Not all water tanks are shaped like cylinders. Suppose a tank has cross-sectiona

Question

Not all water tanks are shaped like cylinders. Suppose a tank has cross-sectional area A(h) at height h. Then the volume of water up to height h is (V = int_{0}^{h}A(u)du) and so the Fundamental Theorem of Calculus gives dV/dh = A(h). It follows that

(dV/dt = (dV/dh) *(dh/dt) = A(h) dh/dt)

and so Torricelli's Law becomes

(A(h) dh/dt = -asqrt{2gh})

(a) Suppose the tank has the shape of a sphere with radius 2 meters and is initially half full of water. If the radius of the circular hole is 1 cm and we take g = 10 m/s^2, show that h satisfies the differential equation

((4h-h^2)dh/dt = -0.0001sqrt{20h})

(b) How long will it take for the water to drain completely?

Explanation / Answer

(a) By Bernoulli's law p+rho g h + (1/2) rho v^2 = constant. The pressure at the surface of the water is the same (atmospheric pressure) as the pressure on the water coming out of the hole. On the surface of the water the speed is zero and the height above the hole is h. At the hole, the height is zero and the speed is v so we must have
rho g h = (1/2) rho v^2 or, dividing through by rho
g h = (1/2) v^2, or, taking square roots of both sides and noting that v<0 as h decreases with time
(1) v = - sqrt (2gh)
The volume displaced by unit of time at the free surface has to be the same as the volume coming out of the hole per unit of time, as water is incompressible and mass is constant. If R is the radius of the sphere (R=2m) and ro is the radius of the hole (r0=1cm) then the radius of the circle which constitutes the free surface is r = sqrt (R^2 - (R-h)^2) = sqrt(2Rh-h^2) = sqrt(4h-h^2). The area of the free surface is therefore pi r^2 = pi (4h-h^2) and the volume displaced by unit of time is
pi (4h-h^2) (dh/dt). At the hole, if v is the velocity of the water, the volume displaced is pi ro^2 v so we must have
pi (4h-h^2) (dh/dt) =pi ro^2 v and therefore
v = (4h-h^2) (dh/dt) / ro^2
Replacing in formula (1) above we have
(4h-h^2) (dh/dt) / ro^2 = - sqrt(2gh). as ro=10^(-2) m and g = 10m/s2 we have therefore
(4h-h^2) (dh/dt) = - 10^(-4) sqrt(20h) as was required
(b) Dividing both sides of the equation above by sqrt(h) we have
(4 h^(1/2) - h^(3/2)) dh = - 10^(-4) sqrt(20) dt. Integrating between t=0 (when h=2) and t=T (when h=0) we have
4(2/3) 2^(3/2) - (2/5) 2^(5/2) = 10^(-4) sqrt(20) T, from where we can calculate that the time for the water to drain completely is
T = 11805.84 seconds or 3 hours 16 minutes and 45.84 seconds

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