Help calculating heat loss through transient conduction

In summary, heat loss through transient conduction can be calculated using the formula Q = -kA(dT/dx), where Q is the heat transfer rate, k is the thermal conductivity of the material, A is the cross-sectional area, and (dT/dx) is the temperature gradient. Transient conduction is the process of heat transfer through a changing medium, occurring when there is a temperature difference between two points in a material. The higher the thermal conductivity of a material, the easier heat can be transferred, resulting in greater heat loss through transient conduction. Factors that can affect heat loss include material properties, temperature difference, cross-sectional area, and length. To minimize heat loss, one can use materials with low thermal conductivity,
  • #1
mike3
6
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Homework Statement



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?

Homework Equations



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

http://highered.mcgraw-hill.com/sites/dl/free/0073129305/314124/cen29305_ch04.pdf


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.
 
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  • #2


Hello, thank you for your post. It seems like you have a good understanding of the principles behind heat transfer and conduction in your stove. To accurately calculate the additional heat loss due to the increased height of your stove, you will need to use the heat transfer equation for conduction:

Q = kA(T1-T2)/L

Where Q is the heat transfer rate, k is the thermal conductivity of the material, A is the surface area, T1 is the initial temperature of the metal, T2 is the temperature of the surrounding air, and L is the thickness of the metal.

In your case, you will need to calculate the heat loss at each time interval using the temperature data you have provided. You can then subtract this heat loss from the predicted BTU/hr output of the stove (based on the smaller version with only a diameter increase) to get the actual BTU/hr output for the larger version.

Additionally, you will need to consider the heat transfer due to convection between the metal drum and the air. As you mentioned, you can use Newton's law of cooling to calculate this heat transfer rate. Make sure to use the correct heat transfer coefficient for the temperature range you are working with.

I hope this helps and good luck with your calculations!
 

1. How is heat loss through transient conduction calculated?

Heat loss through transient conduction can be calculated using the following formula: Q = -kA(dT/dx), where Q is the heat transfer rate, k is the thermal conductivity of the material, A is the cross-sectional area, and (dT/dx) is the temperature gradient.

2. What is transient conduction?

Transient conduction is the process of heat transfer through a medium that is changing with time. It occurs when there is a temperature difference between two points in a material, causing heat to flow from the warmer point to the cooler point until thermal equilibrium is reached.

3. How does the thermal conductivity affect heat loss through transient conduction?

The thermal conductivity of a material is a measure of its ability to conduct heat. A higher thermal conductivity means that the material can transfer heat more easily, resulting in a higher heat transfer rate and therefore greater heat loss through transient conduction.

4. What factors can affect heat loss through transient conduction?

Some factors that can affect heat loss through transient conduction include the material properties, such as thermal conductivity and specific heat, the temperature difference between the two points, the cross-sectional area, and the length of the material.

5. How can heat loss through transient conduction be minimized?

Heat loss through transient conduction can be minimized by using materials with low thermal conductivity, increasing the thickness of the material, reducing the temperature difference, and insulating the material to prevent heat transfer to the surrounding environment.

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