fluid mechanics case report
TRANSCRIPT
DRAGOS VASILE MITROFAN & DAVID OGDEN
5/7/2012
FLUID MECHANICS
CASE REPORT WATER ROCKET
Page 1
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
Page 2
FLUID MECHANICS
WATER ROCKET CASE REPORT
Teacher: Jens Brusgaard Vestergaard
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
π1 + 1
2πππ£1
2 + ππππ1 = π2 + 1
2πππ£2
2 + ππππ2 + βππππ π
where
βππππ π = (ππΏ
π+ π1 + π2)
1
2πππ£2
2 and π = 1
2 ππππ π π
2.51
2
v1 = Speed of water inside the bottle (v1 β 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
ΟW = Density of water (Ο β 1000 kg/m3)
L = length of the nozzle
d = internal Diameter of nozzle
ΞΆ = 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
π£2 = 2π1
π 1 + π + ππΏπ
Now that we found V2 we can calculate the mass flow of air:
π π = π΄πππ§π§ππ β π£2 β ππ
Page 3
FLUID MECHANICS
WATER ROCKET CASE REPORT
Teacher: Jens Brusgaard Vestergaard
Having the mass flow now we can start working on the variation of water per time unit :
βππ€ = π πβπ‘
βt = time step
From this we can deduce that mass of the water per time unit is:
ππ€ = ππ€π β βππ€
Mwi = Initial mass of water
Mw = Mass of the water per time unit
Total mass of the rocket is
ππ πππππ‘ = ππ΅ππ‘π‘ππ + ππ€ππ‘ππ
Next step is to calculate the variation of air per time unit:
βππππ = βππ€ = βππ€
ππ
By having the volume of air variation we can calculate the new volume of air in the bottle
ππππ = π0πππ + βππππ ,
π0πππ - 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:
π0 β π0π = π1 β π1
π
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.
Page 4
FLUID MECHANICS
WATER ROCKET CASE REPORT
Teacher: Jens Brusgaard Vestergaard
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
ππππ’π π‘ = π π€π£π
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
πΉπ = ππ π πΉπ· = 1
2πππππ΄πππ‘π‘ππ πΆπ·π£π
g = Gravity force (g = 9.82 m/s2)
Abottle = Area of bottle section
CD = Coefficient of drag of the bottle
Ξ‘air = Density of air (Ο β 1.25 kg/m3)
Now we can find the Force acting on the rocket
πΉ = πΉπ‘πππ’π π‘ β πΉπ β πΉπ· β ππ = πΉπ‘πππ’π π‘ β πΉπ β πΉπ·
ππ
aR = Acceleration of the rocket
We also know that the speed is variable so we have a variable speed at a certain time (βv):
π£π = π£0π + βπ£ where βπ£ = ππ βπ‘
V0R = Initial rocket speed
Knowing the Thrust and speed we can start calculating the height that the rocket can reach:
π = π0 + βπ where βπ = π£π βπ‘
Page 5
FLUID MECHANICS
WATER ROCKET CASE REPORT
Teacher: Jens Brusgaard Vestergaard
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
Page 6
FLUID MECHANICS
WATER ROCKET CASE REPORT
Teacher: Jens Brusgaard Vestergaard
IV. CARRY OUT A SIMULATION USING EXCEL AND EULERβS 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.