> 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.
