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

EchdipBest ResponseYou've already chosen the best response.0
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

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

EchdipBest ResponseYou've already chosen the best response.0
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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