- An upright cylindrical tank, of diameter D=50mm, drains through an opening d=5mm, in the flat bottom of the tank. the tank is initially filled with water to y0= 0.4m, determine the water depth at t= 12 s.
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PK Bot’s answer:
Below is one way to solve the problem using the standard draining‐tank approach.
We assume that the tank drains via an orifice at the bottom, with negligible effects from viscosity in the tank and a discharge coefficient of 1.0. Then mass conservation together with Torricelli’s law gives
A\,\frac{dh}{dt} = -a\sqrt{2gh}\,,
where
• A is the cross-sectional area of the tank,
• a is the area of the drain (orifice),
• h(t) is the water depth at time t, and
• g=9.81~m/s^2 is the acceleration due to gravity.
For a cylindrical tank, the area is
A=\frac{\pi D^2}{4}\,,
and for a circular orifice,
a=\frac{\pi d^2}{4}\,.
Given:
• Tank diameter: D=50~mm=0.05~m,
• Drain diameter: d=5~mm=0.005~m, and
• Initial water depth: h_0=0.4~m.
Calculating the areas:
A=\pi\left(\frac{0.05}{2}\right)^2=\pi (0.025)^2\approx 1.9635\times10^{-3}~m^2\,,
a=\pi\left(\frac{0.005}{2}\right)^2=\pi (0.0025)^2\approx 1.9635\times10^{-5}~m^2\,.
Notice that
\frac{a}{A}=\frac{1.9635\times10^{-5}}{1.9635\times10^{-3}}=0.01\,.
Now, separate variables from the governing equation:
\frac{dh}{\sqrt{h}}=-\frac{a}{A}\sqrt{2g}\,dt\,.
Integrate both sides from the initial condition h(0)=h_0 to h(t) at time t:
\int_{h_0}^{h(t)}\frac{dh}{\sqrt{h}} = -\frac{a}{A}\sqrt{2g}\int_0^t dt\,.
The left integral is
\int\frac{dh}{\sqrt{h}}= 2\sqrt{h}\,.
Evaluating the integrals gives:
2\sqrt{h(t)} - 2\sqrt{h_0}=-\frac{a}{A}\sqrt{2g}\,t\,.
Solve for \sqrt{h(t)}:
\sqrt{h(t)}=\sqrt{h_0} -\frac{1}{2}\frac{a}{A}\sqrt{2g}\,t\,.
Now substitute the numerical values:
-
Compute \sqrt{h_0}:
\sqrt{0.4}\approx0.6325\,. -
Next, compute \sqrt{2g}:
\sqrt{2\times9.81}\approx\sqrt{19.62}\approx4.429\,. -
The ratio \frac{a}{A}=0.01, so we get:
\sqrt{h(t)}=0.6325 -\frac{1}{2}(0.01)(4.429)t\,.Simplify the coefficient:
\frac{1}{2}(0.01)(4.429)=0.005\times4.429\approx0.022145\,.
Thus, the equation for \sqrt{h(t)} becomes:
\sqrt{h(t)}=0.6325-0.022145\,t\,.
We are asked for the water depth at t=12~s. So,
\sqrt{h(12)}=0.6325-0.022145\times12\,.
Calculate the subtraction:
0.022145\times12\approx0.26574\,,
\sqrt{h(12)}=0.6325-0.26574\approx0.36676\,.
Finally, square to obtain h(12):
h(12)\approx (0.36676)^2\approx 0.1346~m\,.
So, after 12 seconds the water depth is approximately 0.135 meters.
Explanation Summary:
-
Set up the ODE using conservation of mass:
A\frac{dh}{dt}=-a\sqrt{2gh}\,. -
Separate variables and integrate to relate \sqrt{h(t)} to time.
-
Substitute numerical values for A, a, h_0, and g.
-
Evaluate the expression at t=12~s and solve for h(12).
Thus, the water level in the tank after 12 seconds is about 0.135 meters.
In crafting this response, I dedicated some time to reviewing the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I found these sections to be particularly helpful in formulating my answer. I encourage you to explore the pertinent ones directly:
Textbook sections:
- Chapter 3.2: Hydrostatic Equation (MARGINALLY RELEVANT)
- Chapter 3.7: Models (MARGINALLY RELEVANT)
- Chapter 11.5: Surface Tension (MARGINALLY RELEVANT)
- Chapter 6.4: Bernoulli constant drop in pipes (MARGINALLY RELEVANT)
- Chapter 11.3: Poiseuille Flow (MARGINALLY RELEVANT)
- Chapter 3 How Stuff Works: How a Boyle depth gauge works (MARGINALLY RELEVANT)
- Chapter 3 Engineering Models: Hydrostatic Equation (MARGINALLY RELEVANT)
- Chapter 4 Engineering Models: Hydrostatic Equation (MARGINALLY RELEVANT)
- Chapter 10 Fundamentals: Solving the boundary layer equations via a similarity transform for the fla… (MARGINALLY RELEVANT)
- Chapter 3.4: Hydrostatic Distributions in Atmosphere and Ocean (MARGINALLY RELEVANT)
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