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anonymous
 one year ago
Can anyone please help me with this mass balance differential problem
anonymous
 one year ago
Can anyone please help me with this mass balance differential problem

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If it helps I believe the lower bound is h0 and upper is h(t)

TheCatMan
 one year ago
Best ResponseYou've already chosen the best response.0theres no way for me to help its to advance for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alrity thanx anyway :) @TheCatMan

TheCatMan
 one year ago
Best ResponseYou've already chosen the best response.0whpalme4 is an expert he can help better

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What are \(\rho\) and \(A\)? Just constants?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A is the cross sectional area of the tank top but it's a constant here and p crosses out @SithsAndGiggles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0After 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops, 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: \[h^{1/2}=\frac{k}{2A}t+{h_0}^{1/2}~~\implies~~h=\left({h_0}^{1/2}\frac{k}{2A}t\right)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OOh I totally forgot about that constant thank you so much
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