# What Size Solar Panel for Well Pump: Comprehensive Sizing Guide

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The size of the solar panel system required to power a well pump depends on several factors, including the pump’s horsepower rating and daily energy needs. As a rule of thumb, approximately five solar panels are often needed to run a 1 hp solar pump.

Following this comprehensive sizing guide, you can accurately determine the solar array size needed to match your well pump’s demands.

We’ll walk through critical calculations, discuss how sizing differs for AC versus DC pumps, and examine various pump types. Equipped with this knowledge, you’ll be ready to size, select, and install solar panels for your off-grid well pump.

## Step-by-Step Solar Panel Sizing Guide to Run a Well Pump

Follow these comprehensive steps to accurately size your solar panels based on your well pump’s specific requirements:

### Step 1. Conversion from Horsepower to Watts

Regardless of its type, every well pump often has a power rating of horsepower (hp). This power requirement and other operational factors determine how much power and solar panel capacity you need.

Solar panels, however, provide power in watts (or kilowatts). Thus, the first task in sizing solar panels for your well pump is to convert the pump’s horsepower into a comparable unit, typically watts or kilowatts.

Let’s use a 3hp motor as an example to explain the process. Using the conversion factor of 0.746, the calculation becomes:

$$3hp * 0.746 = \textbf{2.238 kW} or\ \textbf{2238 watts}$$

### Step 2. Accounting for the Starting Current

Well pumps, particularly induction motors, draw a higher current during startup briefly. This surge in demand can be substantial, and it’s crucial to account for it to avoid under-sizing our solar setup.

For our 3hp motor example, the demand may inflate to about 3.0 kW to cater to this starting current.

### Step 3. Find the Inductive Power Factor

At this juncture, we must touch upon the concept of the inductive power factor. Essentially, it’s the ratio of the actual power (that does the valuable work) to the apparent power (the total power in the system). Due to the inherent characteristics of motors, they don’t use all the power provided to them efficiently.

To supply the 3.0 kW of actual power, we need to generate even more to account for this inefficiency. With an inductive power factor of 0.85 for our example, the calculation is:

$$\displaystyle {\frac {3.0\ kW}{0.85}} = \textbf{3.53 kW} of\ total\ power$$

### Step 4. Calculate the Average Daily kWh Requirement

This is based on how long and frequently your well pump runs daily. Suppose the pump runs for 4 hours daily and requires 3.53 kW (as we found previously considering inverter losses), then:

$$Average\ Daily\ kWh\ Usage = 3.53\ kW * 4\ hours = \textbf{14.12 kWh/day}$$

### Step 5. Determine Peak Sunlight Hours

This depends on your location and the time of year. The average peak sunlight hours vary from place to place, but as an example, let’s assume 5 peak sunlight hours for our location.

### Step 6. Incorporate the Design Factor (DF)

This factor accounts for inverter losses and potential degradation of solar panel efficiency, especially in hotter environments. Usually, we take

$$DF = 1.35\ for\ hot\ climates\ and\ critical\ applications$$

$$DF = 1.2\ for\ moderate\ climates\ and\ non-critical\ applications$$

If we consider a hot climate with critical applications, DF = 1.35.

Given the formula:

$$Solar\ System\ Required\ Power = \displaystyle {\frac {Average\ Daily\ kWh\ Usage}{Peak\ Sunlight\ Hours}} * DF$$

Our required solar system size is:

$$Solar\ System\ Required\ Power = \displaystyle {\frac {14.12 kWh}{5\ hours}} * 1.35 = \textbf{3.81 kW}$$

This indicates that our solar system should ideally produce around 3.8124 kW to cater to average daily usage, peak sunlight hours, inverter losses, and efficiency degradation in hot climates.

### Step 7. Battery Storage Sizing and Its Effect on Solar System Size

Incorporating a battery bank into the solar system requires adjusting the system size for energy losses during battery charging and discharging.

The previous calculation determined that you needed a solar system with a power of 3.8124 kW for your well pump. With battery storage in the mix, you’d need to consider the battery’s round-trip efficiency. Assuming an average round-trip efficiency of 0.85:

$$Adjusted\ Solar\ System\ Power = \displaystyle {\frac {3.81\ kW}{0.85}} = \textbf{4.48 kW}$$

You’d need a slightly larger solar system for battery-related energy losses. Depending on the wattage of the panels you choose, this might require an additional panel or two.

### Step 8. Calculating How Many Solar Panels You Need

The number of solar panels you need is then calculated using the following formula:

$$No.\ of\ Solar\ Panels = \displaystyle {\frac {Solar\ System\ Required\ Power}{Power\ Output\ of\ a\ Single\ Panel}}$$

Let us consider that we have already selected a 300-watt solar panel. In an ideal world, a 300-watt solar panel would deliver 300 watts.

However, most solar panels deliver slightly less due to factors like sun angle, temperature, and potential obstructions. A typical 300-watt panel might realistically provide up to 250 watts. Therefore, to determine the number of panels needed:

$$\displaystyle {\frac {3.81\ kW}{250\ watts}} = \textbf{18 panels}$$

Based on our calculations and real-world conditions, you would need approximately 18 solar panels, each rated at 300 watts, to sufficiently power your well pump while accounting for various efficiency losses.

## What Size Solar Panel for Well Pump for Different Power Sources

Understanding the energy needs of your water pump is critical. Different power sources and designs can impact how much energy your pump needs and, consequently, the number of solar panels required.

Let us apply the above steps to AC and DC pumps.

### Using an AC Pump

AC well pumps are common in residential settings, offering the advantage of working with most household electrical systems. They typically have a power factor of 0.9.

#### Calculations

• Convert horsepower to AC power in watts:

$$1hp * 0.746 = 746\ watts\ or\ \textbf{0.746 kW}$$

$$\displaystyle {\frac {0.746\ kW}{0.9}} = \textbf{0.828 kW}$$

• Calculate daily kWh requirement (using 4 hours a day as an example):

$$0.828\ kW * 4 = \textbf{3.312 kWh/day}$$

• Incorporate the design factor using a hot climate factor of 1.35:

$$\displaystyle {\frac {3.312\ kWh}{5\ hours}} * 1.35 = \textbf{0.896 kW} or\ 896\ watts$$

• The number of solar panels:

$$\displaystyle {\frac {0.896\ kW}{250\ watts}} = 3.584 \ or\ \textbf{4 panels}\ (rounded\ up)$$

• Adjusted Solar System Power to account for battery losses (using round-trip battery efficiency of 0.85):

$$\displaystyle {\frac {0.896\ kW}{0.85}} = 1.054\ kW\ or\ \textbf{1054 watts}$$

• Number of solar panels with battery backup: 1.054 kW / 250 watts = 4.216 or 5 panels (rounded up)

$$\displaystyle {\frac {1.054\ kW}{250\ watts}} = 4.216\ or\ \textbf{5 panels}\ (rounded\ up)$$

### Using a DC Pump

DC water pumps are designed to work directly with solar panels. That means inverters do not need to change the current type, making the whole setup simpler and often more efficient. Also, the power factor for DC systems is equal to 1.

#### Calculations

• Convert horsepower to DC power in watts: 1hp x 0.746 = 746 watts or 0.746 kW

$$1hp * 0.746 = 746\ watts\ or\ \textbf{0.746 kW}$$

• Adjust for power factor: 0.746 kW / 1 = 0.746 kW

$$\displaystyle {\frac {0.746\ kW}{1}} = \textbf{0.746 kW}$$

• Calculate daily kWh requirement: 0.0.746 kW * 4 = 2.984 kWh/day

$$0.746\ kW * 4 = \textbf{2.984 kWh/day}$$

• Incorporate the design factor using a hot climate factor of 1.35: (2.984 kWh/5 hours) x 1.35 = 0.806 kW or 806 watts

$$\displaystyle {\frac {2.984\ kWh}{5\ hours}} * 1.35 = \textbf{0.806 kW} or\ 806\ watts$$

• Number of solar panels: 0.806 kW / 250 watts = 3.22 or 4 panels (rounded up)

$$\displaystyle {\frac {0.806\ kW}{250\ watts}} = 3.22 \ or\ \textbf{4 panels}\ (rounded\ up)$$

• Adjusted Solar System Power to account for battery losses: 0.806 kW / 0.85 = 0.948 kW or 948 watts

$$\displaystyle {\frac {0.806\ kW}{0.85}} = 0.948\ kW\ or\ \textbf{948 watts}$$

• Number of Panels with battery backup: 0.948 kW / 250 watts = 3.79 or 4 panels (rounded up)

$$\displaystyle {\frac {0.948\ kW}{250\ watts}} = 3.79\ or\ \textbf{4 panels}\ (rounded\ up)$$

## Different Types of Well Pumps and Solar Panel Sizing

Different pump designs serve various purposes, each with characteristics that can influence energy requirements. Let us look that on average how many 300-watt solar panels are required for each pump type.

### Jet Pump

Commonly found in homes, these pumps draw water from surface sources like ponds or shallow wells. Their energy behavior is similar to single-phase AC pumps.

So, for a typical 1hp jet pump, you’ll require around 5 panels, as calculated earlier.

### Centrifugal Pump

These are versatile and used in many applications, from agriculture to homes. They work by spinning water using centrifugal force.

For a 1hp centrifugal pump, you’ll be looking at around 5 panels.

### Submersible Pump

Submersible pumps are unique as they’re placed underwater, directly in the source. Being submerged offers some benefits, like cooling, which can make them slightly more efficient.

For a 1hp submersible well pump, you might need only 4 panels if it’s a DC design.

### Floating Pump

Floating pumps are designed to float on the surface of the water source, often used for large ponds, lakes, or reservoirs.

Their primary advantage is the flexibility in drawing water from varying depths without requiring extensive installation procedures. They are often used for larger applications, so their power requirements vary.

It would generally align with the energy behavior of a single-phase AC pump, so you’d be looking at around 5 panels.

## Factors to Consider When Sizing Solar Systems for Well Pumps

This section will explore many factors influencing the solar panel sizing process.

### Geographical Location

Places nearer the equator generally receive more consistent solar energy. Similarly, certain regions may experience frequent cloudy or rainy days, which can reduce solar panel efficiency.

### Panel Tilt and Orientation

The angle and orientation at which your panels are set can significantly influence their efficiency. Ideally, panels should be oriented toward the equator and tilted to an angle approximately equal to the location’s latitude.

### Seasonal Variations

The sun’s position varies throughout the year. This means the energy output of your panels may be different in summer versus winter.

Ensure your system is sized to handle the needs of the least productive months.

Nearby trees, buildings, or other obstructions can cast shadows on your solar panels, reducing efficiency. Considering the entire day’s sun trajectory is crucial when positioning panels.

Solar panels degrade over time, typically losing about 0.5% to 1% of their yearly efficiency. While this might seem minimal initially, it can accumulate over a decade or more.

Future-proofing by oversizing slightly can help account for this.

### Temperature

Solar panels are tested at a specific temperature, usually 25°C (77°F). However, in real-world conditions, especially in hot climates, the panel temperature can rise significantly, reducing efficiency.

### System Losses

Apart from inverter efficiency, other system components like cables, connectors, and charge controllers also introduce losses.

### Pump Efficiency

The solar pump’s efficiency can change over time due to wear and tear, scaling, or other issues.

If a pump becomes less efficient, it might require more solar power to offer the same output, impacting the number of panels needed.

### Backup and Storage

Consider energy storage and retrieval efficiency if you plan to store energy in batteries for nighttime or cloudy day operations. Battery efficiency can vary, and their storage capacity may diminish as they age.

### Future Needs

Always consider any future needs or expansions you might be planning. Whether adding more solar panels, using a different solar well pump kit, or accounting for potential future needs.

Planning for these changes during the initial setup is often more cost-effective.