A jet on a water feature shoots a column of water H = 80 ft high. The nozzle is located just above the surface of a reflecting pool and has a diameter of 1 in. A submersible pump with a 4 in dia outlet pipe and an 6 in. diameter inlet pipe located D = 2.5 ft below the surface of the pool operates the jet. For this configuration,
a) Derive an expression for the velocity of the water at the nozzle exit and simplify accordingly
b) Calculate the velocity of the water at the nozzle exit
c) Derive an expression for pressure at a point just inside the pump intake pipe and simplify accordingly
d) Calculate the pressure at a point just inside the pump intake pipe(γwater = 62.4 lb/ft^3)
a) To derive an expression for the velocity of the water at the nozzle exit, we can use the principle of conservation of energy. At the nozzle exit, the water has reached its maximum height, so its potential energy is at a maximum. This means that all of the pump's work goes into raising the water to that height.
The work done by the pump is given by:
Work = Force * Distance
The force is equal to the pressure at the nozzle exit (P) multiplied by the area of the nozzle exit (A). The distance is the height of the water column, H.
Work = P * A * H
Since the work done is equal to the change in potential energy, we can equate the two expressions:
P * A * H = m * g * H
Where m is the mass of the water and g is the acceleration due to gravity.
The mass of the water is equal to its density (ρ) multiplied by its volume (V). The volume of the water column is equal to the cross-sectional area of the column (π * radius^2) multiplied by its height (H).
m = ρ * V = ρ * (π * r^2 * H)
Plugging this back into the equation:
P * A * H = (ρ * π * r^2 * H) * g
We can simplify this expression by canceling out the height:
P * A = ρ * π * r^2 * g
Finally, we solve for the velocity (v) at the nozzle exit by dividing both sides by the area (A):
v = (ρ * π * r^2 * g) / A
b) To calculate the velocity of the water at the nozzle exit, we need to substitute the values into the derived expression. The radius of the nozzle exit is half of the diameter, so r = 0.5 in = 0.0417 ft. The area of the nozzle exit is:
A = π * (0.0417 ft)^2 = π * 0.001739 ft^2
The density of the water is given as 62.4 lb/ft^3. The acceleration due to gravity is approximately 32.2 ft/s^2.
v = (62.4 lb/ft^3 * π * (0.0417 ft)^2 * 32.2 ft/s^2) / (π * 0.001739 ft^2)
Simplifying the expression, we find:
v ≈ 53.79 ft/s
c) To derive an expression for the pressure at a point just inside the pump intake pipe, we can use Bernoulli's equation. Bernoulli's equation states that the pressure at any point in a fluid is a combination of the static pressure (P), the dynamic pressure (0.5 * ρ * v^2), and the gravitational potential pressure (ρ * g * h), where ρ is the density of the fluid, v is the velocity, and h is the height above a reference level.
Since the point is just inside the pump inlet pipe, we can assume that the velocity is negligible and the height above the reference level is zero. Therefore, the pressure at this point is equal to the static pressure.
P = ρ * g * h
d) To calculate the pressure at a point just inside the pump intake pipe, we need to substitute the values into the derived expression. The density of the water is given as 62.4 lb/ft^3 and the acceleration due to gravity is approximately 32.2 ft/s^2.
P = 62.4 lb/ft^3 * 32.2 ft/s^2 * 2.5 ft
Simplifying the expression, we find:
P ≈ 4981.4 lb/ft^2
a) To derive the expression for the velocity of water at the nozzle exit, we can use the principle of conservation of energy. Assuming there is no energy loss due to friction or other factors, the total energy at the nozzle exit will be equal to the total energy at the pump intake.
The total energy of a fluid can be represented as the sum of its kinetic energy and its potential energy. The kinetic energy per unit mass is given by the formula v^2/2, where v is the velocity of the fluid. The potential energy per unit mass is given by the formula g*z, where g is the acceleration due to gravity and z is the elevation of the fluid above a reference point.
At the nozzle exit, the elevation z is equal to the height of the water column H. The velocity v can be determined by the flow rate Q and the cross-sectional area of the nozzle A. The flow rate Q can be calculated as the product of the velocity v and the cross-sectional area of the nozzle A, given by Q = v*A.
b) To calculate the velocity of the water at the nozzle exit, we need to determine the flow rate Q and the cross-sectional area of the nozzle A. The cross-sectional area can be calculated as the product of pi (π) and the square of the radius of the nozzle (diameter/2).
c) To derive an expression for the pressure at a point just inside the pump intake pipe, we can use the Bernoulli's equation, which relates the pressure, velocity, and elevation of a fluid at two different points. Bernoulli's equation states that the sum of the pressure, velocity head, and elevation head at one point is equal to the sum of the pressure, velocity head, and elevation head at another point in a constant streamline flow.
In this case, we can treat the nozzle exit and the pump intake as two points. The velocity head at a point is given by the formula v^2/2g, where v is the velocity of the fluid and g is the acceleration due to gravity. The elevation head at a point is given by the formula z, where z is the elevation of the fluid above a reference point. Since the nozzle exit has a higher elevation than the pump intake, the elevation head at the nozzle exit is greater than the elevation head at the pump intake.
d) To calculate the pressure at a point just inside the pump intake pipe, we need to determine the velocity and elevation of the fluid at that point. The velocity can be determined by the flow rate and the cross-sectional area of the pump intake pipe, using the same formula as in part b. The elevation can be determined by subtracting the depth D of the pump inlet pipe from the water surface elevation. The water surface elevation is assumed to be the reference point for calculating elevations. The pressure at the pump intake can be calculated using the Bernoulli's equation, as described in part c.
a) To derive an expression for the velocity of the water at the nozzle exit, we can use the principle of conservation of energy. The water column gains potential energy as it rises to a height of 80 ft. This gain in potential energy is equal to the loss of kinetic energy at the nozzle exit.
The potential energy gained by the water column can be calculated using the formula: PE = m * g * h, where m is the mass of water, g is the acceleration due to gravity, and h is the height gained.
The kinetic energy lost by the water at the nozzle exit can be calculated using the formula: KE = 0.5 * m * v^2, where v is the velocity of the water at the nozzle exit.
Since the mass of water is the same in both cases, we can equate the potential energy gained to the kinetic energy lost:
m * g * h = 0.5 * m * v^2
Simplifying and canceling out the mass, we get:
g * h = 0.5 * v^2
Now, we can solve this equation for the velocity of the water at the nozzle exit.
b) To calculate the velocity of the water at the nozzle exit, we need to substitute the given values into the derived equation and solve for v.
Using the given values:
g = acceleration due to gravity = 32.2 ft/s^2
h = height gained by water column = 80 ft
Substituting these values into the equation, we get:
32.2 ft/s^2 * 80 ft = 0.5 * v^2
Simplifying and solving for v, we get:
v^2 = 32.2 ft/s^2 * 80 ft * 2
v^2 = 5152 ft^2/s^2
v = √5152 ft/s
v ≈ 71.82 ft/s
Therefore, the velocity of the water at the nozzle exit is approximately 71.82 ft/s.
c) To derive an expression for the pressure at a point just inside the pump intake pipe, we can use Bernoulli's equation. Bernoulli's equation relates the pressure, velocity, and elevation of a fluid flowing in a streamline.
Bernoulli's equation is given as:
P + 0.5 * ρ * v^2 + ρ * g * h = constant
Where P is the pressure, ρ is the density of the fluid, v is the velocity of the fluid, and h is the elevation of the fluid.
Since we are interested in the pressure at a point just inside the pump intake pipe, where the elevation is D (2.5 ft below the surface of the pool), we can set the reference elevation to be the surface of the pool.
So, the equation becomes:
P + 0.5 * ρ * v^2 + ρ * g * (D + h) = constant
Since we are looking for the pressure just inside the pump intake pipe, where the water velocity is low (almost stagnant), we can neglect the velocity term in the equation, making it:
P + ρ * g * (D + h) = constant
Simplifying and rearranging the equation, we get the expression for pressure:
P = constant - ρ * g * (D + h)
d) To calculate the pressure at a point just inside the pump intake pipe, we need to substitute the given values into the derived expression for pressure.
Using the given values:
ρ (density of water) = 62.4 lb/ft^3
g (acceleration due to gravity) = 32.2 ft/s^2
D (elevation below surface) = 2.5 ft
h (height gained by water column) = 80 ft
Substituting these values into the expression, we get:
P = constant - 62.4 lb/ft^3 * 32.2 ft/s^2 * (2.5 ft + 80 ft)
Simplifying the expression, we get:
P ≈ constant - 211777.28 lb/ft^2
Since the constant term is not given in the problem, we cannot calculate the exact pressure value. However, you can use this expression and the value of the constant in subsequent calculations