Well pump electricity cost depends on the pump’s electrical input, how long it runs, and the electricity rate that applies when it operates. A hypothetical 1,000-watt pump running 30 minutes per day uses 15 kWh in a 30-day month and costs USD 3.00 at an assumed U.S. rate of USD 0.20/kWh. Water use helps explain that runtime, but gallons alone cannot determine electricity consumption: lift, delivery pressure, friction losses, and system efficiency also matter.

Well Pump Running Cost at Different Daily Runtimes

The following comparison uses an assumed constant electrical input of 1,000 watts and an illustrative U.S. electricity rate of USD 0.20/kWh. These are calculation examples, not typical pump ratings or local utility prices.

Daily pump-on time	Electrical input	Electricity per 30-day month	Monthly energy cost, USD	Applicable conditions
15 minutes	1,000 W	7.5 kWh	1.50	Constant running power; no separate standby load
30 minutes	1,000 W	15 kWh	3.00	Same pump input and assumed rate
1 hour	1,000 W	30 kWh	6.00	Same pump input and assumed rate
2 hours	1,000 W	60 kWh	12.00	Same pump input and assumed rate

Use total runtime, not the number of hours that the water system is available. A pump connected to power all day does not necessarily pump all day; a conventional pressure-tank system starts and stops as system pressure changes.

For a national reference, EIA reported a U.S. residential average electricity price of USD 0.1831/kWh for July 2026, released September 24, 2026—the latest monthly release available as of October 11, 2026. This is an average calculated from electricity revenues and sales, not a quoted utility tariff or a prediction of your next bill. Use it only as a starting reference; replace it with your applicable local rate.

Calculate Cost From Watts and Runtime

The electrical method is the most direct approach when you have measured energy use or reliable running-power data. It avoids estimating hydraulic efficiency and accounts for the actual installation when the measurement covers the complete pump system.

For approximately constant running power:

[
E=\frac{P_{\text{input}}\times t}{1{,}000}
]

[
C=E\times r
]

Where:

• (E) = electrical energy, in kWh.
• (P_{\text{input}}) = electrical input power, in watts.
• (t) = total pump-on time, in hours.
• (r) = applicable electricity rate, in USD/kWh.
• (C) = energy cost, in USD.

For variable operating conditions, calculate energy for each power-and-time interval and add the results. DOE’s pumping guidance uses this operating-condition approach when estimating total pumping energy costs.

Inputs to Collect

Input	Unit	Preferred source	Important condition
Electrical energy	kWh	Dedicated energy monitoring	Include the equipment you intend to evaluate
Running input power	W or kW	Measured real power or manufacturer operating data	Match the actual operating point
Pump-on time	Minutes or hours	Controller logs or suitable monitoring	Do not substitute elapsed calendar time
Pumped water volume	U.S. gallons	Water meter or a documented usage estimate	Include irrigation and other well-supplied uses
Delivered flow	U.S. gallons per minute, gpm	Measured flow or the model’s performance curve	Flow depends on operating head
Electricity rate	USD/kWh	Current utility tariff and bill	Account for applicable time periods or tiers
Standby power	W	Measurement or manufacturer documentation	Add separately if excluded from running-energy data

A horsepower label is not an electrical wattage measurement. Horsepower describes mechanical output capacity; electrical input also reflects motor losses and the load imposed by the pump. Likewise, voltage multiplied by current is not necessarily real power for an AC motor unless power factor is included. Manufacturer motor specifications distinguish horsepower, efficiency, and power factor for this reason.

Have a qualified professional install any monitoring that requires access to fixed wiring or an electrical enclosure. Do not open energized equipment to collect cost inputs.

Calculation Procedure

1. Establish the measurement boundary: pump only, pump plus drive, or the entire water-supply system.
2. Obtain measured kWh, or collect running watts and total pump-on hours for a representative period.
3. Convert minutes to hours and watts to kilowatts.
4. Multiply kilowatts by operating hours to calculate kWh.
5. Apply the electricity rate for the corresponding billing period.
6. Add separately measured standby consumption or additional pump loads that were outside the original boundary.

You can reproduce the constant-power calculation with the Appliance Cost Calculator using your running wattage, equivalent daily operating hours, and local electricity rate. For a variable-speed system, use measured kWh or calculate separate operating intervals rather than entering the drive’s maximum rating as a constant load.

How Water Use Becomes Electricity Consumption

Moving more water requires more hydraulic work at the same head. However, two homes using the same number of gallons can have different pumping costs because their systems require different lift, pressure, and friction head. DOE distinguishes static head from flow-dependent friction losses and emphasizes evaluating the complete pumping system.

Estimate Runtime From Water Volume

If delivered flow remains approximately constant:

[
t_{\text{day}}=\frac{V_{\text{day}}}{Q\times60}
]

Where (V_{\text{day}}) is daily pumped volume in U.S. gallons and (Q) is delivered flow in gpm.

For an assumed 240 gallons per day at an actual delivered flow of 10 gpm:

[
t_{\text{day}}=\frac{240}{10\times60}=0.40\text{ hours}
]

That equals 24 minutes of pumping per day. The calculation assumes all counted water passes through the pump and that 10 gpm represents its operating flow—not simply the flow category printed in a product listing.

Calculate Total Dynamic Head

For a well supplying a pressurized destination, a practical head breakdown is:

[
H_{\text{TDH}}=
H_{\text{lift}}+
H_{\text{pressure}}+
H_{\text{friction}}
]

Total dynamic head includes the lift from the pumping water level to the delivery elevation, the required delivery pressure expressed as head, and losses through piping and equipment. This matches the head components identified in Oklahoma State University’s pumping-efficiency guidance.

For water, pressure head is approximately:

[
H_{\text{pressure}}(\text{ft})=2.31\times p(\text{psi})
]

An illustrative system with 75 ft of lift, 50 psi delivery pressure, and 9.5 ft of friction loss has:

[
H_{\text{TDH}}=75+(2.31\times50)+9.5=200\text{ ft}
]

Use the water level while pumping, not automatically the well’s drilled depth or the pump’s installation depth. For a pressure-tank system, pressure changes through the pumping cycle, so one assumed pressure is an approximation rather than a complete operating profile.

Estimate Energy From Gallons and Head

For water, the hydraulic-work relationship can be expressed in U.S. units as:

[
E_{\text{input}}(\text{kWh})
\approx
\frac{3.14\times10^{-6}\times V_{\text{gal}}\times H_{\text{ft}}}
{\eta_{\text{overall}}}
]

Here, (\eta_{\text{overall}}) is the combined electrical-to-water efficiency expressed as a decimal. It must include pump, motor, and drive losses within your calculation boundary; motor efficiency alone is not sufficient. Pumping-system efficiency depends on the complete system and its operating point.

For an explicitly hypothetical 240 gallons per day, 200 ft total dynamic head, and 40% overall efficiency:

[
E_{\text{day}}

\frac{3.14\times10^{-6}\times240\times200}{0.40}

0.3768\text{ kWh}
]

[
E_{\text{30 days}}=0.3768\times30=11.304\text{ kWh}
]

At an assumed USD 0.20/kWh:

[
C_{\text{30 days}}=11.304\times0.20
\approx\text{USD }2.26
]

The 40% efficiency is an example assumption, not a residential well-pump benchmark. This method is useful for understanding the effect of volume and head, but it is less dependable for billing estimates when efficiency is unknown. It excludes separate standby consumption and does not establish pump suitability or guaranteed performance.

Horsepower and Flow: Manufacturer Reference Examples

There is no universal “horsepower equals gallons per minute” conversion. Franklin Electric’s FS Series includes pumps with the same horsepower but different flow categories and shutoff heads, illustrating why the exact model and performance curve matter.

The examples below are manufacturer-listed 60 Hz, 230 VAC, two-wire thermoplastic models, checked October 11, 2026.

Franklin FS model	Rated horsepower	Listed flow category	Shutoff head at 60 Hz
5FS05P4-2W230	1/2 hp	5 gpm	312 ft
10FS05P4-2W230	1/2 hp	10 gpm	282 ft
16FS05P4-2W230	1/2 hp	16 gpm	167 ft
5FS07P4-2W230	3/4 hp	5 gpm	491 ft
22FS07P4-2W230	3/4 hp	22 gpm	186 ft
10FS1P4-2W230	1 hp	10 gpm	526 ft
26FS1P4-2W230	1 hp	26 gpm	221 ft

Source: Franklin Electric’s FS Series order information. These are model-specific reference values, not a generic sizing table.

Shutoff head is the zero-flow end of a pump curve; it is not the head at which the pump delivers its listed flow category. Use the manufacturer’s performance curves to determine delivered flow at your system head. Do not run a pump at shutoff to test these values.

Worked Example: Electricity Cost for a Four-Person Household

This example assumes a U.S. household with four occupants using 60 gallons per person per day, all supplied by one well pump. The water-use allowance, running power, flow, and electricity rate are hypothetical inputs—not official household averages.

Example input	Assumed value
Occupants	4
Well-supplied water use	60 U.S. gallons/person/day
Delivered pump flow	10 gpm
Electrical input while pumping	1,000 W
Electricity rate	USD 0.20/kWh
Monthly calculation period	30 days
Separate standby and auxiliary loads	Excluded

1. Calculate daily water volume:

[
V_{\text{day}}=4\times60=240\text{ gallons}
]

2. Calculate daily pumping time:

[
t_{\text{day}}=\frac{240}{10\times60}=0.40\text{ hours}
]

3. Calculate monthly electricity:

[
E_{\text{month}}=\frac{1{,}000}{1{,}000}\times0.40\times30
=12\text{ kWh}
]

4. Calculate monthly energy cost:

[
C_{\text{month}}=12\times0.20=\text{USD }2.40
]

5. Calculate a 365-day annual estimate:

[
C_{\text{year}}=1.00\times0.40\times365\times0.20
=\text{USD }29.20
]

This result covers pumping electricity only. It excludes water heating, irrigation beyond the assumed volume, maintenance, treatment equipment, and a separate booster pump. Systems with intermediate storage may use both a well pump and a pressure pump, so counting only one motor can omit part of the water-supply energy use.

The result differs slightly from the hydraulic example because this example assumes 1,000 W electrical input, while the hydraulic example calculates energy using an assumed 40% overall efficiency. They are separate estimates, not interchangeable measurements.

Fixed-Speed Versus Variable-Speed Well Pumps

Variable-speed control can match pump output to changing demand, but constant pressure does not establish a particular electricity saving. DOE’s guidance explains that savings depend on the system curve, static head, operating profile, and the control method being replaced.

Comparison	Fixed-speed pressure-tank system	Variable-speed constant-pressure system
Operating behavior	Pump starts and stops within a pressure range	Drive adjusts motor speed to regulate pressure
Practical energy estimate	Running watts × total pump-on hours	Measured kWh or the sum of power-and-time intervals
Main comparison condition	Actual flow and head during pumping	Actual flow, head, speed, and drive losses
Basis for evaluating savings	Baseline kWh for a documented water volume	Comparable kWh at the same delivered volume and service pressure

Pressure-tank operation is described by Penn State Extension; variable-speed operation and its energy limitations are covered by DOE and Franklin Electric.

For a concrete manufacturer example, Franklin’s SubDrive QuickPAK model 25SDQP-1.5HP-N4 lists 1.5 hp, a 25 gpm flow category, a 30–80 Hz submersible output-frequency range, and maximum input power of 2,400 W. These specifications apply to that package; maximum input power is not its constant operating consumption.

A common calculation mistake is applying the pump affinity-law cube relationship directly to a deep-well system’s electricity bill. DOE warns that substantial static head can make simplified speed-based energy predictions seriously inaccurate. Use the actual pump and system curves or measured energy instead.

Compare systems using:

[
\text{Specific energy}

\frac{\text{Measured kWh}}{\text{Pumped gallons}}
\times1{,}000
]

The result is kWh per 1,000 U.S. gallons. Compare periods with similar lift and delivery pressure; otherwise, a change in pumping conditions may be mistaken for an efficiency improvement.

Apply the Local Tariff Without Misreading the Bill

For a flat energy rate, multiply pump kWh by the applicable USD/kWh charge. For time-of-use service, calculate each period separately:

[
C_{\text{energy}}=\sum_i E_i r_i
]

For tiered service, the additional pump consumption may fall into a higher-priced usage tier. PG&E’s official explanations illustrate both time-dependent and usage-tier pricing, but its California schedules are not substitutes for another utility’s tariff.

Use your utility’s current tariff to identify applicable usage-based delivery charges, adjustments, and taxes. An unchanged fixed customer charge should not be presented as an additional cost caused by the pump; distinguish incremental pumping cost from an allocated share of the whole bill.

For commercial or agricultural accounts with demand charges, kWh cost alone may be incomplete. Determine whether pumping increases the tariff’s billed peak demand. PG&E, for example, describes certain demand charges based on the highest 15-minute demand interval—not simply a motor’s momentary starting current. Your tariff’s measurement rules control.

Investigate Changes in Pumping Cost

Start by separating an energy-use change from a rate change. Compare pump kWh, pumped gallons, and the applicable tariff over matching periods rather than attributing a higher whole-house bill to the well pump.

• Increased water volume raises hydraulic work when head and efficiency remain comparable.
• Greater lift or delivery pressure increases the required head.
• More friction loss changes the system operating point and can reduce delivered flow.
• Repeated short cycles warrant inspection of the pressure tank, controls, and plumbing; cycle count alone does not quantify the electricity penalty.

Do not diagnose a major energy increase from starting amperage alone. The relevant quantity is accumulated real energy over time.

Well Pump Cost Self-Check

• I used electrical input watts or measured kWh, not horsepower as wattage.
• I converted operating minutes to hours.
• I used total pump-on time rather than 24-hour availability.
• I used U.S. gallons consistently.
• I used delivered flow at the operating head, not a product flow category alone.
• I included every pump and controller within the intended measurement boundary.
• I used the current local rate and checked time-of-use, tiers, or demand charges.
• I labeled assumed water use and efficiency as assumptions.
• I treated unusual cycling as an inspection issue, not permission to bypass protection.

Use the applicable utility tariff for billing rules, the exact manufacturer documentation for equipment limits, and qualified on-site judgment for measurement and diagnosis. These calculations and the appliance-cost calculator do not replace NEC requirements, local AHJ decisions, permits, manufacturer instructions, or professional pump selection.