1

CEE170: Introduction to Fluid Mechanics Water Jet Cart Design Project

Group 18: Top Crew

Chosita Sribhibhadh

Kevin Krik

Patricia Kharazmi

Megan Hanrahan

Saman Shaolian

Ali Behbahani

Toni Lynch

2

Table of Contents Section Title Page Number

Introduction…………………………………………………………… 3

Predicting the Velocity of the Water Jet………..…………………….. 3

Predicting the Rate of Change of Tank Water Height over Time….…. 6

Predicting the Speed of the Cart............................................................. 7

Predicting the Distance the Cart Will Travel…….…………………… 8

The Matlab Code……………………………………………………… 8

Tank Options and Specifications……………………………..……...... 9

Cart Materials and Design……………....…………………………..... 10

Preliminary Cart Testing Results…………………………….……….. 11

Final Cart Race Results……………………………………………….. 12

Race Videos and Pictures.…………………………………………….. 13

Appendix…………………………………………………………….… 14

Resources……………………………………………………………… 17

3

Introduction The purpose of this project is to design a cart that can travel a 50 foot distance along a 2% slope

with a high speed. Each group should be able to predict the speed of their cart based on Matlab

modeling and calibration of these models due to preliminary testing. The goal of the project is to

design a cart with the fastest design speed or have the closest prediction of time using the test

data and Matlab model. Provided is a pressurized tank filled with water. The tank has a nozzle of

interchangeable diameter that will shoot out a jet of water. The teams’ job is to design a chamber

to attach to the cart that will catch the water jet and use it to propel the cart forward and to

choose certain tank specifications that will lead to a successful run.

Predicting the Velocity of the Water Jet The energy equation, which comes from conservation of energy in the Reynolds Transport

Theorem is used to find the velocity of the water jet. Here is how the equation used is derived:

Sketch of pressurized tank:

𝑃1𝛾+𝑉12

2𝑔+ 𝑍1 + ℎ𝑝 =

𝑃2𝛾+𝑉22

2𝑔+ 𝑍2 + ℎ𝑡 + ℎ𝐿

4

P1 is the pressure in the tank, which can be denoted as P

γ is the density of water (ρ=998 kg/m3) multiplied by the acceleration of gravity (g=9.81 m/s2)

V1 is the velocity of the water in the tank, which is such a small number, it can be considered zero

Z1 is the height of water in the tank

P2 is the pressure at point 2 and since it is open to the atmosphere, it can also be consider zero

V2 is the velocity of the water jet which can be denoted as VJ

Z2 is the height of the centerline of the nozzle

hp is the height of energy gained due to a pump, but since there is no pump, it equals zero

ht is the height of energy removed from the system by a turbine, this term also equals zero

hL is the height of energy lost due to friction in the nozzle

A new variable h is used to represent the height of the water in the tank to the centerline

of the nozzle where: ℎ = 𝑍1 − 𝑍2

Taking all these factors into account, the equation is then simplified to:

𝑃

𝛾+ ℎ =

𝑉𝐽2

2𝑔+ ℎ𝐿

where ℎ𝐿 = (𝐾𝑣 + 𝑓𝐿

𝑑)𝑉𝐽2

2𝑔

Kv is the minor loss coefficient

𝑓 is the friction factor

L is the length of the nozzle

d is the diameter of the nozzle

The simplified energy equation now becomes:

𝑃

𝛾+ ℎ =

𝑉𝐽2

2𝑔(1 + 𝐾 + 𝑓

𝐿

𝑑)

Finally, solving for VJ results in the equation:

𝑉𝐽 = √2𝑔 (

𝑃𝛾 + ℎ)

1 + 𝐾 + 𝑓𝐿𝑑

5

The equation that accounts for the change in pressure in the tank over time is:

𝑃 = 𝑃0 (𝑠

𝑠 + ℎ0 + ℎ)𝑘

P0 is initial pressure in the tank

s is the initial height from the top of the tank to the water in the tank

h0 is the initial height of water in the tank

k is the ratio of specific heats inside and outside of the tank

This equation must be plugged into the equation for VJ for accuracy:

𝑉𝐽 =

√

2𝑔(𝑃0 (

𝑠𝑠 + ℎ0 + ℎ

)𝑘

𝛾 + ℎ)

1 + 𝐾 + 𝑓𝐿𝑑

6

Predicting the Rate of Change of Tank Water Height over Time

Now that the velocity of the water jet has been estimated, it can be used to estimate the rate at

which the height of the water in the tank changes over time. The volume of water changing in the

tank over time can be expressed as:

𝑑𝑉𝑡𝑎𝑛𝑘𝑑𝑡

= −𝑄𝑜𝑢𝑡

Qout is the flow of water out of the tank (A2V2) which can be written as:

𝑄𝑜𝑢𝑡 =𝜋

4𝑑2𝑉𝐽

Since the diameter of the tank always remains constant, the expression for the change in volume

of the tank can be written as:

𝑑𝑉𝑡𝑎𝑛𝑘𝑑𝑡

=𝜋

4𝐷2

𝑑ℎ

𝑑𝑡

D is the diameter of the tank

dh/dt is the change in height of the water over time

Substituting these two equations into the original results in:

𝜋

4𝐷2

𝑑ℎ

𝑑𝑡= −

𝜋

4𝑑2𝑉𝐽

Solving for dh/dt gives:

𝑑ℎ

𝑑𝑡= (−

𝑑2

𝐷2𝑉𝐽)

Plugging the jet velocity equation into this equation supplies the first ordinary differential

equation (ODE) that will be solved in Matlab:

𝑑ℎ

𝑑𝑡= −

𝑑2

𝐷2∗

(

(

2∗(𝑔∗ℎ+(𝑃𝑜∗(𝑠𝑜

𝑠𝑜+ℎ𝑜−ℎ)𝑘))

𝑝)

1 + 𝐾𝑣 + 𝑓 ∗ (𝐿𝑑)

)

0.5

7

Predicting the Speed of the Cart

Once the velocity of the water jet has been predicted, that value, along with the dimensions of

the cart design can be used to predict the speed of the cart. The prediction of the cart’s speed

comes from Newton’s Second Law of Motion Equation (Σ𝐹 = 𝑚𝑎). Accounting for all the

forces acting on the cart gives the equation:

𝐹𝐽 − 𝐹𝐺 − 𝐹𝑅 = 𝑚𝑎

Solving the equation for acceleration gives:

𝑎 =𝐹𝐽 − 𝐹𝐺 − 𝐹𝑅

𝑚

FJ is the force exerted on the cart by the jet, which can be written as:

𝐹𝐽 = 𝛼𝜌𝐴𝐽𝑉𝑅2

o α is the momentum transfer coefficient (calibration parameter from 0-2)

o AJ is the area of the water jet (π/4*d2)

o VR is the relative velocity between the jet and the cart (VJ-VC)

FG is the gravitational force on the cart which can be written as:

𝐹𝐺 = 𝑚𝑔𝑠𝑖𝑛𝜃

o θ is the angle of incline

o m is the mass of the cart

FR is the force due to rolling friction and air resistance which can be ignored in further

calculations

Plugging all of these forces into Newton’s Equation solved for acceleration gives:

𝑎 = 𝛼𝜌𝐴𝐽𝑉𝑅

2

𝑚− 𝑔𝑠𝑖𝑛𝜃

Knowing that acceleration is the change in velocity over time (𝑎 =𝑑𝑉𝑐

𝑑𝑡), this equation can be

rewritten as:

𝑑𝑉𝐶𝑑𝑡

=𝛼𝜌𝐴𝐽𝑉𝑅

2

𝑚− 𝑔𝑠𝑖𝑛𝜃

Accounting for deceleration due to drag and friction (Cdf) supplies the second ODE to be solved

in Matlab:

𝑑𝑉𝐶𝑑𝑡

=𝛼𝜌𝐴𝐽𝑉𝑅

2

𝑚− 𝑔𝑠𝑖𝑛𝜃 − 𝐶𝑑𝑓

8

Predicting the Distance the Cart Will Travel

Now that the speed of the cart has been predicted, the distance it will travel can also be predicted.

The total distance the cart will travel is simply the change in the cart’s velocity over time. This

can be expressed as:

𝑑𝑥 = 𝑉𝑐𝑑𝑡

Solving for velocity supplies the third ODE to be solved in Matlab:

𝑑𝑥

𝑑𝑡= 𝑉𝑐

The Matlab Code

Using an initial set of parameters, the system of three ODEs was used to solve for the unknown

variables. These initial conditions were set for the dependent variables:

ht=0 = 0

Vc t=0 = 0

xt=0 = 0

Next, a script was created to solve the system of ODEs using the built-in MATLAB function

ode45. (The actual script can be found in the appendix) Here, the design parameters and the

calibration parameters were defined and given values were input in order to solve for the

dependent variables in the ODEs. Also, an initial conditions vector and a timespan vector were

defined so that the MATLAB program would perform ode45 over a given time interval. Lastly, a

new function file was created for use in the ode45 built-in function to help evaluate the three

differential equations.

The calibration parameters used in the MATLAB script were chosen based on the data and times

we recorded during the test run of our cart. In order to accurately predict the time our cart took to

cross 50ft at a 2% grade, we had to adjust our calibration parameters accordingly. These

calibration parameters included the drag and friction coefficients, which helped reduce the

velocity of our cart to zero. By only taking into account the effects of gravity, our cart would

have taken a very large amount of time for velocity to reach zero; therefore we manipulated the

drag and friction coefficients.

9

After the program solved the ODEs, a plot of time versus each dependent variable was produced,

along with a plot of the jet velocity versus time:

From these results, we predicted that our cart would take 2.799 seconds to cross the finish line on

race day.

Tank Specifications

Diameter of tank (D) 12 in.

Z1 8.25 in.

s + h 25 in.

L 7.75 in.

Tank Options (Chosen Prior to Race)

d - diameter of nozzle 1 in.

P – pressure 55 psi

h – water height 4in.

t – time prediction 2.79 sec.

10

Cart Materials and Design

The chamber attached to the cart consists of a cone made of aluminum steel, riveted together on

opposite sides of the lateral area. The cone was attached to a typical skateboard with two

bearings and four wheels using four metal legs. The metal legs are bolted to the skateboard and

welded to the cone. This specific design was chosen because the cone shape has a low level of

aerodynamic resistance. Also, the goal was to have a chamber with a large volume so that

maximum water tank pressure could be used, which ideally would result in the fastest speed.

Cross section views of cart:

Profile view of cart:

11

Preliminary Cart Testing Results (Friday 11/15)

Pressure (psi) Water Height (in.) Nozzle (in.) Time (seconds)

Test 1 60 3.25 1/2 None*

Test 2 55 4 1 2.71

Test 3 55 4 1 2.79

Test 4 55 4 1 None*

Test 5 45 6 1 None*

* Cart crashed before crossing finish line

During the preliminary cart test, we completed 5 test runs before the 15 minute time limit was

reached. Each of the 5, except tests 2 and 3, varied from one another. Test 1 was performed at a

pressure of 60 psi and a corresponding water height of 3.25 inches. The first test was done with

the 2nd smallest nozzle and failed to reach the finish line. The problem with the first test was

mainly attributed to the nozzle size and for further tests; the largest nozzle size was used.

For the second and third tests, the group decided on a pressure of 55 psi and a corresponding

water height of 4 inches. After the pressurized water was released, the cap detached from the end

of the cone. This helped release some pressure and water and helped the cart stay on track

without tipping over, but it did not help us achieve the fastest time. After two successful runs, the

group decided to tape the cap of the cone shut to achieve optimal time. Both test 4 and 5 were

unsuccessful due to the cart tipping over.

We realized the reason why our cart kept tipping over was because so much weight from the

water was building up in the tip of our cone and the creating a moment. That is why our cart was

only successful when the end cap popped off. This, along with the fact that the cart’s center of

gravity was closer to the front end, caused a moment at the front end of the cart and caused the

cart to tip over whenever the cap stayed on. In order to counteract this, we decided to try adding

a bowl to our final design. The bowl would be attached to the inside of the cone, causing more

weight to be added toward the back of the cone and causing the water jet to hit the cart farther

back.

12

Final Cart Race Results (Wednesday 11/27)

Pressure (psi) Water Height (in.) Nozzle (in.) Time (seconds)

Test 1 55 4 1 None*

Test 2 55 4 1 3.09

Test 3 55 4 1 None*

Test 4 55 4 1 None*

Test 5 45 6 1 3.90

Test 6 45 6 1 None*

* Cart crashed before crossing finish line

Test 1 was performed with a metal bowl attached to the inside of our cart. We lined the inside of

our cart with a trash bag and taped it to the outer edge and to the bowl to help keep the bowl in

place and make sure no water could get inside the cone. Our cart tipped over during test 1

because the pressure we specified wasn’t meant for the bowl design and it was too high.

For test 2 we took the bowl out and the cap popped off of the cone like it did in the test runs. The

cart made it across the finish line in 3.09 seconds which is only 0.3 seconds off from our

predicted time of 2.79 seconds. For tests 3 and 4 we left the bowl out again. Our cart wasn’t

tipping over this time, but it kept veering off to the side instead of staying on track.

For the 5th test, we decided to add the bowl again and reduce the pressure to 45 psi. During test 5

the cart made it across the finish line in 3.9 seconds, but during test 6 it flipped over again.

Overall, our cart only worked when the cap popped off because of the flaws in our design.

Because the cart was too heavy at the front end and because the skateboard was so small

compared to the cone, our cart was very unstable. Even if we had fixed the problems with our

cart and got it to go straight across the finish line without tipping over or having the cap pop off,

we still most likely wouldn’t have gotten the fastest time since the materials of the cart were so

heavy.

13

Race Videos and Pictures

Race Video:

http://www.youtube.com/watch?v=lZN4Dj4QmcQ&feature=c4-

overview&list=UUY_Dpnug5Jui2w-NLSP1JNQ

Figure of cart with bowl taped inside and trash bag attached:

14

Appendix

Matlab Code:

15

16

17

Resources

1. Sanders, Prof. B.F.. Water Jet Cart Analysis, CEE170 Class Website. 2012.

2. White, F.M.. Fluid Mechanics, 7th Edition, McGraw Hill 2011.