A large open tank is filled with water to a depth of 15 m.
A spout located 10.0 m above the bottom of the tank is then opened as shown in the drawing. What is the volume flow rate from the spout? Assume that the radius of the spout is 2 cm.
To find the volume flow rate from the spout, we can use the principle of Torricelli's law, which states that the speed of fluid flowing out of an opening at the bottom of a tank is equal to the speed of an object in free fall from the same height.
Given:
- Depth of water in the tank, h = 15 m
- Height of spout from the bottom of the tank, H = 10.0 m
- Radius of spout, r = 2 cm = 0.02 m
First, let's find the speed of the fluid flowing out of the spout using Torricelli's law:
Speed of fluid = √(2gh)
Here, g is the acceleration due to gravity, which is approximately 9.8 m/s².
Substituting the values:
Speed of fluid = √(2 * 9.8 * 15) ≈ 19.80 m/s
Now, to find the volume flow rate, we can use the equation:
Volume flow rate = Area of spout * Speed of fluid
The cross-sectional area of the spout can be calculated using the formula:
Area = π * r²
Substituting the values:
Area = π * (0.02)² ≈ 0.00125664 m²
Finally, we can find the volume flow rate:
Volume flow rate = Area * Speed of fluid
Volume flow rate = 0.00125664 m² * 19.80 m/s ≈ 0.0249 m³/s
Therefore, the volume flow rate from the spout is approximately 0.0249 cubic meters per second.
To find the volume flow rate from the spout, we can use Torricelli's law, which states that the speed of fluid flowing out of a small hole is given by the formula:
v = √ (2gh)
Where:
v is the speed of fluid out of the spout
g is the acceleration due to gravity (9.8 m/s^2 on Earth)
h is the height of the water above the spout
In this case, the height of the water above the spout is 15 m, since the tank is filled to a depth of 15 m.
Using Torricelli's law, we can calculate the speed of the fluid coming out of the spout:
v = √ (2 * 9.8 * 15)
v ≈ 17.29 m/s
To find the volume flow rate (Q), we use the equation:
Q = Av
Where:
Q is the volume flow rate
A is the cross-sectional area of the spout
v is the speed of fluid out of the spout
The radius of the spout is given as 2 cm, which is equivalent to 0.02 m. The cross-sectional area of the spout can be calculated using:
A = πr^2
A = π * (0.02)^2
A ≈ 0.0012566 m^2
Now we can calculate the volume flow rate:
Q = 0.0012566 * 17.29
Q ≈ 0.0217 m^3/s
Therefore, the volume flow rate from the spout is approximately 0.0217 cubic meters per second.