CEE170: A University of California, Irvine upper division civil engineering course

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
1

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
2

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:
2
𝑃1
𝑉12
𝑃2
𝑉2
+
+ 𝑍1 + ℎ 𝑝 =
+
+ 𝑍2 + ℎ 𝑡 + ℎ 𝐿
𝛾 2𝑔
𝛾 2𝑔
3

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:
𝑉𝐽2
𝑃
+ℎ =
+ℎ𝐿
𝛾
2𝑔
𝐿
where ℎ 𝐿 = (𝐾𝑣 + 𝑓 )
2
𝑉𝐽
𝑑 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:
𝑉𝐽2
𝑃
𝐿
+ℎ=
(1 + 𝐾 + 𝑓 )
𝛾
2𝑔
𝑑
Finally, solving for VJ results in the equation:
𝑃
2𝑔 ( 𝛾 + ℎ)
𝑉𝐽 = √
𝐿
1+ 𝐾+ 𝑓
𝑑
4

5. The equation that accounts for the change in pressure in the tank over time is:
𝑘
𝑠
)
𝑃 = 𝑃0 (
𝑠 + ℎ0 + ℎ



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:
𝑘
𝑠
)
𝑃0 (
𝑠 + ℎ0 + ℎ
2𝑔 (
+ ℎ)
𝛾
𝑉𝐽 =
√
1+ 𝐾+ 𝑓
𝐿
𝑑
5

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:
𝑑𝑉 𝑡𝑎𝑛𝑘
= −𝑄 𝑜𝑢𝑡
𝑑𝑡

Qout is the flow of water out of the tank (A2V2) which can be written as:
𝑄 𝑜𝑢𝑡 =
𝜋 2
𝑑 𝑉𝐽
4
Since the diameter of the tank always remains constant, the expression for the change in volume
of the tank can be written as:
𝑑𝑉 𝑡𝑎𝑛𝑘
𝜋
𝑑ℎ
= 𝐷2
𝑑𝑡
4
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 2 𝑑ℎ
𝜋
𝐷
= − 𝑑 2 𝑉𝐽
4
𝑑𝑡
4
Solving for dh/dt gives:
𝑑ℎ
𝑑2
= (− 2 𝑉𝐽 )
𝑑𝑡
𝐷
Plugging the jet velocity equation into this equation supplies the first ordinary differential
equation (ODE) that will be solved in Matlab:
𝑘
𝑠𝑜
2∗(𝑔∗ℎ+(𝑃𝑜∗( 𝑠𝑜+ℎ𝑜−ℎ) ))
(
𝑑ℎ
𝑑2
=− 2∗
𝑑𝑡
𝐷
𝑝
0.5
)
𝐿
1 + 𝐾𝑣 + 𝑓 ∗ ( )
𝑑
(
)
6

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 − 𝐹 𝐺 − 𝐹 𝑅 = 𝑚𝑎
Solving the equation for acceleration gives:
𝑎=

𝐹𝐽 − 𝐹 𝐺 − 𝐹 𝑅
𝑚
FJ is the force exerted on the cart by the jet, which can be written as:
𝐹𝐽 = 𝛼𝜌𝐴 𝐽 𝑉 2
𝑅
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 (V J-VC)

FG is the gravitational force on the cart which can be written as:
𝐹 𝐺 = 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:
𝛼𝜌𝐴 𝐽 𝑉 2
𝑅
𝑎=
− 𝑔𝑠𝑖𝑛𝜃
𝑚
Knowing that acceleration is the change in velocity over time (𝑎 =
rewritten as:
𝑑𝑉 𝑐
𝑑𝑡
), this equation can be
𝛼𝜌𝐴 𝐽 𝑉 2
𝑑𝑉 𝐶
𝑅
=
− 𝑔𝑠𝑖𝑛𝜃
𝑑𝑡
𝑚
Accounting for deceleration due to drag and friction (Cdf) supplies the second ODE to be solved
in Matlab:
𝛼𝜌𝐴 𝐽 𝑉 2
𝑑𝑉 𝐶
𝑅
=
− 𝑔𝑠𝑖𝑛𝜃 − 𝐶 𝑑𝑓
𝑑𝑡
𝑚
7

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 = 𝑉𝑐 𝑑𝑡
Solving for velocity supplies the third ODE to be solved in Matlab:
𝑑𝑥
= 𝑉𝑐
𝑑𝑡
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.
8

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)
Z1
s+h
L
12 in.
8.25 in.
25 in.
7.75 in.
Tank Options (Chosen Prior to Race)
d - diameter of nozzle
P – pressure
h – water height
t – time prediction
1 in.
55 psi
4in.
2.79 sec.
9

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:
10

11. Preliminary Cart Testing Results (Friday 11/15)
Pressure (psi) Water Height (in.)
Test 1
60
3.25
Test 2
55
4
Test 3
55
4
Test 4
55
4
Test 5
45
6
* Cart crashed before crossing finish line
Nozzle (in.)
1/2
1
1
1
1
Time (seconds)
None*
2.71
2.79
None*
None*
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.
11

12. Final Cart Race Results (Wednesday 11/27)
Pressure (psi) Water Height (in.)
Test 1
55
4
Test 2
55
4
Test 3
55
4
Test 4
55
4
Test 5
45
6
Test 6
45
6
* Cart crashed before crossing finish line
Nozzle (in.)
1
1
1
1
1
1
Time (seconds)
None*
3.09
None*
None*
3.90
None*
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.
12

13. Race Videos and Pictures
Race Video:
http://www.youtube.com/watch?v=lZN4Dj4QmcQ&feature=c4overview&list=UUY_Dpnug5Jui2w-NLSP1JNQ
Figure of cart with bowl taped inside and trash bag attached:
13

14. Appendix
Matlab Code:
14

15. 15

16. 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.
17