Heat Transfer from metal wall

In summary, to estimate the average temperature the wall will attain at equilibrium, you first need to calculate the hydraulic diameter of the wall, then use it to calculate the Reynold's number, and finally use the appropriate Nusselt number equation to calculate the heat transfer coefficient. With the heat transfer coefficient, you can then calculate the average temperature of the wall using the heat transfer equation.
  • #1
JSBeckton
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A light breeze of 10mi/hr blows across a metal building. The buildings wall is 3.7m tall and 6.1m wide. A net energy flux of 347w/m^2 from the sun is absorbed by the wall and then dissipated to the air through convection. Assuming the air is 27c and 1 atm and blows across the wall as a flat plate, estimate the average temp the wall will atain at equlibrium.

I took the flux and divided it by the area to get the heat transfer. I was thinking that i need to work backwards through the Nussalt number to find h, then I can solve from the convection eqn but I have a sheet with a bunch of Nussalt equations that all depend on the Renolds number for this situation, but that depends on the temp of the wall? Where do I begin?

Thanks in advance for any feedback.
 
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  • #2
You can start by calculating the characteristic length of the wall, which is the hydraulic diameter. The hydraulic diameter is given by: Dh = 2(lw)/ (l+w) Where l is the length and w is the width of the wall. Substituting the values you have, you get a hydraulic diameter of Dh = 2(3.7)(6.1) / (3.7 + 6.1) = 3.3 m Now you can calculate the Reynold's number for this flow, which is given by: Re = (vDhρ)/μ where v is the velocity of the air, Dh is the hydraulic diameter, ρ is the density of the air and μ is the dynamic viscosity of the air. Substituting the values you have, you get a Reynold's number of Re = (10 x 3.3 x 1.2)/1.8 x 10^-5 = 0.23 x 10^5 Now you can use the appropriate Nusselt number equation to calculate the heat transfer coefficient. Once you have the heat transfer coefficient, you can then calculate the average temperature of the wall using the following equation: T_avg = (Q/hA) + T_air where Q is the net energy flux, h is the heat transfer coefficient, A is the area of the wall and T_air is the temperature of the air. Substituting the values you have, you get an average temperature of T_avg = (347/h x 3.7 x 6.1) + 27 = 37.3°C
 
  • #3


I would approach this problem by first understanding the principles of heat transfer and the factors that affect it. In this case, the heat transfer is occurring through convection, which is the transfer of heat through the movement of a fluid (in this case, air). The rate of heat transfer is dependent on the temperature difference between the wall and the air, as well as the properties of the fluid and the surface area of the wall.

To begin, we can use the heat transfer equation, Q = hA∆T, where Q is the heat transfer rate, h is the convective heat transfer coefficient, A is the surface area of the wall, and ∆T is the temperature difference between the wall and the air. We know that the heat transfer rate is 347 W/m^2, and we can assume that the temperature difference (∆T) is the same as the wall temperature, which we are trying to find.

Next, we need to determine the convective heat transfer coefficient (h). This can be done using the Nusselt number (Nu), which is a dimensionless number that relates the convective heat transfer to the thermal conductivity of the fluid. The Nusselt number can be calculated using different correlations or equations, depending on the flow conditions and geometry of the system.

In this case, we can use the correlation for a flat plate in cross-flow, which is given by Nu = 0.664 * Re^0.5 * Pr^0.33, where Re is the Reynolds number and Pr is the Prandtl number. The Reynolds number can be calculated using the air velocity (10 mi/hr or 4.47 m/s), the characteristic length (3.7 m), and the kinematic viscosity of air at 27°C (1.9*10^-5 m^2/s). The Prandtl number can also be calculated using the air properties at 27°C.

Once we have calculated the Nusselt number, we can rearrange the heat transfer equation to solve for the wall temperature. This will give us an estimated average temperature that the wall will reach at equilibrium.

However, it's important to note that this is a simplified approach and there may be other factors at play that could affect the heat transfer, such as thermal radiation from the sun or other sources. It's also important to consider the assumptions made in the calculations and the accuracy of the data used. Overall, this approach can
 

1. How does heat transfer occur in metal walls?

Heat transfer in metal walls occurs through conduction, where heat energy is transferred from molecule to molecule within the metal. This is due to the close proximity of metal atoms, allowing for efficient transfer of heat energy.

2. What factors affect the rate of heat transfer in metal walls?

The rate of heat transfer in metal walls can be affected by several factors, including the type of metal, thickness of the wall, temperature difference between the two sides of the wall, and the presence of any insulation or surface coatings.

3. How does the shape of a metal wall affect heat transfer?

The shape of a metal wall can affect heat transfer through conduction by altering the surface area available for heat transfer. For example, a flat wall will have a higher surface area and therefore allow for more efficient heat transfer compared to a curved or corrugated wall.

4. What is the difference between heat transfer in conductive and non-conductive metal walls?

Conductive metal walls allow for direct transfer of heat energy through the metal, while non-conductive metal walls may have a layer of insulation or coating that slows down the rate of heat transfer. This can result in a longer time for the metal wall to reach thermal equilibrium with its surroundings.

5. Can the rate of heat transfer in metal walls be controlled?

Yes, the rate of heat transfer in metal walls can be controlled through various methods such as using different types of insulation or surface coatings, adjusting the temperature difference between the two sides of the wall, and selecting the appropriate thickness and shape of the wall for the desired heat transfer rate.

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