Question

Water is flowing into the top of an open cylindrical tank (which has a diameter D) at a volume flow rate of Qi and the water flows out through a hole in the bottom at a rate of Qo. The tank is constructed from wood that is old and has become very porous so that water leaks out through the walls at a rate of q per unit of wetted surface area. The initial depth of the water in the tank is Z1. a) Derive an equation for the depth of water in the tank at any time.
b) If Qi = 10 gpm, Qo = 5 gpm, D = 5 ft, q = 0.1 gpm/ft2, and Z1 = 3 ft, is the level in the tank rising or falling?

Answer 1

Answer:

Z = 3 + 0.23t

The water level is rising

Explanation:

Please see attachment for the equation

Answer 2

Answer:

a) the expression is $z=\frac{(Q_{1}-Q_{0})t }{q\pi D+\frac{\pi }{4} D^{2} } +z_{1}$

b) z = 0.00057t + 3

the water level in the tank is rising

Explanation:

a) Given

Q₁ = flow rate into the tank

Q₀ = flow rate out of the tank

$q\frac{dA}{dt}$ = rate of water leakage

V = volume

$\frac{dV}{dt}$ = rate of accumulation

the expression is

$Q_{1} =Q_{0} +q\frac{dA}{dt} +\frac{dV}{dt} \\Q_{1} -Q_{0}=q\frac{d(\pi Dz)}{dt} +\frac{d(\frac{\pi }{4} D^{2}z) }{dt} \\(Q_{1} -Q_{0})dt=(q\pi D+\frac{\pi }{4} D^{2} )dz$

we apply the integral on both sides

$\int\limits^t_0 {(Q_{1} -Q_{0})dt=\int\limits^a_b {q\pi D+\frac{\pi }{4} D^{2} } \, } \, dz$

$z=\frac{(Q_{1}-Q_{0})t }{q\pi D+\frac{\pi }{4} D^{2} } +z_{1}$

b) Replacing values in the expression

$z=\frac{(0.022-0.011)t}{0.022*\pi *5+\frac{\pi }{4} *5^{2} } +3\\z=0.00057t+3$

the water level in the tank is rising