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anonymous
 4 years ago
Find the BVM of θ(x,t):
by Deriving the equation ∂φ/∂t=k/ρc (∂^2 φ)/(∂x^2 )  (h/ρAc) φ , where φ(x,t) = (θ(x,t) θ_0)
anonymous
 4 years ago
Find the BVM of θ(x,t): by Deriving the equation ∂φ/∂t=k/ρc (∂^2 φ)/(∂x^2 )  (h/ρAc) φ , where φ(x,t) = (θ(x,t) θ_0)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1356836792021:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how to prove \[\theta(x,t)\] ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0form this qs: Find the BVM of θ(x,t): by Deriving the equation ∂φ/∂t=k/ρc (∂^2 φ)/(∂x^2 )  (h/ρAc) φ , where φ(x,t) = (θ(x,t) θ_0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0please jsu ignore the BVM, just fint the partial derivative of (θ(x,t)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The new BC at L/2 would be: ∂θ(L/2, t)/∂x = 0, which means for all time t the slope of the temperature curve versus x is zero at x = L/2, because the temperature profile is symmetric and continuous about the midpoint. To prove this is true take your heat solution equation where θ(x,t) = θ0 + Sum terms and take the partial derivative of that eqn. with respect to x.
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