water flows at 5 gal/min through a small circular hole in the bottom of a large tank. assuming that the water in the tank approaces the hole radially, find te velocity in the tank at 2, 4, and 8 in from the hole.
This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is one way to analyze the problem. We assume that the small orifice acts like a hemispherical point sink on the bottom wall of the tank. In that case, the streamlines in the tank will be those of a hemispherical sink. By applying conservation of mass over a hemispherical surface of radius r (with area A=2\pi r^2), the radially inward (area‐averaged) velocity is
u(r)=\frac{Q}{2\pi r^2}\,.
Because the given flow rate is 5 gal/min, we first convert it to ft³/s. Using
1~\text{gal}=0.13368~\text{ft}^3\quad \text{and}\quad 1~\text{min}=60~\text{s}\,,
we have
Q=5\times0.13368/60\approx0.01114~\text{ft}^3/\text{s}\,.
The distances from the orifice are given in inches and should be converted to feet:
• 2 in =2/12\approx0.167 ft
• 4 in =4/12\approx0.333 ft
• 8 in =8/12\approx0.667 ft
Now, we compute the velocity at each specified location.
- For r=0.167 ft:
u(0.167)=\frac{0.01114}{2\pi (0.167)^2}\,.
Compute the denominator:
(0.167)^2\approx0.0278\quad\Longrightarrow\quad 2\pi (0.0278)\approx0.1745\,.
Thus,
u(0.167)\approx \frac{0.01114}{0.1745}\approx0.0639~\text{ft/s}\,.
- For r=0.333 ft:
(0.333)^2\approx0.1111,\quad 2\pi (0.1111)\approx0.6981\,,
so
u(0.333)\approx \frac{0.01114}{0.6981}\approx0.0160~\text{ft/s}\,.
- For r=0.667 ft:
(0.667)^2\approx0.4444,\quad 2\pi (0.4444)\approx2.7925\,,
thus
u(0.667)\approx \frac{0.01114}{2.7925}\approx0.0040~\text{ft/s}\,.
Assumption: The analysis uses the hemispherical sink model because the orifice is small and lies on the bottom wall. In that scenario the radially converging flow occupies a half‐space (a hemisphere) so that the conservation of mass over a hemispherical surface yields the factor 2\pi r^2.
The answers are:
• At 2 in (0.167 ft): u\approx0.064~\text{ft/s}
• At 4 in (0.333 ft): u\approx0.016~\text{ft/s}
• At 8 in (0.667 ft): u\approx0.0040~\text{ft/s}
To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked their significance below. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 3 Part 1 (HIGHLY RELEVANT)
- Chapter 9.3: Why We Use Irrotational Flow Analysis Part 5 (HIGHLY RELEVANT)
- Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 2 (HIGHLY RELEVANT)
- Chapter 11.3: Poiseuille Flow Part 3 (RELEVANT)
- Chapter 4 Engineering Models: Hydrostatic Equation (RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.