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P(kW) = E(kWh) / t(hours)
Where:
This calculation helps determine the average power consumption of an appliance or system. For example, if your electricity bill shows 720 kWh for a month (30 days × 24 hours = 720 hours), your average household power consumption is 1 kW.
| kWh | 1 hour | 2 hours | 4 hours | 8 hours | 24 hours |
|---|---|---|---|---|---|
| 1 kWh | 1 kW | 0.5 kW | 0.25 kW | 0.125 kW | 0.042 kW |
| 5 kWh | 5 kW | 2.5 kW | 1.25 kW | 0.625 kW | 0.208 kW |
| 10 kWh | 10 kW | 5 kW | 2.5 kW | 1.25 kW | 0.417 kW |
| 24 kWh | 24 kW | 12 kW | 6 kW | 3 kW | 1.000 kW |
| 50 kWh | 50 kW | 25 kW | 12.5 kW | 6.25 kW | 2.083 kW |
| 100 kWh | 100 kW | 50 kW | 25 kW | 12.5 kW | 4.167 kW |
| 500 kWh | 500 kW | 250 kW | 125 kW | 62.5 kW | 20.830 kW |
The kWh to kW conversion calculates the average power consumption (in kilowatts) from a known amount of energy consumed (in kilowatt-hours) over a specific time period. While kWh measures the total energy used—the quantity that appears on your electricity bill—kW measures the instantaneous rate at which that energy is consumed. Think of kWh as the distance traveled and kW as the speed: dividing total energy by time gives average power, just as dividing distance by time gives average speed. This conversion is used daily by energy managers analyzing utility bills, engineers sizing generators and solar systems, and homeowners evaluating appliance efficiency. Utilities track both metrics because kWh determines your energy charges while peak kW demand determines infrastructure sizing and demand charges per commercial rate schedules aligned with ANSI C12 metering standards.
Find the total kWh consumed from your electricity bill, energy monitor, smart meter, or equipment sub-meter. Monthly residential bills typically show 500–1,500 kWh depending on climate and household size.
Identify the time span in hours over which the energy was consumed. A 30-day billing cycle = 720 hours. A 24-hour day = 24 hours. For appliance-level analysis, use the actual run time of the device.
Divide the total energy by the time period. Example: 900 kWh used over 720 hours (one month) = 900 ÷ 720 = 1.25 kW average power consumption.
Remember this is average power, not peak. Your actual peak demand is typically 2–4 times higher. For sizing generators or backup systems, add a 25–50% safety margin above the calculated average to handle load surges and startup currents.
Understanding average kW from kWh data helps identify opportunities to shift loads and reduce peak demand charges, which can account for 30–50% of commercial electricity bills.
Converting daily kWh usage to average kW helps determine the solar panel array size needed to offset consumption, factoring in local peak sun hours and system efficiency losses.
Comparing average kW across time periods reveals consumption trends, efficiency improvements, and anomalies that indicate equipment malfunction or energy waste.
| Appliance | Monthly kWh | Daily kWh | Avg kW (24h) |
|---|---|---|---|
| Central Air Conditioning | 300–600 | 10–20 | 0.42–0.83 |
| Electric Water Heater | 150–400 | 5–13 | 0.21–0.56 |
| Refrigerator | 40–80 | 1.3–2.7 | 0.06–0.11 |
| Electric Dryer | 60–100 | 2–3.3 | 0.08–0.14 |
| Lighting (LED, whole home) | 30–60 | 1–2 | 0.04–0.08 |
| EV Charging (Level 2) | 200–500 | 7–17 | 0.29–0.69 |
| Entire US Household (avg) | 886 | 29.5 | 1.23 |
kW (kilowatts) measures the rate of energy consumption at any given instant—it is power. kWh (kilowatt-hours) measures the total amount of energy consumed over time. A 2 kW space heater running for 5 hours consumes 10 kWh of energy. Your electricity bill charges you for kWh consumed, while demand charges are based on peak kW.
Divide the total kWh on your bill by the number of hours in the billing period. For a typical 30-day month: hours = 30 × 24 = 720. If your bill shows 900 kWh, your average power consumption is 900 ÷ 720 = 1.25 kW. This means your home uses an average of 1,250 watts continuously.
The kWh ÷ hours formula gives average power. Your actual peak demand (measured in 15-minute intervals by utilities) is typically 2–4 times higher than the average. For example, a home averaging 1.25 kW may peak at 5–8 kW when the AC, dryer, and oven run simultaneously.
Divide your daily kWh consumption by your location's peak sun hours (not 24 hours). If you use 30 kWh/day and receive 5 peak sun hours, you need 30 ÷ 5 = 6 kW of solar panels. Add 20–25% for system losses (inverter efficiency, wiring, shading) for a final size of 7.2–7.5 kW.
kWh charges cover the cost of generating and delivering energy. kW demand charges cover the infrastructure the utility must build and maintain to meet your peak power needs. A factory using 10,000 kWh spread over 720 hours (14 kW average) costs less to serve than one using 10,000 kWh in 100 hours (100 kW peak), because the latter requires much larger transformers, feeders, and generation capacity.