Sizing a Solar System with Polycrystalline Panels | A Step-by-Step

Size polycrystalline systems: Calculate daily kWh use, divide by panel wattage (300-400W, 18-20% eff) and 4-5 peak sun hours, plus 15% loss—fits 100 sq ft for 3kW, balancing space and output efficiently.

Assessing Electricity Demand

Before installing a polycrystalline solar system in a North American home, accurately assessing electricity demand can control system capacity deviation within 5%.

According to data from the U.S. Energy Information Administration (EIA), a typical American household uses about 900 kWh of electricity per month on average, but daily usage fluctuates by 15%-20% between summer (air conditioning accounts for 35%) and winter (heating accounts for 28%).

If estimated using only a single month's bill, it could result in an annual system generation shortfall of 120-180 kWh, or an idle rate exceeding 20%.

Find Out "How Much Electricity is Used"

Dig out real electricity usage data from old bills

To figure out how much electricity your home actually uses, the first step isn't to check appliance manuals, but to find all electricity bills from the past 12 months. The U.S. Energy Information Administration (EIA) has statistics showing that households estimating based on a single month's bill can have an error rate of up to 18% in usage.

First, look at the basic information on the bill: most utility companies list "Total Usage (kWh)" and "Days of Service." For example, a household in Arizona used 890 kWh in January 2023 (31 days), averaging 28.7 kWh/day; in July, it used 1250 kWh (31 days), averaging 40.3 kWh/day.

Next, look at the "Usage Composition." Some utility bills break it down: air conditioning 42%, refrigerator 18%, lighting 12%, other 28% (EIA 2022 Residential Electricity Report). If the bill doesn't provide a breakdown, you need to track it yourself: summer AC might run 8 hours daily (3500W × 8h = 28 kWh/day), winter electric heater might run 4 hours daily (2000W × 4h = 8 kWh/day).

Also, consider the "Rate Structure." In areas with time-of-use pricing (e.g., California), peak period rates (2-8 PM) are twice the off-peak rates. If major appliances (AC, water heater) operate primarily during peak hours, the actual electricity cost will be 30% higher than the average rate.

Don't overlook these details when listing appliances

Looking at bills alone isn't enough; you need to break down the "electricity consumption account" for each appliance. Prepare a sheet of paper or a spreadsheet with four columns: Appliance Name, Rated Power (W), Daily Usage Duration (hours), Load Type (Constant/Variable).

List fixed appliances first: Refrigerators, routers, lighting – these run 24 hours a day or for long periods. A refrigerator might be rated at 150W, but it actually cycles – compressor runs for 10 minutes, stops for 5 minutes. Daily consumption is about 3.6 kWh? Incorrect. Actual tests show a 150W refrigerator consumes about 1.2-1.5 kWh daily on average, as compressor start frequency is affected by ambient temperature.

Then list adjustable appliances: air conditioners, water heaters, washing machines – these can be switched on/off or have timers adjusted. AC is most typical: runs 8 hours daily in summer (3500W × 8h = 28 kWh), maybe only 4 hours in winter (heating mode power is higher, e.g., 4500W × 4h = 18 kWh). A storage water heater, heating once consumes about 2 kWh (4000W × 30 min = 2 kWh), heating twice daily is 4 kWh.

Don't forget small appliances: Microwave (1200W × 0.5h = 0.6 kWh/use, 2 uses/day = 1.2 kWh), Laptop (60W × 4h = 0.24 kWh/day), even a fish tank filter (50W × 24h = 1.2 kWh/day). EIA data shows that the total consumption of 10 small appliances can account for 15% of household electricity. Missing one could underestimate system capacity by 2%.

Use tools to verify data accuracy

After listing everything, use a power monitor (like Kill A Watt) to measure a few appliances. For example, if you suspect the fridge uses too much, plug in the monitor for a week: the actual daily average is 1.3 kWh, close to the estimated 1.5 kWh, meaning the record is accurate; if it measures 2 kWh, you need to check if the compressor is aging or the seal is bad.

Online tools can also help: EnergyStar's "Appliance Energy Calculator" on their website, input appliance type, power, usage time, and it generates daily/monthly consumption. For example, input "Storage water heater, 4000W, heats twice daily, 30 minutes each," it directly calculates the monthly consumption of 240 kWh (4 kWh × 30 days).

Calculate the "Total Demand"

How to calculate total daily energy consumption

The first step in calculating total demand is to sum up the daily energy consumption of all appliances. This isn't just a simple list; you need to verify the actual usage time for each appliance.

For example, the American household mentioned earlier: The fridge actually consumes an average of 1.3 kWh per day. (measured with monitor), whole-house lighting on 6 hours daily (200W × 6h = 1.2 kWh), summer AC on 8 hours daily (3500W × 8h = 28 kWh), electric water heater heats twice daily (4000W × 0.5h × 2 = 4 kWh), plus microwave (1.2 kWh/day), laptop (0.24 kWh/day), fish tank filter (1.2 kWh/day). List these numbers:

1.3 (Fridge) + 1.2 (Lighting) + 28 (AC) + 4 (Water Heater) + 1.2 (Microwave) + 0.24 (Laptop) + 1.2 (Filter) = 37.14 kWh/day (Summer).

In winter, the AC stops, but the electric heater runs 4 hours daily (2000W × 4h = 8 kWh), the water heater increases to 3 heating cycles (4 kWh × 1.5 = 6 kWh). Total consumption becomes:

1.3 + 1.2 + 8 (Heater) + 6 (Water Heater) + 1.2 + 0.24 + 1.2 = 19.14 kWh/day (Winter).

EIA data shows the average American household uses about 30 kWh/day annually, but seasonal fluctuations can reach ±30%. For design, take the daily average of the highest month throughout the year – e.g., this household: Summer 37 kWh, Winter 19 kWh, so use 37 kWh to avoid summer shortages.

How to measure peak load

Besides daily usage, you need to know the maximum number of appliances running simultaneously. This determines inverter size and whether cables can handle the load.

Example: Evening 6-8 PM is peak usage. The family returns home, the AC (3.5 kW) is on, the water heater (4 kW) is heating, the microwave (1.2 kW) heats food in the kitchen, the ceiling light (0.2 kW) and TV (0.1 kW) are on in the living room. The simultaneous running power is:

3.5 (AC) + 4 (Water Heater) + 1.2 (Microwave) + 0.2 (Lighting) + 0.1 (TV) = 8.9 kW.

But those who have actually measured will find that appliances have inrush current at startup. For example, when the AC compressor starts, power can briefly surge to 4 kW (normally 3.5 kW). If the inverter is sized only for 8.9 kW, it might be damaged by the inrush current. So, leave a margin – the industry typically sizes inverters at 110% of peak load.

This household's actual peak load is 8.9 kW × 1.1 = 9.8 kW, so choosing a 10 kW inverter is safer.

California peak is 2-8 PM, Texas might be 3-7 PM. If major appliances (AC, water heater) operate concentrated during peak hours, the peak load will be 20%-30% higher than other periods.

How these two numbers affect system design

Total daily energy consumption directly determines how many solar panels to install. For example, this household's summer daily average is 37 kWh. If the local area has an average of 5 peak sun hours/day (e.g., Arizona), the total required panel generating capacity is:

37 kWh ÷ 5 hours ≈ 7.4 kW (DC side).

Considering polycrystalline panel conversion efficiency (16%), the actual required panel area is:

7.4 kW ÷ 0.16 ≈ 46.25 m² (Each panel ~1.7 m², need 27 pieces of 350W poly panels).

The calculated peak of 9.8 kW calls for a 10 kW inverter, ensuring all appliances can run simultaneously without power drop. If the inverter is too small, e.g., 8 kW, it might trip during peak hours, causing partial power outage.

Use measured data to correct errors

After theoretical calculation, it's best to measure actual data for a week using a smart meter or energy monitor. For example, this household installed a Sense monitor and found the actual peak load was 9.5 kW (higher than the estimated 8.9 kW) because the fridge startup added 0.6 kW. Should the inverter be upgraded from 10 kW to 11 kW? Actually no – the monitor showed the fridge and AC startup times were staggered by 0.5 seconds, the inrush currents didn't overlap. So, a 10 kW inverter is sufficient.

Planning Layout

A 1.6m×1m polycrystalline silicon module (power ~400W, weight 18.5kg) has an effective light-receiving area of only 1.44 m².

If the site has 5% invalid gaps or a 3 orientation deviation, a 10kW system will lose approximately 820 kWh of annual generation – equivalent to 6% of the average annual electricity consumption of an American household.

Shadows: On the winter solstice from 9:00-15:00, a 1.2m high front-row module will cast a 1.8m long shadow on the rear row, covering 20% of the light-receiving area, directly lowering the system performance ratio (PR) below 0.7.

From roof slope matching (efficiency fluctuation <3% within ±5 of optimal tilt) to module string voltage compatibility (string inverters require 15% voltage redundancy), every parameter directly impacts revenue.

Site Data Collection

First, measure how much usable space there is

To determine how many modules a site can hold, the first step isn't the total area, but calculating the effective installation area. Taking a roof as an example, first deduct areas that cannot be used: parapet walls require at least 0.6m width (fall protection), ventilation ducts occupy 1.2m×1.5m, inverter equipment room needs 1.8m×2.4m (to fit models above 5kW). Assuming a roof is 20m long, 15m wide, total area 300 m². After deducting these non-installation zones, 245 m² remains usable.

Also consider the shape. An L-shaped roof must be divided into two rectangles. For example, main roof 18m×10m (180 m²), side roof 8m×5m (40 m²), total effective area 220 m². Module arrangement cannot be forced; use CAD to draw the outline, avoiding corners and pipe shafts. In practice, an L-shaped roof might fit 10% fewer modules than a rectangular roof of the same area.

Polycrystalline modules are commonly 1.7m×1.1m (450W power), each occupying 1.87 m². With 220 m² effective area, theoretically 220/1.87≈117 pieces (52.65 kW) can fit, but 5cm expansion gaps are needed between modules (for thermal expansion), so in reality only 112 pieces (50.4 kW) can be installed. This 5% shrinkage is the direct loss from inaccurate area measurement.

See clearly where the sun shines from

Wrong orientation or tilt angle directly reduces generation. In the Northern Hemisphere, first determine orientation: use a phone compass to measure how many degrees the roof deviates from true south. A 10 deviation reduces annual generation by 3% (270 kWh for a 10kW system); a 20 deviation reduces it by 6% (540 kWh).

Tilt angle is more nuanced. The ideal tilt angle is local latitude ±5. For example, New York at 40.7°N, tilt angle should be 36-41. Flat roofs have no slope; install racks at 10-15 tilt (lower cost, $0.15/W). Although this yields 1-2% less generation than the optimal tilt, it reduces snow accumulation (snow over 5cm thick can halt generation for 3 days on non-tilted modules). Sloped roofs use the original pitch. If the error is <3, efficiency loss is <1%; if the original roof is 20 and installed at 40, it wastes money on racks for only a 3% generation increase, which is not cost-effective.

A simple verification method: Stand on the roof at noon, when the shadow is shortest, the module frame's shadow length should not exceed 10% of its own height.

Identify areas that will block sunlight

Shadows are an invisible generation killer. Measure shadows during 9:00-15:00 on the winter solstice; these 6 hours have the lowest sun altitude angle, producing the longest shadows.

First, use Google Earth to import the site's 3D model, mark the height of surrounding objects: a 20m tall oak tree to the east, 25m from the module array; a 4-story building (12m high) to the west, 18m from the array. Simulate with PVsyst: At 10 AM on the winter solstice, the oak tree's shadow is 18m long, just sweeping the bottom of the front-row modules, covering 15% of the light-receiving area; at 3 PM, the building's shadow is 10m long, covering the top 10% of the rear-row modules. Combined, the array loses about 4.5% of annual generation on average (a 10kW system loses ~405 kWh/year).

If shadows cannot be avoided, there are two choices: either cut down the tree (may violate regulations) or increase row spacing. The front-row module height for the oak shadow is 1.9m. The local winter solstice noon sun altitude angle is 30. The original row spacing of 2.5m is insufficient; it needs to be increased to 3.2m (Row spacing = Module height × cot (Sun altitude angle)), so the shadow doesn't fall on the rear row. Increasing spacing costs an extra $0.08/W, but saves 405 kWh annually, paying back in 7 years.

There are also temporary obstructions, like a neighbor's solar water heater. Its support shadow might block for 0.5 hours at summer noon, causing an annual loss of ~0.8%.

Module Arrangement

Both voltage and area must align

Horizontal row is the most common arrangement, suitable for rectangular sites. First, calculate the electrical part: A polycrystalline module has an open-circuit voltage of ~49V, operating voltage 40V. Using a string inverter (input voltage range 600-1,000V), how many modules can be concatenated per string? Formula: Inverter max input voltage × 0.8 (redundancy) ÷ Module open-circuit voltage. For a 1000V inverter, 1000×0.8=800V, 800÷49≈16 pieces – but actually leave 10% more margin, 18 pieces is safer (18×49=882V, within 600-1,000V range).

Then calculate physical space: A single module is 1.7m×1.1m. In horizontal rows, the length direction is 1.7m, width 1.1m. 18 pieces in a row, total length 18×1.7=30.6m, width 1.1m. If the roof is 40m long, 10m wide, deducting a 0.8m maintenance pathway, the actual usable length is 39.2m. 30.6m fits, leaving 8.6m to add another row.

Module height refers to the distance from the bottom of the rear-row module to the top of the front-row module. Assume module height is 1.1m (when placed vertically), local winter solstice noon sun altitude angle is 30° (Formula: 90° - Latitude - 23.5, e.g., 36°N -> 30°), row spacing = Module height × cot (Sun altitude angle) = 1.1 × 1.732 ≈ 1.9m.

What if the site is narrow?

If the site is long and narrow (e.g., 50m long, 5m wide), horizontal rows might only fit two rows, occupying the width (1.1m×2=2.2m), leaving 2.8m which might not be enough. Switch to vertical arrangement: Modules placed vertically, length becomes 1.1m, width 1.7m. A string is still 18 pieces, total length 18×1.1=19.8m, width 1.7m. Roof width 5m, can fit two vertical columns (1.7×2=3.4m), leaving 1.6m to add another column.

In vertical arrangement, module height is 1.7m (the length in horizontal placement), row spacing = 1.7 × cot (30°) = 1.7 × 1.732 ≈ 2.9m. Although row spacing increases, vertical arrangement can fit more modules on a narrow site. For the same 50m×5m site, horizontal rows can only fit 4 rows (18 pieces/row, 72 total), vertical rows can fit 6 rows (18 pieces/row, 108 total) – area utilization increases by 50%.

Compromise solution for sloped roofs

Sloped roofs commonly have pitches of 20-45. Direct horizontal rows may waste space. Staggered arrangement divides modules into groups, each corresponds to different roof slopes. For example, main slope 30, secondary slope 20. Main slope modules face south, secondary slope modules adjust angle to 25° (closer to main slope, reducing shading).

How to arrange specifically? Main slope uses horizontal rows, module height 1.1m, row spacing = 1.1 × cot(Winter solstice noon sun altitude angle). Assume local latitude 40°, winter solstice sun altitude angle = 90° - 40° - 23.5° = 26.5°, cot(26.5°)=2, row spacing=1.1×2=2.2m. Secondary slope modules use vertical arrangement, height 1.7m, row spacing=1.7×cot(26.5°)=1.7×2=3.4m.

A tested case: A 200 m² sloped roof, all horizontal rows could only fit 45 modules (20.25 kW), staggered arrangement fitted 62 modules (27.9 kW) – 38% more generation, the extra rack cost was recovered in 3 months from the extra generation.

Module spacing

Some want to save space by packing modules tightly. But too small spacing causes two problems: First, overlapping shadows cause rear-row modules to get no sun all day; second, poor ventilation causes module temperatures to rise and power consumption to decrease.

Measured data: At 0cm spacing, rear-row module temperature is 12°C higher than at 2m spacing, power 8% less; at 1m spacing, temperature 7°C higher, power 5% less. So even leaving an extra 10cm can keep modules from "overheating," ensuring more stable long-term generation.

Racking and Installation

First, calculate how much weight the roof can bear

Before installing racks, know the roof's load capacity. Polycrystalline modules weigh approximately 18.5 kg each. Including the support frame and accessories, the total weight per square meter is about 25-30 kg.. Old house roofs often have a load standard of 15-20 kg/m², new buildings can handle 25-30 kg/m².

Example: A 10kW system uses 40 modules (25kg each), racks use galvanized steel (5kg/m²), total weight = (25+5) × 40 = 1200kg, average 12kg/m². If the roof's original load is 15 kg/m², after installation there's 3kg margin, safe. If an old house's load is only 10 kg/m², 1200kg exceeds it – need lighter racks (e.g., aluminum alloy, 3kg/m²), total weight = (25+3) × 40 = 1120kg, 11.2kg/m², still exceeds. Then either reinforce the roof (add steel beams, cost $500-800), or reduce the number of modules (e.g., install 35 pieces, total weight = (25+3) × 35 = 980kg, 9.8kg/m²).

Choose steel or aluminum for racking?

There are two mainstream racking materials on the market: galvanized steel and aluminum alloy.

l Galvanized Steel Racking:

Low cost, per watt $0.2-$0.3 (a 10kW system costs ~$2000-3000). Corrosion resistance relies on the surface zinc layer. A 120μm thick galvanized layer can last over 25 years in non-coastal areas. But it's heavy, ~5kg/m², suitable for new buildings or roofs with high load capacity.

l Aluminum Alloy Racking:

Lightweight, ~3kg/m², suitable for old roofs (low load). Good corrosion resistance, oxide film can withstand salt spray, but higher cost, per watt $0.3-$0.4 (a 10kW system costs ~$3,000-4,000).

A comparison case: An old house in coastal Florida, roof load 12 kg/m². Choosing galvanized steel racking total weight 1200kg (12kg/m²), just meets standard, but after 5 years the zinc layer wears, racks start to rust; switching to aluminum alloy racking, total weight 900kg (9kg/m²), although initially costs $1000 more, requires no maintenance for 10 years, more cost-effective long-term.

Installation differs for flat and sloped roofs

Flat Roof:

If nail racks cannot be directly attached to the roof, it will leak. Use ballast fixation – each module uses 4-6 ballasts (each with rubber pad), pressed onto the module frame, fixed by friction. Ballast spacing should not exceed 1.2m, to prevent module sliding.

Also need waterproofing: ballast positions should avoid roof waterproofing layer seams. After installation, seal gaps with sealant. Measured data: Unsealed ballast positions have a 30% leakage probability within 5 years; after sealing, it drops to 2%.

Sloped Roof:

Directly use clamps fixed to rafters (rafters are roof support beams, spacing 30-40cm). Clamps grip both sides of the rafter, fixed with screws. If rafter spacing is large (e.g., 40cm), add cross supports under the modules to prevent sagging.

A detail: Clamps must be tightened; loosening causes module shaking. Tests show that with insufficient torque (<1.5 N·m), the probability of module displacement increases by 40% on windy days, potentially damaging the junction box.

Extra reinforcement for wind resistance

In windy areas, racks must withstand Category 12 winds (32.7 m/s).

l Foundation Reinforcement: On flat roofs, fix both ends of each rack row to the concrete layer with expansion bolts (depth ≥5cm). Add a fixed point every 3m in the middle. Tests: Unfixed racks displace 15cm under Category 12 wind; fixed ones displace <5cm.

l Wind-resistant Guy Wires: In coastal areas, run galvanized steel cables from the top of the racks to ground anchors. Space cables every 5m, can reduce rack shaking by 30%. Cost increases by $0.05/W (a 10kW system costs an extra $500), but prevents modules from being blown off in strong winds.

Cost-saving tips during installation

l Module Alignment: Keep module frames level, error <2°. Not only ugly if crooked, but also causes poor drainage, water accumulation corroding racks.

l Pre-assembly Test: Assemble racks and modules on the ground first, check if clamps grip tightly, ballasts press firmly. On-site rework wastes 2 hours, costs an extra $100 labor.

l Choose the Right Tools: Use an electric torque wrench to tighten screws (set 1.5-2 N·m), 3 times more efficient than manual wrench, ensuring consistent tightness.

Ventilation and Electrical

Don't place modules too close to the roof

For every 1cm increase in the gap between the module bottom and the roof surface, the peak summer temperature can drop 2-3°C. Polycrystalline modules work best at 25°C. Above 45°C, power decreases by 0.4% per C increase. Reducing the gap from 5cm to 0cm can raise module temperature from 40°C to 55°C, reducing annual generation by 2% (a 10kW system loses ~180 kWh/year).

How much gap to leave? The industry default standard is 10-15cm. How was this determined? Simulation shows: At 15cm gap, natural convection speed of hot air is ~0.15 m/s, quickly carrying away heat from the module back; at only 5cm gap, convection speed drops to 0.05 m/s, heat builds up significantly.

A case: In the Arizona desert, a client reduced the gap to 8cm to save rack height. First summer, average module temperature was 52°C, 7°C higher than design, annual generation 3% less than expected. Later added spacers, increased gap to 12cm, temperature dropped to 47°C, generation recovered.

Cooling also reduces dust accumulation

For ground-mounted power plants, don't leave the area between module arrays bare concrete. Planting low grass (height ≤15cm) can help cool modules by 3-5°C.

Test data: Module back temperature in arrays on concrete ground is 48°C, under grass is 43°C, front temperature also 2°C lower. More surprisingly, grass reduces dust accumulation by 60%. Dust 0.5mm thick reduces module power by 1.5%. The cost of cleaning twice a year is basically saved by planting grass.

Route cables correctly

How DC and AC cables are routed directly affects loss and lifespan.

l DC Cables: From modules to inverter, run along the back of the racks, through PVC conduit (diameter ≥20mm). Don't take shortcuts across the roof; exposure to sunlight accelerates cable jacket aging (40% higher probability of insulation cracking after 5 years UV exposure). PVC conduit blocks 90% of UV, extending cable life to over 25 years.

l AC Cables: From inverter to grid connection point, the shorter the path, the better. Line loss per meter can be calculated: 4mm² copper cable resistance 0.0027 Ω/m, 10kW system current ~4.5A, loss per meter = I²R = 4.5² × 0.0027 ≈ 0.055W, 15m total loss 0.8W – although small, 100m is 8W, consuming an extra 70 kWh annually.

A project shortened AC cables from 18m to 12m, annual line loss dropped from 100 kWh to 60 kWh, equivalent to earning back the cost of 50 kWh of electricity.

Junction boxes face downward

Don't install module junction boxes randomly.uniformly facing down (≥0.5m from ground), later maintenance doesn't require ladders to reach the roof. Tests: Arrays with boxes facing up take an average of 15 minutes more per module for fault finding; facing down, 5 minutes done.

With boxes facing down, rainwater flows away along the frame, less likely to seep inside; facing up, water accumulation may corrode internal terminals.. Tests show: Boxes facing up have a 25% probability of water ingress within 5 years; facing down only 5%.

Don't mount combiner boxes high

1. Combiner boxes collect current from multiple module groups. Where to install? Don't mount them high on the roof for esthetics. Install under the eaves or on the ground (height 1.2-1.5m). Three major benefits:

2. Reading data doesn't require ladders, measure directly with a handheld meter, high efficiency;

3. Rainwater is less likely to splash in (Under eaves, which are 10cm lower than the roof plane, the probability of water splashing is reduced by 70%.);

4. Better heat dissipation – every 10°C increase in combiner box temperature shortens internal module life by 15%.

A case: A project moved combiner boxes from the roof to under the eaves. Monthly inspection time for maintenance personnel dropped from 2 hours to 40 minutes, equipment failure rate also dropped from 2 times/year to 0.

Seal against oxidation

All cable connections must be waterproof sealed. Use heat shrink tubing, then apply silicone grease. Exposed wire connections have a 50% oxidation probability within 1 year, contact resistance increases 10% (equivalent to 10% more line loss); properly sealed ones don't need touching for 5 years.

Test comparison: Unsealed connections, voltage drop increases by 0.5V after 3 years operation (affects inverter input); sealed connections, voltage drop almost unchanged.

Considering Efficiency Loss

A polycrystalline silicon module labeled 350W can only stably output 260-290W in real environments.

Data from the International Renewable Energy Agency (IRENA) shows that systems without corrected efficiency losses have an average first-year actual generation 22%-28% lower than theoretical values.

Taking a 10 kW residential project in Arizona, USA as an example, if designed using only nominal power, first-year test generation was 14,200 kWh, 27% less than the theoretical value (19,400 kWh), extending the investment payback period by 11 months.

The superposition of six factors – temperature fluctuation (-0.4%/°C), dust accumulation (annual average 6%), DC line loss (4%), etc. – is the main cause of generation deviation.

Sources of Efficiency Loss

Power drop due to temperature

The labeled 350W nominal value is measured at an ideal 25°C temperature. Outdoors, under the sun, module surface temperature can soar to 50-70°C. For every 1°C temperature increase, power drops by 0.3%-0.5%.

A practical example: A mainstream polycrystalline silicon module in Arizona, USA, has a temperature coefficient of -0.4%/°C. Summer noon ground temperature 40°C, module surface can reach 68°C (28°C higher than ambient). The power loss is then 28°C × 0.4% = 11.2%. In desert areas, extreme summer heat can reach 75°C, module surface 85°C, 60°C difference from ambient, loss directly 60 × 0.4% = 24%.

How to check your module's temperature coefficient? Look in the manual for "Temperature Coefficient (Pmax)", a negative value. The smaller the absolute value, the better (e.g., -0.3%/°C is better than -0.5%/°C).

Dust and bird droppings building up?

Dust, bird droppings, leaves covering the module block light and may cause local heating, burning out cells (hot spot effect). How much loss? Depends on location:

l Middle East desert: Little rain year-round, dust like cement powder, annual accumulation loss can reach 8%;

l Northern Germany: More rain, dust washed off frequently, loss 3%-5%;

l Industrial building roofs: Nearby factories emit exhaust, fine dust particles stick to glass, hard to remove, loss may be up to 7%.

A project in Texas, USA actual measurement: Two identical module sets, one cleaned monthly, one not cleaned for six months. Year-end generation differed by 12%. The cleaned set had an average annual loss of 4%; the uncleaned set, 9%.

Solution? Don't wait until it's dirty black to clean it. For dry areas, it's recommended to clean it quarterly using a soft brush and diluted water. Avoid using hard objects to scrape the glass, as this can easily scratch it.

Wires too thin or too long?

DC electricity from modules travels through cables to the inverter. Wires have resistance, high current causes heating, electricity is consumed this way. The loss formula is I²×R (current squared times resistance). Sounds complicated, actual data is clearer:

Each MC4 connector also hides loss, ~0.5% per connector. A string of 20 modules might have 5 connectors, adding an extra 2.5% loss. Therefore, minimize connectors during wiring and use whole cable adapters.

Conversion efficiency fluctuates

The inverter converts DC to AC, consuming electricity itself. Its efficiency isn't fixed; low load (e.g., only a few lights on at night) has low efficiency, full power operation has high efficiency.

Data for mainstream inverters:

l Half load (50% power): Efficiency 95%-96%;

l Full load (100% power): Efficiency 97%-98%;

l Small systems (e.g., home 3kW): Since usually only 30% load is used, annual average efficiency may drop to 94%, loss 6%.

A case: A 5kW home system in Australia, inverter only at full load from 10 AM to 3 PM daily, half load or light load rest of the time. Annual calculation shows inverter average efficiency 95%, loss 5%.

Mismatch loss is easily overlooked

In the same string, if modules have large power differences, the entire string's output is dragged down by the weakest one. For example, a string of 10 modules, 9 are 350W, 1 is 330W. The string can only output 330W×10=3300W max, not 350×10=3500W, directly losing 200W.

Where does this difference come from? It could be due to different production batches (manufacturers allow ±2% power deviation), or some modules were shaded by branches, causing them to age faster.

How to avoid? During design, require modules in the same string to be from the same batch, factory power difference controlled within ±1%. If impossible, leave 5% capacity buffer per string.

Power quietly decreases every year

Polycrystalline modules age internally over time, power decreases year by year. Industry standard is first-year degradation ≤2%, thereafter ≤0.5% per year, total degradation after 25 years ≤20% (i.e., 80% power remains).

Actually measured data:

l A project in Northern Europe: Average annual degradation 0.6%, 85% remaining after 25 years;

l A project in the Middle East: High temperature accelerates aging, average annual degradation 0.7%, 84% remaining after 25 years;

l First-year degradation: Some modules can reach 1.5% (e.g., damaged during transport), most are 1%-1.2%.

Design must reserve space for this. For a 25-year target, initial capacity needs to account for 20% total degradation, or install additional small modules year by year.

Total Loss Factor and Capacity Correction Formula

Calculate a clear account

We've broken down individual losses like temperature, dust, line loss. Now sum them up to get the total loss factor. But don't just add percentages directly – some losses may overlap (e.g., dust and temperature both affect module efficiency). Engineering typically uses empirical ranges or measured values.

Take the Arizona, USA example:

l Temperature loss: Module temp coefficient -0.4%/°C, summer extreme ΔT 60°C, loss 24%;

l Dust loss: Arid, little rain, annual average 7%;

l DC line loss: 4mm² cable, 15m wiring, loss 4%;

l Inverter loss: System average load factor 60%, efficiency 95.5%, loss 4.5%;

l Module mismatch: Same batch but production deviation, loss 1.5%;

l First-year degradation: Module first-year degradation 1.2%.

Sum these: 24%+7%+4%+4.5%+1.5%+1.2%=42.2%. But in actual projects, engineers don't add them all – some losses are long-term cumulative (e.g., aging), some are short-term fluctuations (e.g., temperature). A more common method is to take weighted averages of the losses, e.g., temperature, dust, line loss account for 70% weight, aging and mismatch account for 30%. Final total loss factor ~35% (42.2%×0.7 + (first-year 1.2% + annual 0.6%)×25 years×0.3 ≈ 35%).

How to determine the total loss factor?

Total loss factors vary greatly by region and system type. Compiled reference data (based on statistics from 200 cases by the International PV Association NREL):

For example, a home project in Berlin, Germany: High annual precipitation, dust loss 3%; low winter temperatures (-5°C to 15°C), small temperature loss (modules actually have higher efficiency at low temps, loss -3% → actually a gain); line loss using 4mm² cable, 5m distance, loss 2%; inverter annual efficiency 96%, loss 4%; total loss factor ≈ 3%+2%+4%=9%.

Use the formula to calculate actual required module power

Knowing the total loss factor, you can calculate how many modules to install. The formula is:

Required Total Nominal Module Power (kW) = Target Annual Generation (kWh) / (Local Annual Equivalent Sun Hours × (1 - Total Loss Factor) × Inverter Annual Efficiency)

Break it down:

l Target Annual Generation: How much electricity the user needs the system to generate annually, e.g., 15,000 kWh;

l Equivalent Sun Hours: Locally, the number of hours per year where solar irradiance equals 1000W/m² (e.g., Arizona 5.5 hours, Germany 1.2 hours);

l (1 - Total Loss Factor): The actual efficiency modules can output after deducting all losses;

l Inverter Annual Efficiency: The average efficiency of the inverter converting electricity throughout the year (e.g., 96%).

Practical Case: A household in Texas, USA wants to install a system, target annual generation 12,000 kWh. Local data: Equivalent sun hours 4.8, total loss factor 30%, inverter annual efficiency 95.5%.

Plug into formula:

Required Module Power = 12,000 / (4.8 × (1 - 30%) × 95.5%) = 12,000 / (4.8 × 0.7 × 0.955) ≈ 12,000 / 3.21 ≈ 3739W ≈ 3.7 kW.

But module nominal power comes in integers, so choose 4 kW (e.g., 10 pieces of 370W modules). If losses are ignored,Press directly 12,000/(4.8×0.955)≈2625W≈2.6kW, actual generation will be much less.

Loss changes over different years

In early years, new modules have small aging loss; after 5 years, the aging loss increases year by year; after 10 years, it may need to be recalculated.

For the Texas project above, first-year total loss factor 28% (aging 1%), year 5 aging total 5%, total loss factor 32%; year 10 aging total loss factor 10%, total loss factor 37%. If generation is insufficient then, two methods:

1. Initially install 5%-8% more modules (e.g., install 4.2 kW first year, no need to add later);

2. Add 0.5 kW modules in year 5 to balance the loss.

Use software to automatically calculate losses

Manual calculation is error-prone. Now there's PV design software that can automatically calculate total loss. For example, PVsyst. Input module model, local weather data, installation method, the software automatically calculates temperature, dust, line loss, etc., and outputs the total loss factor and required capacity.

An Australian project used PVsyst simulation: 30 pieces of 350W modules, total nominal power 10.5 kW. Software calculated total loss factor 29%, actual usable power 7.4 kW.