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Echdip
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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
 one year ago
Echdip Group Title
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
 one year ago

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Echdip Group TitleBest ResponseYou've already chosen the best response.0
dw:1356836792021:dw
 one year ago

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how to prove \[\theta(x,t)\] ?
 one year ago

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form 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
 one year ago

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Boundasy Value Mean
 one year ago

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please jsu ignore the BVM, just fint the partial derivative of (θ(x,t)
 one year ago

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The 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.
 one year ago
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