I have a stove, basically a metal drum which is the fuel chamber. Think of a hobo stove, just a metal drum with a fire in it. This stove creates convection heat via exhaust, and conductive heat to the metal drum itself which "slowly" releases the heat from there....now, I have the BTU/hr output of this stove at a smaller size, but I altered the dimensions disproportionally. Now, the new, larger version has a much greater height than the previous version. Greater height means more metal therefore, more heat loss through transient conduction...and a lower BTU/hr output of the stove. So, If I only increased the diameter of the stove by some percentage, the BTU/hr output would be exactly that percent greater. The surface area of that stove would be 1464.2448"squared. Now that I've increased the height as well, the surface area is 4560.2848"squared. And, now we don't know the BTU/hr output of this stove. If we subtract the SA of the increased diameter stove (predictable BTU/hr output) from the SA of the increased diameter + height stove (unpredictable BTU/hr output) we have: 4560.2848 - 1464.2448 = 3096.0400"squared This accounts for the additional, unaccounted SA of conductive heat loss which this new stove has. So, I will be able to subtract the amount of heat loss of this surface area from the BTU/hr output the stove would have had if I didn't make it taller. Then I will have an accurate BTU/hr output for this new stove.
So, I think what I need is to calculate the heat loss through transient conduction.
The metal is 304 Stainless steel and has a heat transfer coefficient of 16-24 W/mk.
SA we are working with = 3096.0400"squared
Volume of the new section of fuel chamber = 17957.032"cubed
Density of 304 stainless steel = .029 lbs/in cubed
Mass = 520.753928
Specific Heat, BTU/lb/°F (kJ/kg•K)
32 - 212°F (0 - 100°C) – 0.12 (0.50)
More stats on thermal conductivity:
Thermal Conductivity, BTU/hr/ft2/ft/°F(W/m•K)
at 212°F (100°C) – 9.4 (16.2)
at 932°F (500°C) – 12.4 (21.4)
Initial temperature of medium = 50
Initial temperature of the mass (metal) = 50
After 90 minutes the temperature of the mass peaks @ 600F
The temperature plateaus for 240 minutes
Over the next 30 minutes the temperature goes down to 400F
Then the reaction in the stove is finished and the metal cools down over time to meet an equilibrium with the outside air temperature which is @ 60F. Maybe this takes 1 hr, but I do not know.
How much additional heat loss am I getting?
I think this one:
T(t) - T∞ -bt
_________ = e
T(i) - T∞
This gets the temperature of body at time, or time required to get body to temperature
Then Newton's law of cooling to see the rate of convection heat transfer between the metal drum and the air.
Q(t)=hAs[T(t) - T∞]
These equations and explanations thereof can be found at this link
The Attempt at a Solution
I'm not even sure I'm using the right equations...I can't be sure I know how to use them anyway, I need some assistance.