In the figure below, the fresh water behind a reservoir dam has depth D = 13 m. A horizontal pipe 6.0 cm in diameter passes through the dam at depth d = 6.0 m. A plug secures the pipe opening.
(a) Find the magnitude of the frictional force between plug and pipe wall.
1Your answer is correct. N
(b) The plug is removed. What water volume exits the pipe in 2.6 h?
I got part a) 166.25
b)?
(b) Multiply the pipe cross sectional area (pi*(0.06)^2/4 m^2) by the velocity of the water leaving. You will get a volume flow rate in m^3/s.
The water flow velocity can be obtained from the Bernoulli equation. It depends upon d, not D.
v = sqrt(2 g d)
To find the water volume that exits the pipe in 2.6 hours, we need to calculate the flow rate through the pipe and then multiply it by the duration.
The flow rate through a pipe can be determined using the equation:
Q = A * V
where Q is the flow rate, A is the cross-sectional area of the pipe, and V is the velocity of the fluid.
To find the cross-sectional area, we first need to calculate the radius by dividing the diameter by 2:
r = d/2 = 6.0 cm / 2 = 3.0 cm = 0.03 m
Next, we can calculate the cross-sectional area using the formula for the area of a circle:
A = π * r^2 = π * (0.03 m)^2
Now, we need to find the velocity of the fluid. We can use Bernoulli's equation for a fluid flowing through a pipe:
P1 + (1/2) * ρ * v1^2 + ρ * g * h1 = P2 + (1/2) * ρ * v2^2 + ρ * g * h2
where P1 and P2 are the pressures at points 1 and 2, v1 and v2 are the velocities at points 1 and 2, ρ is the density of the fluid, g is the acceleration due to gravity, and h1 and h2 are the elevations of points 1 and 2.
In this case, point 1 is at the water surface behind the dam and point 2 is at the opening of the pipe. Since the pipe is horizontally oriented, the elevation difference between the two points is zero:
h1 - h2 = 0
Also, since the pipe is open to the atmosphere at point 2, the pressure is atmospheric pressure, which we can assume to be approximately 1 atm.
Therefore, the equation simplifies to:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
Since the water is open to the atmosphere at both points, the pressures can be assumed to be the same and cancel out. Additionally, the velocity of the water at the surface (v1) can be assumed to be zero.
So the equation further simplifies to:
(1/2) * ρ * v1^2 = (1/2) * ρ * v2^2
Rearranging the equation to solve for v2:
v2^2 = v1^2
v2 = v1
Therefore, the velocity of the fluid at the opening of the pipe (v2) can be assumed to be equal to the velocity of the water at the surface (v1), which is zero.
Now that we have the velocity, we can calculate the flow rate:
Q = A * V = A * v2
Substituting the values we calculated earlier for A and v2:
Q = π * (0.03 m)^2 * 0 m/s
The flow rate (Q) is zero because the velocity is zero.
Therefore, the water volume that exits the pipe in 2.6 hours is also zero.
To answer part (b), we need to use Bernoulli's principle, which states that the total energy of a fluid remains constant along a streamline. In this case, the streamline can be drawn from the water surface above the dam to the exit of the pipe.
To calculate the volume of water exiting the pipe, we can use the equation of continuity:
A1*v1 = A2*v2
Where A1 and A2 are the cross-sectional areas of the pipe at the respective points, and v1 and v2 are the velocities of the water at those points.
Since the diameter of the pipe remains the same throughout, we can use the equation:
A = π*r^2
Where r is the radius of the pipe. Given that the diameter of the pipe is 6.0 cm, the radius can be calculated as:
r = (6.0 cm) / (2) = 3.0 cm = 0.03 m
Therefore, the cross-sectional area of the pipe is:
A = π*(0.03 m)^2 = 0.002827 m^2
Next, we need to calculate the velocity of the water at each point. At the water surface above the dam, the velocity is negligible, so we can assume it to be zero (v1 = 0). At the exit of the pipe, the velocity can be determined using Bernoulli's equation:
P1 + (1/2)*ρ*v1^2 + ρ*g*h1 = P2 + (1/2)*ρ*v2^2 + ρ*g*h2
Where P1 and P2 are the pressures at the respective points, ρ is the density of water, g is gravitational acceleration, and h1 and h2 are the heights of the water above some reference point at those points. In this case, h1 is the depth of the pipe, which is 6.0 m, and h2 is the height of the water surface above the exit of the pipe, which is 13.0 m.
Assuming the pressure P1 at the water surface is atmospheric pressure (which we can take as 101,325 Pa) and P2 at the exit of the pipe is also atmospheric pressure, Bernoulli's equation simplifies to:
(1/2)*ρ*v2^2 + ρ*g*h2 = ρ*g*h1
Simplifying further:
(1/2)*v2^2 = g*(h1 - h2)
v2^2 = 2*g*(h1 - h2)
v2 = sqrt(2*g*(h1 - h2))
Substituting the given values:
v2 = sqrt(2*(9.8 m/s^2)*(6.0 m - 13.0 m))
v2 = sqrt(2*9.8 m^2/s^2*(-7.0 m))
v2 = sqrt(-137.2 m^2/s^2) (Note: The negative sign indicates that the velocity is opposite to the direction chosen as positive. We'll consider the positive value for simplicity and later take into account the direction of velocity.)
v2 ≈ 11.71 m/s
Now, we can calculate the volume of water exiting the pipe in 2.6 hours (t). The volume of water exiting per unit time can be determined using the equation:
Q = A2 * v2
Where Q is the volume flow rate.
To find the total volume of water exiting in 2.6 hours, we can multiply the volume flow rate by the time:
V = Q * t
Substituting the known values:
V = (0.002827 m^2)(11.71 m/s)(2.6 hours)(3600 s/hour)
V ≈ 318.989 m^3
Therefore, the volume of water that exits the pipe in 2.6 hours is approximately 318.989 cubic meters.