Can anyone please help me with this mass balance differential problem

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Can anyone please help me with this mass balance differential problem

Mathematics
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If it helps I believe the lower bound is h0 and upper is h(t)
theres no way for me to help its to advance for me

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alrity thanx anyway :) @TheCatMan
whpalme4 is an expert he can help better
What are \(\rho\) and \(A\)? Just constants?
A is the cross sectional area of the tank top but it's a constant here and p crosses out @SithsAndGiggles
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.
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?
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: \[h^{1/2}=-\frac{k}{2A}t+{h_0}^{1/2}~~\implies~~h=\left({h_0}^{1/2}-\frac{k}{2A}t\right)^2\]
OOh I totally forgot about that constant thank you so much
yw

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