Heat and Temperature Equations-Separation of variables

In summary, the function θ(x,t) is given by θ_0+∑_(n=1)^∞▒〖B_n e^(-k/ρc ((n^2 π^2)/l^2 +h/KA)t ) Sin (nπx/l)〗.
  • #1
ra_forever8
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I really don't how to start this question. Please help me.

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  • #2
grandy said:
I really don't how to start this question. Please help me.

Hi grandy, :)

Welcome to MHB! I think you have been given how to start this problem in the heading; Separation of Variables. So initially what you have to do is write the function \(\psi\) as,

\[\psi(x,t)=A(x)B(t)\]

A good introduction about the method of separation of variables can be found >>here<<.

Kind Regards,
Sudharaka.
 
  • #3
Consider an element of the rod having length dx and situated at x:
1. Rate of heat loss by this element to the surroundings = h.φ.dx ; so that
Heat loss to surroundings in time dt = h.φ.dx.dt
Change in temperature due to heat loss to surroundings in dt = h.φ.dx.dt/(ρ.A.c.dx)

2. Rate of heat loss by conductivity along the rod = [dφ(x+dx)/dx – dφ(x)/dx].k.A
Change in temperature of the element in time dt = [dφ(x+dx)/dx – dφ(x)/dx].dt.k.A/(ρ.A.c.dx)]

3. Total change in temperature dφ = [dφ(x+dx)/dx – dφ(x)/dx].dt.k.A/(ρ.A.c.dx)]-h.φ.dt/(ρ.A… giving:
∂φ/∂t = [dφ(x+dx)/dx – dφ(x)/dx].k/(ρ.c.dx)]-h.φ/(ρ.A.c)
Now [dφ(x+dx)/dx – dφ(x)/dx]/dx = ∂²φ/∂x² giving finally:
∂φ/∂t = (k/ρc).∂²φ/∂x² - φ.h./(ρ.A.c)

I did the first part a) would please check my answer and confirm me the result, Now would you please help me with the second part. I looked at the separation of variables but I was unable to do this one because it is tough for me. Your help is really appreciated.
 
  • #4
I try to show that θ(x,t)= θ_0+∑_(n=1)^∞▒〖B_n e^(-k/ρc ((n^2 π^2)/l^2 +h/KA)t ) Sin (nπx/l)〗by looking at paul online notes but I was unsucessful. And also i was unable to find the value of Bn too. This qs is impossible tough., Please help me.
 
  • #5
now i have finished the first and second of this qs. for the third part of qs, I have got the new boundary condition but again I am not sure . would please help me.In that case the BC at L/2 would become
∂θ(L/2,t)/∂x = 0, meaning it is like an insulated end with no heat flow across it.

After that I dnt know how to do?
 

FAQ: Heat and Temperature Equations-Separation of variables

What is the difference between heat and temperature?

Heat and temperature are often used interchangeably, but they are actually two different concepts. Heat is the transfer of energy from one object to another, while temperature is a measure of the average kinetic energy of the particles in a substance.

What is the separation of variables method used for in heat and temperature equations?

The separation of variables method is a mathematical technique used to solve partial differential equations, including those related to heat and temperature. It involves separating the variables in the equation and solving for each one separately.

What are the common heat and temperature equations that use the separation of variables method?

Some common heat and temperature equations that use the separation of variables method include the heat equation, the diffusion equation, and the wave equation. These equations are used to model the transfer of heat and temperature in various systems.

How does the separation of variables method help in solving heat and temperature equations?

The separation of variables method simplifies the solution of complex heat and temperature equations by breaking them down into smaller, more manageable equations. This allows for easier computation and a better understanding of the behavior of heat and temperature in a given system.

Are there any limitations to the separation of variables method in heat and temperature equations?

While the separation of variables method is a powerful tool in solving heat and temperature equations, it does have some limitations. It is not always possible to use this method for more complex equations, and it may not always provide an exact solution. In these cases, other numerical or analytical methods may be needed.

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