Sizing an on-grid system for a home is mostly one division. The skill lies in picking the right yield figure, rounding sensibly, and explaining the monthly pattern so the customer is not surprised in the monsoon.
The core formula
System size (kWp) = annual units to cover ÷ specific yield (kWh per kWp per year)
Specific yield is how many units one kWp of modules produces in a year at that location, after normal losses. For a south-facing rooftop at a good tilt, the Sunlectric city data (Global Solar Atlas PVOUT less 5% for rooftop losses, October 2026) gives:
| City | kWh/kWp/yr | City | kWh/kWp/yr |
|---|---|---|---|
| Jodhpur | 1,659 | Hyderabad | 1,510 |
| Ahmedabad | 1,616 | Bengaluru | 1,498 |
| Jaipur | 1,546 | Chennai | 1,486 |
| Indore | 1,526 | Kochi | 1,491 |
| Pune | 1,513 | Lucknow | 1,390 |
If the array faces east or west or sits at a poor tilt, multiply the yield by the orientation factor from module m02.
Performance ratio: what is hidden inside the yield
Performance ratio (PR) compares what the system actually delivers with what the sunlight on the modules could deliver with no losses:
PR = actual AC units ÷ (kWp × plane-of-array irradiation in kWh/m²)
Example: a 1 kWp array receives 2,000 kWh/m² on its plane in a year and delivers 1,540 kWh. PR = 1,540 ÷ (1 × 2,000) = 0.77.
A well-built Indian rooftop runs at a PR of about 0.75–0.80. Heat, dust, wiring and inverter losses make up the rest (lesson 3). If a commissioned system shows a PR below 0.70 over a year, something is wrong: shade, soiling, a failed string, or an inverter clipping hard.
You will use PR in two ways:
- Estimating from sunlight data when no yield figure is available: yearly units ≈ kWp × POA irradiation × PR.
- Checking a running system: compare measured units with irradiation (module m09).
Worked example A: a Pune home using 300 units a month
- Annual units: 300 × 12 = 3,600
- Size: 3,600 ÷ 1,513 = 2.38 kWp
- With 550 Wp modules: 2.38 ÷ 0.55 = 4.3 modules. Choose 4 (2.20 kWp) or 5 (2.75 kWp).
| Option | Yearly generation | Share of use | What happens |
|---|---|---|---|
| 4 modules, 2.20 kWp | 2.20 × 1,513 = 3,329 | 92% | About 270 units a year still bought from the grid |
| 5 modules, 2.75 kWp | 2.75 × 1,513 = 4,161 | 116% | About 560 surplus units, credited at a low rate |
How to choose:
- In Maharashtra, surplus units at the annual settlement are credited at only about ₹2.82 (MERC, FY27; check the current order), while each unit bought in the 101–300 slab costs the customer about ₹14. Undersizing costs more than slight oversizing.
- Is consumption growing (children, a new AC, an EV planned)? Then 5 modules.
- The PM Surya Ghar CFA rises up to 3 kW (₹60,000 for 2 kW plus ₹18,000 per kW for the third kW, 2024 guidelines), so the fifth module also earns extra subsidy.
For most families here, 2.75 kWp is the better answer. Write the reason in the proposal.
Monthly values use Global Solar Atlas monthly data for Pune, scaled to 1,513 kWh/kWp/yr.
The chart shows why monthly patterns matter. From October to May the system makes more than the house uses; from June to September it makes less. Under annual net metering the summer surplus is banked and used in the monsoon. If the state settles monthly (Telangana, Andhra Pradesh and Karnataka do, for example), surplus months are paid out at the export rate and deficit months are bought at retail, so the sizing logic changes (lesson 4).
Worked example B: a Lucknow home using 450 units a month
- Annual: 450 × 12 = 5,400 units
- Size: 5,400 ÷ 1,390 = 3.88 kWp
- 7 × 550 Wp = 3.85 kWp, generating 3.85 × 1,390 = 5,352 units (99%)
The same 450 units in Ahmedabad need only 5,400 ÷ 1,616 = 3.34 kWp, so 6 modules (3.30 kWp). Location changes the size by a whole module.
Worked example C: sizing from daily units and peak sun hours
When you have only daily figures, use:
kWp = daily units ÷ (peak sun hours on the module plane × PR)
A Bengaluru customer wants to cover 12 units a day. With about 5.5 peak sun hours on a 15° south-facing plane and PR 0.75:
kWp = 12 ÷ (5.5 × 0.75) = 12 ÷ 4.13 = 2.91 kWp, so 5 or 6 modules of 550 Wp. Check with specific yield: 12 × 365 = 4,380 units ÷ 1,498 = 2.92 kWp. The two methods agree closely; use the city yield where you have it.
Choosing the inverter size
For on-grid systems, the inverter's AC rating is usually close to the DC kWp: a DC/AC ratio of about 1.0–1.2 is common for homes (module m04). In the Pune example, a 2.75 kWp array works well with a 2.5–3 kW single-phase inverter.
Key points
- kWp = annual units ÷ specific yield; use the city's own yield, not a national rule of thumb.
- Specific yield already includes normal losses; PR of 0.75–0.80 is typical for good Indian rooftops.
- Round to whole modules and explain why: low export rates, future loads and the subsidy steps all matter.
- Show the monthly generation profile; the monsoon dip is normal.
- Never size from a single month's bill.
Check yourself
- A Hyderabad home uses 4,800 units a year (1,510 kWh/kWp/yr). What size do you need, and how many 550 Wp modules?
- A 4 kWp system produced 5,600 units in a year with 1,850 kWh/m² on the module plane. What is its PR?
- Why might you round up rather than down in Maharashtra?
Answers
- 4,800 ÷ 1,510 = 3.18 kWp; 6 modules (3.30 kWp).
- 5,600 ÷ (4 × 1,850) = 5,600 ÷ 7,400 = 0.76.
- Each unit still bought from the grid costs far more (about ₹14–19 in upper slabs) than the export credit for a surplus unit (about ₹2.82), so a small oversize costs less than a small undersize.