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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

2. anonymous

If it helps I believe the lower bound is h0 and upper is h(t)

3. TheCatMan

theres no way for me to help its to advance for me

4. TheCatMan

@whpalmer4

5. anonymous

alrity thanx anyway :) @TheCatMan

6. TheCatMan

whpalme4 is an expert he can help better

7. anonymous

What are $$\rho$$ and $$A$$? Just constants?

8. anonymous

A is the cross sectional area of the tank top but it's a constant here and p crosses out @SithsAndGiggles

9. anonymous

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

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

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

OOh I totally forgot about that constant thank you so much

13. anonymous

yw

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