Heat transfer derivation problem

In summary, to solve this problem, you will need to use Newton's law of cooling and Fourier's law to derive an equation for the rate of heat transfer through a wall, assuming steady state and neglecting end effects. A diagram and labeling of variables will be helpful in this process.
  • #1
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Homework Statement



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Homework Equations


Newton's law of cooling
Fourier's law


The Attempt at a Solution



Besides writing down both the laws and perhaps drawing a diagram, I don't really know how to go about this derivation.

Any help would be really appreciated
 
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  • #2
The diagram will probably be very helpful, but be sure to label it with all of the variables which will affect the rate of heat transfer i.e. the wall width, material conductivity, convection coefficients, far-stream temperatures etc.

If the system is in steady state, then you can conclude that the rate of heat transfer is uniform throughout the wall. You should maybe also assume that the wall is large enough in two dimensions so that "end effects" are negligible, thus reducing the problem to one in only 1-dimension.

If the far-stream temperatures are T1 and T2 and the wall surface temperature are Ts1 and Ts2, then the rates of convection on either side of the wall are (Newtons law of cooling):

[tex]
q'' = h_1 (T_1 - T_{s1} )
[/tex]

[tex]
q'' = h_2 (T_{s2} - T_2 )
[/tex]

Since the system is in steady state and the temperature gradient through the wall will be linear, the heat flux through the wall will be (fouriers law):

[tex]
q'' = \frac{k}{L} (T_{s1}-T_{s2})
[/tex]

You can use these three equations to derive the equation you need.
 

1. How do you derive the heat transfer equation?

The heat transfer equation can be derived using the principles of conservation of energy and Fourier's law of heat conduction. It involves considering the rate of heat transfer, the temperature gradient, and the thermal conductivity of the material. The final equation is a differential equation that describes how heat flows in a given material.

2. What is the significance of the boundary conditions in a heat transfer derivation problem?

Boundary conditions are essential in a heat transfer derivation as they provide the necessary information to solve the differential equation. They specify the temperature at specific points on the material's surface or interface, which helps determine the rate of heat transfer. Without boundary conditions, the solution of the heat transfer equation would be incomplete.

3. How does the material's thermal conductivity affect the heat transfer equation?

The thermal conductivity of a material is a crucial factor in determining the rate of heat transfer. It is a measure of how easily heat can flow through a material. Materials with higher thermal conductivity will transfer heat more quickly than those with lower thermal conductivity. Therefore, the value of thermal conductivity is directly incorporated into the heat transfer equation.

4. Can you explain the difference between steady-state and transient heat transfer?

Steady-state heat transfer occurs when the temperature of a material remains constant over time. In this case, the rate of heat transfer does not change, and the temperature gradient is constant. Transient heat transfer, on the other hand, occurs when the temperature of a material changes over time due to a changing heat source or change in the material's properties. The rate of heat transfer and temperature gradient are not constant in transient heat transfer.

5. What are some common methods used to solve heat transfer derivation problems?

There are several methods used to solve heat transfer derivation problems, including analytical methods, numerical methods, and experimental methods. Analytical methods involve using mathematical equations and techniques to solve the heat transfer equation. Numerical methods use computer algorithms to approximate the solution of the equation. Experimental methods involve conducting physical experiments to measure the heat transfer rate and validate the theoretical solutions.

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