<div dir="ltr"><div><br></div><div>kg/(m2-s)*(86400 s/day)*(30 day/month)==> kg/(m2-month)</div><div><br></div><div>kg/(m2-month)*area(m2) ==> (kg/month)<br></div><div><br></div><div>1 kg = 0.001 tonnes.</div><div><br></div><div>(kg/month)*0.001 (tonnes/kg) ==> tonnes/month</div><div><br></div><div><br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, Apr 25, 2020 at 10:51 AM Jayant via ncl-talk <<a href="mailto:ncl-talk@ucar.edu">ncl-talk@ucar.edu</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr"><div>Dear,</div><div>I have monthly gridded CO2 emissions due to fires in units of kg/m2/s. I want to compute the annual emissions (in tonnes). I am confused how to handle the units of time and area. For time, is it correct to multiply by a factor 30days x 24x60x60?</div><div>and for area, multiply with corresponding gridarea? and then add together. Is this correct?</div><div>Best regards,</div><div>Jayant<br></div></div>
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