How Can You Calculate Your Solar Energy Needs | Power Usage, Roof Space, System Size

Based on electricity bills, divide monthly electricity consumption by the product of daily average sunshine hours and 30 days to derive installed power.

Every 1 kW needs about 7 square meters of roof, reserve 20% margin to offset system loss, then can precisely calculate required photovoltaic panel quantity and area.

Power Usage

Look at past years' electricity consumption

The first step to accurately estimating solar energy demand is to retrieve past 12 months' complete electricity bills. Monthly electricity consumption exists significant seasonal fluctuation, usually in summer because air conditioning operation will lead to monthly readings soaring to 1,500 degrees (kWh) or even 2,200 degrees, while spring and autumn two seasons might fall back to 600 degrees to 800 degrees. To get a reliable daily average benchmark, you must add the total degrees of the whole year, then divide by 365 days. Taking a typical family annual electricity consumption of 11,000 degrees for example, daily average electricity consumption is about 30.1 degrees. The degree electricity unit price on the bill usually fluctuates between 0.12 USD and 0.35 USD, through calculating annual total expenditure, can preliminary lock system investment recovery expectation.

In addition to observing the total electricity amount, also need to analyze the electricity habit's time distribution. Morning and evening peak period's electricity price under Time-of-Use (TOU) mode is often 2 times to 3 times of usual, if the system is not equipped with energy storage cells, the daytime produced surplus electricity amount is usually sold to the grid at an extremely low price (about 0.03 USD to 0.08 USD), while at night, it needs to be bought back at a high price. Therefore, counting high price electricity consumption proportion from 4 PM to 9 PM, is crucial for deciding whether to increase Lithium Iron Phosphate cells with per kilowatt-hour cost about 400 USD.

Calculate appliance wattage

Central air conditioning system's surge current at startup instant might reach 45 amperes (A), daily operation power is usually between 3,500 watts (W) to 5,000 watts. If running 6 hours every day, a single unit air conditioner will consume 21 degrees to 30 degrees of electricity, occupying over 70% of many families' daily electricity consumption. An electric water heater is also a large electricity consumer. Its power is usually fixed at around 4,500 Watts, heating 2 hours every day, namely producing 9 degrees of electricity load.

l Refrigerator: Power is between 150 Watts to 400 Watts, although seemingly not high, but because of its 24-hour operation, deducting compressor stop time, daily average electricity consumption is usually 1.2 degrees to 2.5 degrees.

l Dishwasher and washing machine: Standard washing mode single time electricity consumption about 1.2 degrees to 1.8 degrees, if using 5 times every week, monthly cumulative increase 30 degrees load.

l Dryer: Power as high as 5,000 watts, every drying one drum of clothes consumes about 3 degrees to 5 degrees.

l Standby equipment: Such as routers, set-top boxes, and game consoles, combined standby power year-round maintains at 50 Watts to 100 Watts, every day has 1.2 degrees to 2.4 degrees of electricity volume in unconsciousness wasting away.

Find electricity consumption peak

If a family turns on a 5,000-watt dryer, a 4,500-watt electric water heater, and a 1,500-watt microwave oven at evening time, the total load will instantly rush toward 11,000 watts. At this time, if the selected inverter rated power is only 7.6 kilowatts (kW), the system will because of overload protection power off. Usually suggest inverter's rated capacity to be 15% to 20% safety margin higher than family's common maximum concurrent load.

Monitoring data shows that ordinary families' electricity consumption curve usually presents "double peak" characteristics, namely morning 7 AM to 9 AM as well as evening 6 PM to 10 PM. Solar panels reach 100% output efficiency at around 12 PM noon, having about 8 hours time difference with evening 8 PM electricity peak. This kind of mismatch, if not performing electricity habit adjustment (such as scheduling washing clothes at noon), about 60% of solar power generation cannot be immediately consumed when produced, this part of energy flow will affect the overall 25-year life cycle investment return rate.

Reserve growth space

When calculating system size, you cannot only stare at current demand, but must consider future 5 to 10 years' electricity increment. Buying one pure electric vehicle (EV) will completely change the electricity structure. An ordinary electric car every driving one mile (about 1.6 kilometers) consumes about 0.3 degrees of electricity. If annual driving mileage is 12,000 miles, annual electricity demand will surge 3,600 degrees, 10 kilowatts system needs to extra add about 8 to 10 pieces of 400 Watts panels to offset this part of expenditure.

l Expansion cost: After installation, adding panels' per watt cost is usually 40% to 60% higher than the initial installation, because you need to re-pay permit fees, lifting fees, and possible inverter upgrade fees.

l Family members: Adding one member usually daily electricity consumption increases 2 degrees to 4 degrees.

l Efficiency degradation: Photovoltaic panels' power generation efficiency every year will drop 0.5% to 0.8%, in 25th year, system's output usually only has 80% to 85% of initial value.

l Redundancy design: In initial design, reserving 20% power margin is an effective means of hedging future energy price rising 5% to 10% annual growth rate.

Deduct loss rate

A system labeled as 10 kilowatts in reality will never be able to stably output 10 kilowatts of power. Direct current to alternating current inverter loss is between 3% and 5%. Besides, when current passes through cables, it will produce resistance loss. If wiring length exceeds 30 meters, loss might reach 2%. Dust, bird droppings, or slight shading on the panel surface will bring a 3% to 7% efficiency drop, and after the air temperature exceeds 25 degrees Celsius (77 degrees Fahrenheit), every temperature rise 1 degree, monocrystalline silicon panel's power generation power usually will drop 0.3% to 0.4%.

Considering these negative factors, the whole system's comprehensive efficiency coefficient (Derate Factor) is usually set between 0.75 and 0.82. To ensure every day actually gets 30 degrees of available electricity, the system's theoretical output under ideal sunshine must be set at around 37 degrees. If the local average daily peak sunshine duration is 4.5 hours, then the required system power calculation method should be: 30 degrees divided by 4.5 hours, then divided by 0.8 efficiency coefficient, deriving finally needing to install an 8.33 kilowatt system.

Determine daily units

Currently, mainstream monocrystalline module specifications are 400 Watts to 450 Watts, single piece size is about 1.7 square meters. If calculated needing 8.33 kilowatts of system, selecting 415 Watts panels, then need to install 21 pieces. These 21 pieces of panels will occupy about 36 square meters of continuous roof space.

On the budget level, the current per watt installation price is between 2.5 USD and 3.8 USD. One system around 8 kilowatts total budget is about between 20,000 USD and 30,000 USD. If local government provides a 30% tax credit (ITC), actual expenditure can drop to 14,000 USD to 21,000 USD. Calculating by saving 2,500 USD electricity bills every year, the cost recovery cycle is about 6 years to 8.5 years. Through these precise data deconstructions, we can clearly see every cent how to transform into future 25 years' free energy flow.

Roof Space

Measure accurately with a ruler

If you plan to install a 10 kW system, you probably need 23 to 25 pieces of panels of this specification. Just these modules tiled out need 45 to 50 square meters of net area. But in actual operation, you absolutely cannot strictly stick to this area for the roof, because between panels must leave 20 mm to 25 mm installation gaps for fixing middle clamps and end clamps, meanwhile also must consider edge's wind-resistant and earthquake-resistant spacing.

Roof's shape directly decides panels' arrangement efficiency. Rectangular flat roofs' utilization rate is highest, can reach 80% above; while that kind with multiple dormer windows, chimneys, or polygonally cut roofs, the actual available area often only has 50% to 60% of the total area. When measuring must be precise to centimeters, because in wiring planning, having 10 centimeters more might be able to arrange one whole extra row of panels, directly increasing 2kW installed capacity, while 10 centimeters less might lead to the whole row plan being scrapped.

One set standard 5 kW system, on a pitched roof, usually needs about 30-35 square meters of installation space, a flat roof because of needing to consider spacing left for front and back row shading, area demand will double to 60-70 square meters.

Avoid obstacles

Even if it's only a 5-centimeter-wide vent pipe shadow, as long as it falls on series-connected modules, it might trigger bypass diode operation, leading to whole panel even whole string module output dropping 30% to 90%. When planning space, must according to the local winter solstice day (the day with the lowest sun altitude angle) shadow trajectory to define the "prohibited installation zone". Usually require reserving at least 2 times the obstacle height distance outside the obstacle shadow range.

In addition, fire regulations (such as the International Fire Code IFC) usually require roof edges, two sides of the roof ridge, to leave 45 centimeters to 90 centimeters (about 18-36 inches) of emergency pedestrian walkways, convenient for firefighters to evacuate or cut off power during a fire. If your roof total width is 10 meters, deducting 0.9 meters for fire passages on each end, the remaining 8.2 meters is where you can truly place panels.

l Shadow loss coefficient: Even if only 3% of the area is long-term shaded, because of the hot spot effect, not only power generation shrinks, but also will accelerate encapsulation material aging, making module life shorten 5-10 years.

l Avoidance distance: For a 1-meter high chimney, in Northern Hemisphere middle latitude regions, it is suggested not to lay panels within at least 2.5 meters on the North side.

Calculate roof slope

In the Northern Hemisphere, true South orientation is the golden position, power generation efficiency is defined as 100%. If the roof is biased East or West 30 degrees, annual total power generation will drop about 5% to 8%; if it's pure East-West orientation, the loss will then expand to 15% to 20%. This does not represent East-West cannot be installed, through increasing 20% installed capacity can offset this orientation disadvantage, and East-West arrangement's performance during morning and evening electricity peak periods (cooking/AC peak) often fits family electricity consumption curve better than true South.

Regarding tilt angle, the best angle usually equals local geographical latitude. When installing on a flat roof, usually, the angle will be raised through aluminum alloy brackets, but pay attention that the higher the bracket is raised, the larger the wind load area, the requirement for expansion bolts' pull-out resistance will lift from normal 2 kN to 5 kN above.

Weigh the load-bearing capacity

Photovoltaic modules themselves weigh about 18 kg to 25 kg, plus aluminum alloy brackets, ballast blocks or stainless steel hooks, converted per square meter will add 15 kg to 30 kg extra weight to the roof. For color steel tile roofs, must confirm purlin spacing (usually require within 1.2 meters to 1.5 meters) and thickness (not lower than 0.6 mm) to prevent crushing deformation.

If it's an asphalt shingle roof with age over 15 years, forced installation might lead to trampling during the installation process, causing shingles to break, triggering late-stage water leakage. Because photovoltaic system's warranty period is usually 25 years, if the roof itself only has 5 years left, the most stable practice is first spending 5,000 to 10,000 USD to re-lay the roof, then perform photovoltaic construction. Otherwise, 10 years later for repairing the roof, the fee for dismantling and re-installing the photovoltaic system will occupy over 40% of the original installation cost. This is extremely uneconomical.

Structural check items:

1. Roof material remaining life whether > 20 years.

2. House beams whether have decay, cracking or obvious deflection (sagging).

3. Wind load design: Under local maximum gust wind speed, module fixing point's tensile strength.

Leave foot space

The photovoltaic system is not finished after installation and no need to worry, you need to consider later maintenance costs. Dust, bird droppings, or fallen leaves accumulated on the module surface, if not regularly cleaned, monthly power generation will because of light transmittance drop produce 10% to 15% linear landslide. Therefore, when arranging panels, every two rows need to leave one about 30 centimeters to 40 centimeters maintenance passage, convenient for personnel standing and using a high pressure water gun or cleaning brush.

In addition, the inverter's installation position should also be close to the roof access point, to reduce the direct current cable length. The longer the DC line is, the more electrical energy is dissipated by internal resistance. Taking commonly used photovoltaic dedicated cable for example, if the line round-trip length exceeds 50 meters, voltage drop loss produced during line full load might reach 1% to 2%.

System Size

Calculate total wattage

If your home's daily average consumption is 30 kWh of electricity, and the region where you are located has 4.5 hours of peak sunshine time, you cannot simply use 30 divided by 4.5 to get 6.6 kW. Because the solar system in the energy conversion process will produce various losses, usually need to introduce a comprehensive efficiency coefficient between 0.75 and 0.85. If calculating by 0.8 efficiency, the actual required installed capacity should be 30 divided by (4.5 multiplied by 0.8) equals 9.375 kW.

If your goal is 100 percent off-grid, then system size usually needs to increase 20% to 30% than average demand. This is to cope with continuous cloudy and rainy days or winter sunshine duration shrinking. In a 10 kW system, every increase in 1 kW capacity will result in annual power generation in sunshine-abundant regions (such as California or parts of Australia) being roughly 1,450 kWh to 1,650 kWh. This part of extra generated electricity if cannot be real-time consumed, and no energy storage cell configured, then in regions without Net Metering tariff policy, its economic value will be greatly discounted.

Pick inverter

Inverter's size usually does not need to perfectly 1:1 match with solar panels' total power, here involves a professional DC-AC ratio concept. In the photovoltaic industry, a reasonable DC-AC ratio is usually between 1.2 and 1.25. If you install 10 kW solar panels, matching one 8 kW inverter is often the most efficient. This is because solar panels can only reach peak power during that short one or two hours at noon, and also must be under conditions that temperature, angle, and air transparency are all perfect. In most months of the whole year, panels' actual output might only be around 80% of nominal value.

Although at midday, the midsummer noon might appear slight clipping phenomenon, namely, the inverter forcibly limits output to 8 kW, losing that small part of electrical energy, but according to industry statistics, this kind of clipping loss is usually less than 1% of annual total power generation. In contrast, purchasing more expensive high-power inverters increased hardware expenditure, often needs over 15 years to earn back through extra generated electricity. For a 10kW DC system, choosing an 8kW inverter can save about 15% of the initial inverter purchase cost, while maintaining 97% above average conversion efficiency.

Deduct loss

Photovoltaic module surface temperature every rising 1 degree Celsius, its output power will drop about 0.3% to 0.4%. If the ambient temperature is 35 degrees Celsius, the black panel surface temperature might soar to 65 degrees Celsius, this 40-degree temperature difference will lead to power directly shrinking 12% to 16%. Plus direct current cable during transmission process's voltage drop loss (usually 1% to 2%), inverter's conversion loss (2% to 4%), and matching loss between modules (1% to 2%), your 10 kW system's real-time display power at the meter end is hard to exceed 8.5 kW.

Dust and dirt accumulation is also a variable, long-term uncleaned panels' power generation will drop 5% to 10%, in arid and sandy regions, this number will even reach 20%. If weight-calculating these data, one standard 10 kW system's annual total derate coefficient is roughly 0.78. When calculating system size, must put this coefficient into the denominator. If you need 15,000 kWh of electricity every year, and the local per kilowatt every year can generate 1,600 degrees of electricity, you actually need an installed capacity of 15,000 divided by (1,600 multiplied by 0.78) equals 12 kW, not 9.3 kW.

System derate details:

l Inverter efficiency loss: 3.0%

l Cable resistance loss: 1.5%

l Module dirt loss: 5.0%

l Temperature rise loss: 10.0%

l Power degradation and matching loss: 2.0%

Leave some margin

Considering photovoltaic modules' 25-year long life, system size design must include anticipation of power degradation. Mainstream monocrystalline silicon PERC modules or N-type TOPCon modules, first year degradation is usually between 1% to 2%, after that, every year will linearly drop at a speed of 0.4% to 0.55%. When your system runs for 20 years, its power generation capability is only about 88% of its installed time. If your system size design is just enough to use, then by the 10th year, you have to again face the awkwardness of buying electricity from the grid.

Life electrification trend also requires you to reserve more 15% to 20% redundancy when determining size. If you don't have a new energy vehicle yet, but plan to buy an electric car with a cell capacity of 80 kWh within 5 years, running 300 kilometers every week, then you need to extra supplement about 60 kWh of electricity every week. This equals annually adding 3,120 kWh load, needing extra 2 kW to 2.5 kW installed capacity to cover. If the initial period did not reserve this margin, late-stage adding two or three pieces of panels' labor cost and filing fee, will be more than 3 times the cost of installing a well at once now.

Calculate payback period

System size directly decides your investment return rate. Currently, photovoltaic system's average installation cost is between 2.5 to 3.5 USD per watt (including labor and equipment). One 10 kW system total investment is roughly around 30,000 USD. If the local average electricity price is 0.2 USD per degree, your system annually generates 15,000 degrees of electricity and all for self-use, then every year you can save 3,000 USD in electricity bills. Under the condition of not calculating any government subsidies, the simple payback period is 10 years. If counting a 30% federal or local tax credit, the payback period will shorten to around 7 years.

The larger the system scale is, the per watt unit price actually is decreasing. Installing one 5 kW system's fixed cost (such as design fee, permit fee, truck dispatch fee) is almost the same as a 10 kW system. For one 5kW system, its soft cost proportion might be as high as 40%, while for a 15kW system, the soft cost proportion will be diluted to below 25%. Under the premise of roof area allowing, try to choose a larger system capacity, which can effectively reduce the Levelized Cost of Electricity (LCOE). Currently, large systems' electricity cost can be as low as 0.05 to 0.08 USD per degree, far lower than most regions' commercial grid electricity price.