I want to calculate watt hour per square meter to watt per square meter.
Irradiance is understood as instantaneous density of solar radiation falling on unit area per unit time, typically expressed in W/m^2. This unit time is a hour.
Irradiation is the sum of irradiance over a time period (e.g. 1 hour, day, month, year, etc.), or with other words, the amount of solar energy falling on unit area over a stated time interval, it is expressed in Wh/m^2.
Usually you can't compare irradiance and irradiation. Irradiation is a sum of irradiance values, hour after hour (or other short period). This is like to compare speed (irradiance) with traveled distance (irradiation).
From other hand, if both values are given per hour, they are equivalent and you don't need to convert. It is like to know that someone is moving at 10 km / h OR to know that he travels 10 km per hour. It's the same.
May be you don't need to convert Wh/m^2 to W/m^2. Could you give more details?
As we know that Wh is unit of energy and W is unit of power so divide Wh by given time in hour then you will get Wh/m^2 into W/m^2.
Irradiance is understood as instantaneous density of solar radiation falling on unit area per unit time, typically expressed in W/m^2. This unit time is a hour.
Irradiation is the sum of irradiance over a time period (e.g. 1 hour, day, month, year, etc.), or with other words, the amount of solar energy falling on unit area over a stated time interval, it is expressed in Wh/m^2.
Usually you can't compare irradiance and irradiation. Irradiation is a sum of irradiance values, hour after hour (or other short period). This is like to compare speed (irradiance) with traveled distance (irradiation).
From other hand, if both values are given per hour, they are equivalent and you don't need to convert. It is like to know that someone is moving at 10 km / h OR to know that he travels 10 km per hour. It's the same.
Be careful, however, that in some applications (like climatology, but also sometimes solar energy), irradiation results are expressed as W/m2 with the hidden convention that it is applied to a 24 h period. With such a confusing convention, 100 W/m2 actually means 2400 Wh/m2.
radiation input for 1 sq.m area expressed in Joules/sec = W/m2
hourly radiation input (for 1 hour) for 1 sq.m. area expressed in Joules/sec = Wh/m2
Divide Wh/m2 by 3600, it will be converted into W/m2
Anbu, you are just repeating what has been said earlier. Please check the whole thread before responding!
OK Mr Chris. I am new to this research group.
Shall take care in future
As already mentioned by me, taking your question "by the word" leads to nonsense. However, measuring power [W] directly is seldom possible. Mostly energy E [Ws=J] over a period deltat [s] is recorded. Power P [Ws/s=W] is then approximated by E/deltat.
Clearly, it is possible only on the contrary path when you know the length of time during which a constant (mean) power is acting, otherwise it is a doubtful problem as long as solar irradiance is a time dependent value.
- 1 Watt = 1 joule en une seconde
- 1 KW = 1000 W
- 1 MW = 1000000 W
- 1 WH = 1 watt pendant 1 heure donc au total 3600 watts (3600 secondes)
- 1 KWH = 1000 watts pendant 1 heures donc 3,6 MW (3600 * 1000)
- 1 KWH/m2 = 3,6 MW sur 1 m2
- 1 KWH/an = 3,6 MW sur une période de 1 an donc il faut diviser par 365 jours puis 24 heures puis 3600 secondes pour avoir les watts (moyenne généralement)
Donc 1300 kwh/an = 4680 MW/an (1300 * 1000 * 3600)
Donc 4680 MW / 365 jours / 24 heures / 3600 secondes = 148 W environ (moyenne indiquant les joules que l'on peut avoir en une seconde pendant un an)
I too have the same doubt. I received the Insolation data in average sense i.e, 6.5kWh/m^2/day. This value how can i convert it into W/m^2. Can you please help me. Also in data provided by the resources, Global Horizontal Irradiance, Direct Normal Irradiance and Diffuse Horizontal Irradiance. Among these parameters which value I have to use for understanding of solar farm in simulation
How to convert kJ/m2 to W/m2 and Wh/m2 to W/m2
Please clear the doubt.
http://www.fao.org/docrep/X0490E/x0490e0i.htm
watt per m2 (W m-2)
1 W = 1 J s-1
1 W m-2 = 0.0864 MJ m-2 day-1
1W/m2 = 24hrs*60minutes*60seconds/1000000 MJ/m2/day
1 W h/m^2= 1*(1 J/s) * 1 h / m^2=1*(1 J/s)*3600 s / m^2=3600 J / m^2=3.6 KJ/ m^2
Yes, it's possible to convert back to power from energy as long as you know the time magnitude. You can even assume the power == the energy if your data are given with the time interval of ONE hour. Simply, apply the definition of power and energy.
Good luck.
to convert 6.5 kWh/m2-day to W/m2
you need the length of the day (not 24hr but sunshine hours )or the average if the data for a week or a month.
e.g. if the day length is 10 hrs
then you get 650W/m2 which is the average of the day from sunrise to sunset .
The question is similar to this one:
How can I convert km to km/h?
I recommend 2 bellow sites:
https://barani.biz/apps/solar/
https://en.wikipedia.org/wiki/Solar_irradiance
How to convert the Wm-2 to KWhm-2
Again how to calculate the daily average value of solar radiation
1 kwh/m2/d =( 1x1000 w)/(24 m2) =41.666 w/m2
ex. 4.29 kwh/m2/d = 178.75 w/m2
A lot of explanations have been written here. I want to clear something.
If annual solar radiation 5.3509 KWh / m2 / day is given on the basis of hourly estimation of global horizontal radiation (GHI) or direct normal radiation (DNI) radiation and we are taking 10 hours of sunshine for the length of the day. so,
1 KWh/m2-day = (1*1000*3600)/(10*3600) W/m2 =100 W/m2 ( if day length is 10 hours) or
1 W/m2 = 0.01 KW/m2-day So, 5.3509 KWh/m2/day = 5.3509*100=535.09 W/m2.
also
1W/m2 = 24hrs*60minutes*60seconds/1000000 MJ/m2/day = 0.0864 MJ m-2 day-1
1MJ/m2-day=1000000/(24hrs*60minutes*60seconds) W/m2=11.574 W/m2 if day length is 24
1 KWh/m2-day = (1*1000 W*3600)/(24 m2*3600) =41.666 w/m2. if day length is 24
1 KWh / year = (1*1000*3600) W/year = 3.6 MW/year = (3.6*1000000)/(365*24*3600)=0.1141 W or J/s yearly ( used for average usually)
So 5 kWh/year = 18 MW/year (365 * 24* 3600)
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