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Calculate the cost of Photovoltaic Systems (Home Solar Electricity)

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.

Table of Contents

1. Introduction

2. Approach 

3. Cost of inverter as function of peak power required 

4. Cost of solar panels as a function of energy usage  

5. Cost of batteries as a function of energy usage 

6. Calculation of upfront cost 

7. Calculation of life-cycle cost per kilowatt-hour 

8. Summary of definitions and formulas 

9. Examples

Introduction

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? 

Approach

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. 

Cost of inverter as function of peak power required 

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}

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

For example, if we need 2 kilowatts of peak power used, the cost of the inverter will be about $2000 dollars.

Cost of solar panels as a function of energy usage 

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,

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.

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 of batteries as a function of energy usage

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

Because we have included the factor of two, then we are probably safe to assume at least a six year lifetime on the batteries:

Calculation of upfront cost

Adding up the costs of the inverter, panels and batteries, we find:

Calculation of life-cycle cost per kilowatt-hour

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

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 

Summary of definitions and formulas

Examples

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:

  1. The panels are the lion's share of the cost.

  2. The upfront cost covers most of the life-cycle cost.

  3. 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!)

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