A calculation of the total electricity output of a solar panel system in kilowatts hours (kWh)

A calculation of the total electricity output of a solar panel system in kilowatts hours (kWh)

Use the following steps as a guide to making this calculation:

• How many panels fit on the roof, assuming the building is rectangular? (To make this determination, determine the number of panels that can fit along one side, and the number of rows of panels that can fit along the opposite side. Round down to the nearest whole number of panels in each direction and multiply to obtain the required number of panels.)
  • What are the dimensions of the building’s roof in meters?
    • 60m x 30m
  • What are the dimensions of each solar panel in centimeters? (To convert from inches to centimeters, multiply the dimensions in inches by 2.54 and do not round until the last step of the calculation.) Find the number of meters for each dimension
    • Solar panel dims – 80 in x 40 in x 2 in =
    • 80 in x 2.54 = 203.2 cm
      • 2 cm / 100 = 2.03 m in length
    • 40 in x 2.54 = 101.6 cm
      • 6 cm / 100 = 1.016 m in width
    • 2 in x 2.54 = 5.08 cm
      • 08cm / 100 = 0.0508 m in height
    • How many panels will fit on the roof in each direction? Round down to the nearest panel. When calculating the number of panels, be sure that when you change from length to width on the roof, you also change from length to width on the panels.
      • Panels fitting lengthwise on long side of roof
        • Length of roof – 60 m / 2.03 cm = 29.55 = 29 panels
        • Width of roof – 30 m / 1.016 cm = 29.52 = 29 panels
        • 29 panels lengthwise x 29 panels widthwise = 841 panels
      • Panels fitting widthwise on long side of roof
        • Length of roof – 60 m / 1.016 m = 59.05 = 59 panels
        • Width of roof – 30 m / 2.03 m = 14.76 = 14 panels
        • 59 panels lengthwise x 14 panels widthwise = 826 panels
      • More panels will fit on the roof by installing them lengthwise on the long side of the roof = 841
    • Find the area of the panels in meters to make sure that the area of the panels is less than the area of the roof.
      • Area of roof = 60 m x 30 m = 1,800 sq. m.
      • Area of panels = 29 panels lengthwise by 29 panels widthwise
        • Length – 29 panels x 2.032 m = 58.93 m
        • Width – 29 panels x 1.016 m = 29.46 m
        • Area = 58.93 m x 29.46 m = 1736.08 sq. m.
      • 841 panels have an area of 1,736.08 square meters that will fit inside the area of the 1,800 square meters of the roof.
    • What is the total amount of electricity that could be produced by adopting a solar panel system that covers the entire roof based on the average monthly sunlight? (1kW = 1000W, and 1 hour of sunlight produces 400 watts per panel)
      • How many kW per hour of sunlight could be produced per solar panel? Per entire system (based on how many panels could fit on the roof)?
        • 1 panel = 400 watts per hour
          • 1 kW = 1000 watts = kW ratio = 1:1000
          • Panel ratio X/400 watts = 1/1000 watts
            • X1000=400(1) = X1000=400
            • X = 400/1000 = X = .40 = .40 kW/hour
            • One panel produces .4 kW/hour
          • 841 panels x .40 kW/hour = 336.4 kW/hour
        • How many hours of sunlight are expected on average per month? Calculate the average hours based on the monthly data provided. Round down to the nearest tenth of an hour.
          • Average monthly sunlight = total hours of sunlight of each month divided by 12 months
            • Jan 163 hours
            • Feb 168
            • Mar 214
            • Apr 227
            • May 276
            • Jun 287
            • Jul 301
            • Aug 277
            • Sep 237
            • Oct 206
            • Nov 143
            • Dec 142
            • Total 2,641 hours
          • 2,641 hours of sunlight / 12 months = 220.1 monthly average of sunlight hours
        • What is the likely average amount of kWh produced per panel based on the average amount of sunlight per month? Per year? Round to the nearest hundredth kWh.
          • Per month – per panel per hour = .40 kWh
            • Average hours of sunlight per month = 220.1
              • 1 x .40 kWh = 88.04 kWh per panel per month
            • Per year – 2,641 hours of sunlight
              • 2,641 x .40kWh = 1,056.40 kWh per panel per year
            • What is the total amount of kWh that is produced by the entire solar panel system per month based on the average monthly sunlight? Per year?
              • Per month full system –
                • 841 panels x 220.1 hours of sunlight per month = 185,104.1 hours
                  • 185,104.1 hours x .40 kWh = 74,041.64 kWh per month
                • Per year full system –
              • 74,041.64 kW x 12 months = 888,499.68 kWh per year

A calculation of the difference between the current electricity usage of the building and the electricity generated by a solar panel system in kilowatt hours and in dollars

Use the following steps as a guide to making this calculation:

• How much electricity does the building use on average annually?
• 271,253 kWh
  • What is the estimated total amount of kWh produced by the entire solar panel system on this building per year, based on the average monthly sunlight for the area?
    • 888,500 kWh
      • Based on the average cost of electricity in the area, about how much is the annual electricity cost for the SNHU building?
    • 271,253 kwh x $0.165/kWh = $44,756.75 annual electricity cost
  • Is the amount of electricity generated by the solar array sufficient to cover SNHU’s yearly electricity usage?
    • Yes, the solar energy is sufficient to cover the yearly electricity usage.
  • If not, what is the remaining energy needed in kW? How much would this cost in dollars? In other words, what is the remaining utility bill?
  • If the energy generated is more than the energy needed to run the building, how much additional savings is there for energy that can be channeled to other buildings on campus or sold back to the energy company?
    • Annual solar kWh = 888,500
    • Annual electric kWh = 271,253
      • Solar – electric = 888,500 kWh – 271,253 kWh = 617,247 kWh overage
        • 617,247 kWh overage x $0.165/kWh = $101,845.76 savings

A determination of the likelihood of receiving a damaged panel

SNHU expressed some concerns about receiving damaged solar panels from the manufacturer. You would like to be transparent and address these concerns by illustrating the likelihood of a damaged panel based on the size of the system SNHU would be purchasing.

The manufacturer has reported that since solar panels are complex and evolving technology, 1 out of every 1,000 manufactured solar panels is defective.

• How many panels fit on the roof?
• 841 panels
  • What is the probability or likelihood that SNHU will receive a damaged solar panel, based on the number of panels it would be purchasing?
• P(damaged panel) = 1/1000 panels = 0.001 x 841 panels = 0.84 = 84% chance of receiving a damaged panel

A determination of how long it would take to pay back the cost of buying the system in years

Use the following steps as a guide to making this calculation—you can assume there will not be any required maintenance during the first 10 years:

• What would be the upfront cost to purchase and install the solar panel system?
  • How much does each panel cost? How much does the entire system cost?
    • $560 per panel
    • $560 x 841 panels = $470,960 total panel cost
  • How much does installation cost?
    • $0.75 per watt
    • 400 watt x $0.75 = $300 per panel
    • $300 x 841 panels = $252,300 installation cost
  • Total system cost
    • $470,960 + $252,300 = $723,260
  • What are the government incentives? How does that affect the cost?
    • Discount of 30% of the panels and installation
    • $723,260 x 30% = $723,260 x 0.30 = $216,978
      • The discount will lower the price by $216,978
      • Total system Cost = $723,260 – $216,978 = $506,282
    • What is the remaining utility cost, if there is one?
      • Based on the system generating more power than needed, there is no remaining utility cost.
    • How much will your solar panels save SNHU per year?
      • $101,846 per year

How long would it take to pay back the cost of purchasing the solar panel system in years? (Years = Cost to Purchase and Install Solar Panel System / Savings per Year) The time in years should take into account all energy savings, not just those for the building on which the solar array is installed.

• $506,282 / $101,846 savings/year = 4.97 = 4.9 years to pay back the cost of purchase

A determination of whether there is a cost savings over 10 years for leasing the solar panel system

Use the following steps as a guide to making this calculation:

What is the total cost without solar for 10 years, in dollars?

• Annual average use = 271,253 kWh
• $0.165 per kWh
• 271,253 x $0.165 = $44,756.75 per year
• $44,756.75 x 10 years = $447,567.45
• The cost for 10 years without solar equals $447,567.45

What is the total cost with solar for 10 years, in dollars?

• How much does it cost to rent the entire solar panel system?
• Annual cost to rent = $97.00 per panel x 841 panels x 12 months
• = $978,924
• 10 year rental cost = $978,924 x 10 = $9,789,240 10 year cost

What is the total remaining utility bill for 10 years?

• $20 per month x 12 months = $240 per year
• $240 x 10 years = $2,400
• Total 10 year total remaining utility bill is $2,400.

What are the 10-year savings? (Cost Without Solar for 10 Years – (Cost of Solar Panel Rental for 10 Years + Remaining Utility Bills for 10 Years) = Total 10-Year Saving)

• 10 year cost without solar = $447,567.45
• 10 year savings = $447,567.45 – $9,789,240 + $2,400 = -$9,339,272.55
• There is no cost savings for leasing over 10 years.

Hint: leasing costs should be a substantial fraction of purchase costs without any government incentives. Costs for leasing are typically listed per panel.

A recommendation for whether SNHU should install solar energy panels on its buildings based on your calculations

Currently, the annual cost using electricity is $44,756.75. By using solar power, the energy produced by the panels exceeds the required energy needed for this building. The savings from this energy overage is $101,846 which can be spread over to offset costs in other buildings. Based on the savings, we are recommending installing the solar panel system.

An explanation of whether SNHU should invest in a solar energy system by purchasing it upfront or by leasing it (Base your response on your calculations.)

The cost of purchasing the solar energy system after discount is $506,282 which is less than the yearly cost of leasing the system of $978,924. By purchasing the system upfront, the system will start showing savings in less than 5 years. If leased, there is no savings at any point. Based on these totals, we recommend purchasing the solar panels upfront rather than leasing.