fluid mechanics case report

DRAGOS VASILE MITROFAN & DAVID OGDEN

5/7/2012

FLUID MECHANICS

CASE REPORT WATER ROCKET

FLUID MECHANICS

WATER ROCKET CASE REPORT

Teacher: Jens Brusgaard Vestergaard

I. GENERAL CONSIDERATIONS, PHYSICAL, THERMODYNAMIC AND FLUID

DYNAMIC PROCESSES ACTING ON THE WATER ROCKET

A water rocket is a type of model rocket using water as its reaction mass. The pressure vessel in

our case the PET container is the engine of the rocket. The water inside is forced out by a pressurized

gas, typically compressed air. which force the water inside the container throughout the nozzle. In our

case the water acts as a fuel... This type of model demonstrate the working principles of Newton`s third

Law.

In this report the water rocket is subjected to a 600 kPa pressure and we want to see what is the

maximum altitude that can reach considering also different parameters like nozzle diameter, water

content, pressure loss etc...

Basic principle of reaction of a jet

Whenever the momentum of a stream of fluid is increased in a given direction in passing from

one section to another, there must be a net force acting on the fluid in that direction and by the Newton`s

third law there will be an equal and opposite force exerted by the fluid on the system which is producing

the change of momentum.

In order to start calculate first we have to make a FBD(free body diagram ) of the water rocket with all the

forces that are acting on our model

http://en.wikipedia.org/wiki/Model_rocket

http://en.wikipedia.org/wiki/Water

http://en.wikipedia.org/wiki/Reaction_mass

http://en.wikipedia.org/wiki/Compressed_air

FLUID MECHANICS



II. MATHEMATICAL FORMULATION OF THE PROCESSES

Knowing all the forces that are present we can start formulate a mathematical model of our rocket

starting with the expelled water throughout the nozzle by using Bernoulli`s equation on the surface of

contact between water and air and the surface at the end of the nozzle

𝑃1 + 1

2𝜌𝑊𝑣1

2 + 𝜌𝑊𝑔𝑕1 = 𝑃2 + 1

2𝜌𝑊𝑣2

2 + 𝜌𝑊𝑔𝑕2 + ∆𝑃𝑙𝑜𝑠𝑠

where

∆𝑃𝑙𝑜𝑠𝑠 = (𝑓𝐿

𝑑+ 𝜁1 + 𝜁2)

1

2𝜌𝑊𝑣2

2 and 𝑓 = 1

2 𝑙𝑜𝑔𝑅𝑒 𝑓

2.51

2

v1 = Speed of water inside the bottle (v1 ≈ 0)

v2 = Exhaust velocity of water

P1 = Air pressure inside the bottle

P2 = Pressure at the end of the nozzle (Atmospheric pressure)

h1 = h2= Altitude of points 1 and 2

f = Friction factor from Moody`s Diagram

ρW = Density of water (ρ ≈ 1000 kg/m3)

L = length of the nozzle

d = internal Diameter of nozzle

ζ = Secondary looses according to the local water speed

Based on these formulas the expelled water speed v2 can be calculated as a function of the pressure inside

the bottle

𝑣2 = 2𝑃1

𝜌 1 + 𝜏 + 𝑓𝐿𝑑

Now that we found V2 we can calculate the mass flow of air:

𝑚 𝑓 = 𝐴𝑛𝑜𝑧𝑧𝑙𝑒 ∙ 𝑣2 ∙ 𝜌𝑊

FLUID MECHANICS



Having the mass flow now we can start working on the variation of water per time unit :

∆𝑀𝑤 = 𝑚 𝑓∆𝑡

∆t = time step

From this we can deduce that mass of the water per time unit is:

𝑀𝑤 = 𝑀𝑤𝑖 − ∆𝑀𝑤

Mwi = Initial mass of water

Mw = Mass of the water per time unit

Total mass of the rocket is

𝑀𝑅𝑜𝑐𝑘𝑒𝑡 = 𝑀𝐵𝑜𝑡𝑡𝑙𝑒 + 𝑀𝑤𝑎𝑡𝑒𝑟

Next step is to calculate the variation of air per time unit:

∆𝑉𝑎𝑖𝑟 = ∆𝑉𝑤 = ∆𝑀𝑤

𝜌𝑊

By having the volume of air variation we can calculate the new volume of air in the bottle

𝑉𝑎𝑖𝑟 = 𝑉0𝑎𝑖𝑟 + ∆𝑉𝑎𝑖𝑟 ,

𝑉0𝑎𝑖𝑟 - initial volume

We know that inside the bottle there is a pressure of 600 kPa and when lunched there will be a

pressure drop so these means that we will have a variation of the pressure. We can calculate this with the

relation between pressure and volume:

𝑃0 ∗ 𝑉0𝑘 = 𝑃1 ∗ 𝑉1

𝑘

V0 = Initial volume of air

V1 = Volume per time unit

P0 = Initial pressure of air

P1 = Pressure of air per time unit

and we have a K value and this is equal with 1.4.

FLUID MECHANICS



Having all the masses and the pressure forces now we can start working on the Technical parameters such

as : Thrust, Speed, Acceleration and the Altitude

The thrust is given by the force of the running water at the nozzle

𝑇𝑕𝑟𝑢𝑠𝑡 = 𝑚 𝑤𝑣𝑅

vR = Speed of rocket per time unit

Also there are other 2 forces present : the Drag force and the gravity and they are expressed as follow

𝐹𝑔 = 𝑀𝑅𝑔 𝐹𝐷 = 1

2𝜌𝑎𝑖𝑟𝐴𝑏𝑜𝑡𝑡𝑙𝑒 𝐶𝐷𝑣𝑅

g = Gravity force (g = 9.82 m/s2)

Abottle = Area of bottle section

CD = Coefficient of drag of the bottle

Ρair = Density of air (ρ ≈ 1.25 kg/m3)

Now we can find the Force acting on the rocket

𝐹 = 𝐹𝑡𝑕𝑟𝑢𝑠𝑡 − 𝐹𝑔 − 𝐹𝐷 → 𝑎𝑅 = 𝐹𝑡𝑕𝑟𝑢𝑠𝑡 − 𝐹𝑔 − 𝐹𝐷

𝑀𝑅

aR = Acceleration of the rocket

We also know that the speed is variable so we have a variable speed at a certain time (∆v):

𝑣𝑅 = 𝑣0𝑅 + ∆𝑣 where ∆𝑣 = 𝑎𝑅∆𝑡

V0R = Initial rocket speed

Knowing the Thrust and speed we can start calculating the height that the rocket can reach:

𝑕 = 𝑕0 + ∆𝑕 where ∆𝑕 = 𝑣𝑅∆𝑡

FLUID MECHANICS



III. WIND TUNNEL MEASUREMENTS

By the use of wind tunnel we can find out what is the Drag Coefficient for our bottle we can also use the

following graphic( FLUID mechanics 1 semester) for rapid approximation of the Drag coefficient and it s

around Cd = 0.5

FLUID MECHANICS



IV. CARRY OUT A SIMULATION USING EXCEL AND EULER’S METHOD

(the Excel file is attached to this report)

By the use of Microsoft Excel we have carried out a simulation of all the mathematical formulas

written above and we came with the following conclusion:

For this experiment, the water rocket consisted of a recycled 2 litre plastic bottle, fitted with a 9

mm nozzle in its lid. The bottle was filled up with 1litre of water, placed onto a rig and attached to a

container of compressed air. The compressed air created high pressure around 600 [kPa] in the bottle, and

when the bottle was disconnected from the air and released from the rig, the air pressure forced the water

through the nozzle to create a water jet, and propel the bottle vertically.

The bottle was propelled to around 23.814 metres high, before it ran out of water. The bottle was

highly unstable throughout its flight, before falling back to the ground.

fluid mechanics case report

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