- #1
LennoxLewis
- 129
- 1
Here is the problem. Skip the first two paragraphs to get to the pure math part.
There are two cubes of water, both with area of 2 x 2 dm^2. Via the bottom they are connected by a tube, and the flow of water is (surprise, surprise) proportional to the difference in water level between the two, by 0.1 * delta h. In addition to that, a constant amount of 1.0 liter / min is pumped from one to the other. At t=0, the pump is turned on when h1 = h2 = 10 dm. Derive a formula that describes delta h as function of t.
I started with: Dv1 / dt = A * dh1 / dt = 0.1*delta h - 1.0, and a similar equation for h2.
Substracting these two [i know this is risky, but i didn't know what else to do], i get:
A* d(h1-h2) / dt = A * d(delta h)/dt = 0.2*delta h - 2.0. Substracting A gives:
The math part:
d(delta h) / dt - 0.05 delta h = 0.5
Let's make things easier by renaming delta h to "x":
dx/dt - 0.05x = 0.5
I'm using a standard solution-formula, which says that for:
dx/dt + a*x = F(t), the solution is:
x(t) = exp(-a*t) * integrand (F(t')*exp(a*t')) + c*exp(-a*t)
Using this formula, i get: x(t) = -10 + c* exp(0.05*t).
Obviously, this problem would lead to a negative exponential as solution, or the difference in water level would go to infinity as time goes by!
Where do i miss a minus sign?
There are two cubes of water, both with area of 2 x 2 dm^2. Via the bottom they are connected by a tube, and the flow of water is (surprise, surprise) proportional to the difference in water level between the two, by 0.1 * delta h. In addition to that, a constant amount of 1.0 liter / min is pumped from one to the other. At t=0, the pump is turned on when h1 = h2 = 10 dm. Derive a formula that describes delta h as function of t.
I started with: Dv1 / dt = A * dh1 / dt = 0.1*delta h - 1.0, and a similar equation for h2.
Substracting these two [i know this is risky, but i didn't know what else to do], i get:
A* d(h1-h2) / dt = A * d(delta h)/dt = 0.2*delta h - 2.0. Substracting A gives:
The math part:
d(delta h) / dt - 0.05 delta h = 0.5
Let's make things easier by renaming delta h to "x":
dx/dt - 0.05x = 0.5
I'm using a standard solution-formula, which says that for:
dx/dt + a*x = F(t), the solution is:
x(t) = exp(-a*t) * integrand (F(t')*exp(a*t')) + c*exp(-a*t)
Using this formula, i get: x(t) = -10 + c* exp(0.05*t).
Obviously, this problem would lead to a negative exponential as solution, or the difference in water level would go to infinity as time goes by!
Where do i miss a minus sign?