You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work on water resources project. The department will be overseeing the construction of a dam to create a large fresh water lake that will be approximately 18 meters deep. A horizontal pipe 1.2 meters long and 4 cm in diameter will pass through the dam at a depth of 7 meters to allow for release of the water in emergencies and for sampling. In normal situations, a plug will secure the pipe opening.
a) What will be the total force on the left side of the plug?
FLEFT = N
b) What is the total force on the right side of the plug?
FRIGHT = N
c) What is the gauge pressure on the plug?
Pgauge = Pa
d) If the plug were to be removed, how much water would flow out in 3 hours?
ΔVOL = m3
a) A*[Po + (rho) g H]
Po is atmospheric pressure
rho is the mass density of water, 1000 kg/m^3
H is the water height above the hole, 7 m
A is the pipe inner area, pi*D^2/4, in square meters
b) A* Po
c) (rho) g H
d) A*(10,800 s)*V
where V is the speed of wather leaving the pipe.
Use Bernoulli's principle to get V.
http://en.wikipedia.org/wiki/Bernoulli's_principle
To answer these questions, we need to consider the hydrostatic pressure exerted by the water at different depths within the lake.
a) The total force on the left side of the plug can be calculated using the formula: F = P * A, where P is the pressure and A is the cross-sectional area of the pipe.
First, let's find the pressure at a depth of 7 meters. The hydrostatic pressure is given by the equation P = ρ * g * h, where ρ is the density of water, g is the acceleration due to gravity, and h is the height.
Density of water (ρ) ≈ 1000 kg/m^3
Acceleration due to gravity (g) ≈ 9.8 m/s^2
Height (h) = 7 m
P = (1000 kg/m^3) * (9.8 m/s^2) * (7 m) = 68,600 Pa
Now, let's calculate the cross-sectional area of the pipe:
Radius (r) = diameter / 2 = 0.04 m / 2 = 0.02 m
Area (A) = π * r^2 = 3.14159 * (0.02 m)^2 ≈ 0.00126 m^2
FLEFT = P * A = 68,600 Pa * 0.00126 m^2 = 86.436 N
Therefore, the total force on the left side of the plug is approximately 86.436 N.
b) The total force on the right side of the plug will be equal to the force on the left side, as the pressures on both sides of the plug are equal when it is secured.
FRIGHT = FLEFT = 86.436 N
c) The gauge pressure on the plug is the pressure difference between the inside and outside of the pipe. The gauge pressure is zero when the plug is secured. So:
Pgauge = 0 Pa
Therefore, the gauge pressure on the plug is zero.
d) If the plug were removed, we can calculate the volume of water that would flow out using the formula:
ΔVOL = A * h * t, where A is the cross-sectional area of the pipe, h is the height difference (depth), and t is the time.
Height difference (h) = 18 m - 7 m = 11 m
Time (t) = 3 hours = 3 * 60 * 60 seconds
Using the same area (A) calculated earlier:
ΔVOL = 0.00126 m^2 * 11 m * (3 * 60 * 60 s) = 166.32 m^3
Therefore, if the plug were removed, approximately 166.32 m^3 of water would flow out in 3 hours.
To answer these questions, we need to consider the principles of fluid pressure and hydrostatics.
a) To determine the total force on the left side of the plug, we can use the formula F = P x A, where F is the force, P is the pressure, and A is the area. The area of the pipe can be calculated using the formula A = πr^2, where r is the radius of the pipe. Given the diameter of the pipe (4 cm), we can calculate the radius as 2 cm or 0.02 m.
First, let's calculate the area of the pipe:
A = πr^2 = π(0.02)^2 = 0.001256 m^2
Now, to find the force on the left side of the plug, we need to calculate the pressure at that depth. The pressure at a certain depth in a fluid is given by the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
The density of water is approximately 1000 kg/m^3, and the depth at which the plug is located is 7 meters. The acceleration due to gravity can be approximated as 9.8 m/s^2.
P = ρgh = (1000 kg/m^3)(9.8 m/s^2)(7 m) = 68,600 Pa
Now, we can calculate the force on the left side of the plug:
FLEFT = P x A = (68,600 Pa)(0.001256 m^2) = 86.3 N
Therefore, the total force on the left side of the plug is approximately 86.3 N.
b) The total force on the right side of the plug will be equal to the force on the left side since the system is in equilibrium. Therefore, the total force on the right side of the plug is also approximately 86.3 N.
c) The gauge pressure on the plug can be calculated by subtracting the atmospheric pressure from the absolute pressure. Gauge pressure is the pressure measured relative to atmospheric pressure.
The atmospheric pressure at sea level is approximately 101,325 Pa. Since the gauge pressure is relative to the atmospheric pressure, the gauge pressure can be calculated as:
Pgauge = P - Patm = 68,600 Pa - 101,325 Pa = -32,725 Pa
Therefore, the gauge pressure on the plug is approximately -32,725 Pa.
d) To determine the amount of water that would flow out if the plug were removed in 3 hours, we need to calculate the volume of water that would flow through the pipe.
The volume of water flowing through a pipe in a given time can be calculated using the formula ΔVOL = A x v x Δt, where ΔVOL is the change in volume, A is the cross-sectional area of the pipe, v is the velocity of the water, and Δt is the time.
First, we need to find the velocity of the water. The velocity can be calculated using Bernoulli's equation, assuming no energy losses due to friction:
P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2
Since the section where the water flows (pipe diameter) is constant, the heights (h1 and h2) cancel out, and the equation simplifies to:
P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2
At the left side of the plug (P1), the pressure is equal to the atmospheric pressure (101,325 Pa). At the right side of the plug (P2), the pressure is equal to the gauge pressure (-32,725 Pa). Neglecting the kinetic energy change (v1 = v2) and canceling the height term, the equation simplifies to:
101,325 Pa = -32,725 Pa + 1/2ρv^2
Solving for v^2, we get:
1/2ρv^2 = 134,050 Pa
v^2 = (134,050 Pa) / ρ
Now, we can calculate the velocity of water:
v^2 = (134,050 Pa) / (1000 kg/m^3) = 134.05 m^2/s^2
v ≈ 11.58 m/s
Now, we can calculate the volume of water that would flow through the pipe in 3 hours:
ΔVOL = A x v x Δt = (0.001256 m^2)(11.58 m/s)(3 hours x 3600 seconds/hour) = 167.65 m^3
Therefore, if the plug were removed, approximately 167.65 m^3 of water would flow out in 3 hours.