A water tank is filled by an inflow x(t), the tank is emptied by the outflow y(t)
The outflow if controlled by a resistance R
The water depth in the tank is represented by d(t)
The surface area of the water is A, independent of depth
The tank is 1.5m high with a 1m diameter and a valve resistance of 10s/m^2
We are being asked to write a differential equation in terms of tank dimensions and valve resistance
The outflow is related by y(t) = d(t)/R
d/dt(Volume of tank) = x(t)-y(t)
dh/dt = -a*e^(10t) --> I pulled this from another example online
The Attempt at a Solution
I don't understand how to express this without using the terms in regards to outflow and inflow. From the given equation we can calculate...
d(t) = y(t)R --> From here I assume the following
d(t ) = x(t) - y(t)R , can we assume that x(t) is the initial volume of the tank?
I honestly don't know where to go from here, the only times I have seen a problem like this has been in regards to inflow and outflow, not tank dimensions and valve resistance.
d(t) = pi(.5^2)(1.5) - y(t)*10
I feel like I am overlooking something very simple here, I haven't slept in about a day and a half so that might be it :p