Note: Students should first become familiar with the material in Explore Photovoltaics first.
Objective: To introduce students to upfront and life-cycle cost calculation and the relative costs of solar electricity.
A commonly asked question is "How much does solar electricity cost?" There are really two questions here, which must be answered separately. The first question is
What is the upfront cost to install solar? In other words, how much do I have to pay today to have a system installed that delivers a given peak power and a given amount of energy storage.
The second question is:
What is the life-cycle cost per kilowatt-hour of solar energy? In other words, how does solar compare to the cost of grid power?
In this exercise, the overall approach will be:
First calculate the cost of each major component in terms of user specified variables. The major components are:
Inverter
Solar Panels
Batteries
We will ignore adding in the cost of the charge controller, since this is only a few hundred dollars (whereas the whole system cost will be in the thousands of dollars).
The user specified variables will be:
Peak power required to power appliances
Total energy produced/consumed per day
Hours of sunshine (average)
After calculating the component costs, we add them up to create simple formulas with which to answer each of the questions above.
Note: The numbers used below are fairly conservative and include costs such as installation, and so the results should be taken to indicate upper bound estimates. Moreover, we have tried to round up to simple numbers to make the calculations easier.
Recall that power is defined to be the rate at which energy is delivered (or captured), that is, energy per unit time:
Power = Energy / Time
For electrical applications, Power is usually specified in kilowatts (kw) which means in thousands of watts. One kilowatt is enough power to light ten 100 watt light bulbs, or one typical hair dryer (hair dryers use alot of power!)
There are two types of power requirements one needs to know when designing a solar system: The peak power delivered to the load, and the peak power produced by the solar panels by the system.
The peak power delivered to the load is the total maximum power level one expects to be drawn by appliances in the home. For example, if one expects to run, at most, a 1 kilowatt hairdryer, five 100 watt light bulbs, and a 500 watt refridgerator, then the peak power would be:
P_{peak, usage} = 1 kw + 5 x .1 kw + .5 kw = 2 kilowatts
Two kilowatts is a probably a good peak power target for small energy efficient solar home. Some people may require significantly more (say, up to 5 kilowatts).
The amount of peak power the system can deliver will be determined by the size of the system's inverter, the inverter being the device which converts the dc battery power to ac:
P_{peak,}_{ usage} = P_{peak, inverter}
As determined by surveying current market prices for inverters, the costs of an inverter are about $1 per watt, or (multiplying by 1000): Cost_{inverter} = $1000/kilowatt |
Thus, the cost of the inverter, as a function of the peak power used, is therefore:
Cost_{inverter }(P_{peak, usage}) = P_{peak, usage} x Cost_{inverter }
or
Cost_{inverter }= P_{peak, usage} x $1000/kilowatt |
For example, if we need 2 kilowatts of peak power used, the cost of the inverter will be about $2000 dollars.
The peak power produced by the solar panels is determined by the type and number of solar panels one uses:
P_{peak panels} = # of panels x power per panel
Although the energy used by the appliances will of course be produced by the solar panels, it is not necessary that the peak output of the solar panels be equal the peak power used:
P_{peak, usage} : NOT NECESSARILY EQUAL TO: P_{peak panels}
This is because the power generated by the solar panels is stored up over time by batteries, so more peak power (but not energy!) can be delivered by the inverter than is produced by the panels.
Instead, we should calculate the peak power of the solar panels, and hence the number of solar panels, from the total amount of energy we want them to produce each day.
Calling the energy produced E_{produced}, we want this to equal the amount of energy used each day,
Important connection: E_{produced} = E_{used} |
We will specify energy in units of kilowatt-hours:
Energy = Power (in kilowatts) x Time (in hours) = # of kilowatt-hours
A good target for E_{used} for an energy efficient home is 10 kilowatt-hours. Electrical energy from the grid in the United States typically costs between 6 to 12 cents per kilowatt-hour. So, for example, if you use 10 kilowatt-hours a day, and the cost of power is about 10 cents per kilowatt-hour, then you electrical costs would be about $1 per day (ten times 10 cents), or $30 per month.
Also, we need to know how long the sun shines each day on average. Let this be denoted by T_{sun},
T_{sun} = Hours of Sunshine on average.
Using the formula for power and energy (Power = Energy / Time), we have
P_{peak panels }= E_{used} / T_{sun}.
Note that the fewer hours of sunshine available, the more peak power from the panels will be needed.
As determined from a survey of current market prices, it costs about $600 to purchase and install a 75 watt panel. Therefore, the upfront cost of the solar panels per watt are Cost_{panels }= $600/75 watts = $8/watt Or, by multiplying numerator and denominator by 1000, Cost_{panels }= $600/75 watts = $8000/kilo-watt |
Thus, as a function of Energy use, the cost of the solar panels will be
Cost_{panels }= P_{peak panels } x_{ }Cost_{s.p.} =_{ }(E_{used} / T_{sun }) x Cost_{s.p.}
or
Cost_{panels }= (E_{used} / T_{sun }) x $8000/kilo-watt |
The amount of energy stored (by batteries) determines how much energy can be used after dark, or on a rainy day.
The number of kilowatt-hours we can store will be determined by the number and type of batteries we have:
E_{stored }= Energy per battery x number of batteries
The lifetimes of deep cycle batteries are fairly short (3 - 10 years), and depend on how well they are maintained (for example, one needs to avoid overcharging, and overdrawing, and in many cases to keep the water levels up). Typically, if a battery is discharged to 50% every day, it will last about twice as long as if it is cycled to 80%. If cycled only 10%, it will last about 5 times as long as one cycled to 50%.
We will assume, in order not to discharge the battery more than 50%, that the batteries will be able to store twice the amount of energy we use:
E_{stored }= 2 x E_{used }
Presently, the cost of batteries is about $100 per kilowatt-hour of storage:
Cost_{batteries} = $100/kilowatt-hour
The cost of batteries, therefore, as a function of energy used, is
Cost_{batteries} = 2 x E_{used }x $100/kilowatt-hour |
Because we have included the factor of two, then we are probably safe to assume at least a six year lifetime on the batteries:
Lifetime_{batteries} = 6 years |
Adding up the costs of the inverter, panels and batteries, we find:
Cost_{upfront }= Cost_{inverter }+ Cost_{panels }+ Cost_{batteries} = P_{peak, usage} x $1000/kw + (E_{used} / T_{sun }) x $8000/kw + 2 x E_{used }x $100/kwh |
As mentioned above, today's solar panels are estimated to last at least 25 years. We will therefore use 25 years as our lifetime with which to calculate the life-cycle cost:
T_{system }= 25 years
Note that this figure is somewhat arbitrary: using a longer lifetime will tend to decrease the life-cycle cost calculated, and vice versa.
The total life-cycle cost per kilowatt hour is given by
Cost_{kwh} = (Total life-cycle cost)/(Total kilowatt-hours used).
To calculate the total life-cycle cost, we need to account for periodic replacement of the batteries. Assuming a lifetime of six years for the batteries (which we helped insure by sizing the battery to twice the daily energy usage), The number of times we have to replace the batteries is
N_{batteries} = T_{system} / Lifetime_{batteries} = 25/6 = (approximately) 4
The total life-cycle cost of the batteries will therefore be
Cost_{batteries, life-cycle} = 4 x Cost_{batteries }= 8 x E_{used }x $100/kwh
The total life-cycle cost of the system will therefore be
Cost_{life-}_{cycle }= Cost_{inverter }+ Cost_{panels }+ Cost_{batteries, life-cycle} = P_{peak, usage} x $1000/kw + (E_{used} / T_{sun }) x $8000/kw + 8 x E_{used }x $100/kwh |
Note that this is similar to the upfront cost formula, except for the extra factor of four in the last term.
Because we defined the quantity E_{used} to be the number of kilo-watt hours used per day, the number of kilowatt-hours used over the lifetime of the system will be:
Total kilowatt-hours used = 25 years x 365 days x E_{used} = 9125 x E_{used}.
We therefore have
Cost_{kwh} = Cost_{life-cycle} / (9125 x E_{used}) |
P_{peak, usage} = Peak power usage in kilowatts |
E_{used} = Total daily energy usage in kilowatt-hours |
T_{sun }= Hours of sunshine (average) |
Cost_{inverter }= P_{peak, usage} x $1000/kilowatt |
Cost_{panels }= (E_{used} / T_{sun }) x $8000/kilo-watt |
Cost_{batteries} = 2 x E_{used }x $100/kilowatt-hour |
Cost_{batteries, life-cycle} = 4 x Cost_{batteries }= 8 x E_{used }x $100/kwh |
Cost_{upfront }= Cost_{inverter }+ Cost_{panels }+ Cost_{batteries} |
Cost_{life-cycle }= Cost_{inverter }+ Cost_{panels }+ Cost_{batteries, life-cycle} |
Cost_{kwh} = Cost_{life-cycle} / (9125 x E_{used}) |
We now give some concrete examples using the formula above:
Small Cabin: Choose
P_{peak, usage} = 1 kw.
E_{used }= 5 kwh,
T_{sun }= 6 hours.
From these we find that:
Inverter: $1000
Panels: $6666
Batteries (upfront): $1000
Batteries (life-cycle): $4000
Cost_{upfront }= $8666.
Cost_{life-cycle }= $11,666
Cost_{kwh} = $.255 per kwh - (26 cents/kwh)
One can see that:
The panels are the lion's share of the cost.
The upfront cost covers most of the life-cycle cost.
That solar is two to three times the price of grid power
Small Home: Choose
P_{peak, usage} = 2 kw.
E_{used }= 10 kwh,
T_{sun }= 6 hours.
From these we find that:
Inverter: $2000
Panels: $13333
Batteries (upfront): $2000
Batteries (life-cycle): $8000
Cost_{upfront }= $17,333.
Cost_{life-cycle }= $23,333
Cost_{kwh} = $.256 per kwh - (still 26 cents/kwh!)
One can see that increasing the size of the system had little impact on the life-cycle cost per kilowatt-hour.
Large Home: Choose
P_{peak, usage} = 5 kw.
E_{used }= 20 kwh,
T_{sun }= 6 hours.
From these we find that:
Inverter: $5000
Panels: $26,666
Batteries (upfront): $4000
Batteries (life-cycle): $16,000
Cost_{upfront }= $35,666.
Cost_{life-cycle }= $47,666
Cost_{kwh} = $.261 per kwh - (still 26 cents/kwh!)