On Oct 18, 2011, at 10:50 PM, Colin Hercus wrote:
Hi Horace,
I find your posts quite interesting and you seem to have a rational
rather than emotional approach which makes for good reading.
I just read your reply to Dave and as it seemed to make the ECat
(and my kettle) impossible I thought I'd double check some of your
calculations and I think you've made a mistake on the heat flow
from the reactor:
R = (0.002 m)/(16 W/(m K)*(1.8x10^-2 m^2) = 1.78 °C/W
By my calculations:
R = 0.002/(16 * 0.018)
= 0.002/.288
= 0.007 °C/W
Yes you are right! Another one of my clerical mistakes. The above
should be written:
R = (0.002 m)/((16 W/(m K)*(1.8x10^-2 m^2)) = 6.94x10^-3 °C/W
Producing a heat flow of 4.39 kW, or 4390 W then requires a delta T
given as:
delta T = (6.94x10^-3 °C/W) * (4390 W) = 30.46 °C
Using Fourier's law to check, I get
q = (16 W/(m K))*(1.8x10^-2 m^2)*(30.47 K)/(0.002 m) = 4387 W
which is well within tolerance.
I did not put the review up on my site. I should correct it and put
it there.
Thanks for the correction! I wish my calculations were checked more
often.
From engineeringtoolbox.com
Fourier's Law express conductive heat transfer as
q = k A dT / s (1)
where
A = heat transfer area (m2, ft2)
k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr
oF ft2/ft))
dT = temperature difference across the material (K or oC, oF)
s = material thickness (m, ft)
So A = 180 CM^2 = 0.018 M^2
K = 16 W/(m K)
s = 0.002m
Then q = 16*0.018*dT/0.002
= 144 * dT
So for 2500W we'd have a temperature difference of 2500/144 = 17 C
which is quite reasonable.
This is all way out of my area of expertise so I could be messing
up units somewhere.
Best Regards, Colin
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
