water rockets

[Sasso SP]

Bye to all,
I wanted to ask for your help in designing a water rocket.
The rockets I saw on the internet are very rudimentary so I thought I could use the principle but to review it as a whole by giving a somewhat more technical approach. What do you say? ?
I should like to thank you for your comments. Hi.

 

[zeigs]

What do you mean "more technical"? remote control? greater autonomy? Trajectory control? load capacity? application to a saturn? Pilot on board? nanotechnological materials?

more seriously, if you want a technical approach, I'd say start by defining what you want to make him do to this rocket. his because to this world, in practice. As much as I know you can try to go into space. then we sell the project to the Chinese.. .

 

[Sasso SP]

What do you mean "more technical"? remote control? greater autonomy? Trajectory control? load capacity? application to a saturn? Pilot on board? nanotechnological materials?

more seriously, if you want a technical approach, I'd say start by defining what you want to make him do to this rocket. his because to this world, in practice. As much as I know you can try to go into space. then we sell the project to the Chinese.. .

hello zeigs my intent is not to go on the moon but to perform a design exercise.
the end of the project must be to realize a water rocket that reaches a speed of 300 km/h:biggrin:
the study will focus on that given: choice of materials, pressure calculation, calculation of water and room volume, water output speed, mechanisms to be used for the vent, aerodynamics of the rocket, effective possibility to achieve this purpose....ecc
I understand that it might sound stupid but I repeat that it's just an exercise, what do you say? ? ?

 

[maxopus]

I know all your rockets

 

[Exatem]

I know all your rockets

We could install a battery on the laodice. the raw material is not missing...

 

[zeigs]

I guess the judge is a boat. But I think rock wants to use a pressure tank
I'd say that anyway:
-stim the total mass of the rocket (dry rocket plus full tank)
-calculating that speed must have the outgoing water to give the boost to the rocket (it is a budget of energy, in fact not trivial but since it is an exercise you can beat a little to reason above)
-calculating which pressure must be in the tank to have sufficient potential energy
-calculate the nozzle to transform water pressure into speed (and here I want you)
-size everything (I deliberately missed the aerodynamics because at this stage I see it a secondary problem) and try to estimate more or less the various weights
-Recommend with new data
once the "motor" is baptised in principle, keeping a certain margin for not having to start over every time I add a fine, I start looking at details
regarding the materials I use steel and aluminum, so I can't give you advice about it

 

[Sasso SP]

thanks zeigs thanks for the advice,
Just one thing, I don't know what you mean by "calculating the nozzle to turn water pressure into speed"?
I thought I would use a simple valve mounted on a "tape" calibrated according to the required vent pressure.

 

[stevie]

the nozzle converts the piezometric energy of a fluid in pressure into kinetic energy.
In the case of water I think reasonable the use of a simple convergent nozzle, unlike that of the "real" rockets where a convergent-divergent nozzle is used, this because the exhaust gases reach the speed of the sound and in order to be further accelerated it is necessary to use a divergent trait.
nozzle and push calculation is not complicated. the problem lies in the fact that your rocket does not ensure a constant drive, but by downloading water the pressure in the tank drops and consequently also the water output speed. to this add that even the mass of the rocket decreases therefore the dynamics of the rocket must be evaluated in numerical way, with a nice program that at very small intervals evaluate the generated thrust and the mass of the object.
However, your idea is not so crazy if you consider that there are those who have experienced nuclear propulsion rockets. . .

 

[zeigs]

the nozzle, from what I know, is not exactly immediate to draw, from its shape depends on the yield of the transformation from potential energy in kinetics, and a simple conicity is not enough, while in the form enter roughness, viscosity, etc.
if you want to reach 300 km/h, the water will have to go out at a much greater speed: 300 km/h are 83 m/s, with a hole of 15 mm we beat from 15 liters per second up, and 83 m/s are many for a liquid

 

[myface]

Bye to all,
I wanted to ask for your help in designing a water rocket.
The rockets I saw on the internet are very rudimentary so I thought I could use the principle but to review it as a whole by giving a somewhat more technical approach. What do you say? ?
I should like to thank you for your comments. Hi.

You want to assault a seed bank?

 

[Sasso SP]

You want to assault a seed bank?

I don't understand? ? ! ! !

I have not yet had time to make some calculations, just to start but in the meantime I have recovered some material that I think will be very useful.

 

[Sasso SP]

Bye to all,
I finally started putting down some calculations and as always the thing is more complicated than expected.
I started by establishing the mass of water and dry rocket (3 kg - 1.5 kg) and then starting to calculate the amount of motion (q) agent in the system.
calculation the amount of the rocket motion:
q=m*v=1.5*83,33=125 kg*m/s
knowing that the total speed of the system does not change,
and considering that q of the rocket is equal to q of the water calculates the speed of the water:
m*v (water) = 125*3= 41,66 m/s (about 150km/h)

as you will have noticed I set a mass of the water greater than that of the rocket this to make it not to have a speed of the outgoing water too high.
question : the speed of water found (150) do I have to assign it as an initial, final, medium speed? ?
look your comments, hi

 

[maca75]

I think that speed you must consider it instantaneous when you have water mass equal to 3kg, probably at the start of the rocket

 

[Sasso SP]

I think that speed you must consider it instantaneous when you have water mass equal to 3kg, probably at the start of the rocket

ok, but in my calculations the speed of water is in relation to the speed of the final rocket!

 

[Pastorman75]

I would say missing 2 data and 1 hypothesis:

given1) the time necessary to reach 300 km/h (83 m/s)
given2) how many liters of water are in the tank

hyp1) the sebator ends its charge at the same time when it comes to the required speed

for example we put:

given1= 5 sec
given2= 10 litres

to get from 0 to 83 m/s in 5 sec you need a uniform acceleration equal to:

a = 3.32 m/s^2

if the serrbatio has 10 liters and the dry razo weighs 2 then the initial total mass is 12 kg therefore:

f = ma = 12x3.32 = 40 n (approx)

therefore the thrust that serves is 40 n and must be obtained by expelling (hyp1):

mw = 10 [litri] / 5 [sec] * 1 [kg/l] = 2 kg/sec

with mw: water mass coming out from the nozzle to the second

Now, the thrust is equal to the variation of the amount of motion (of water obviously). the plugt s must be equal to the calculated f force to obtain the acceleration "a".
then with:

vi: initial water speed=0 m/s (it is stopped in the tank)
vf:water output speed (incognita)

(vf-vi) = 2 [kg/s] (vf-0) [m/s] = 40 n [kgm/s^2]give to whom:

vf = 40 / 2 = 20 m / s


Go on.....

 

[Pastorman75]

We said:

20 m/s
2 kg/s

are the speed and flow characteristics to keep for 5 seconds out of the rocket to make it reach 300 km/h.

until now we have considered only the initial conditions for which the mass remains constant throughout the time of thrust. this is a strong approximation mitigated only partly by the fact that the pressure drops during the thrust itself. for now let's settle.

Now, what pressure is in the tank and what volume should it have?
beautiful domende.... .

we can use the perfect gas formula:

e=p*vs*ln(p/patm)

where:

e= energy possessed by the perfect gas
p= pressure in the tank
vs= volume tank
atmospheric pressure

had, the energy necessary and calculated by:

e=power*time=spinta*speed=40*20*5=4000 joule

Note: 4000j for 5 seconds are a power of 800w

solving the perfect gas energy formula by attempting to fix vs= 5 liters, you have:

p=5 atm

then the project parameters are:

water tank: 10 litres
air tank: 5 liters
dry rocket weight: 2 kg
push time: 5th century
theoretical pressure air tank: 5 atm


we make two considerations on the nozzle

continue...

 

[Pastorman75]

the nozzle, as they have rightly observed, transforms potential energy into kinetics.

if well done allows to obtain good yields however of the non isoentropic transformation losses are inevitable and you will have to take into account.

we want 20 m/s and 2 kg/s for water.
If I consider water incomprehensible then I have:

mass flow / density = speed * section

2 [kg/s] / 1000 [kg/m^3] = 20 [m/s] * amin

from which:

amin = 1 cm^2

This is the theoretical minimum air of the nozzle.

now you should do tests at 5 seconds considering the mass of the rocket equal to 2 kg (water is over!) and the volume of the air tank equal to 15 liters.
the pressure is determined by:

pv=constant (perfect gas law)

It is necessary to see if with that energy the thrust is sufficient to accelerate to 3.32 m/s^2 the residual mass of the rocket (2kg)

Finally, considering a 60% yield for the nozzle I would say to increase the energy figure similarly:

4000/0,6=6665

from which for the formula of the energy of the perfect gases:
(air tank 5 liters)

p=7 atm (approx)


Hopefully decolli!! ! ! !
greetings