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Echdip

  • 2 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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  1. Echdip
    • 2 years ago
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    |dw:1356836792021:dw|

  2. Echdip
    • 2 years ago
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    how to prove \[\theta(x,t)\] ?

  3. Echdip
    • 2 years 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

  4. abb0t
    • 2 years ago
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    whats BVM?

  5. Echdip
    • 2 years ago
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    Boundasy Value Mean

  6. Echdip
    • 2 years ago
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    please jsu ignore the BVM, just fint the partial derivative of (θ(x,t)

  7. Echdip
    • 2 years 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.

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