One Dimensional Heat Equation

In summary, the heat energy necessary to raise the temperature of a thin slice of thickness \Deltax from 0^o{} to u(x,t) is not c(x)u(x,t). but instead \int_0^uc(x,\overline{u})d\overline{u}.
  • #1
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Homework Statement



Show that the heat energy per unit mass necessary to raise the temperature of a thin slice of thickness [itex]\Delta[/itex]x from [itex]0^o{}[/itex] to [itex]u(x,t)[/itex] is not [itex]c(x)u(x,t)[/itex]. but instead [itex]\int_0^uc(x,\overline{u})d\overline{u}[/itex].

Homework Equations



According to the text, the relationship between thermal energy and temperature is given by

[itex]e(x,t) = c(x)p(x)u(x,t)[/itex],

which states that the thermal energy per unit volume equals the thermal energy per unit mass per unit degree times the temperature time the mass density.

When the specific heat [itex]c(x)[/itex] is independent of temperature, the heat energy per unit mass is just [itex]c(x)u(x,t)[/itex].


The Attempt at a Solution



The only hint really is that this is related to the area, from the solution. How can I go about this geometrically and/or algebraically?

Any help/pointers will be much appreciated. Thank you!
 
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  • #2
that integral doesn't make sense based on what you've posted, can you check it... c has changed to a function of x only or x and u (position and temperature)...?
 
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  • #3
so i think you do need to assume c = c(x,u), then try and find the change in energy de, for a small change in temp du and integrate.
 
  • #4
Argh of course! Thank you very much! :approve:
 
  • #5
I have used your advice and went about it in the following way:

For a small slice of thickness [itex]\Delta{x}[/itex] a small change in energy will be given by

[itex]de = c(x,u)du[/itex]

Dividing by [itex]du[/itex] I obtain [itex]e_{u} = c(x,u)[/itex].

From the Fundamental Theorem of Calculus, this really says that

[itex]e(x,t) = \int_0^uc(x,t)dt[/itex].

Am I correct in my thinking? It feels a bit messy somehow...
 
  • #6
you can just start from differentials
[tex] de = c(x,u)du [/tex]
[tex] \delta e = \int_{e_0}^{e_f}d\bar{e} = \int_{0}^{u} c(x,\bar{u})d\bar{u} [/tex]
 
  • #7
Your equation did not even occur to me but simplifies it a lot. Thank you very much for your help. I can now get some sleep again :-)
 

What is the one dimensional heat equation?

The one dimensional heat equation is a partial differential equation that describes the flow of heat in a one dimensional space. It is commonly used in physics and engineering to model heat transfer in various systems.

What are the key components of the one dimensional heat equation?

The one dimensional heat equation includes the following components: the heat flux, the thermal conductivity, the temperature, and the heat source/sink. These components are used to describe the rate of heat transfer in a specific direction.

What are the boundary conditions for the one dimensional heat equation?

The boundary conditions for the one dimensional heat equation include the initial temperature distribution, the boundary temperatures, and the heat transfer coefficients at the boundaries. These conditions help determine the behavior of the system over time.

What are some real-world applications of the one dimensional heat equation?

The one dimensional heat equation has many practical applications, including predicting the temperature distribution in a metal bar as it is heated, modeling the cooling of an object in a room with varying temperatures, and analyzing the thermal behavior of electronic devices.

What are some numerical methods used to solve the one dimensional heat equation?

Some commonly used numerical methods for solving the one dimensional heat equation include the finite difference method, the finite element method, and the method of lines. These methods involve discretizing the one dimensional space and solving for the temperature at each point in time.

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