How Many Solar Panels To Run A 5-ton AC/heat Pump?

The solar power needed to operate an AC or heat pump primarily depends on two key factors:

The energy consumption of the unit: This is measured in Watt-hours (Wh) or more commonly, kilowatt-hours (kWh). It represents the electricity usage of your AC or heat pump, and your solar panels should be capable of generating this amount of electricity.
The Peak Sun Hours in your location: This indicates the amount of sunlight available for your solar panels to convert into electrical energy. The more Peak Sun Hours you have, the fewer solar panels you’ll need.

In this article, I will begin by discussing the electricity usage of 5-ton ACs and heat pumps, providing you with estimates and explaining how to accurately determine this energy usage.

Next, I will delve into the concept of Peak Sun Hours, guiding you on how to determine this value in your specific location and how to use it to calculate the required solar power to run your 5-ton unit.

For those considering an Off-grid system, I will also cover the various components necessary for a complete system.

Let’s get started.

How many solar panels to run a 5-ton air conditioner or a heat pump?

If you’re planning an off-grid system that can run your 5-ton unit, you’ll need to ensure that the daily energy production of the solar panels exceeds the daily energy consumption of the AC or heat pump.

For off-grid systems, a general rule of thumb is that you typically require between 700 to 900 Watts of solar panels for each hour of daily operation of a 5-ton (60,000 BTU) air conditioner or heat pump during the cooling season.

For instance, if you plan to run the unit for 8 hours a day, you would need 5500 to 7000 Watts (5.5 – 7 kW) of solar power. Depending on the wattage of the solar panels, a system of this size would consist of 18 to 28 individual panels.

However, during the heating season, when heat pumps require more energy to operate, and sunlight is relatively scarce, a 5-ton heat pump would require around 1300 to 1700 Watts of solar panels for each hour of operation.

For example, if you plan to operate the heat pump for 12 hours each day during the heating season, you would need 15,000 to 20,000 Watts (15 – 20 kW) of solar power just to offset the heat pump’s daily energy consumption.

In this case, the system would consist of 45 to 60 solar panels, depending on their wattage.

Here’s a table providing initial estimates of the required solar power to operate a 5-ton unit in both cooling and heating seasons based on the daily runtime:

Daily Runtime	Required Solar Power in the cooling season (Watts)	Required Solar Power in the heating season (Watts)
1 hour	700 – 900 Watts	1300 – 1700 Watts
2 hours	1400 – 1800 Watts	2600 – 3400 Watts
4 hours	2800 – 3600 Watts	5200 – 6800 Watts
6 hours	4200 – 5400 Watts	7800 – 10200 Watts
8 hours	5600 – 7200 Watts	10400 – 13600 Watts
10 hour	7000 – 9000 Watts	13000 – 17000 Watts
12 hour	8400 – 10800 Watts	15600 – 20400 Watts
16 hour	11200 – 14400 Watts	20800 – 27200 Watts

Solar Power (Watts) required to run a 5-ton (60,000 BTU) AC/heat pump during both cooling and heating seasons

These figures should provide you with a general understanding of what to expect.

However, it’s important to note that the exact solar power requirement will depend on your 5-ton unit’s precise energy consumption, and the amount of sunlight typically available in your location.

With these two variables accounted for, you could calculate the size of the solar array that you need as follows:

Required Off-Grid Solar Power (kW) = (Daily Energy Consumption (kWh) ÷ Daily Peak Sun Hours) x 1.25

If you are planning a grid-tied (on-grid) solar installation with net metering, you’ll need to ensure that the annual energy production of the solar panels matches the annual energy consumption of your 5-ton unit.

For a grid-tied system to offset the annual energy consumption of a cooling-only 5-ton air conditioner, you would typically require between 1.5 kW and 2.8 kW (1,500 – 2,800 Watts) worth of solar panels.

However, if you are looking to offset the annual energy consumption of a 5-ton heat pump that provides both heating and cooling, you would need between 7 kW and 11.5 kW (7,000 – 11,500 Watts) worth of solar panels.

The formula for sizing a grid-tied solar system capable of offsetting the energy consumption of your 5-ton AC or heat pump can be expressed as follows:

Required On-Grid Solar Power (kW) = (Annual Energy Consumption (kWh) ÷ Annual Peak Sun Hours) x 1.25

Let’s see how to determine the daily or annual energy consumption of your 5-ton unit.

How much energy does your 5-ton AC/heat pump use per day?

As a general rule of thumb, a 5-ton air conditioner or heat pump will typically use between 3 and 4.5 kWh (kiloWatt-hours) of energy per hour of operation during the cooling season. However, in the heating season, a 5-ton heat pump’s energy consumption can range from 5 to 7.5 kWh per hour.

The exact energy consumption of a 5-ton AC or heat pump depends on the specific model’s efficiency, as well as various operating factors including outdoor temperature, indoor temperature settings, insulation, and daily usage patterns.

Learn more about this here: How much electricity does a 5-ton AC/heat pump use?

The most accurate method to determine the energy consumption of your unit is to actually measure it, using an electricity monitoring device.

But it’s important to note that, unlike smaller air conditioners that operate at 120 Volts, most central air conditioners, including 5-ton units, require 240 Volts to operate. This means that your 5-ton air conditioner has a dedicated circuit, and you can’t simply use a plug-and-play monitor to assess its electricity usage.

To measure the energy consumption of your 5-ton unit, you’ll need a clamp meter such as the Emporia Monitor. However, since this device needs to be installed inside your breaker box, it is advisable to seek the assistance of a professional electrician for installation.

You can watch the following video for a detailed explanation of how this monitoring process works:

If you’re comfortable with estimates for now, a less precise but quicker method involves using the efficiency rating of your HVAC unit. Here’s how it works:

The 5-ton (60,000 BTU) rating on your AC or heat pump represents the amount of heat that the unit can either remove or deliver within an hour. However, the energy required by the unit to achieve this depends on its energy efficiency.

Energy Efficiency = BTUs/hr ÷ Hourly Energy Consumption

Air conditioners typically have a SEER (Seasonal Energy Efficiency Ratio) rating, which reflects their efficiency in removing heat during the cooling season.

Heat pumps, on the other hand, have both a SEER rating (for cooling season efficiency) and an HSPF rating (Heating Seasonal Performance Factor) for heating season efficiency.

You can use these energy efficiency ratings to estimate the hourly energy consumption in kWh per hour (kilowatt-hours per hour) of your unit with the following simplified formulas:

Hourly Energy Consumption in the Cooling Season (kWh per hour) = 60 ÷ SEER rating

Hourly Energy Consumption in the Heating Season (kWh per hour) = 60 ÷ HSPF rating

To illustrate, let’s consider an example with a 5-ton heat pump:

The manufacturer specifies a SEER rating of at least 15.5. Let’s use this rating to estimate the hourly energy consumption of the unit in the cooling season:

Hourly Energy Consumption in the Cooling season (kWh per hour) = 60 ÷ SEER rating

Hourly Energy Consumption in the Cooling season (kWh per hour) = 60 ÷ 15.5

Hourly Energy Consumption in the Cooling season (kWh per hour) = 3.87 kWh/hour

Now, let’s calculate the energy consumption of the unit in the heating season using its HSPF rating (8.5):

Hourly Energy Consumption in the Heating season (kWh per hour) = 60 ÷ HSPF rating

Hourly Energy Consumption in the Heating season (kWh per hour) = 60 ÷ 8.5

Hourly Energy Consumption in the Heating season (kWh per hour) = 7.05 kWh/hour

Notice that the unit consumes almost twice as much energy in the heating season. This is typical as heat pumps require more electrical energy to bring in heat than to remove it.

How much energy does your 5-ton AC/heat pump use annually?

The exact annual energy use of your 5-ton unit depends on its energy efficiency, as well as operating factors such as climate, insulation, and usage patterns. However, by using rough estimates and rules of thumb, we can arrive at a reasonably close estimate of the annual energy consumption for HVAC systems of this capacity.

According to the U.S. DOE’s test procedures for measuring the energy use of central ACs and heat pumps, these systems usually undergo an average annual cooling cycle of 1000 hours per year and an average annual heating cycle of approximately 1600 hours per year.

By incorporating these annual usage cycles, which are based on the U.S. DOE’s guidelines, along with the previously calculated hourly energy consumption values, you can calculate the annual energy consumption of your 5-ton unit accurately.

To find the annual energy consumption of a cooling-only 5-ton air conditioner, you can use the following formula:

Annual Energy Consumption (kWh/year) = Hourly Energy Consumption (kWh/hour) x 1000 hours/year

To determine the annual energy consumption of a 5-ton heat pump that provides both cooling and heating, you can use this formula:

Annual Energy Use (kWh/year) = (Hourly Energy Consumption (cooling) x 1000 hours/year) + (Hourly Energy Consumption (Heating) x 1600 hours/year)

For example, using the hourly energy usage estimates calculated earlier, the annual energy consumption of that specific heat pump can be calculated as follows:

Annual Energy Use (kWh/year) = (Hourly Energy Consumption (cooling) x 1000 hours/year) + (Hourly Energy Consumption (Heating) x 1600 hours/year)

Annual Energy Use (kWh/year) = (3.87 kWh/hour x 1000 hours/year) + (7.05 kWh/hour x 1600 hours/year)

Annual Energy Use (kWh/year) = (3,870 kWh/year) + (11,280 kWh/year)

Annual Energy Use (kWh/year) = 15,150 kWh/year

Now, we’ve been discussing the energy consumption of your heat pump. However, if you plan to install an off-grid solar system, it’s crucial to consider the energy consumption of your other appliances as well.

To do this, you’d need to list your appliances, note their wattages, and calculate their daily usage to determine your total energy consumption.

To simplify this process, I’ve created an Appliance Energy Consumption Calculator that does the heavy lifting for you and provides energy consumption estimates.

Once you have these estimates, the next step is to determine the amount of sunlight available in your location.

How much sunlight energy do you receive each day?

The amount of sunlight that an area receives at a given moment is measured in W/m² (Watts per square meter) and is referred to as solar irradiance. For a solar panel to produce 100% of its rated power, it requires exactly 1000W/m² of solar irradiance.

For example, a 100W solar panel could generate 100 Watts of power if it receives 1000W/m² of solar irradiance. However, if the same solar panel only receives 800W/m² of sunlight, it would produce only 80 Watts of power.

However, it’s important to keep in mind that watts measure power, not energy.

Related: Power (kW) vs. Energy (kWh)

The amount of “sunlight energy” that an area receives over a specific timeframe is measured in kWh/m² (kiloWatt-hours per square meter) and is referred to as Peak Sun Hours (PSH). (1 PSH = 1 kWh/m²)

For instance, if our 100W solar panel receives 5 kWh/m² of sunlight energy per day, we could say that it experiences 5 Peak Sun Hours per day. In this scenario, the solar panel would produce 500 Wh (Watt-hours) or 0.5 kWh (kilowatt-hours) per day.

The relationship between Peak Sun Hours, solar panel power rating, and energy production can be expressed as follows:

Energy Production (Watt-hours) = Solar Power Rating (Watts) x Peak Sun Hours

Since we’re determining the system size, the formula becomes:

Solar Power Rating (Watts) = Energy Production (Watt-hours) ÷ Peak Sun Hours

Solar Power Rating (Watts) = Energy Consumption (Watt-hours) ÷ Peak Sun Hours

However, due to typical system imperfections, energy losses are bound to occur. To account for these inefficiencies and losses, it’s advisable to incorporate a multiplier of 1.25 when determining the size of your solar system:

Solar Power Rating (Watts) = (Energy Consumption (Watt-hours) ÷ Peak Sun Hours) x 1.25

We’ve already discussed the energy consumption of your 5-ton AC unit and additional appliances; now we need to know the average daily Peak Sun Hours in your location.

So how many Peak Sun Hours do you get per day?

The Peak Sun Hours that your solar panels would receive depend on various factors, primarily your location and local weather conditions.

Fortunately, you can make use of historical data to estimate the average daily Peak Sun Hours for your location by using the PVWatts Calculator, a free tool provided by the National Renewable Energy Laboratory (NREL).

To get your estimate, simply provide your address in the calculator.

As an example, when I input an address in Phoenix, AZ, into the PVWatts Calculator, it generates the following information in the “Results” section:

The amount of Peak Sun Hours you receive will naturally vary throughout the year. In the example image provided, you can see that this location in Phoenix, AZ, receives 8.11 Peak Sun Hours per day in June but only 4.46 PSH/day in December.

The tool also calculates an annual average, which in this case is 6.57 Peak Sun Hours per day.

It’s important to note that the default settings of the tool assume your solar panels will be installed on a surface that faces due South (Azimuth angle = 180 degrees) and has a 20-degree tilt (4-5/12 roof pitch).

However, the actual orientation angles of your solar panels will impact the amount of sunlight they receive and, consequently, the amount of energy they produce.

You can adjust these parameters in the “System Info” section of the PVWatts calculator to match the actual orientation of your future solar panel setup, which will provide a more accurate estimate of Peak Sun Hours.

For instance, if you plan to install solar panels on your roof but don’t know the exact Azimuth angle of the roof section to be used, you can use apps like Commander Compass Go for iOS or Azimuth Compass for Android to determine the Azimuth angle.

Similarly, if you’re uncertain about the Tilt Angle of the roof, apps such as Measure for iOS or Bubble Level for Android can help you measure the tilt angle of the roof section.

Once you have the correct details to describe your future solar panel setup, you’ll get more accurate estimates for monthly and yearly Peak Sun Hour averages.

If you’re planning a grid-tied (on-grid) solar system, you don’t need to be overly concerned about the fluctuations in Peak Sun Hours throughout the year.

Grid-tied systems, with net metering or similar arrangements, use the grid as a large battery to store excess energy for later use. Excess energy generated during the sunny months goes to the grid and can be used for little to no cost during the less sunny winter months when energy production is lower.

This means you can use the annual average Peak Sun Hours to calculate the system size.

However, if you’re planning an off-grid system, your solar panels need to consistently generate enough energy to meet your daily consumption. To ensure your off-grid system can reliably cover your energy needs, you should calculate its size using the lowest average Peak Sun Hours.

For example, if you’re sizing a solar system for an off-grid setup to power your 5-ton heat pump during the winter, you should base your calculations on the Peak Sun Hours in December.

This value typically represents the lowest daily sunlight historically received during the winter and ensures your system can provide enough energy even during the least sunny times.

To better understand this, let’s look at a couple of examples.

Example 1 (Grid-Tied System):

For this example, let’s assume you want to install a grid-tied solar system to offset the energy consumption of your 5-ton (60,000 BTU) heat pump, which uses approximately 11,000 kWh of energy per year for both cooling and heating.

Now, let’s also assume that based on the PVWatts Calculator, your solar panels would, on average, receive 6 Peak Sun Hours per day throughout the year.

Since this is an annual average, you can multiply this value by 365 (the number of days in a year) to determine the annual Peak Sun Hours your system would receive:

Annual Peak Sun Hours = Daily Peak Sun Hours x 365

Annual Peak Sun Hours = 6 Peak Sun Hours x 365

Annual Peak Sun Hours = 2190 Peak Sun Hours

Now, let’s calculate the size of the solar system required to offset this annual energy consumption:

Required On-Grid Solar Power (kW) = (Annual Energy Consumption (kWh) ÷ Annual Peak Sun Hours) x 1.25

Required On-Grid Solar Power (kW) = (11,000 kWh ÷ 2190 PSH) x 1.25

Required On-Grid Solar Power (kW) = (5.02 kW) x 1.25

Required On-Grid Solar Power (kW) = 6.27 kilowatts

For this setup, you would need 6.27 kilowatts (6270 Watts) of solar panels. If you’re using 400W solar panels, you can calculate the number of panels required:

Number of solar panels = Required Solar Power (kW) ÷ Individual Solar Panel Rating (kW)

Number of solar panels = 6.27 kW ÷ 0.4 kW

Number of solar panels = 15.69

Rounding up to the nearest whole number, you would need 16 solar panels, resulting in a 6.4 kW solar system.

Example 2 (Off-Grid System):

Now, let’s consider an off-grid system where you have a 5-ton cooling-only air conditioner with a peak energy consumption of 40 kWh per day in July. Additionally, you have other appliances (such as a refrigerator, freezer, lights, etc.) that collectively consume about 15 kWh per day.

In this off-grid scenario, we’ll use the Peak Sun Hours for the month of July, assuming 7 Peak Sun Hours per day.

To calculate your total daily energy consumption:

Daily Energy Consumption (kWh/day) = Air conditioner’s energy consumption (kWh/day) + Other appliances energy consumption (kWh/day)

Daily Energy Consumption (kWh/day) = 40 kWh/day (air conditioner) + 15 kWh (other appliances)

Daily Energy Consumption (kWh/day) = 55 kWh/day

Now, let’s determine the size of the solar system needed to offset this daily energy consumption:

Required Solar Power (kW) = (Daily Energy Consumption (kWh) ÷ Daily Peak Sun Hours) x 1.25

Required Solar Power (kW) = (55 kWh ÷ 7 PSH) x 1.25

Required Solar Power (kW) = (7.85 kW) x 1.25

Required Solar Power (kW) = 9.82 kilowatts

Using 400-watt solar panels, you can calculate the number of solar panels required:

Number of solar panels = Required Solar Power (kW) ÷ Individual Solar Panel Rating (kW)

Number of solar panels = 9.82 kW ÷ 0.4 kW

Number of solar panels = 24.55

Rounding up, you would need 25 solar panels, resulting in a 10 kW solar system.

However, it’s important to note that we’ve only discussed solar panels so far. If you’re planning an off-grid system, there are several more components you need to consider.

Off-grid system components

A complete off-grid solar system comprises several essential components:

Solar Panels
Battery Bank
Solar Charge Controller
Inverter
Wires and Over-Current Protection Devices (Fuses/Circuit Breakers)

Ideally, these components should be sized in the following sequence:

1. Battery Bank:

The battery bank is the heart of an off-grid system. It stores surplus energy generated by the solar panels during the day, providing power during the night or when sunlight is unavailable.

To ensure its effectiveness, the battery bank should have an energy capacity equal to or greater than your daily energy consumption. But there are other factors to be taken into consideration, such as battery type and days of autonomy.

Please refer to our guide on calculating battery capacity for your off-grid system here.

2. Solar Charge Controller:

A solar charge controller connects the solar panels to the battery bank. Its primary functions are to maximize solar array power production and protect both the solar array and the battery.

Our MPPT Charge Controller Calculator can help determine the required charge controller specifications.

3. Inverter:

In off-grid solar systems, the inverter plays a vital role in converting low-voltage DC power (12V, 24V, 48V) from the battery bank into the higher voltage (115V, 230V) needed for your 5-ton AC/heat pump and other household appliances.

When sizing an off-grid inverter, ensure its specifications align with your system’s requirements and the power needs of your appliances. Learn more about sizing an inverter for your system here.

4. Wires and Over-Current Protection:

After sizing the aforementioned components, the final step is to determine the appropriate wire sizes and over-current protection devices (fuses or circuit breakers) that connect these components.

To assist you in this process, I’ve put together a library of guides on this specific topic:

These guides will help you ensure a well-functioning and efficient off-grid solar system.

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How many solar panels to run a 5-ton AC/heat pump?