- #36
Jason-Li
- 119
- 14
BvU said:That is not the idea for this exercise, no. You now have $$ {d\,(h_1-h_2)\over dt} =
-{D\,g\over R_f} {A_1+A_2\over A_1 A_2} \,(h_1-h_2)$$Which is of the form ##\ y'= -k\,y\ ##, so easily solved if the initial ##h_{1,0}## and ##h_{2,0}## are given .
When I re-read your post #3 I understand where your interest in ##h_2## comes from. Well, at least now you have the answer for part (b) .
For part (a): Do you see a path to express ##h_2## when ##h_{1,0}## and ##h_{2,0}## are given and ##h_1-h_2## is known as a function of time ?
##\ ##
For Part (b) I can basically follow the other thread as we are at a point now that I can bring the equation to meet the equation here although I don't really see how they did the step shown by the red arrow
For part (a) could I integrate the left hand side between limits of h1-h2 and h1,0-h2,0 to give:
$$(h_1-h_2)-(h_{1,0}-h_{2,0}) = -{D\,g\over R_f} {A_1+A_2\over A_1 A_2} \,(h_1-h_2)t$$
$$-h_2 = -{D\,g\over R_f} {A_1+A_2\over A_1 A_2} \,(h_1-h_2)t+(h_{1,0}-h_{2,0})-h_1$$
Along the right lines here?