Mr Youkee Kong 
Solar Water Pump Solution in Agricultural Irrigation
The application of solar-powered water pumps in agricultural irrigation represents a profound transformation. By integrating traditional farming practices with renewable energy technologies, this innovation not only significantly reduces production costs and frees up labor resources but, more importantly, enables agricultural development in arid and remote regions. This advancement drives the evolution of agriculture towards water conservation, smart management, and sustainable development.
| Crop Name | Growth stage | Daily water demand range (mm/day) | Total water requirement during the entire growth period (m³/mu) | Critical water demand period |
|---|---|---|---|---|
| paddy | Entire growth period | 4.0 - 8.0+ (Excluding field evaporation) | 300 - 500 | Tillering stage, booting and flowering stage |
| wheat | Overwintering period | 0.5 - 2.0 | ||
| Jointing - Anthesis | 4.0 - 7.0 | 250 - 400 | The jointing stage, the heading and flowering stage, the filling stage | |
| corn | Seedling stage | 1.5 - 3.0 | ||
| Anthesis - Filling | 5.0 - 8.0 | 200 - 350 | "Flowering and Filling Period" "Flowering and Ear Development Period" |
|
| Cotton | Seedling bud stage | 2.0 - 4.0 | 300 - 500 | Flowering and Boll Development Period |
| Bloom - Podding | 5.0 - 8.0 | |||
| Beans | Bloom - Pod filling | 4.5 - 7.5 | 250 - 400 | Flowering period, pod formation and grain filling stage |
| Leafy vegetables (lettuce, spinach) | Whole growth period | 3.0 - 6.0 | 200 - 300 | Rapid growth stage |
| Fruit vegetables (tomato, cucumber) | Seedling stage | 2.0 - 3.0 | 300 - 450 | From flowering and fruit setting stage to peak fruiting stage |
| Bearing stage | 5.0 - 8.0+ | |||
| Root vegetables (carrots, potatoes) | Root/stem expansion stage | 4.5 - 7.0 | 250 - 400 | Root and tuber expansion period |
| Citrus fruits | Spring and summer shoot growth stage | 3.0 - 6.0 | 400 - 700 | Spring shoot emergence period, fruit expansion period |
| Apples | New shoot growth / Fruit expansion | 3.5 - 6.5 | 350 - 600 | New shoot rapid growth period, fruit expansion rapid period |
| Grapes | Bloom - Fruit expansion | 4.0 - 7.0 | 400 - 650 | Bud emergence period, flowering period, fruit expansion period |
"How much water does my land need every day?"
Calculate the total daily water demand (Q_day):
Core Formula:
Q_day (m³/ day) = irrigated area (A, unit: hectares) × crop water demand (ET, unit: mm/day) × 10
Parameter analysis：
-Irrigation area (A):
the total area of land to be irrigated, which can be measured. (Note unit conversion: 1 hectare = 15 mu.)
-Evapo Transpiration (ET):
This is the most critical data point, representing the daily water consumption of crops under specific growth conditions. It is influenced by multiple factors, including crop type (e.g., corn, vegetables, fruit trees), growth
stage (with varying requirements during seedling, flowering, and fruiting phases), local climate conditions (such as temperature, humidity, wind speed, and sunlight exposure), and soil composition.
How to obtain this information? Initially, consult the local agricultural technology extension station or review the local agricultural irrigation manual to gather region-specific data. If such data is unavailable, one may refer to the experience-based values (for instance, 4-7 mm/day for field crops, 5-8 mm/day for vegetables and orchards), yet it is essential to prioritize local data as the benchmark.
-Constant 10:
Unit Conversion Coefficient (1 mm of Water Depth = 10 Cubic Meters of Water).
Example:
Suppose you have 3 hectares of vegetable land, local data shows that the average
daily water requirement for vegetables in summer is 6 mm.
Q = 3 ha × 6 mm/day × 10 = 180 m³/day
Q_day (m³/ day) = irrigated area (A, unit: hectares) × crop water demand (ET, unit: mm/day) × 10
Parameter analysis：
-Irrigation area (A):
the total area of land to be irrigated, which can be measured. (Note unit conversion: 1 hectare = 15 mu.)
-Evapo Transpiration (ET):
This is the most critical data point, representing the daily water consumption of crops under specific growth conditions. It is influenced by multiple factors, including crop type (e.g., corn, vegetables, fruit trees), growth
stage (with varying requirements during seedling, flowering, and fruiting phases), local climate conditions (such as temperature, humidity, wind speed, and sunlight exposure), and soil composition.
How to obtain this information? Initially, consult the local agricultural technology extension station or review the local agricultural irrigation manual to gather region-specific data. If such data is unavailable, one may refer to the experience-based values (for instance, 4-7 mm/day for field crops, 5-8 mm/day for vegetables and orchards), yet it is essential to prioritize local data as the benchmark.
-Constant 10:
Unit Conversion Coefficient (1 mm of Water Depth = 10 Cubic Meters of Water).
Example:
Suppose you have 3 hectares of vegetable land, local data shows that the average
daily water requirement for vegetables in summer is 6 mm.
Q = 3 ha × 6 mm/day × 10 = 180 m³/day
How to calculate Lift/Head?
Total Head (Total Lift):
Total Head = Vertical Lift Height + Friction Loss
Vertical Lift Height:
The vertical height from the driven water level to the highest point of the outlet. This is the most important part.
Pipeline Friction Loss (Friction Loss):
The pressure lost by water flowing through a pipeline due to friction. The smaller the diameter, the longer the pipe, and the more bends, the greater the loss. Typically, an additional 10% -20% head is added to compensate for this loss.h1: Lift underwater (the vertical distance between the water pump and the water surface)
h2: Lift above water (the vertical distance between water surface and the wellhead)
h3: The horizontal distance between the well and the water tank
h4: Tank height
Actual lift required: H=h1+h2+h3/10+h4
| Model | Motor Power | Rated Flow | Rated Head | The best use the flow head | |||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Three phase | kW | HP | m³/h | m | L/min m³/h |
0 |
333 |
500 |
666 |
750 |
833 |
| 6SP40-41S-8-2.3 | 3 | 4 | 40 | 18 | Head (m) | 26 |
20 | 30 | 40 | 45 | 50 |
| 6SP40-18-3-5 | 5.5 | 7.5 | 40 | 27 | 40 |
37 | 31 | 18 | 13 | ||
| 6SP40-27-3-5.5 | 5.5 | 7.5 | 40 | 27 | 40 |
37 | 31 | 27 | 25 | 20 | |
| 6SP40-36-4-7.5 | 7.5 | 10 | 40 | 36 | 53 |
50 | 43 | 36 | 32 | 26 | |
| 6SP40-46-5-7.5 | 7.5 | 10 | 40 | 46 | 67 |
63 | 54 | 46 | 41 | 33 | |
| 6SP40-55-6-9.2 | 9.2 | 12.5 | 40 | 55 | 80 |
75 | 65 | 55 | 49 | 40 | |
| 6SP40-64-7-11 | 11 | 15 | 40 | 64 | 94 |
88 | 73 | 64 | 57 | 47 | |
| 6SP40-74-8-13 | 13 | 17.5 | 40 | 74 | 107 |
101 | 86 | 74 | 65 | 54 | |
| 6SP40-92-10-15 | 15 | 20 | 40 | 92 | 134 |
126 | 108 | 92 | 82 | 67 | |
| 6SP40-110-12-18.5 | 18.5 | 25 | 40 | 110 | 161 |
151 | 130 | 110 | 98 | 80 | |
| 6SP40-129-14-22 | 22 | 30 | 40 | 129 | 188 |
176 | 151 | 129 | 115 | 94 | |
| 6SP40-156-17-26 | 26 | 35 | 40 | 156 | 228 |
214 | 184 | 156 | 139 | 114 | |
| 6SP40-184-20-30 | 30 | 40 | 40 | 184 | 268 |
252 | 216 | 184 | 164 | 134 | |
| 6SP40-221-24-37 | 37 | 45 | 40 | 221 | 322 |
302 | 259 | 221 | 196 | 161 | |
Rough Calculation Formula:
Pump Power (kW) ≈ [Flow Rate (m³/h) × Head (m) × Water Density (kg/m³) × Acceleration Due
to Gravity (9.8 m/s²)] / (360,000 × Pump Efficiency × Inverter Efficiency).
Typically, the pump efficiency and inverter efficiency are estimated at 0.5 and 0.95 respectively.
This formula is relatively complex to calculate. We strongly recommend using parameters directly provided by the pump manufacturer.
Calculate the required pump flow rate (Q_pump)
Core formula:
Q_pump (m³/h) = Q_day (m³/day) ÷ [daily average peak sunshine hours (T, unit: hours) × irrigation efficiency (η)]
Q_pump (m³/h) = Q_day (m³/day) ÷ [daily average peak sunshine hours (T, unit: hours) × irrigation efficiency (η)]
Parameter analysis：
-Daily peak sunshine hours (T): A standardized reference value, not the duration of sunshine. It can be understood as "effective power generation hours", usually 4-6 hours. You can check the specific value in your location through NASA database or weather software.
-Irrigation efficiency (η): must be considered! Different irrigation methods result in huge differences in water loss. This is an efficiency coefficient less than 1.
-Flood irrigation: η ≈ 0.5
-Furrow irrigation: η ≈ 0.6-0.7
-Sprinkler irrigation: η ≈ 0.7-0.8
-Drip irrigation (highly recommended): η ≈ 0.85-0.95
This formula means that the pump needs to draw out the water required for a full day (and consider losses) within T hours.
Example:
Take T=5 hours and use drip irrigation η=0.9.
Q_pump = 180 m³/ day ÷ (5 h/ day × 0.9) = 40 m³/h
Conclusion: You need to choose a pump with a flow rate of no less than 40 m³/h at a lift of 60 meters.
Pump selection
according to head and flow type:
-High head, small and medium flow (such as deep well): submersible pump (most common).
-Low head, large flow (such as water from rivers and ponds): surface centrifugal pump is selected (note that the suction head should not exceed 8 meters).
Selection Guidelines:
Based on the calculated H_total and Q_pump values, consult the performance curve (H-Q curve) provided by the pump manufacturer. The required operating point (H=60m, Q=40m³/h) should fall within the pump's high-efficiency zone rather than the edge. Confirming Pump Power: Locate the corresponding operating point on the performance curve to determine the required input power (kW).
Based on the calculated H_total and Q_pump values, consult the performance curve (H-Q curve) provided by the pump manufacturer. The required operating point (H=60m, Q=40m³/h) should fall within the pump's high-efficiency zone rather than the edge. Confirming Pump Power: Locate the corresponding operating point on the performance curve to determine the required input power (kW).
How to choose solar water pump controller
The solar water pump controller is the brain of the entire irrigation system and must be strictly matched with the pump.
-Type matching: select the corresponding DC controller or AC inverter according to whether the pump is DC or AC.
-Parameter matching: The rated voltage, power and maximum current of the controller must be greater than or equal to the rated value of the pump. Considering the extension cable, we usually recommend an inverter model one step higher than the rated power of the pump.
-Functional requirements: must have MPPT (maximum power point tracking) function and perfect protection functions (such as dry turn protection, over voltage, over current, under voltage protection, etc.).
The solar water pump controller is the brain of the entire irrigation system and must be strictly matched with the pump.
-Type matching: select the corresponding DC controller or AC inverter according to whether the pump is DC or AC.
-Parameter matching: The rated voltage, power and maximum current of the controller must be greater than or equal to the rated value of the pump. Considering the extension cable, we usually recommend an inverter model one step higher than the rated power of the pump.
-Functional requirements: must have MPPT (maximum power point tracking) function and perfect protection functions (such as dry turn protection, over voltage, over current, under voltage protection, etc.).
How to calculate photovoltaic array power (P_pv)
Core formula:
P_pv (kW) = [Pump input power (kW) ÷ total system efficiency (η_sys)] × redundancy coefficient
o Parameter analysis:
-Total system efficiency (η_sys): includes controller efficiency (MPPT efficiency, about 95%) and line loss (about 95%), which is usually taken as 0.9.
-Redundancy coefficient: In order to deal with the power attenuation of photovoltaic panels, dust shading, temperature rise and other factors, it is usually increased by 1.2-1.3 redundancy.
Example:
P_pv = (11 kW ÷ 0.9) × 1.3 ≈ 15.9 kW
Conclusion: You need to install solar panels with a total power of about 16 kW.
| Steps | Core tasks | Key formula/method | Example result |
|---|---|---|---|
| 1. Requirement Analysis | Determine the daily total water demand Q_day | Q_day = A × ET × 10 | 180 m³/day |
| Determine the total head H_total | H_total = H_vertical + H_friction | 60 m | |
| 2. Pump Selection | Calculate the required pump flow Q_pump | Q_pump = Q_day ÷ (T × η) | 40 m³/h @ 60m |
| Determine the pump model and power | Examine the H-Q performance curve of the water pump | ~11 kW | |
| 3. Photovoltaic array | Calculate the total power P_pv of the photovoltaic panels | P_pv = (Pump Power ÷ η_sys) × 1.3 | ~16 kW |
| 4. System integration | Select the controller and construct water storage facilities | Match parameters, build the pool |
For large projects, we strongly recommend consulting ZRI's professional sales team, who have extensive experience and expertise to provide you with optimal configuration to ensure the stable operation of the system for decades to come, so that your one-time investment will yield long-term returns.
Suggestion and optimization
Energy Storage vs. Water Storage:
Agricultural irrigation prioritizes water storage solutions.
Constructing a sufficiently large reservoir or water tower (with capacity equivalent to 1-2 days' water demand, i.e., 180-360 m³) over using expensive batteries allows full-capacity pumping during sunny hours to store water for nighttime or cloudy conditions. This approach significantly enhances system reliability and operational efficiency.
Installation details:
-Solar panels: face south (northern Hemisphere) with a tilt Angle similar to the local latitude to maximize total annual power generation.
-Lightning protection and grounding: must be installed to protect expensive equipment.
-Professional tools: It is strongly recommended to use the free online selection software provided by major brands (such as Lorentz, Grundfos, etc.).
As long as you input the location, headway, daily water demand and other parameters, the software can automatically recommend a complete solution, greatly reducing the difficulty of design and error risk.
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