anonymous
  • anonymous
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)
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
how to prove \[\theta(x,t)\] ?
anonymous
  • anonymous
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

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abb0t
  • abb0t
whats BVM?
anonymous
  • anonymous
Boundasy Value Mean
anonymous
  • anonymous
please jsu ignore the BVM, just fint the partial derivative of (θ(x,t)
anonymous
  • anonymous
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.

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