Math Project by me.

1. SMJM 1023
ENGINEERING MATHEMATICS 2
PROJECT
•MUBAREK KURT
GROUP
X+Y=0
Water Flow from Plastic Bottles through Pin-Hole T

2. INTRODUCTION
• We performed an experiment on elementary hydrodynamics.
The basic system is a cylindrical bottle from which water flow
through a pin-hole located at the bottom of its lateral surface.
We measured the speed of the water leaving the pin-hole, as a
function of both the time and current level of water still inside
the bottle.
• We use the concept of conservation of energy that lead to a
differential equation and Toricelli`s Law .

3. OBJECTIVES
To investigate the relationship between the speed of the flow (𝑣) and the
height of the level of water(ℎ) .
To apply the Torricelli’s law in this problem cases.
To apply differential equation in our daily life.
To increase our knowledge on differential equation by solving the problem
cases given.

4. PROBLEM
SOLVING
Construct a differential equation for the speed of the flow
and height of the level of water, and find its solution.
water exiting the hole:
𝑑𝑉
𝑑𝑡
= −𝑎𝑣
Conservation of energy:
1
2
𝑚𝑣2 = 𝑚𝑔ℎ
Relate the velocity of the fluid leaving the hole to the height of the water
in the Tank :
𝑣2
= 2𝑔ℎ
The speed of the fluid is related to the height of the water :
𝑣 = 2𝑔ℎ

5. Recall that the volume of the water in the tank, V (t) is related to the height
of fluid h(t) by
𝑉 𝑡 = 𝐴ℎ(𝑡)
where A > 0 is a constant, the cross-sectional area of the tank. Thus, we can
simplify as follows:
𝑑𝑉
𝑑𝑡
=
𝑑(𝐴ℎ 𝑡 )
𝑑𝑡
= 𝐴
𝑑(ℎ 𝑡 )
𝑑𝑡
𝑑𝑉
𝑑𝑡
= −𝑎𝑣 = −𝑎 2𝑔ℎ
𝐴
𝑑(ℎ 𝑡 )
𝑑𝑡
= −𝑎 2𝑔ℎ
𝑑ℎ
𝑑𝑡
= −
𝑎
𝐴
2𝑔ℎ = −𝑘 ℎ
where k is a constant that depends on the size and shape of the cylinder and its
hole:
𝑘 =
𝑎
𝐴
2𝑔

6. The height h(t) of water in the tank at time t satisfies the following differential equation:
𝑑ℎ
𝑑𝑡
= −𝑘 ℎ
Integration of the formula would give the time and height:
𝑑ℎ
ℎ
= −𝑘 𝑑𝑡
ℎ−
1
2 = −𝑘 𝑑𝑡
2 ℎ 𝑡 − ℎ0 = −𝑘𝑇

7. Thus the decreasement of the height over time is:
ℎ 𝑡 = ( ℎ0 − 𝑘
𝑡
2
)2 , 𝑘 =
𝑎
𝐴
2𝑔
In order the empty the tank, the ℎ 𝑡 = 0:
( ℎ0 − 𝑘
𝑡
2
)2
= 0
𝑡 𝑒 =
2 ℎ0
𝑘
Where 𝑡 𝑒 is the time taken for the tank to empty, ℎ0 is the initial height of the water in the
container.

8. Sketch of the graph of the solutions
0
5
10
15
20
25
0 2 4 6 8 10 12 14 16 18 20
Graph of solution

9. Data for the level and speed of water leaving the
pin-hole through the range X
TIME/s Height/m X/m Speed of water/ms-
1
0 0.135 0 0
20 0.128 0.0230 0.3222
40 0.123 0.0225 0.3152
60 0.122 0.0220 0.3082
80 0.118 0.0210 0.2942
100 0.106 0.0205 0.2871

10. EXPERIMENT VS THEORY
y = 2E-05x2 - 0.004x + 0.2
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
0 20 40 60 80 100 120 140
Experimental Vs Theoretical

11. REASON FOR THE SMALL HOLE
 With the bigger cross sectional area on the
top, there will be a very high atmospheric
pressure act on the surface of the water.
The pressure is transmitted throughout the
liquid equally. As the high atmospheric
pressure pushing the water to the bottom,
the water will run through the small hole
with high velocity, inflicting greater
distance of X. If bigger hole is made, the
velocity would not be as high as the small
one.

12. What will happen if the change the bottle to rectangular form?
 NOTHING CHANGE BUT IT DEPENDS ON THE
HEIGHT OF THE WATER IN THE CONTAINER
 BECAUSE HEIGHT INFLUENCES THE VELOCITY OF
THE WATER FLOW

13. IF H=60M, WHAT IS MAX SPEED OF WATER EMERGE?
𝑣 = 2𝑔ℎ
𝑣 = 2(9.81)(60)
𝑣 = 34.31 𝑚/𝑠

14. CONCLUSION
The experiment values is different with
theoretical values because of some error .
Height is directly proportional to the speed
of water emerge.

15. THE
END