Solar PV Water Pumping: Sizing and When It Pays Off

Executive summary

A solar PV water pumping system turns sunlight into lifted water, with no fuel and no electricity bill. Sizing it is no mystery: it reduces to a chain of basic physics. You begin with the daily water demand and the total dynamic head (TDH), compute the hydraulic energy with E = ρ·g·Q·H, correct it for pump and system efficiency to obtain the electrical energy, and divide that by the site's peak sun hours to reach the array power in watts-peak (Wp). In this article we walk that chain with a verified worked example and close with what matters most: when solar pumping pays off versus grid or diesel, and why storing water almost always beats storing batteries.

Who this is for

For project engineers, agronomists, irrigation and livestock managers, rural water-supply staff, and anyone who has to decide whether to electrify a well with panels or keep a diesel pumpset. If you are asked to justify a PV array with a credible number — how many Wp, how many panels, what flow to expect — this builds the calculation from scratch. You do not need proprietary software: the water demand, the depth and the site irradiation are enough.

The real field problem

The classic mistake is buying from a spec sheet: "this solar pump delivers so many liters per hour", without tying that figure to the real head or the site's solar resource. The result is systems that look fine on paper and fall short at a cloudy noon, or oversized arrays that cost too much. The other trap is reaching for batteries: reproducing at night what the sun gives by day inflates the project unnecessarily, because water is already a cheap battery when you store it in an elevated tank.

Engineering fundamentals

The heart of the calculation is the hydraulic energy: the minimum work to lift a volume of water a given height. It does not depend on the pump brand, only on physics:

E = ρ · g · Q · H

where ρ = 1,000 kg/m³ (water density), g = 9.81 m/s² (gravity), Q = daily volume (m³) and H = total dynamic head, TDH (m). The result in joules is divided by 3,600,000 to convert to kWh. That is the energy in the water; the pump always needs more, because nothing is perfect.

The TDH is not just the well depth. It is the sum of the vertical lift (water level with drawdown up to the discharge point), the friction losses in the pipe (typically 5-15% of the lift in a sensibly sized pipe) and any required discharge pressure (a sprinkler, a tank height). To turn hydraulic energy into electrical energy, you divide by the pump and system efficiency:

E_electrical = E_hydraulic / (η_pump · η_system)

The η_pump is the wire-to-water efficiency (motor + pump); in typical solar submersibles it runs 35-55%, use 45% if you do not have the manufacturer curve. The η_system (derate) bundles controller, wiring, temperature and soiling losses: 85% is a reasonable value, 80% in dusty environments or long cable runs. Finally, the PV array is sized by dividing the electrical energy by the site's peak sun hours (PSH):

P_array (Wp) = E_electrical (Wh/day) / PSH (h/day)

PSH are the equivalent hours of irradiance at 1,000 W/m². A 12-hour day may yield only 4 to 6 PSH depending on latitude and climate. By definition, 1 Wp delivers 1 Wh per peak sun hour: that is why the division gives the array peak power directly.

How to apply it step by step

• 1. Water demand: define the daily volume Q (m³/day) for irrigation, livestock or supply, in the most demanding month.
• 2. Total dynamic head: add vertical lift + friction + discharge pressure to obtain the TDH.
• 3. Hydraulic energy: apply E = ρ·g·Q·H and convert it to kWh/day.
• 4. Electrical energy: divide by η_pump · η_system.
• 5. Peak sun hours: take the site's PSH (irradiation databases) for the critical month.
• 6. PV array: divide the electrical energy by the PSH → power in Wp, and round to real panels.

Worked example with numbers

An irrigation well must deliver 40 m³/day. The TDH is made up of 25 m of vertical lift (level with drawdown), 3 m of friction and 2 m of discharge height to the tank → TDH = 30 m. Assumptions: submersible pump at η = 45%, system derate 85% and 5.5 peak sun hours.

Step 1 — Hydraulic energy:
E = ρ·g·Q·H = 1,000 × 9.81 × 40 × 30 = 11,772,000 J
11,772,000 / 3,600,000 = 3.27 kWh/day

Step 2 — Electrical energy: combined efficiency = 0.45 × 0.85 = 0.3825
E_electrical = 3.27 / 0.3825 = 8.55 kWh/day

Step 3 — PV array:
P = 8.55 kWh / 5.5 h = 1.55 kWp ≈ 1,555 Wp
In real panels, four 400 Wp modules (1,600 Wp) cover the demand with a small margin.

Consistency check: pumping 40 m³ in 5.5 h equals 7.27 m³/h ≈ 2.02 L/s. The mean hydraulic power is 3.27 kWh / 5.5 h = 0.59 kW, and the mean electrical power 0.59 / 0.3825 = 1.55 kW: exactly the array size. Everything closes. Change any input — more depth, less sun — and the power moves proportionally and traceably.

When it applies and when it does not

Solar pumping has high upfront investment and almost zero operating cost: no fuel, no tariff. That is why it pays off where energy is expensive or unavailable: remote off-grid wells, diesel systems with costly fuel logistics, dispersed irrigation and livestock. Case studies report solar pumping costs per kWh well below diesel and grid in those contexts (World Bank, 2018; IRENA, 2016).

It pays off less against a stable, cheap grid already connected at the wellhead, or when you need water on constant demand day and night with no room for a tank. And here is the underlying decision: to provide autonomy, storing water in a tank is far cheaper than storing energy in batteries. A tank sized for 2-3 days of consumption (80-120 m³ in the example) costs and lasts more than an equivalent battery bank, and irrigation tolerates that flexibility well (FAO, 2018).

Common mistakes

• Confusing depth with TDH: forgetting friction, drawdown and discharge pressure undersizes the pump.
• Using annual average PSH: size with the critical month (least sun, most demand), not the average.
• Ignoring the real pump efficiency: assuming 70% when it is 45% leaves the array short.
• Defaulting to batteries: it inflates the project; the tank usually provides the needed autonomy.
• Skipping margin and derate: dust, temperature and aging subtract; the 85% is not optional.

Decision checklist

• Did you define the daily water demand in the most demanding month?
• Did you compute the full TDH (lift + friction + discharge)?
• Did you apply E = ρ·g·Q·H and convert it to kWh?
• Did you divide by η_pump · η_system with realistic values?
• Did you use the site's critical-month PSH?
• Did you compare storing water (tank) against batteries?
• Did you weigh the life-cycle cost against grid and diesel?

Frequently asked questions

How do you size a solar PV water pumping system?

You start from the daily water demand and the TDH, compute the hydraulic energy E = ρ·g·Q·H, divide by the pump and system efficiency to obtain the electrical energy, and divide that energy by the site's peak sun hours to get the array power in Wp.

How many solar panels do I need to pump water?

It depends on the daily water, the TDH and the peak sun hours. For 40 m³/day at 30 m of TDH, with a pump at 45% and 5.5 PSH, the array is about 1,555 Wp: four 400 Wp panels. More depth or less sun raises that figure.

When is solar pumping worth it versus diesel or grid power?

When energy is expensive or unavailable: off-grid wells, high-tariff diesel, remote irrigation. Against a stable, cheap grid it usually loses, except for autonomy. The key economic move is to store water in a tank instead of energy in batteries.

SEMHYS tools

Before sizing the array you need a reliable TDH and operating point: if you get the head wrong, the panel count carries the error. Our free pump calculator gives you the TDH, the friction losses and the operating point from the real curve, so the hydraulic energy of your solar pumping starts from data and not from guesses. And if you are looking for engineering tools and templates to run the project end to end, find them in the shop.

References

1. World Bank (2018). Solar Pumping: The Basics. Water Global Practice.
2. IRENA (2016). Solar Pumping for Irrigation: Improving Livelihoods and Sustainability. International Renewable Energy Agency.
3. FAO (2018). The Benefits and Risks of Solar-Powered Irrigation — A Global Overview. Food and Agriculture Organization.
4. NYSERDA (2019). Guide to Solar-Powered Water Pumping Systems in New York State.