eli5: How is the use of watt-hours useful?

598 views

I’m confused why the term watt-hours (Wh) is used. Watts is joules per second. So watt-hours is joules per second per hour. There are 2 time measurements within the same term.

Why not just use joules for total amount of energy used and watts for consistent/average output over a specific time? I assume watt-hours is how many watts are consistently produced over an hour period but it is used often in unhelpful scenarios. When talking about say solar generation, someone could say “my solar array produces 12kWh every day”. So 12kWh for 24 hours means your solar array produces 0.5kW of power for the entire day. How was watt-hours helpful in anyway to describe the solar array’s power output?

Or when talking about a cities power output, the reports are measured in Gigawatt-hours over the entire year. Why is quantifying a cities yearly output over an hour long period helpful?

Now if we compare the 2 given examples, it becomes even more confusing. If I had a solar array that produces 12kWh every day, how many solar arrays would I need to power a city the needs 5000 GWh every year? 5000 GWh every year is around 570.78 MW, so if I just used the standard watts over watt-hours I would have a simple convertion between scenarios while still having the option to say “0.5kWs each day” or “570.78MWs for a year”.

Sorry if this is sounding like a rant post but I’m really annoyed at this term.

In: 2

18 Answers

Anonymous 0 Comments

>I’m confused why the term watt-hours (Wh) is used. Watts is joules per second. So watt-hours is joules per second per hour.

The last part “watt-hours is joules per second per hour. ” is incorrect it is joules per second times hours. One hour = 3600 seconds so it all works out to 1Wh= 3600J

You measure electric usage over time as energy not as power, Power is just the current rate of consumption

Watt-hour or more often Kilowatt-hour is useful when typical usage time is in hours.

How much energy do a 40W device you run 8 hours per day use each day? 40*8 = 320 Wh =0.32kWh

If you do not use the Wh the calculation is 40*8*3600 = 1 152 000 J

The energy cost starts to be reasonable at kWh too. In the US the average residential energy cost in Feb 2023 was 23 cents. that is 0.00000638cent per joule.

Ther is a reason that the electrical energy is measured in kWh, the number gets reasonable.

For solar power systems, it is important to know that the mount of energy is produce is not constant but changes with the elevation and direction of th sun. Clouds that bloc the sun has also a huge effect

>”So 12kWh for 24 hours means your solar array produces 0.5kW of power for the entire day.”

The last part is incorrect, the output was not 0.5kW during the day it was 0kW during the night. It varies during the day and if you orient the panel to the south the peek will be at noon. 0.5kW is the average power generation for the day not what is produced during the day.

For the city example 5000 GWh every year with power generation of 12kWh every day. The number of generation systems you need is 5000/(12*365/1000000) = 1141000 systems. Ths assumes the power generation is the same all of the year, that it is quite far from current if you are away from the equator.

>5000 GWh every year is around 570.78 MW,

This is the on average power but because power usage is not contend during the day the peak power usage will be a lot higher, you have both daily and yearly variation. It is also not just 570.78 MW you need to say it is “570.78 MW on average during the year” to get energy, the time it is for is important.

The statements “0.5kWs each day” or “570.78MWs for a year” are not correct compared to the rest of the text. a kWs is kilo Watt second = 1000 joule. That means “0.5kWs each day” is an energy generation of 500 joules in a day or 0.14kWh of energy

“570.78MWs for a year” will just mean you generate 570.75 *10^6 Joule each year which is equal to 158 kWh

You are viewing 1 out of 18 answers, click here to view all answers.