A black wood stove with surface area 4.6 m^2 is made from cast iron 4.0 mm thick. The interior wall of the stove is at 650 C, while the exterior is at 647 C.
Part A
What is the rate of heat conduction through the stove wall?
Part B
What is the rate of heat loss by radiation from the stove?
Part C
Use the results of parts A and B to find how much heat the stove loses by a combination of conduction and convection in the surrounding air.
To find the rate of heat conduction through the stove wall (Part A), we can use Fourier's Law of Heat Conduction, which states that the rate of heat conduction (Q) is given by the equation:
Q = (k * A * ΔT) / d
Where:
- Q is the rate of heat transfer
- k is the thermal conductivity of the material
- A is the surface area
- ΔT is the temperature difference across the wall
- d is the thickness of the wall
Given:
- Surface area (A) = 4.6 m^2
- Thickness (d) = 4.0 mm = 0.004 m
- Temperature difference (ΔT) = 650°C - 647°C = 3°C
To solve Part A, we need to determine the thermal conductivity of cast iron. You can find this value in a materials properties reference, such as a handbook or database specifically for engineering materials. The thermal conductivity is usually given in units of W/(m*K) or W/(m°C).
Once you have the thermal conductivity value (k) for cast iron, you can substitute the given values into the equation and calculate the rate of heat conduction.
For Part B, to find the rate of heat loss by radiation, we can use the Stefan-Boltzmann Law, which states that the rate of heat loss (Q) due to radiation is given by:
Q = ε * σ * A * (T^4 - Ts^4)
Where:
- Q is the rate of heat transfer
- ε is the emissivity of the material
- σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2*K^4))
- A is the surface area
- T is the temperature of the interior wall
- Ts is the temperature of the surroundings
Given:
- Surface area (A) = 4.6 m^2
- Emissivity (ε) for cast iron can be assumed to be around 0.7-0.8
- Temperature of the interior wall (T) = 650°C + 273 = 923 K (convert to Kelvin)
- Temperature of the surroundings (Ts) = 647°C + 273 = 920 K (convert to Kelvin)
Substituting the values into the equation, you can calculate the rate of heat loss by radiation.
For Part C, you can add the rates of heat conduction (Part A) and heat loss by radiation (Part B) to get the total rate of heat loss by conduction and radiation combined.