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

Echdip
 one year 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

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

Echdip
 one year 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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