How Long Can Solar Panels Power a House in a Day

Solar panels power homes based on sunlight duration and household usage. A 5kW system with 4-6 peak sun hours generates 20-30kWh daily. For a typical 25kWh home, it covers the full day, with excess stored in batteries or fed to the grid for later use, ensuring reliability.

Solar System Basics

At its core, a standard grid-tied system includes the photovoltaic (PV) panels themselves, which typically convert 15% to 22% of incoming sunlight into direct current (DC) power. Modern string or microinverters operate at 96% to 99% conversion efficiency. For backup power during grid outages or at night, a cell storage unit, like a lithium-ion cell with a usable capacity of 10 kWh to 20 kWh, can be added. All of this is managed by a system controller and is connected to the public utility grid through a bi-directional meter, allowing you to draw power when needed and send excess back, often for a credit.

A standard residential panel today has a power rating between 370 watts (W) and 450 W. Its physical size is roughly 68 inches long by 40 inches wide, with an area of about 18-20 square feet, and it weighs 40 to 50 pounds. The efficiency difference—say, a 20% efficient panel versus a 17% one—means the higher-efficiency model will generate more power in the same limited roof space. For example, on a south-facing roof with a 30-degree pitch in Phoenix, a 400W panel at 20% efficiency might produce an average of 2.2 kWh per day over a year, while the same panel in Seattle might average 1.5 kWh per day due to lower solar irradiance.

A 7.6 kW AC inverter is a common pairing for a 8.0 kW DC panel system, a ratio of about 1.0 to 1.1. Any energy lost here is pure waste; an inverter with 98% peak efficiency versus one with 96% can save you 20-30 kWh of lost electricity per year for a medium-sized system. If you have a complex roof with multiple directions or shading, microinverters (one per panel, often with 25-year warranties) can optimize output at the individual panel level, preventing a single shaded panel from dragging down the performance of the entire string, which can cause energy losses of up to 30% in suboptimal conditions.

For cell storage, the key metrics are usable capacity (kWh) and power rating (kW). A 13.5 kWh lithium-ion cell can typically deliver a continuous power output of 5 kW. This means it could theoretically run ten 500-watt refrigerators simultaneously, but only for as long as its capacity lasts. At a 5 kW continuous draw, a 13.5 kWh cell would be depleted in 2.7 hours. Batteries are also rated by their cycle life—a good quality home cell is rated for 6,000 cycles to 70% of its original capacity. The round-trip efficiency (the energy you get out versus what you put in) for modern systems is between 90% and 95%.

Sunlight in Your Area

One peak sun hour equals one hour of sunlight at an irradiance of 1,000 watts per square meter, the standard test condition for panels. However, real-world sunlight intensity varies dramatically. A home in sun-drenched Phoenix, Arizona, averages about 6.0 peak sun hours per day annually.

In contrast, a similarly equipped home in Seattle, Washington, receives closer to 3.5 peak sun hours. This 71% difference in solar fuel means the Phoenix system will generate nearly twice the daily kilowatt-hours over a year, fundamentally impacting system sizing, financial returns, and how long your panels can power your home. Local climate patterns, including rainy seasons (e.g., the Pacific Northwest's 150+ cloudy days per year), humidity, and even air pollution levels can reduce irradiance by 10-25% on affected days compared to clear-sky models.

To estimate production, you first need your location's average daily peak sun hours. This is not the same as daylight hours; a 14-hour summer day might only yield 5.5 peak sun hours because the sun's angle and intensity are low in the morning and evening. This annual average is a composite of seasonal extremes. For instance, in Denver, Colorado, a July day might provide 6.8 peak sun hours, while a December day may only offer 3.2. This seasonal variance of over 100% is a major planning consideration. Resources like the National Renewable Energy Laboratory (NREL) PVWatts Calculator use 30-year historical weather data to model these averages with an error margin of less than 5% for most locations.

City, State	Avg. Daily Peak Sun Hours (Annual)	Seasonal Range (Summer High / Winter Low)	Estimated Daily Output for a 5 kW System (kWh)
Albuquerque, NM	6.5	7.8 / 5.1	30.0 - 35.0
Boston, MA	4.0	5.5 / 2.5	17.5 - 22.5
Miami, FLFL	5.0	5.8 / 4.2	22.5 - 26.0
Seattle, WAWA	3.5	5.2 / 1.8	14.0 - 20.5
Chicago, ILIL	4.2	5.9 / 2.3	18.0 - 24.0

A coastal community may experience regular morning fog that reduces morning generation by 40-60% until it burns off, while a home 30 miles inland does not. Tree cover is another major local factor; a dense deciduous tree shading your roof from 11 AM to 2 PM during summer can cut production during those peak hours by over 80%, which are the most productive hours of the day.

Power Output Estimate

A common mistake is to simply multiply your system's size in kilowatts by a few hours of sun, which can lead to a 10-20% overestimate and financial disappointment. For a typical 7 kW DC system, an overestimate of just 10% equals roughly 300 fewer kilowatt-hours (kWh) produced per month, which at the U.S. average electricity rate of 0.16 perk Wh ,translates to about 50 less in monthly savings.

A system with twenty 400-watt panels has an 8,000-watt or 8.0 kW DC capacity. Next, multiply this by the average daily peak sun hours for your specific location and roof orientation that you researched. For our example, let's use a home in Atlanta, Georgia, with a south-facing roof. Atlanta receives about 4.8 average peak sun hours per day. The initial, ideal calculation is: 8.0 kW x 4.8 hours = 38.4 kWh.

However, this 38.4 kWh number is unachievable. Total system losses will reduce it. You must apply a system loss factor, typically between 14% and 25%, to this initial figure. A standard default used for preliminary estimates is 18%. The formula becomes: Estimated Daily Output = System Size (kW) x Peak Sun Hours x (1 - Loss Factor). For our Atlanta home: 8.0 kW x 4.8 hours x (1 - 0.18) = 8.0 x 4.8 x 0.82 = 31.5 kWh. This 31.5 kWh is a much more realistic daily average. This loss factor is a composite of several specific, quantifiable inefficiencies:

· Inverter Efficiency: Modern inverters convert DC to AC with about 97% efficiency, meaning a 3% loss.

· Temperature-induced Losses: Panels lose power as they heat up. Depending on your climate and roof ventilation, annual losses can range from 5% to 15% of potential output.

· Soiling (Dirt/Dust): Light buildup on panels can cause a 2% to 5% reduction in yield. In very dry, dusty areas, this can spike above 7% without cleaning.

· Shading & Mismatch: Even small shadows from a vent pipe can disproportionately impact a string of panels. For systems with partial shading or multiple roof planes, losses can be 3% to 10%.

· DC/AC Wiring Resistive Losses: Electricity loses a small amount of power traveling through wires, typically 1% to 3%.

· Age-related Degradation: In the first year, panels can lose about 1% to 2% of their rated power, then degrade at a slower rate of roughly 0.4% to 0.7% per year thereafter.

If your 8.0 kW system is on a West-facing roof instead of South, it might only receive the equivalent of 4.1 peak sun hours in Atlanta. Your new estimate becomes: 8.0 kW x 4.1 hours x 0.82 = 26.9 kWh. This is a 14.6% decrease from the south-facing estimate, highlighting how crucial orientation is. The annual production is what matters for your utility bill. Multiply your daily average by 365 days. Our south-facing Atlanta example: 31.5 kWh/day x 365 days = 11,497.5 kWh per year.

Real-World Example

Consider a single-family home in Austin, Texas, with a 4,200-square-foot roof and an annual electricity consumption of 12,500 kWh. The homeowners installed a 7.6 kW DC solar system in March 2023. The system uses nineteen 400-watt panels with 20.5% efficiency and a single 7.6 kW string inverter with 97.5% peak efficiency. Their roof is south-facing with a 22-degree pitch, and they have no shade after 9:00 AM. According to the pre-installation estimate from NREL's PVWatts calculator, the system was projected to produce 12,100 kWh in its first year, offsetting about 97% of their usage.

Parameter	Specification / Value
Location	Austin, Texas
System Size (DC)	7.6 kW (19 x 400W panels)
Total Installed Cost	22,800(3.00/W)
Net Cost after 30% ITC	$15,960
Average Daily Peak Sun Hours	5.2 hours (annual average)
Estimated Annual Production	12,100 kWh (PVWatts)
Actual First-Year Production	11,860 kWh
Percentage of Home Usage Covered	95%
Average Daily Output (Annual)	32.5 kWh

The average daily output was indeed 32.5 kWh, but this average masks massive seasonal fluctuations. The peak production month was July, with a total of 1,380 kWh generated, averaging a stellar 44.5 kWh per day. The highest single-day output recorded was 48.2 kWh on a clear, cool day in early June. Conversely, the lowest production month was December, with only 620 kWh total, averaging just 20.0 kWh per day. The range from peak to trough is over 120%, a critical point for homeowners considering energy independence.

For comparison, a utility-scale solar farm in West Texas might achieve a capacity factor near 25%. The variance comes from the residential system's fixed roof angle and unavoidable local shading patterns. The financial outcome for the first year was calculated precisely. The home's utility has a net metering agreement that provides a 1:1 bill credit for excess generation. The system produced 11,860 kWh, and the home consumed 12,500 kWh.

Weather and Seasons

In a city like Denver, Colorado, a 7 kW system might produce an average of 42 kWh on a clear June day but only about 19 kWh on a clear December day—a 55% reduction due solely to the sun's lower altitude and shorter daylight duration, which drops from about 15 hours to just 9.5 hours. Weather events superimpose on this cycle. A passing thunderstorm can cut instantaneous power output by over 80% in under 60 seconds.

The shift is caused by the sun's maximum altitude in the sky, which in mid-latitudes can differ by over 40 degrees from summer to winter. This lower winter angle means sunlight passes through a thicker layer of the atmosphere, increasing scattering and absorption, and strikes your panels at a more oblique angle, reducing energy density per square foot.

A day with high, thin cirrus clouds might only reduce total daily yield by 15-25%, while a solid stratus overcast can cause a 70-90% reduction. The most disruptive pattern is rapidly shifting cumulus clouds, which cause a "rollercoaster" effect on inverter output, with power swinging from 100% to 20% and back over 2-minute intervals, stressing the system and making self-consumption planning difficult.

For a concrete example, take the same 7 kW system on two different days in Atlanta. On a pristine, cool spring day (55°F) in April, it produced a total of 48.1 kWh, with a near-perfect bell curve peaking at 6.4 kW at solar noon. On a hot, hazy, and partially cloudy day in August (95°F), with high humidity and intermittent cumulus clouds, it produced only 34.7 kWh, a 28% reduction, despite the longer summer day. The peak output was clipped at 5.9 kW due to thermal losses, and the graph was a jagged series of peaks and valleys.

High ambient humidity can scatter sunlight, reducing peak irradiance by 5-10%. Wildfire smoke or sustained air pollution creates a persistent haze, which can lower daily production by a consistent 10-30% for weeks. Snow cover is binary: a light dusting that melts quickly has minor impact, but 1 inch of accumulated snow can block 100% of production until it slides off or is removed.

These models use 30-year climatological data to account for the probability of rainy seasons, monsoon patterns, and typical cloud cover. A homeowner should expect their system to hit the annual estimate within ±10%, but any single month can deviate wildly from the 30-year average. For instance, a June with 50% more rainfall than the historical mean can result in a monthly production figure that is 20-25% below the expected average for that month.

Using Cell Storage

A typical 10 kilowatt-hour (kWh) usable capacity home cell, like a Tesla Powerwall 2 or comparable unit, can store enough energy to power essential home circuits for 6 to 12 hours, depending on consumption.

On a standard day, a 7 kW solar system might produce 35 kWh, with 18 kWh used immediately during daylight and 17 kWh exported to the grid. With a cell, the system logic changes. The 17 kWh of excess is first used to charge the cell until it's full; only the remainder is exported. Later, when the home's evening load spikes to 3 kW from 6 PM to 11 PM (a 5-hour period drawing 15 kWh), the cell discharges to cover that demand. This increases your self-consumption rate—the percentage of your own solar energy you use directly—from about 50% without a cell to 80% or more with one. The precise benefit depends on your load profile. A home with a high daytime baseload will have a lower self-consumption rate without a cell than a home where everyone is out until evening.

Cell Consideration	Key Metrics & Typical Ranges	Real-World Implication for a 10 kWh Cell
Daily Cycling for Savings	Round-Trip Efficiency: 90-94%; Depth of Discharge (DoD): 90-100%; Cycles: 6,000+	You put in 10.6 kWh to get 10 kWh out. Daily use degrades capacity by ~0.05% per cycle.
Backup Power Duration	Usable Capacity: 10-20 kWh; Continuous Power: 5-11 kW; Surge Power: 7-15 kW	A 10 kWh cell running a 2 kW essential load (lights, modem, fridge, fan) lasts ~5 hours.
Financial & Longevity	Installed Cost: 900−1,500 per kWh; Warrantied Capacity Retention: 70% after 10 years	A 13 kWh unit costs ~$12,000. It may hold only 9.1 kWh in year 10, a 30% total capacity loss.

A 13.5 kWh cell powering a carefully managed essential load panel with a constant draw of 0.8 kW could theoretically last for nearly 17 hours. However, if the outage occurs on a cloudy day, the solar array's recharge potential may be reduced by 70% or more, severely limiting the cell's ability to recharge for the next night.