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anonymous

  • one year ago

Can anyone please help me with this mass balance differential problem

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    If it helps I believe the lower bound is h0 and upper is h(t)

  3. TheCatMan
    • one year ago
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    theres no way for me to help its to advance for me

  4. TheCatMan
    • one year ago
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    @whpalmer4

  5. anonymous
    • one year ago
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    alrity thanx anyway :) @TheCatMan

  6. TheCatMan
    • one year ago
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    whpalme4 is an expert he can help better

  7. anonymous
    • one year ago
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    What are \(\rho\) and \(A\)? Just constants?

  8. anonymous
    • one year ago
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    A is the cross sectional area of the tank top but it's a constant here and p crosses out @SithsAndGiggles

  9. anonymous
    • one year ago
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    Okay, so you have the differential equation \[\frac{d\rho Ah}{dt}=\rho A\frac{dh}{dt}=\rho q_e=\rho k h^{1/2}\] which, as you said, simplifies to \[A\frac{dh}{dt}=k h^{1/2}\] This equation is separable; we can rearrange accordingly to get \[h^{-1/2}\,dh=\frac{k}{A}\,dt\] then integrate both sides.

  10. anonymous
    • one year ago
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    After I integrated I got h(t)= (-Kt/A2)^2 +h0 the solution says h(t)= (h0^(1/2) - Kt/2a)^2 not sure what I'm doing wrong, also how did you remove the negative on the right side there?

  11. anonymous
    • one year ago
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    Oops, I didn't see that at first. Should be \[h^{-1/2}\,dh=-\frac{k}{A}\,dt\] Integrating yields \[2h^{1/2}=-\frac{k}{A}t+C\] We're told that the initial height at \(t=0\) is \(h_0)\), so \[2{h_0}^{1/2}=-\frac{k}{A}(0)+C~~\implies~~C=2{h_0}^{1/2}\] The particular solution is then \[2h^{1/2}=-\frac{k}{A}t+2{h_0}^{1/2}\] Divide by 2 and square both sides: \[h^{1/2}=-\frac{k}{2A}t+{h_0}^{1/2}~~\implies~~h=\left({h_0}^{1/2}-\frac{k}{2A}t\right)^2\]

  12. anonymous
    • one year ago
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    OOh I totally forgot about that constant thank you so much

  13. anonymous
    • one year ago
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    yw

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