First, you have draw the figure, and figure volume in the tank as a function of height (radius=1.5*height)
Volume=1/3 PI (1.5h)^2 *h
the rate of volume change (dV/dt) is given as constant (which ignores Bernoulli principle), but anyway
dV/dt= PI/3 ( 2.25h^2 + 3h^2) dh/dt
check that.
Now consider evaporation:
Ve=constant*PIr^2=k PI*(1.5h)^2
dVe/dt=constant*PI*3h dh/dt
so in the end, total change in volume/time is
deltaV/dt=-PI PI/3 ( 2.25h^2 + 3h^2) dh/dt - constant*PI*3h dh/dt
but dh/dt=.01m/day
so now you have some function for dV/dt.
V=INT dV=INT from o to V = INT function dt from zero to time T. Solve for T after you put in the limits.
A tank for storing water is conical in shape, with maximum radius 20 m and maximum depth 15 m. The large, circular end of the tank points vertically upwards and is open to the air. It is known that the volume of water in the tank will decrease at a rate proportional to the area of the water's surface, due to evaporation. Show that the water-level in the tank decreases at a constant rate. If the tank originally stores 1000 $\mathrm{m}^3$ of water and the water-level drops 1cm/day, how long will it be before the tank is empty?
2 answers
If the area of the water is A, and its depth is h, then
V = 1/3 Ah
But, due to the conical shape, r = kh, so V = kr^3
Taking derivatives,
dV/dt = kr^2 dh/dt
But, we are told that dV/dt = kA, so
(dV/dt)/A = k = dh/dt
So, dh/dt = k
Note that all the k's used are different, but each is constant.
V = 1/3 Ah
But, due to the conical shape, r = kh, so V = kr^3
Taking derivatives,
dV/dt = kr^2 dh/dt
But, we are told that dV/dt = kA, so
(dV/dt)/A = k = dh/dt
So, dh/dt = k
Note that all the k's used are different, but each is constant.
2 answers