From Sunlight to Units: The Mathematics of Solar Energy Yield and Performance Ratio

A quotation says your roof will make so many units a year. Behind that sentence sits a short chain of physics: sunlight reaching the panels, the panels turning it into DC at a given temperature, and the DC surviving wiring, the inverter and dust. Each link is an equation, and once you can write it you can question it.

This article builds that chain from first principles: irradiance and irradiation, plane-of-array transposition, the performance ratio of IEC 61724, temperature loss, the 4 to 4.5 units per kW per day rule of thumb we use in Andhra Pradesh, 25 years of degradation, and a 10 kWp worked example.

All site data here are illustrative, and every result is an estimate that a site audit would refine. Companion articles cover sun path, tilt angle and shading geometry and the module I-V curve, which explains the temperature coefficient.

From power density to energy: irradiance and peak sun hours

Two quantities are easily confused: power and energy. We define both before using either.

• $G$: irradiance, the solar power arriving per unit area, in W/m². It changes from second to second.
• $H$: irradiation, the solar energy arriving per unit area over a period, in kWh/m².
• $t$: time, in hours.
• $G_{\mathrm{ref}}$: the reference irradiance of 1 kW/m² (1,000 W/m²), also the standard test condition value.

Energy is the time integral of power, so irradiation is the time integral of irradiance over the period of interest, usually a day, a month or a year.

If G is in kW/m² and t in hours, H comes out in kWh/m². A pyranometer or reference cell measures G, and a data logger approximates the integral by summing short samples.

Peak sun hours

Peak sun hours (PSH) are a unit conversion, not a count of sunshine. Define the equivalent time $t_{\mathrm{eq}}$ as the hours at the reference irradiance that would deliver the same irradiation:

With $G_{\mathrm{ref}} = 1\,\mathrm{kW/m^2}$, $t_{\mathrm{eq}}$ in hours equals H in kWh/m². That is the whole identity: a day with $H = 5.5\,\mathrm{kWh/m^2}$ has 5.5 peak sun hours. Sanity check: the sun is up for roughly twelve hours, but irradiance averages far below 1 kW/m² over them, so 5 to 6 equivalent hours is plausible.

GHI, DNI and DHI

Sunlight arrives either straight from the solar disc or scattered by the atmosphere. Databases report three components, each on a defined surface.

• $\mathrm{DNI}$: direct normal irradiance, in W/m². Beam light measured on a surface kept perpendicular to the sun's rays.
• $\mathrm{DHI}$: diffuse horizontal irradiance, in W/m². Scattered sky light falling on a horizontal surface.
• $\mathrm{GHI}$: global horizontal irradiance, in W/m². Everything arriving on a horizontal surface.
• $\theta_z$: solar zenith angle, in degrees. The angle between the sun and the vertical.

A horizontal surface sees the beam at an incidence angle equal to the zenith angle, so the beam contributes DNI × cos θz to it. Adding the diffuse part gives the closure relation:

Worked check, with illustrative numbers. At $\theta_z = 25^\circ$, take DNI = 700 W/m² and DHI = 130 W/m². Then $\mathrm{GHI} = 700\times 0.9063 + 130 = 764.4\,\mathrm{W/m^2}$, below the 1,000 W/m² reference, which is reasonable for a hazy or non-peak hour. If a source gives two components, closure recovers the third.

Transposing sunlight to the plane of the array

Panels are tilted, so the number that matters is the plane-of-array irradiance $G_{\mathrm{poa}}$. Databases usually give GHI, so a model must convert it. The simplest is the isotropic sky model of Liu and Jordan. Anisotropic models such as Hay-Davies and Perez add circumsolar and horizon brightening and are used in professional simulation software. At moderate tilts they differ by a few per cent of annual irradiation, and more at steep tilts.

• $\beta$: module tilt from the horizontal, in degrees.
• $\theta$: angle of incidence between the sun's rays and the module normal, in degrees.
• $\gamma_s,\ \gamma_p$: solar azimuth and module azimuth, in degrees.
• $\rho_g$: ground albedo, the dimensionless fraction of GHI that the ground reflects. Values around 0.2 are commonly assumed for ordinary ground and 0.3 or more for a bright, whitewashed roof.

The beam term needs the incidence angle, from the sun-path geometry of the companion article:

When the sun is behind the module plane, cos θ is negative and the beam term is set to zero.

Deriving the sky view factor

Assume uniform sky radiance $L$ in W/m²/sr over the upper hemisphere, which is what isotropic means. A surface with unit normal $\hat n$ receives $\int L\,(\hat n\cdot\hat s)\,d\Omega$ from the directions $\hat s$ it can see, with $d\Omega$ the solid angle in steradians. For a horizontal surface:

Here $\alpha$ is the polar angle from the vertical and $\varphi$ the azimuth. Now tilt the surface by β. The module's front hemisphere (where $\hat n\cdot\hat s\gt 0$) splits into a sky part and a ground part, with contributions per unit radiance $E_{\mathrm{sky}}$ and $E_{\mathrm{gnd}}$. Together they fill the front hemisphere:

For a second relation, integrate $\hat n\cdot\hat s$ over the whole upper hemisphere with no visibility cut. Horizontal components cancel by symmetry and the vertical one integrates to π, so $\int \hat s\,d\Omega = \pi\hat z$ and the result is $\pi\cos\beta$. This counts the visible sky positively and the sky behind the module plane negatively. Reflecting that region through the origin maps it onto the ground region in the module's front hemisphere, so its magnitude is $E_{\mathrm{gnd}}$. Therefore

Adding and subtracting the two relations gives the sky view factor and the ground view factor:

Dividing by $\pi$ (since $\mathrm{DHI}=\pi L$), the module receives the fraction $(1+\cos\beta)/2$ of the diffuse irradiance. Sanity check: at $\beta = 0$ it is 1, and at $\beta = 90^\circ$ it is one half, as for a vertical wall. A uniform diffuse ground of radiance $\rho_g\,\mathrm{GHI}/\pi$ gives, by the same argument, the ground-reflected term $\mathrm{GHI}\,\rho_g\,(1-\cos\beta)/2$. The assumptions are a flat unobstructed horizon, a uniform sky and uniform ground. Nearby walls, parapets and trees break all three, which is why shade surveys matter.

The Liu and Jordan result

The three terms are beam, sky diffuse and ground-reflected. Worked example, illustrative: sun at $\theta_z = 25^\circ$, tilt $\beta = 15^\circ$ facing the sun's azimuth so $\theta = 10^\circ$, DNI = 700, DHI = 130, GHI = 764.4 W/m² and $\rho_g = 0.2$.

The plane receives about 7 per cent more than the horizontal (819.8 against 764.4 W/m²), sensible for a tilt towards the sun. The ground term is tiny at a shallow tilt, so a low-tilt rooftop array gains little from albedo.

Ideal energy and the performance ratio

Module ratings are given at standard test conditions (STC): 1,000 W/m², 25 °C cell temperature and a defined air-mass spectrum. The nameplate kWp is the DC power under those conditions. We ask what energy an ideal plant would deliver, then compare the real plant with it.

• $P_{\mathrm{STC}}$: array rating at STC, in kWp.
• $G_{\mathrm{STC}}$: STC irradiance, 1 kW/m².
• $H_{\mathrm{poa}}$: plane-of-array irradiation over the period, in kWh/m².
• $E_{\mathrm{ideal}},\ E_{\mathrm{actual}}$: ideal and measured energy at the AC meter, in kWh.

Assume an ideal array whose power is proportional to irradiance, with the cell at 25 °C and no other losses. Then $P(t) = P_{\mathrm{STC}}\,G_{\mathrm{poa}}(t)/G_{\mathrm{STC}}$, and integrating gives

Unit check: $\mathrm{kWp}\times\mathrm{kWh/m^2}\div\mathrm{kW/m^2} = \mathrm{kWh}$. Example: 10 kWp on a day with $H_{\mathrm{poa}} = 5.5\,\mathrm{kWh/m^2}$ gives $10\times 5.5\div 1 = 55\,\mathrm{kWh}$, which is 10 kW for 5.5 peak sun hours.

The performance ratio

IEC 61724 defines the monitoring quantities for photovoltaic systems. Two of them are the reference yield $Y_r = H_{\mathrm{poa}}/G_{\mathrm{STC}}$ in hours and the final yield $Y_f = E_{\mathrm{actual}}/P_{\mathrm{STC}}$ in kWh/kWp, which is also hours. Their ratio is the performance ratio:

PR is dimensionless and not an efficiency: it compares output with what the nameplate rating would deliver, which makes plants comparable. PR falls in hot weather because the reference stays at 25 °C, so compare like periods with like. The denominator also needs plane-of-array data from a pyranometer or reference cell, since a satellite-based PR inherits the satellite error.

The multiplicative loss chain

Energy flows through a series of stages: the module, the DC wiring, the inverter and the AC wiring to the meter. Each stage passes on a fraction of what it receives. Let $l_i$ be the fractional loss at stage i, a dimensionless number between 0 and 1. Stage i retains $1-l_i$, so after n stages

This is exact when the stages act in series on the same energy stream, each loss defined against the energy entering that stage. Simulation tools split losses differently, so treat it as a close approximation. Expanding for small $l_i$ shows what simply adding percentages loses:

The first-order estimate $1-\sum l_i$ is slightly low because the pair term is positive: a second loss acts on energy already reduced, so it removes a little less than its face value.

An illustrative loss table

The table lists typical ranges for a rooftop plant in a hot, dusty climate, and the values used in the worked examples. They are illustrative. Confirm each per site, because roof ventilation, cleaning, cable length, inverter model and grid reliability differ. The last column is the running product. An optical, spectral and weak-light row is included because real plants lose a few per cent there too.

The values add up to $\sum l_i = 22.5\%$, so the first-order estimate gives $\mathrm{PR}\approx 0.775$. The exact product is 0.7938. The pair terms sum to 0.0197, and including them gives 0.7947, within 0.1 percentage point. Adding losses understates PR by about 1.9 points here.

Temperature loss and why it must be irradiance-weighted

Silicon cells lose voltage as they warm, so output falls with temperature. In India this is usually the largest single loss, so it gets its own derivation. Notation first.

• $T_c$: cell temperature, in °C.
• $T_a$: ambient air temperature, in °C.
• $\gamma$: power temperature coefficient, in 1/K. It is negative for silicon, and a typical datasheet value is about $-0.0030\,\mathrm{K^{-1}}$ (that is, −0.30 % per kelvin). Use your own module datasheet.
• $\mathrm{NOCT}$: nominal operating cell temperature from the datasheet, in °C. We use 45 °C as an illustrative value.
• $k$: heating coefficient, in K·m²/W.

Assume the cell runs above ambient in proportion to irradiance. NOCT is defined at 800 W/m² and 20 °C ambient, so the rise there is NOCT − 20 and

For NOCT = 45 °C, $k = 25/800 = 0.03125\,\mathrm{K\,m^2/W}$, so a cell at 1,000 W/m² runs about 31 K above ambient. This ignores wind and mounting. Close-mounted modules on a concrete roof run hotter than the datasheet suggests, perhaps by several kelvin. To first order, the power at irradiance G and cell temperature $T_c$ is

Integrating over the period, the energy is $E = (P_{\mathrm{STC}}/G_{\mathrm{STC}})\int G\,[1+\gamma(T_c-25)]\,dt$. Pull the constants out and define the irradiance-weighted mean cell temperature:

The temperature loss is exactly $l_T = -\gamma(\bar T_w - 25)$, and it uses $\bar T_w$, not the simple time average. Energy is the quantity being summed, and it arrives mostly when the sun is high. Cool early-morning and late-afternoon hours contribute little energy, so averaging them in equally would flatter the plant.

A worked clear day

Take an illustrative clear day with $G = 900\sin(\pi(t-6)/12)$ W/m² from 06:00 to 18:00 and $T_a = 26 + 8\sin(\pi(t-9)/12)$ °C. Twelve hourly midpoints give:

The day's irradiation is $H = 6.90\,\mathrm{kWh/m^2}$. The simple mean cell temperature is 47.6 °C, but the irradiance-weighted mean is $\bar T_w = 52.5\,\mathrm{°C}$. With $\gamma = -0.0030\,\mathrm{K^{-1}}$:

The unweighted average understates the loss by about 1.5 percentage points, roughly one-fifth of the true loss. Sanity check: the weighted value lies closer to the midday cell temperature (about 60 °C), as it should. A real annual figure weights all 8,760 hours of measured or modelled weather. The 9 per cent used in the chain is plausible for a hot climate and a close-mounted roof, and your datasheet coefficient and roof airflow could move it by a couple of points.

Specific yield, CUF and the site rule of thumb

Two numbers compare plants of different sizes: specific yield, met above as final yield, and the capacity utilisation factor, which compares energy with what the array would make at full rated power for every hour of the year.

• $Y_f$: specific (final) yield, in kWh/kWp, equal numerically to hours.
• $\mathrm{CUF}$: capacity utilisation factor, dimensionless, often quoted in per cent.
• $8760\,\mathrm{h}$: hours in a non-leap year.

Some reports use AC capacity in the denominator, so check which one a quoted CUF uses. Combining the PR definition with the reference yield gives a forecast formula, $Y_f = \mathrm{PR}\times Y_r$, where $Y_r$ is numerically the annual $H_{\mathrm{poa}}$ in kWh/m².

Reconciling with 4 to 4.5 units per kW per day

Our working rule for this part of Andhra Pradesh is about 4 to 4.5 units per kW per day, or roughly 1,400 to 1,600 units per kW per year. A unit is one kWh. Convert it:

Dividing the annual range by 365 gives 3.8 to 4.4 units per kW per day, matching the quoted daily figures within the rounding that "about" allows. Now work backwards to the implied PR. For an illustrative plane-of-array irradiation of 5.3 to 5.6 kWh/m² per day, the annual $Y_r$ is 1,934.5 to 2,044 hours, and $\mathrm{PR} = Y_f/Y_r$.

The implied PR runs from 0.69 to 0.83 across the grid, and from 0.72 to 0.78 if you pair low yield with low irradiation and high with high. Our clean chain gave 0.794, so the rule sits somewhat below it. That is deliberate: a planning figure should absorb outages, partial shade, poor years, hotter roofs and slow decline.

The 5.3 to 5.6 kWh/m² figure is illustrative. For a real site, pull plane-of-array irradiation from NASA POWER, the NREL National Solar Radiation Database (through PVWatts) or Meteonorm, then transpose for your tilt and azimuth. Our guide on how to size a solar system turns such figures into kW.

Degradation and seasonal variation

Degradation over a lifetime

Modules lose a small fraction of output each year. Let d be the annual degradation rate (dimensionless, per year), $E_1$ the first-year energy in kWh and $E_n$ the energy in year n. Assume a constant rate and the same weather every year. Each year retains $1-d$ of the previous one, so

The lifetime energy over N years is a geometric series with ratio $1-d$:

For small d the binomial expansion $(1-d)^{n-1}\approx 1-(n-1)d$ gives a linear approximation. Summing it with $\sum_{n=1}^{N}(n-1) = N(N-1)/2$:

Worked example. Take an illustrative $d = 0.005$ per year, $N = 25$ years and the first-year energy of the 10 kWp example below, $E_1 = 15{,}936$ kWh.

The geometric sum is $(1-0.995^{25})/0.005 = 23.556$ first-year equivalents, so $S_{25} = 23.556\times 15{,}936 = 375{,}395$ kWh. The linear form gives $25\times(1-0.005\times 12) = 23.5$, or 374,503 kWh, only 0.24 per cent lower. Ignoring degradation gives $25\times 15{,}936 = 398{,}408$ kWh, 6.1 per cent above the geometric result. The 25-year average is 0.942 of year one, about 1,502 kWh/kWp per year, near the middle of the 1,400 to 1,600 rule.

Seasons and monsoon months

Monthly energy follows irradiation, modified by temperature and soiling. Clear spring months give the highest irradiation but hot cells and dusty glass, so PR dips while energy peaks. Monsoon months bring cloud, which cuts the beam and raises the diffuse share, and rain, which washes the glass. Winter has a lower sun but cooler cells and a higher PR. The profile below is illustrative, not site data, and holds PR at 0.794 for simplicity.

The profile averages 5.50 kWh/m² per day, like the worked example, and sums to about 15,922 kWh, within 0.1 per cent of the one-step annual figure. The best month gives about 1.45 times the energy per day of the weakest, so your meter will not show 4.4 units per kW every day. Compare a month with the same month last year.

Worked example: a 10 kWp rooftop

Now put the chain together. The inputs are illustrative: 10 kWp, $H_{\mathrm{poa}} = 5.5\,\mathrm{kWh/m^2}$ per day averaged over 365 days, and the loss values above. The annual irradiation is $5.5\times 365 = 2{,}007.5\,\mathrm{kWh/m^2}$, so

Each loss then multiplies the running energy by its retained fraction.

The final energy is $E_{\mathrm{actual}} = 15{,}936$ kWh in the first year. The headline metrics follow from the definitions above:

Sanity checks: 4.37 units per kW per day lies inside the 4 to 4.5 rule, and 1,594 kWh/kWp sits at the top of the 1,400 to 1,600 range. Over 25 years with 0.5 per cent degradation, lifetime energy is about 375,000 kWh, roughly 15,000 kWh a year.

How uncertain is this?

With every loss at the optimistic end of its table range the PR becomes 0.863 (about 17,300 kWh). At the pessimistic end it falls to 0.710 (about 14,300 kWh). Raising soiling from 3 to 6 per cent alone cuts the result by 3.1 per cent, and a temperature loss of 12 per cent instead of 9 cuts it by 3.3 per cent. The estimate is a range of a few thousand kWh, narrowed only by a site audit, a shade study and local weather data.

Using these numbers on your own roof

To apply this to a real building, follow the same order. Get plane-of-array irradiation for your tilt and azimuth. Replace the illustrative losses with values from your datasheets, cable lengths, cleaning plan and DISCOM supply record. Compute $E = P_{\mathrm{STC}}\,H_{\mathrm{poa}}\,\mathrm{PR}/G_{\mathrm{STC}}$, apply degradation, and only then ask what the units are worth. That belongs to rooftop solar ROI and payback, and it should use a range, not a point.

Ray2Volt can run this analysis for your own site, including a shade survey, a tilt and orientation check and a loss budget for the equipment we propose. Book a free power audit and we will send the estimate with its assumptions written out, so you can check each line.

Frequently asked questions

Ray2Volt Solar Private Limited

We design, install, and service rooftop solar for homes and businesses across Tirupati district and Andhra Pradesh, from PM Surya Ghar residential systems to commercial and industrial plants. Every enquiry starts with a free power audit.