I believe I've ran into a snag with the very first calculation - surface area of the cylinder. Somehow I got 32.42m^2 instead of 36.95m^2 so here we go again!
Hey,
Yes that's $/GJ and I'm unsure as to how to factor in the 80% efficiency of the oven. Thanks for pointing that out - I should definitely leave in the sec/yr calculation. This has been a big help!
Area of the oven exposed to the air;
(pi x d^2)/2 + (pi x d x L) = 32.42 m^2
Heat transfer rate of the system with no insulation:
q=hA(Ts-Tsurr)
q=20 x 32.42 x ( 80 - 23 )
q=36.96 KW
Heat transfer rate of the system with 1" insulation:
Thermal circuits-
Rcond= ln(r2-r1)/(2 x pi x L x K)
Rcond=...
Hi all, I have a unique problem that I need help with
A company's curing oven's exposed surface temperature is measured at 80°c when the surrounding air is 23°c. You think it should be insulated and wager to pay for the cost of this yourself if you can keep the savings incurred. Is this a smart...