# Homework Help: Impact force of a water jet upon a target (fluid mechanics)

1. Oct 1, 2011

### jjack99645

This one is actually for a lab assignment. In the experiment you shoot a jet of water upward vertically at a variety of targets with different geometries (a flat plate, various cups). The target is suspended from a rod, and on top of the rod is a cup where you can add weights, held up by a spring. With the jet impacting the cup you determine the force two ways, first, by measuring the mass of the weight it takes to push the target apparatus back down to y=0 against the force of the water. Second, by calculating the theoretical value of the force given the density of water, nozzle cross-sectional area, volume of water through flowmeter, and time.

I think the formula given by the professor on the lab sheet is incorrect, or perhaps I don't understand it enough and am missing something.

In all we repeated the process 12 times, so I'll just post the data for one repetition. The process should be the same for each one.

1. The problem statement, all variables and given/known data

Problem:
1. Assuming the impact velocity is the same as nozzle exit velocity, calculate the theoretical impact velocity.
2. Using the theoretical equation for the impact force of a jet, calculate the impact force.
3. Compare the theoretical and measured values.

Data:
ρ = 1000 kg/m^3
θ = 90° (flat plate)
m = 159g (mass required to "zero" the target against the force)
vol = 0.018927 m^3
t = 56.40 s
D = 1cm

2. Relevant equations

F = ρA(V^2)cos(θ+1)
F = mg

where V is jet velocity, A is nozzle cross-sectional area, ρ is fluid density, θ is the angle of the target (given by professor), and g is the constant of gravitation.

3. The attempt at a solution

Experimental force = m*g = 159g(.001kg/g)(9.8m/s^2) = 1.5582 N

Jet area: A = ∏r^2 = ∏((1cm*.01 m/cm)/2)^2 = 7.854e-5 m^2

Jet velocity: V = ( vol / t ) / A = (0.018927 m^3 / 56.4 s) / (7.854e-5 m^2) = 4.2728 m/s

Impact force: F = ρA(V^2)cos(θ+1) = (1000kg/m^3)(7.854e-5 m^2)(4.2728 m/s)^2(cos(90° + 1)

- or -

= (1000kg/m^3)(7.854e-5 m^2)(4.2728 m/s)^2(cos(∏/2 + 1))

I have tried it both ways -- radians and degrees, it does not specify in the lab sheet -- and both of them are incorrect. For this one I get a value of -1.2065 N. This in itself wouldn't trouble me, except that for the next experiment, a 0° "dome" target I get a positive 0.46963N theoretical value versus a measured value of 1.7444N. The values alternate between positive and negative as we go across the chart.

I'm guessing the problem lies in the cos(θ+1). +1 what? It doesn't make sense to me. I've tried Google to see if an alternative form of the equation exists but no luck so far. For the 90° target, which is just a flat plate, the jet force should be the full force of the jet...or F = ρA(V^2). That means the cos() would have to go to 1, which would mean cos(∏) or, in this case, cos(θ + ∏/2). Don't see how +1 plays in.

Thanks for any help on this one in advance. I'm trying to get the data done so I can pass it along to my lab partners for the lab write-up but I've been scratching my head on this all day and they have no suggestions either, and, sine it's Saturday, the professor is not in his office. Hopefully I've explained it well enough to pinpoint where I've gone wrong.

2. Oct 1, 2011