production in the state.
Currently, across all types, Michigan produces about 181,147,228,472 KW • h of energy annually .
Of that, 32,413,660,360 KW • h come from renewable resources already.
So, yes our solar roof for I-75 produces a lot of power. But, it’s only about 17% of the state’s entire current production of non-renewable energy. So, some mix of ~6 times more efficient panels covering I-75 or ~6 times more roads being covered would be enough to match all the energy produced by the state of Michigan using non-renewable sources.
That’s encouraging in that we could easily produce far more energy than is needed by just putting a solar panel roof on additional roads. Needing only 6 times the surface area of I-75 also means it’d be a small fraction of the states roads that we would need to cover.
How much would that cost?
On one hand, we may expect an order of 86 million panels to get a sizable discount. On the other hand, an order that large poses serious logistical problems for manufacturers. Let’s just assume a normal retail price wins out. This comes out to $150 per panel for a total of
150.00 $/panel • 85,699,697.34 panel = $12,854,954,601
$13 billion is a lot of money, but not a scary big number when it comes to matching more than 1/6 of a state’s non-renewable energy production.
As a comparison, it’d take 3.3 coal fired power plants rated at 895 MW to achieve parity with this solar output. Simply building those plans and buying the coal to operate them at today’s price for the lifetime of the solar cells is about $11,886,476,671. With my numbers, the solar cells are at a small disadvantage on price, but they have many other positive features that would compel utilities to use them anyway. At least one of those reasons is immunity from volatility in the cost of Coal, which becomes volatility in the prices consumers pay for power.
The Roof Itself
We mentioned earlier how the roads would be protected from snow by this solar roof. Sadly, solar panels covered in snow don’t produce much power , so we’d need a way to clean these automatically. Or at least remove snow.
Let’s presume we’d want a physical mechanism to lift the solar roof like a drawbridge so that all the snow can fall off. There may be much more elegant and cost effective solutions out there, but this one has the advantages of being practical and more importantly having convenient proxies to estimate the cost. Let’s assume we’ll create something about as mechanically complex as the mechanism that opens and closes retractable roofs on sports stadiums.
The Lucas Oil stadium covers 1,800,000 feet^2. It also cost about $83,200,000 in materials and $15,600,000 in labor to construct.
If we know how many Lucas Oil stadium roof equivalents we need, we can estimate the cost to build a massive expressway roof that can lift itself to shed snow.
feet^2 ÷ 1,800,000 feet^2 = 836.17 stadium roof equivalents
So the estimated cost of our fancy snow removing roof is about…
836.17 •($83,200,000 + $15,600,000) = $14,549,437,900.80
Our fancy mechanical roof costs more than the solar cells, but ensures we can use them year round. That stings and brings our total cost to
$14,549,437,900.80 + $12,854,954,601 = $27,404,392,501.82
Well that’s far less attractive. It gets worse, but not dire.
The first problem is energy transmission. Most of the power generated on I-75 is going to be in the vast rural portions of the state, where energy needs are far lesser. Maybe that means those areas have a much more modest carbon footprint, or it means we need to move that power a long way. Transmitting energy a long distance consumes a lot of energy (ironically). If we do this, our array would be far less efficient than it originally looked.
There isn’t a convenient equation for this, but we may also be able to mitigate this by covering all lanes in urban areas and covering only some lanes in rural areas. We won’t calculate the consequences of these choices because I don’t hate myself. The result would more likely be design choices about how much coverage portions of the freeway get based on local energy demand.
The biggest wrinkle here isn’t getting the power, it’s storing it safely and securely. We need power at night, but the sun doesn’t shine at night.
Flywheels provide a nice non-battery solution for energy storage that includes no hazardous chemicals nor much in the way of rare earth metals, so let’s presume that this is a technology we want to use.
Omnes Energy produces flywheel storage arrays already in use for utilities. Let’s assume we use their solution. To ensure continuous power delivery day or night, Omnes offers an array of 20 flywheels, each with a 200 kwh capacity that is capable of delivering 200 kwh for 20 hours. We’ll use this solution and round it’s energy storage capacity to one day, just because it makes my math easier and has a modest practical impact on the result.
Let’s say we want enough energy storage capacity to hold onto one full week of power. We’ll need a lot of flywheels to make this work:
(25,875,968,080.70 KW • h /
year ÷ 365 day / year = 70,893,063.23 KW • h/ day
KW • h / day ÷ 200 kW • h / flywheel array = 354,465.32 flywheel array / day
flywheel array / day × 20 flywheel / flywheel array = 7,089,306.32 (flywheel / day)
At a cost of $150 per Flywheel, that means we’ll spend about one billion dollars for one day of energy storage.
$150 × 7,089,306.32 = $1,063,395,948.52
For one week of power that is $7,443,771,639.65, which brings or total to
$7,443,771,639.65 + $14,549,437,900.80 + $12,854,954,601.02 = $34,848,164,141.47
What’s this mean?
Solar panels are very first efficient.
If we had roads double as solar energy farms, we’d be able to meet all of our energy needs and then some…during the day on most days.
We could easily overcome the variability in seasonal sunlight by buying enough solar panels to account for 120% (or more) of the non-renewable energy being displaced and storing the excess for when it is needed. Even with this increased cost, the solution is still reasonable.
If we accept reduced solar output in winter, then the cost is very low. Even if we need to pay for some fancy self cleaning system to keep snow off of the panels, the cost is still doable, but is ugly.
Energy storage is expensive, but storing almost one week of power costs about half as much as the solar array itself and would provide a good safety net of available power.
To match all energy produced in Michigan using only solar panels could be achieved using only the area atop some roads and would have a cost near $200 billion dollars.
Is that project worth taking seriously?
Maybe not for all of I-75. But it’s attractive enough to perhaps warrant a proof of concept to explore the idea.