i am new here, I am using MERRA monthly solar radiation data. I want to convert w/M^2 to MJ/m^2

I am bit confused, how to convert solar radiation monthly average data W/m^2 to MJ/m^2

so far i understood by reading different sources, Firstly i have to convert w/m^2 to kw/m^2 after kw/m^2 to mj/m^2 ....... Am i doing correctly

Just i am taking one instance:
For may month i have value 294 w/m^2
So 294 * 0.001 = 0.294 kw/m^2 0.294 * 24 (kw to kwh (m^/day)) = 7.056 kwh/m^2/day 7.056 * 3.6 (kwh to mj) = 25.40 mj/day

i am confused i am doing right or wrong .

3 Answers

Not sure why you would take the kWh step in between.

Your panels do 294 Watt per m², i.e. 294 Joule per sec per m². So that's 24*60*60 * 294 = 25401600 Joule per m² per day, or 25.4016 MJ per m² per day.

1

So if:

1 W/m2 = 1 J/m2 s

Then:

294 W/m2 = 294 J/m2 s

if you want it in days then:

1 day = 60s * 60min *24h = 86400s
294 J/m2 s x 86000s/1day = 25284000 J/m2 day
25284000 J/m2 day x 1MJ/1000000J = 25.284 MJ/m2 day

all together:

294 W/m2 = 294/(1000000/86400) = 25.4016 MJ/m2 day

1

A watt is the unit of power and Joules are the units of energy, they are related by time. 1 watt is 1 Joule per second 1W = 1 J/s. So the extension of that equation is that 1J = 1w x 1second. 1J = 1Ws. A loose analogy is if you say Litre is a unit of volume and L/S is a unit of flow. So your calculation needs to consider how long you are gathering the solar energy. So the number of Joules, if the sunlight shines at 90degrees to the solar panel for 1 hour is 294W/m2 x 3600s and would give ~1 x 10^7 joules per square metre. Of course as the inclination [the angle of light] varies away from 90 degrees, this will cause the effective power and hence the energy absorbed to drop, as a function of the sine of the angle to the sun. 90 degrees gives a sine of 1 and is full power.