Mechanics -- Converting volumetric flow rate to nozzle velocity

[moenste]

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?

 

[Doc Al]

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?

 

[BvU]

For part (b) your intuition is correct. You can give it a solid base by realizing that $F = {\Delta p\over \Delta t}$

 

[moenste]

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?

 

[Doc Al]

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!

 

[R4YY4N]

How did you get the force for part a)? You only had the mass and velocity

 

[Doc Al]

How did you get the force for part a)? You only had the mass and velocity

Consider the change in momentum of the water.