2. The Solar Resource

Learning objectives

• Explain what irradiance and irradiation/insolation are and how they differ.
• Define peak sun hours and use it as the bridge between sunlight and energy.
• Describe how sun-earth geometry (tilt, azimuth, latitude, season) drives available energy.
• Locate and interpret solar resource data for a site.

2.1 The source: the sun’s power

Above the atmosphere, sunlight arrives at about 1,361 W/m² (the solar constant). The atmosphere scatters and absorbs some of it. On a clear day at the earth’s surface with the sun overhead, irradiance peaks near 1,000 W/m². That round figure is not a coincidence in this trade. It is the reference condition that modules are rated at (STC, Chapter 4).

2.2 Irradiance vs irradiation (insolation)

Two words that sound alike and mean different things:

• Irradiance: power of sunlight per unit area, in W/m². An instantaneous rate. Noon on a clear day is roughly 1,000 W/m²; an overcast morning might be 150 W/m².
• Irradiation (also insolation): energy of sunlight per unit area over time, in kWh/m²/day or kWh/m²/year. The accumulated total.

The relationship mirrors power-vs-energy from Chapter 1: irradiance is the rate, irradiation is the rate integrated over time. Insolation is what determines how much a site can produce annually.

2.3 Peak sun hours: the installer’s workhorse concept

Real irradiance varies all day: zero at dawn, peaking near noon, back to zero at dusk, and knocked down by clouds. Rather than integrate that messy curve every time, the trade uses an elegant simplification:

Peak Sun Hours (PSH) = the number of hours per day that, at a constant 1,000 W/m², would deliver the same total energy as the actual day.

Numerically, a site’s daily PSH equals its daily insolation in kWh/m²/day. A location receiving 5.5 kWh/m²/day gets 5.5 peak sun hours. This is the single most useful number for quick sizing. Multiply array kW by PSH to estimate daily kWh before losses.

Example 2.A: A 6 kW array at a 5.0 PSH site produces, ideally, 6 × 5.0 = 30 kWh/day. Apply a realistic system derate of ~0.80 (Chapter 14) and you’d estimate ~24 kWh/day. PSH turns the solar resource into a one-line sizing input.

2.4 Sun-earth geometry

How much sun a surface catches depends on geometry:

• Latitude: higher latitudes see lower sun angles and stronger seasonal swings.
• Season: the earth’s axial tilt (23.5°) makes the sun higher in summer and lower in winter; day length changes too.
• Time of day: sun angle sweeps from horizon to peak and back, which is why azimuth (the compass direction a surface faces) and tilt (its angle from horizontal) matter so much.

For fixed arrays in the northern hemisphere, the rule of thumb is to face true south (azimuth 180°) at a tilt roughly equal to the site latitude for balanced annual yield. Real designs trade this against roof geometry, shading, and rate structures (Chapter 18). Southern hemisphere: face true north.

⚠️ True vs magnetic: azimuth is referenced to true (geographic) south/north, not magnetic. Account for local magnetic declination when using a compass.

2.5 Direct, diffuse, and reflected

Irradiance reaches a panel three ways: direct beam (straight from the sun’s disk), diffuse (scattered by atmosphere and clouds, which is why you still get output on overcast days), and reflected (albedo from the ground, the basis of bifacial module gains, Chapter 5). Total irradiance on a tilted plane is the Plane-of-Array (POA) irradiance, the quantity that actually drives a module.

2.6 Finding the data

You never guess insolation. Standard free sources:

• NREL’s NSRDB / PVWatts: typical-meteorological-year data and a built-in production estimate for any US (and many international) locations.
• Global Solar Atlas: worldwide insolation maps.
• SAM: for detailed hourly modeling (Chapter 19).

These give the PSH / insolation figures that feed every sizing calculation in Part IV.

2.7 The peak-sun-hours picture

The PSH concept replaces the real, lumpy daily irradiance curve with an equal-area rectangle 1,000 W/m² tall:

 Irradiance (W/m²)
 1000 ┤        ____                 ┌───────────┐  ← equal AREA (energy)
      │      /      \               │           │     squeezed into a
  750 ┤    /          \      ≈      │  1000 W/m²│     5-hour-wide
      │   /            \            │           │     rectangle
  500 ┤  /              \           │  (= 5 PSH)│
      │ /                \          │           │
    0 ┼/__________________\_        └───────────┘
      6am    noon      6pm           5 "peak sun hours"

The actual curve (left) and the 5-PSH rectangle (right) enclose the same area, i.e., the same daily energy (kWh/m²). That’s why PSH is so convenient: it collapses a day of varying sun into a single multiplier.

2.8 Worked examples

Example 2.B (seasonal swing): A site averages 5.5 PSH annually but only 2.8 PSH in December and 7.4 PSH in June. A 6 kW array (derate 0.82) produces, roughly:

• December: 6 × 2.8 × 0.82 ≈ 13.8 kWh/day
• June: 6 × 7.4 × 0.82 ≈ 36.4 kWh/day The 2.6× swing between months is exactly why off-grid systems size to the worst month (Chapter 11) while grid-tied systems can lean on the annual average, since the grid covers winter shortfalls.

Example 2.C (orientation penalty): The same array faces due west instead of true south, costing ~15% of annual yield. If the south-facing case made 9,900 kWh/yr, the west case makes ~9,900 × 0.85 ≈ 8,400 kWh/yr. That is still useful, and sometimes preferred under time-of-use rates that pay more for late-afternoon production (Chapter 18).

Chapter 2 summary

Irradiance is sunlight’s power (W/m², ~1,000 at STC); insolation/irradiation is its accumulated energy (kWh/m²/day). Peak sun hours conveniently equals daily insolation and converts array kW into daily kWh. Geometry (latitude, season, tilt, azimuth) sets how much a surface captures; face true south (N. hemisphere) at ~latitude tilt as a starting point. Pull real numbers from NREL/PVWatts, never from memory.

Practice Problems: Chapter 2

1. A location receives 6.2 kWh/m²/day of insolation. How many peak sun hours is that?
2. A 5 kW array sits at a 4.5 PSH site. Estimate its daily output at a 0.80 derate.
3. Convert that to an annual estimate (assume the 4.5 PSH is the annual average).
4. Is “850 W/m²” an irradiance or an insolation value? Which quantity (power or energy) is it?
5. A roof faces true south at the optimal tilt and gets 5.0 PSH. A second roof on the same house faces east and effectively gets 4.2 PSH. What percentage of the south roof’s daily energy does the east roof capture?
6. Why must azimuth be measured to true south rather than magnetic south?

Solutions: Chapter 2

1. PSH = insolation numerically, so 6.2 PSH.
2. 5 × 4.5 × 0.80 = 18 kWh/day.
3. 18 kWh/day × 365 ≈ 6,570 kWh/year.
4. Irradiance: it’s a rate of power per area (W/m²), not accumulated energy.
5. 4.2 ÷ 5.0 = 84%.
6. The sun’s path is referenced to the geographic poles; a compass points to magnetic north, which differs from true north by the local declination, so using magnetic bearings mis-aims the array.