Question

Can anyone please help me with this mass balance differential problem

Answer 1

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

Answer 2

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

Answer 3

@whpalmer4

Answer 4

alrity thanx anyway :) @TheCatMan

Answer 5

whpalme4 is an expert he can help better

Answer 6

What are \(\rho\) and \(A\)? Just constants?

Answer 7

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

Answer 8

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.

Answer 9

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?

Answer 10

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:
\[h^{1/2}=-\frac{k}{2A}t+{h_0}^{1/2}~~\implies~~h=\left({h_0}^{1/2}-\frac{k}{2A}t\right)^2\]

Answer 11

OOh I totally forgot about that constant thank you so much

Answer 12

yw