# how can pv = nrT could be used to describe a water rocket experiment ?Watch

Thread starter 10 years ago
#1
as the questions says " how can pv = nrT and F = (mv-mu)/t could be used to describe a water rocket experiment where we had to find the optimum water load at which the rocket covered the longest horizontal distance.
My results are as follows :

Water/ml Range/m
No water 2.21
250 12.74
500 12.3
750 11.94
1000 11.8
1250 9.14
1500 1.73
1750 1.2
2000 0.002

Any help would be really appreciated.
Thanks
-Bharath
0
10 years ago
#2
Oh dear. It sounds like youare way over your head on this one. In order to be able to do the calculations you are going to have to learn calculus.
This is the rocket equation but with some fun changes.
You have to work out the rate of flow of water coming out of the nozzle wrt to pressue.
You then substitute the rate of flow with dM/dt and you have to work out the velocity of the water wrt flow rate (should be pretty much linear), this then is substituted into the v, velocity of the imparted reactionary mass. Then you 'simply' integrate to dv to find the final velocity of the rocket assuming zero gravitational potential. Then you differentiate wrt mass to find the optimum range.

Not a simple degree level calculation.
0
10 years ago
#3
you wont be expected to come up with a complete formaula describing how the distance is related to the amount of water, just to describe / appreciate how a few physical relationships are relevant.
Here's a few things to think about:

the force pushing the water out is proportional to the pressure of the air,
in most water rockets the release pressure is constant so this is unlikely to change each time.
What does change in your experiment is the amount of time for which the force acts on the rocket and the mass of the rocket.

Little water - small mass so big acceleration but force acts for a short time

Lots of Water - big mass so small acceleration but force acts for a longer time

Also bear in mind that the mass of the rocket gets less as time passes AND the pressure will reduce as the remaining air has more volume.

I have a student doing this now for his investigation and his latest problem is having the bottle vertical but getting the water to squirt out horizontally. If the bottle is at an angle then not all the water is ejected before it starts squirting air and all the theory goes to pot anyway.
0
10 years ago
#4
(Original post by Drummy)
I have a student doing this now for his investigation and his latest problem is having the bottle vertical but getting the water to squirt out horizontally. If the bottle is at an angle then not all the water is ejected before it starts squirting air and all the theory goes to pot anyway.
A good way to solve that would be to add extra length to the rocket, to give it a greater moment of inertia. Apart from that I can only suggest you look up 'inverted pendulum' in wikipedia.

Wait...your trying to shoot the rocket horizontally? That should be simple enough. Seperate the air from the liquid. Use a balloon to hold the water inside the bottle and then just pump air into the surroundings. It'll be a little bit complexified due to the extra valves and stuff, but it should work perfectly.
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Cranfield University
Cranfield Forensic MSc Programme Open Day Postgraduate
Thu, 25 Apr '19
• University of the Arts London
Open day: MA Footwear and MA Fashion Artefact Postgraduate
Thu, 25 Apr '19
• Cardiff Metropolitan University
Undergraduate Open Day - Llandaff Campus Undergraduate
Sat, 27 Apr '19

### Poll

Join the discussion

#### Have you registered to vote?

Yes! (394)
37.56%
No - but I will (80)
7.63%
No - I don't want to (73)
6.96%
No - I can't vote (<18, not in UK, etc) (502)
47.86%