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Water Rocket - Rate at which water exits

by NMS09
Tags: exits, rate, rocket, water
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NMS09
#1
Nov26-09, 02:54 PM
P: 2
1. The problem statement, all variables and given/known data

We're currently studying Momentum & Collisions and Work and Energy

Use algebra no Calculus

Id appreciate help in setting it up. I seem to be missing one equation since I keep coming up with two unknown variables.

A water rocket (2-liter half filled with water and pressurized with compressed air) can develop a thrust of 300 N.

Question: At what rate would water have to come out of the rocket to develop that thrust?

The jet of water emerging from the bottle has a diameter of 2.2 cm. (.022 m.).
The water density is 1,000 kg/cu. m.

The answer provided is 10.7 kg/s.


2. Relevant equations

Equations weve been using recently:

Momentum: p = mv

Change of momentum: F = delta p / delta time

Force: F = ma

Force & Impulse: F = mvf - mvi / delta time

Conservation of Momentum: m1v1,i + m2v2,i = m1v1,f + m1v1,f + m2v2,f Note: Those are subscripts after v1 , ve, etc. I need to learn how to use the formulas on this site.



3. The attempt at a solution

I calculated:

The area of the exit hole = 1.52 x 10^-3 m^2
Water in the rocket = 1 kg

I then tried the convservaton of momentum equation but could not get the 300 N into a form to use on one side of the equation. Everything I tried seem to need a "delta t" value, which I don't know and was not given.

The free body diagram has the forward thrust of the rocket created by the rearward explusion of the presurized water. The 300 N must overcome earth's gravity (-9.81 m / sec^2).

Sorry for not being more organized, but I've spent 3 or 4 hours going around on this one.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Andrew Mason
#2
Nov28-09, 11:29 PM
Sci Advisor
HW Helper
P: 6,679
The force depends on the speed of the water being ejected.

Assume that the water comes out at a speed v relative to the container. The rocket thrust is, as you have stated:

[tex]F = \Delta p/\Delta t = v\Delta m/\Delta t[/tex]

where [itex]\Delta m[/itex] is the mass of a volume of water ejected in time [itex]\Delta t[/itex].

Since the water is being ejected at the speed v, [itex]\Delta m[/itex] is [itex]\rho \dV = \rho A\Delta s = \rho Av\Delta t[/itex]. So the rate of mass expulsion is [itex]\Delta m/\Delta t = \rho Av[/itex]

Work out the expression for force and then find the speed that is needed to reach 300 N. From that you can calculate the rate at which mass has to be ejected.

AM
NMS09
#3
Nov29-09, 07:57 PM
P: 2
Andrew --

Thank you for your time and that direction. It helped a lot and I was able to work through the problem to get the correct answer.

The steps made sense and helped me understand what was going on.


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