Introduction
The basic idea behind an off-peak battery is to use cheaper power during an off-peak period to charge up a battery, then use that power during a peak period when such power would be more expensive.
This article is meant to provide back-of-the-envelope calculations to help you decide if such a battery setup makes sense for you.
Marginal Savings – A Litmus Test
Given transfer and battery efficiency losses, we want to make sure that the difference in price makes up for the efficiency losses of storage. In technical terms, we need to know if our setup would save us any money on a per unit of energy stored basis.
If we lost money on every unit of energy we store, then it makes no sense to even begin considering this approach.
Note that we use kilowatt-hours (kWh) because it is a unit regulators and electricity companies use, and because it is a good metric for comparison between different energy storage solutions.
Cost savings per Input kWh
We begin by borrowing the concept of variable cost. We need this to ensure that we can make marginal savings on every kilowatt-hour (kWh) that we store.
Put simply, we want to be sure that if we charge up our battery with 1 kWh of off-peak energy, when we use that 1 kWh, it will be cheaper than if we bought it from the grid during peak hours. in energy markets parlance, this approach is called time-shifting or arbitrage value
We can describe this with a simple profit per kWh equation:
Cost Savings per Input kWh = Peak Price per kWh - Off-peak Price per kWh
Seems pretty straightforward, right? If Off-peak price is higher, then it is an indication that such a project could be worth pursuing.
Not quite.
Factoring in system efficiencies
Electrical systems are not 100% efficient. Energy is usually lost in the process of distribution and storage. We can refer to the amount of energy retained after such losses as system efficiencies.
Though many factors contribute to system efficiency, the biggest factors we will be concerned with are those related to battery and conversion efficiency.
Battery efficiency
The efficiency of a battery decides how much of the battery’s energy is lost in storage. This means that for every 1 kWh of energy we put into a battery, we do not get the full 1 kWh out of it when we use the battery as a power source.
Different batteries have different efficiencies, depending on the kind of battery we select. For e.g. lithium-ion batteries generally have a charge-discharge efficiency of nearly 90-99%. This means that for every 1 kWh of energy put into the battery, we only get 0.9-0.99 kWh of energy out of it when we use the battery as a power source. This value varies depending on various factors like temperature and age, but there are averages we can usually work with for our purposes.
Conversion efficiency
Power distribution system usually transmit electricity via alternating current (AC) as opposed to direct current (DC) because of better transmission efficiencies over long distances. There may however be a need to change this energy from AC to DC or back again when working with energy storage systems because some of batteries do not work with AC power out-of-the-box.
To do this we can use an AC-DC inverter, for which efficiencies generally range from 95-98%.
Combining the two
Incorporating our 2 factors, gives us the resulting equation:
System Efficiency = Battery Efficiency x Conversion Efficiency
Cost Savings per Input kWh = (Peak Price per kWh x System Efficiency) - Off-peak Price per kWh
If our result is positive (larger than 0), we make marginal savings for every kWh stored which is a good thing. If negative (less than 0), we lose money on every kWh stored. In this case, it makes no sense to continue an examination.
Examples – plugging in some numbers
Let’s say we want to work with a lithium-ion battery. Let’s assume it has an efficiency at the lower end of 90%. We also use an AC-DC inverter with an efficiency of 95%. Let’s assume we are in California, and find out through an inquiry with our power provider that the prices in our locality are 45 c/kWh for peak, and 37.5 c/kWh for off-peak. Let’s keep it at summer prices, for simplicity’s sake.
Plugging this into our equation gives us
Cost Savings per Input kWh = (Peak Price per kWh x System Efficiency) - Off-peak Price per kWh
Cost Savings per Input kWh = (0.45 x 90% x 95%) - 0.375
Cost Savings per Input kWh = 0.00975
Because our result is positive (>0) there is an indication that an off-peak energy battery could save us money.
Capital Costs – The Other Side of the Coin
Simply because we have savings per kWh, doesn’t mean we can have savings on the whole without considering the costs of setting up the system.
What are capital costs?
Capital costs are one-time expenses paid for things used in the production of goods or services. Basically the cost of setting up the system for the first time.
Such costs include the cost of buying the battery, the costs of providing the storage space and safety equipment, costs of setting up the wiring and circuitry, and other forms of fixed costs.
Different battery system setups have different capital cost expenditures. Simple setups may consist of simply wiring a group of Li-ion car batteries together in parallel, while more complicated setups may use more specialised batteries with fancy electrical infrastructure.
For our purposes, let’s assume a simple car battery setup, with negligible wiring costs. This consists of a series of car batteries strung together, and connected to the grid via an AC/DC inverter.
An example capital cost calculation
Pulling a random $260 car battery off of Amazon with a rated Voltage of 12V and stated capacity of 100Ah, we can use the battery capacity equation to determine its battery capacity in kWh.
Battery Capacity in kWh = Voltage x Ampere-hours / 1000
Battery Capacity in kWh = 12V x 100Ah / 1000
Battery Capacity in kWh = 1.2 kWh
That means we need 5 / 1.2 ≈ 4 batteries to provide roughly 5 kWh of energy storage, which is enough storage to satisfy 4 hours of daily use for the average American household.
Why 4 hours? Simply because it is the current industry standard for commercial energy storage solutions. This number can be adjusted according to your individual context.
We also pull a random $400 12V DC-to-AC inverter, with a Power rating in the range of our needs.
At a minimum then,
Capital Cost = Cost of Equipment
Capital Cost = $260 x 4 batteries + $400 x 1 DC-to-AC inverter
Capital Cost = $1,440
With marginal savings and capital costs understood, we can now do some proper economic analysis.
Economic metrics
To evaluate if a project like this would actually be useful to build, we can work with many different metrics and financial modelling approaches. Our 3 selected metrics will be
- Breakeven point
- Total cost savings
- Cost per kWh of storage capacity.
Breakeven point
Breakeven point will measure how long it will take for our project to recoup our original investment with our cost-savings. This is also a metric used in commercial projects, like in this Jordanian hydrodam storage project study.
In a traditional finance context, breakeven point measures the number of units we have to sell in order to cover completely the fixed cost we invested in a project. In such a situation, it is represented by the equation: Breakeven point (units) = Fixed Costs ÷ (Sales price per unit – Variable costs per unit).
Units in our context is represented by the number of kilowatt-hours (kWh). And not kWh sold, but kWh consumed at peak vs off-peak, thus earning cost-savings. Remember that we already have cost-savings per input kWh from above:
Cost Savings per Input kWh = (Peak Price per kWh x System Efficiency) - Off-peak Price per kWh
This directly replaces (Sales price per unit – Variable costs per unit) in the breakeven equation. Therefore, we arrive at a breakeven equation of our own:
Breakeven point (kWh) = Capital Cost / Cost Savings per Input kWh
This tells us how much electricity in kWh we have to use, before we breakeven.
If we have an idea how much electricity we use on average daily, we will also be able to calculate the breakeven point in days:
Breakeven point (days) = Capital Cost / (Cost Savings per Input kWh x Daily kWh Usage)
Total cost savings
Total cost savings broadens breakeven point, and asks how much money we will actually save over time.
Naturally, different projects have different time horizons, dependending on many factors like the setup and durability of the equipment, and technological development over time.
A simple way to decide on the time period is to identify how long the equipment will last. There are many ways to do this, all specific to the kind of setup in question. For example, lithium-ion batteries have a rated number of charge-discharge cycles, so a setup involving this can assume each cycle is one day, and calculate the number of days the batteries will last.
We can then apply the total cost savings equation:
Total Cost Savings = (Cost Savings per Input kWh x Daily kWh Usage) x Time Period in days - Capital Cost
This will tell us how much we will save over the period of time our setup is in operation.
Cost per kWh of Storage Capacity
This metric is less directly useful than the others, but allows us to make comparisons between our project and large-scale commercial battery storage projects, along with the published studies that examine these.
This is because many projects measure capital costs in terms of $/kWh of storage costs. What this metric measures is simply how much each unit of storage capacity will cost, for example in kWh. The equation is straightforward:
Cost per kWh of Storage Capacity = Capital Cost / Total kWh of the Battery System
As mentioned, this allows us to make approximate comparisons across projects, maybe as a way to validate the accuracy of our numbers. For example, in the study Cost Projections for Utility-Scale Battery Storage 2023 Update, utility-scale storage stands at roughly $400 per kWh. If we find that our numbers stray too far from this value, we might be worried enough to pay more attention to how we achieved them.
Complicating factors
Note that the above equations fail to account for additional complexities. Some examples are:
- Safety concerns and expenditures to ensure safety
- Off-peak and on-peak pricing might not be constant
- When to switch from charging mode to power source mode
- Seasonal differences in prices
- System efficiency changes due to temperature changes
- And more…
But addressing all of these is too much for a single blog post. Anyway, here we are focused on quickly guess-timating the feasibility of such a project; contextual details have to be tackled on the reader’s own effort.
Scenario & Cheatsheet
Let’s run a scenario to see how we could apply the above equations. To start, let’s summarise the example numbers that we have been working with so far:
Summary of Numbers
Efficiencies
- Battery Efficiency: 85%
- Conversion Efficiency: 95%
Electricity Pricing
- Peak Price: 0.45 $/kWh
- Off-peak Price: 0.375 $/kWh
Lithium-ion Battery Specification
- Capacity: 1.2 kWh
- Rated Cycles: 15,000
- Time period of use in days: 15,000
Setup
- Cost of Equipment: $1,440
- Total Capacity in kWh: 5.0 kWh
- Daily Usage from Battery Setup: 5.0 kWh
Cheatsheet
We can also summarise the equations from above as below into a cheatsheet:
System Efficiency = Battery Efficiency x Inverter Efficiency
Cost Savings per Input kWh = (Peak Price per kWh x System Efficiency) - Off-peak Price per kWh
Capital Cost = Cost of Equipment
Breakeven point (days) = Capital Cost / (Cost Savings per Input kWh x Daily kWh Usage)
Total Cost Savings = (Cost Savings per Input kWh x Daily kWh Usage) x Time Period in days - Capital Cost
Cost per kWh of Storage Capacity = Capital Cost / Total Capacity in kWh of the Battery System
Application
Plugging in these numbers:
System Efficiency = Battery Efficiency x Inverter Efficiency
System Efficiency = 90% x 95%
System Efficiency = 85.5%
Cost Savings per Input kWh = (Peak Price per kWh x System Efficiency) - Off-peak Price per kWh
Cost Savings per Input kWh = (0.45 x 85.5%) - 0.375
Cost Savings per Input kWh = 0.00975
Capital Cost = Cost of Equipment
Capital Cost = $1,440
Breakeven point (days) = Capital Cost / (Cost Savings per Input kWh x Daily kWh Usage)
Breakeven point (days) = 1,440 / (0.00975 x 5.0)
Breakeven point (days) = 29,538 days
Breakeven point (years) = 80.8 years
Total Cost Savings = (0.00975 x 5) x 15,000 - 1,440
Total Cost Savings = $-708.75
Cost per kWh of Storage Capacity = Capital Cost / Total Capacity in kWh of the Battery System
Cost per kWh of Storage Capacity = 1,440 / 5.0
Cost per kWh of Storage Capacity = 288 $/kWh
Pay attention to our breakeven point of 80 years! That’s much longer in fact than the expected life of our system. As such, we actually make a loss on the project of -$708.75 if we run it to completion in 15 years.
Despite the cheaper cost per kWh of storage compared to most commercial-scale battery storage systems (which stand at $400 per kWh), the setup in our scenario doesn’t make ecomomic sense at all.
But perhaps your numbers might look different?
Conclusion
Through this article, we have acquired an understanding of how to decide if a battery system that arbitrages off-peak and peak electricity prices makes sense for you, and the equations necessary to help you determine this.