Water is descending in a vertical pipe of diameter 7 cm and open to the atmosphere. At a lower point the water flows into a smaller pipe of diameter .84 cm. At a certain instant the depth of the water just above the narrowing point is 30 cm and the water is moving at 135 cm/s. How fast will water be going as it exits a small hole just below the narrowing point?
To determine the speed of water as it exits the small hole just below the narrowing point, you can apply the principle of conservation of mass and Bernoulli's equation.
First, let's find the cross-sectional area of the pipe at the wider and narrower points.
The cross-sectional area of a pipe can be calculated using the formula:
A = π * r^2,
where A is the cross-sectional area and r is the radius of the pipe.
For the wider point, the diameter is given as 7 cm, so the radius (R) would be half of the diameter: R = 7 cm / 2 = 3.5 cm = 0.035 m.
For the narrower point, the diameter is given as 0.84 cm, so the radius (r) would be half of the diameter: r = 0.84 cm / 2 = 0.42 cm = 0.0042 m.
Using the formula for the cross-sectional area, we find:
A_wider = π * (0.035)^2
A_wider ≈ 0.00385 m^2
A_narrower = π * (0.0042)^2
A_narrower ≈ 0.000055 m^2
Next, let's calculate the speed of the water just above the narrowing point using the given information.
The depth of the water just above the narrowing point is given as 30 cm, which can be converted to meters: 30 cm = 0.30 m.
Now we can apply Bernoulli's equation, which states that the total pressure in a fluid system is constant along a streamline:
P + (1/2) * ρ * v^2 + ρ * g * h = constant,
where P is the pressure, ρ (rho) is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the height of the fluid.
In this case, we can neglect the pressure term since the pipe is open to the atmosphere. Also, the acceleration due to gravity (g) remains constant. Therefore, the equation becomes:
(1/2) * ρ * v^2 + ρ * g * h = constant.
But since the cross-sectional area changes, the velocity of the fluid (v) will vary. However, the product of the velocity and the cross-sectional area (A * v) will remain constant:
A_wider * v_wider = A_narrower * v_narrower.
Now we can solve for v_narrower, the velocity of the water as it exits the small hole.
v_narrower = (A_wider * v_wider) / A_narrower.
Substituting the given values:
v_narrower = (0.00385 m^2 * 135 cm/s) / 0.000055 m^2.
Note that we need to convert the velocity from cm/s to m/s to maintain consistency in units:
v_narrower = (0.00385 m^2 * 1.35 m/s) / 0.000055 m^2
v_narrower ≈ 93.409 m/s.
Therefore, the water will be flowing at approximately 93.409 m/s as it exits the small hole just below the narrowing point.