The end a And B of arod of length l are maintained temperature 0 degree celcius and 100 degree celcius until steady state condition. Suddenly temperature at the end Ais increased to 20 degree celcius and B decreased to 60 degree celcius find the temperature distance in th rod at time t
as stead state is achieved,
"\\frac{dT}{dt} =0"
the heat equation
"\\frac{d2T}{d2t} =0"
"\\frac{dT}{dx} =A"
T = Ax + B
usin the boundary condition
T=0
0=a(0) +B
B=0
x=100L T=100
100=AL
T(r.t)="\\frac{100u}{L}+0"
after that at t=0 the temp temp at b is reduced to
u=o a + r=0
u=o a + x=0
Ts(r,t) = "\\frac{(0-0)}{L}r=0"
=0
T(r,t) = Ts(r) + Tt(r,t)
T(r,0) = 0 +100r/L + Tt(r,0)
on trial value condition
Tt(r,t) =
"\\frac{2 x100(-1)n+1}{n}"
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