In reply to Alan Fletcher's message of Mon, 5 Aug 2013 13:32:46 -0700 (PDT): Hi, [snip] >> From: "David Roberson" <[email protected]> >> Sent: Sunday, August 4, 2013 11:42:35 PM > >> Alan, I made a calculation of the output flow rate that came out much >> higher than the value you have shown here. Is it possible for you to >> recheck your calculation? I will do the same. > >Q = Total power = 22 kWh = 79.2 * 10^6 joules = 79.2 * 10^3 kJ > http://www.rapidtables.com/electric/kWh.htm > >dT = 25 C -- temperature difference >c = 4.186 kJ / kG C -- specific heat >m = kg -- in 1 hour > >Q = c * m * dT > >m = Q / ( c * dT) = 79.2 * 10^3 / ( 4.186 * 25 ) = 79200 / 104.65 = 756.8 kg/h >= 12.6 kg/sec > >which is close to what I used.
756.8 kg/hr = 0.21 kg/sec (you forgot to divide by 60 to convert minutes to seconds). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
