Mechanics -- Converting volumetric flow rate to nozzle velocity

In summary: The mass of water is m = ρV, where ρ is the density and V is the volume. Since the water falls vertically, the vertical component of the velocity is zero, so the change in momentum is given by Δp = mΔv = ρVΔv = ρ(0.044 m3 s-1)(8.75 m s-1) = 385 N.
  • #1
moenste
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Homework Statement


A hose with a nozzle 80 mm in diameter ejects a horizontal stream of water at a reate of 0.044 m3 s-1. With what velocity will the water leave the nozzle? What will be the force exerted on a vertical wall situated close to the nozzle and at right-angles to the stream of water, if, after hitting the wall, (a) the water falls vertically to the ground, (b) the water rebounds horizontally? (g = 10 m s -2, density of water = 1000 kg m -3)

Answers: velocity = 8.75 m s-1, (a) 385 N, (b) 770 N.

Homework Equations


Mechanics equations.

The Attempt at a Solution


I tried to get velocity with: V*0.08 m = 0.044 m3 s-1 but got only 0.55. I did try some other various calculations using the given diameter and the rate of ejection but do not think that they are of any value. Really lost on finding velocity.

For a and b:
(a) 0.044 ms * 1000 kgm = 44 kg is being ejected in one second.
44 kg * 8.75 ms (velocity from the answer) = 385 N
(b) I suppose that if the force is so strong that water rebounds backwards, the force should be twice as much = 385 N * 2 = 770 N. The answer fits the book answer, but I am not sure whether my logic is correct.

So in sum: any help with finding velocity? Really lost. And logic for part b is correct? Or I should have calculated it in an other way?
 
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  • #2
moenste said:
I tried to get velocity with: V*0.08 m = 0.044 m3 s-1 but got only 0.55.
Check your units. The left side should not be speed times diameter, but speed times what?
 
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  • #3
For part (b) your intuition is correct. You can give it a solid base by realizing that ##F = {\Delta p\over \Delta t}##
 
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  • #4
Doc Al said:
Check your units. The left side should not be speed times diameter, but speed times what?
In a similar problem from the book the author multiplies 2 ms-1 = 2 m and a cross-sectional area of 0.03 m2, so 2 m * 0.03 m2 = 0.06 m3 --- volume of water leaving the pipe in 1 second.

In my case if I take the book answer of 8.75 ms-1 = 8.75 m then: 8.75 m * X = 0.044 m3s-1. X = 5 * 10-3 m2. I googled cross-sectional area and got this video:

So:
80 mm diameter = 80 m / 2 = 40 mm radius.
40 mm / 1000 = 0.04 m.
A = πr2
A = π*0.042 = 5.0265...*10-3 m2

So finding v:
v * 5.0265...*10-3 m2 = 0.044 m3
v = 8.75 ms-1

Correct?
 
  • #5
moenste said:
So:
80 mm diameter = 80 m / 2 = 40 mm radius.
40 mm / 1000 = 0.04 m.
A = πr2
A = π*0.042 = 5.0265...*10-3 m2

So finding v:
v * 5.0265...*10-3 m2 = 0.044 m3
v = 8.75 ms-1

Correct?
Perfect!
 
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  • #6
How did you get the force for part a)? You only had the mass and velocity
 
  • #7
R4YY4N said:
How did you get the force for part a)? You only had the mass and velocity
Consider the change in momentum of the water.
 

1. How do you calculate the nozzle velocity from a given volumetric flow rate?

The nozzle velocity can be calculated by dividing the volumetric flow rate by the cross-sectional area of the nozzle. This will give you the average velocity of the fluid exiting the nozzle.

2. What units are typically used for volumetric flow rate and nozzle velocity?

Volumetric flow rate is usually measured in cubic meters per second (m3/s) or liters per second (L/s). Nozzle velocity is typically measured in meters per second (m/s).

3. How does the nozzle design affect the conversion between volumetric flow rate and nozzle velocity?

The design of the nozzle, specifically the shape and size of the opening, can affect the velocity of the fluid exiting the nozzle. A smaller opening will result in a higher velocity for the same volumetric flow rate.

4. Can the conversion between volumetric flow rate and nozzle velocity be applied to all fluids?

The conversion formula can be applied to most fluids, but it may not be accurate for non-ideal fluids or those with high viscosities. In these cases, the actual nozzle velocity may be lower than the calculated value.

5. How is the conversion between volumetric flow rate and nozzle velocity important in practical applications?

This conversion is important in many engineering and industrial applications, such as in pumps, turbines, and nozzles for spray applications. It allows for the calculation of the velocity and pressure of a fluid as it exits a nozzle, which is crucial for optimizing the performance of these systems.

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