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- #1

- Thread starter grandy
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- #1

- Feb 5, 2012

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Hi grandy,I really don't how to start this question. Please help me.

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.

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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.

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- #5

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?