Water Level in a Tank - 2

Classical Mechanics Level 3

Consider a tank of a cross-sectional area of A (when viewed from the top). Water flows into this tank from the top at a volumetric flow rate of $Q$ . The tank has a hole at the bottom which can be opened or closed by a valve. At time $t=0$ the valve opens allowing water to flow out. At that same instant, inflow into the tank begins. The area of this hole at the bottom is $A_o$ . The objective of this problem is to ensure that the steady-state height of the tank is 1 meter. To accomplish this, the following law governing the inflow is used.

$Q = Q_{out} + 0.1(1 - h)$

Here, $Q_{out}$ is the tank outflow at an instant when its height is $h$ . Enter your answer as the time (in seconds) required for the height to reach 0.9 meters.

Notes and assumptions:

$A = 0.15 \ m^2$ , $A_o = A/100$ , $g = 10 \ m/s^2$
$A_o<<A$ holds true
The cross-sectional area of the tank remains constant throughout its height.
There are no ambient pressure variations throughout the height of the tank.
The height of the tank can always be accurately and noiselessly measured at all instants of time.
The tank is empty at $t=0$

Bonus: Can you see and explain what I did here?

The answer is 3.454.

1 solution

Steven Chase
Dec 15, 2019

It is evident that when $h = 1$ , $Q = Q_{out}$ , meaning that the water height reaches an equilibrium state once that height is reached.

$\frac{dV}{dt } = Q - Q_{out} = 0.1(1-h) \\ \frac{dV}{dh } \frac{dh}{dt} = 0.1(1-h) \\ A \frac{dh}{dt} = 0.1(1-h) \\ \frac{dh}{dt} = \frac{2}{3} (1-h) \\ dt = \frac{3}{2} \frac{dh}{1-h}$

This results in:

$\Delta t = \frac{3}{2} \int_0^{0.9} \frac{dh}{1-h} \approx 3.454$

This result is easily checked numerically as well:

import math

# Constants

A = 0.15
A0 = A/100.0
g = 10.0

dt = 10.0**(-5.0)

#################################

# Initialize variables

h = 0.0  # height

V = 0.0   # volume
Vd = 0.1    # volume rate of change

t = 0.0
count = 0

#################################

# Run simulation

#print "(t/min) h (Vdout/Vdin)"

while h <= 0.9:

    V = V + Vd * dt   # numerical integration

    h = V/A           # water height

    u = math.sqrt(2.0*g*h)  # flow speed out of valve - Toricelli

    Vdout = A0*u   # outward flow rate
    Vdin = Vdout + 0.1*(1.0-h)       # inward flow rate

    Vd = Vdin - Vdout  # net flow rate

    t = t + dt
    count = count + 1

    #if count%10000 == 0:
        #print (t/min),h,(Vdout/Vdin)

#################################

print ""
print ""
print dt
print t
print h

Water Level in a Tank - 2

The answer is 3.454.

1 solution

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