KM20303 FLUID MECHANICS

Session 2019/2020

Group Assignment:

Finding Drag Coefficient for a Parachute

DR. MOHD KAMEL WAN IBRAHIM

GROUP MEMBERS MATRIC NUMBER

ALFERA YAP BK18110169
EVANS GOH BK18110220
NOORINAH BINTI MASA BK18110142
NUR AINA ALWANI BINTI MOHAMAD BK18110162
RONALDO ULISI BK18110241
OBJECTIVE

Using experimental means, determine the coefficient of drag for a parachute.

METHOD

Drop a parachute from a known height and measured the fall time. Assume the
parachute is moving at terminal velocity and calculate terminal velocity as the ratio of height
to drop time. Next, apply the equilibrium principle such as balance weight with drag force and
use algebra to solve for the coefficient of drag.

Note: This method is only valid if the acceleration period is negligible compared to the drop
time.

INTRODUCTION AND THEORY

The main purpose of this experiment is to make a parachute that will have slow velocity and
stable parachute. Before doing that, the graph was plotted as a reference . All the equation
below is used to analysis the factor that should be consider to make a parashute.

Using the drag force or “wind resistance force” equation.

1
𝐹𝐷 = 𝜌𝐶𝑑 𝐴𝑉 2
2
Where,

F = drag force

𝜌= density of the air (1.22 𝑘𝑔/𝑚3 )

𝐶𝑑 =drag coefficient

A = area of the chute

V = velocity through the air

Meanwhile. The weight of the rocket, otherwise known as the force of gravity 𝐹𝑎 , is
computed to be

𝐹𝑎 = 𝑚𝘨
Where,

m = mass of the payload

𝘨 = acceleration of gravity

Solving where drag force equals force of gravity:

𝐹𝑎 = 𝐹𝐷
 1
𝑚𝘨 = 𝜌𝐶𝑑 𝐴𝑉2
2
 𝜋𝐷 2
Chute area, in term of the projected chute diameter is 𝐴 = , therefore coefficient
4
8𝑚𝘨
of drag is 𝐶𝑑 =
𝜋𝜌𝐷2 𝑉 2

Where,

𝐷= projected diameter or chute diameter in meters

𝘨 = acceleration of gravity (9.81 𝑚/𝑠 2 )

𝜌 = density of air (1.22 𝑘𝑔/𝑚3 )

Procedure for the parachute prototype:

1. A big plastic bag with the diameter of 0.8 m is used as the material for the parachute.
Then the plastic bag was cut into square shape with about 100𝑐𝑚 × 100𝑐𝑚 and was
fold from corner 3 of the plastic to its opposite side of the corner 1 which produce a
triangle shape.

1 2

100cm

100cm 3
4

FIGURE 1

2. Then, the triangle shape of the plastic bag again was fold with the same way which is
from corner 2 of the plastic to its opposite corner 4 of the plastic which will produce
triangle shape again.
 3 to 1

4 2

FIGURE 2

3. Then, the triangle shape of the plastic bag again was fold with the same way which is
from corner 2 of the plastic to its opposite corner 4 of the plastic which will produce
triangle shape again.
4. Then, the triangle was fold by meeting the hypotenuse and the adjacent of the triangle
and was fold once again with the same direction or point which is the corner between
the folded hypotenuse and adjacent of the triangle.
5. Next, the folded plastic was cut to make the arc of the plastic which a circle shape was
produced. The diameter of the plastic is 0.8 m
6. The rope was cut into 17 rope with the same length which is 1.5 m
7. Then, the rope was tape with cellotape and the way it is tape was same.
8. The distance between the rope was tape based on the calculation of the calculated
circumstances divided by 17 ropes.
9. In the middle of the plastic parachute, a small hole was cut with its diameter of … 𝑐𝑚.
10. Then, all the 17 ropes are made sure to have same length before it is tides together.
11. The hook with the ‘S’ shape was used and was tide at the bottom of the first tides
rope.
12. At the surface around the small hole, the rope was tape at one point and to the other
point with 180 degree.

Procedure for parachute challenge:

1. The parachute was hanged on a stick.

2. The time given was for 5 minutes which is from hanging on the parachute, letting go
the parachute until it reached the ground.
3. This process was repeat 3 times.
 Result

Diameter 1 meter
Diameter (m) 𝐶𝑑 = 1.2
Velocity (m/s)
𝐶𝑑 = 0.2 𝐶𝑑 = 0.4 𝐶𝑑 = 0.6 𝐶𝑑 = 1.4 𝐶𝑑 = 2.0
𝐶
𝐶𝑑 = 0.8 𝑑 = 1.0 𝐶𝑑 = 1.6𝐶𝑑 = 1.8
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
0.1 22.6253 15.9985 13.0627 11.3127 10.1183 9.2367 8.5516 7.9993 7.5418 7.1548
0.2 11.3127 7.9993 6.5314 5.6563 5.0592 4.6184 4.2758 3.9996 3.7709 3.5774
0.3 7.5418 5.3328 4.3542 3.7709 3.3728 3.0789 2.8505 2.6664 2.5139 2.3849
0.4 5.6563 3.9996 3.2657 2.8282 2.5296 2.3092 2.1379 1.9998 1.8854 1.7887
0.5 4.5251 3.1997 2.6125 2.2625 2.0237 1.8473 1.7103 1.5999 1.5084 1.4310
0.6 3.7709 2.6664 2.1771 1.8854 1.6864 1.5395 1.4253 1.3332 1.2570 1.1925
0.7 3.2322 2.2855 1.8661 1.6161 1.4455 1.3195 1.2217 1.1428 1.0774 1.0221
0.8 2.8282 2.000 1.6328 1.4141 1.2648 1.1546 1.0689 0.9999 0.9427 0.8943
0.9 2.5139 1.7776 1.4514 1.2570 1.1243 1.0263 0.9502 0.8888 0.8380 0.7950
1.0 2.2625 1.5999 1.3063 1.1313 1.0118 0.9237 0.8552 0.7999 0.7542 0.7155
1.1 2.0568 1.4544 1.1875 1.0284 0.9198 0.8397 0.7774 0.7272 0.6856 0.6504
1.2 1.8854 1.3332 1.0886 0.9427 0.8432 0.7697 0.7126 0.6666 0.6285 0.5962

TABLE 1

Graph of Velocity against Diameter

25.00

20.00
Velocity (m/s)

15.00

10.00

5.00

0.00
0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40
Diameter (m)

Cd = 0.2 Cd = 0.4 Cd = 0.6 Cd = 0.8 Cd = 1

Cd = 1.2 Cd = 1.4 Cd = 1.6 Cd = 1.8 Cd = 2

GRAPH 1
 Height = 5m

Trial Time taken

1 2.13

2 2.00

3 2.56

Table 2

Experimental calculation
𝑑
Velocity, 𝑣 = 𝑡
5
= 2.23

= 2.24 𝑚/𝑠

2 8𝑚𝘨
Coefficient of drag, 𝐶𝑑 = √
𝜋𝜌𝐷 2 𝑉 2

2 8(0.05)(9.81)
=√
𝜋(1.225)(0.8)2 (2.24)2

=0.56343
∴ 𝐶𝑑 = 0.6
1
𝐺𝑟𝑎𝑣𝑖𝑡𝑦 𝑓𝑜𝑟𝑐𝑒, 𝐹𝑔 = 𝑚𝑔 𝐹𝑑 = 2 𝜌𝐶𝑑 𝜋𝑣
1
= (0.05)(9.81) = 2 (1.22)(0.6)(𝜋)(2.24)

=0.4905 𝑁 = 0.412096 𝑁

Percentage Error

exp erimental  theory

%error 
theory
2.24  1.63
 100
1.63
 37.42%
DISCUSSION

From the graph in Figure, the graph of the parachute with Cd = 0.2 is plotted the highest
compared to the other graphs. On the other hand, the graph of the parachute with Cd = 2.0
is plotted the lowest compared to the other graphs. It can be deducted that at the same
mass of payload and chute diameter, parachute with larger Cd will have a smaller fall
velocity.

From the graph, the larger the chute diameter, the smaller the parachute fall
velocity. The differences between graphs of parachute of different drag coefficients also
decrease as the diameter of the parachutes increases. In short, it is beneficial to increase
the chute diameter in order to reduce the fall velocity of the parachute. If the mass of the
payload increases, the fall velocity of the parachute will increase.

So based on the experiment, there are many group that use different diameter of parachute.
The different diameter of parachute was effected the result where the value for the velocity
and the coefficient of drag are different. The greater diameter of parachute leads to a greater
air resistance and a tendency to reach a slower terminal velocity.Once the parachute is
opened, the air resistance oppose the downward force of gravity. The net force and the
acceleration on the falling parachute is upward. An upward net force on a downward falling
object would cause that object to slow down. The parachute thus slows down. When the
velocity being slow cause of the large diameter, the coefficient of the drag will become smaller.
The diameter of parachute is inversly proportional to the Cd. Thus , the best parachute is the
large diameter as possible with considering it stability.

In choosing of the designed of parachute, the stability should be consider too. The diameter
of the parachute should be balance with the the weight provided. We can’t just choose the
large one because of the velocity wil be slow. The light object downward force won’t be able
to equal to the upward air resistance when the diameter is too large. So the net force acting
is not stable. When the force of gravity is equal to the air resistance the object will travel at a
constant velocity. But we still consider the large diameter of parachute with the suitable load
applied. This is because of the aim of the experiment is to get the slowest parachute.

In this assignment of the parachute, it is design by cutting a material which is a plastic, into
a circle shape. The diameter chosen for the parachute is 0.8 m. However, to make the
parachute more stable, a small circle is cut at the middle of the parachute. This is because
when a parachute is released, the weight pulls down on the strings. Hence, the large surface
area of the parachute material provides air resistance to slow the parachute down (Telus
World of Science, 2019). The larger the surface area the more air resistance and the slower
the parachute will drop. To make the parachute more stable, the diameter and the load weight
consideration are very important. This is because a light object won’t be able to equal
resistance the way a heavy object will. When the diameter is big while the weight of the load
is small, it is not stable because the upward force and downward force is not stable as it can’t
be equal to the resistance.

The number used for the parachute is 17. The number of the rope must be an odd
number and it is depended on the diameter used. This is to make sure the surface area open
when its fall to the ground is equal. Hence, the distribution of the upward force is stable. The
length of the string also will affect the stability of the parachute. The length of the string must
be long as it will increase the stability of the parachute. When the parachute stable, the
velocity will increase. Thus, there will be less time taken for the parachute to reach the
terminal velocity. This means, the velocity will be increase as the length of the rope increases
and the parachute will be heavier.

The prototype test of the parachute took place on 19th December 2019, at the void shaft
of main stairs, in Block A of Faculty of Engineering of Universiti Malaysia Sabah. There is
something that went wrong during the prototype test.

The parachute was initially dropped from a height of 16.07 m at the highest floor of the
main stairs. This is to calculate the terminal velocity of the parachute as the ratio of height
to drop time. Unfortunately, for every set of experiment the group conducted, it was
observed that the parachute does not fall straight down the void shaft of the main stairs to
the lowest floor or the ground. The parachute always drifted to a side and landed in the
stairs above the ground.

One of the possible reasons why the parachute drifted to a side is due the uneven
parachute shape. The uneven parachute shape may come from the process of cutting the
plastic for the parachute. When the parachute shape is uneven, it will have a high tendency
to be dragged to a side by the wind. Furthermore, the uneven shape of the parachute will
cause uneven air flow around the parachute. Uneven air flow around the parachute will
produced a net force on one side of the parachute pushing it to a side while the parachute is
falling.

Another possible reason why the parachute drifted to a side may due to uneven lengths of
suspension strings attached to the parachute to the payload. When the lengths of the
suspension strings are different, the payload will not be hanging straight below the center of
the parachute. The payload will be hanging on a side under the parachute. When the
payload is hanging on a side under the parachute, an area under the parachute will be more
exposed to wind blow. Like the uneven parachute shape, the uneven lengths of the
suspension string will also cause uneven air flow around the parachute. All of these will have
a high tendency to cause the parachute to be unable to fall straight to the ground.

Furthermore, the drifting of the parachute might also come from the large area of the
parachute. A large chute area will make it difficult for the parachute to fall through the small
void shaft of the main stairs. More importantly, having a large chute area will cause the
parachute to be easily dragged to the side by the wind or by the unsteady and not uniform
air flow around the parachute.

In order to record the drop time to calculate the terminal velocity of the parachute, the
experiment was adjusted whereby now the parachute will be dropped 5 m from the ground.
By this mean, the experiment is successfully conducted as the recorded fall time for set 1,
set 2, and set 3 are 2.13 s, 2.00 s, and 2.56 s respectively.

The velocity of the parachute dropped from the 5 m height with 50 g load is 2.24 m/s
was obtained from the experiment which give the coefficient of drag of 0.6. From table 1 the
when the coefficient of drag is 0.6 and the diameter of the chute is 0.8 m, the velocity is
1.6328 m/s however the velocity that obtained from the experiment is not same with the value
in the table 1. Therefore, the different between the velocity obtained and the theoretical due
to the error occur during the experiment was conducted. The error might come from the
design of the parachute itself and the air from surrounding that effect the parachute to slide
to other direction rather than fall down straightly.

CONCLUSION

The velocity that obtained from the experiment is not the same with the theoretical
velocity even the diameter of the chute and the coefficient of the drag is the same. The
percentage error recorded in this experiment is 37.423 %. So the design of the parachute
should be suitable with the weight and should consider the stability too. The objective of this
experiment was accepted where the coefficient of drag for a parachute was determined.
References

1. Science World at Telus World of Science, 2019. Retrieved from :

https://www.scienceworld.ca/resource/parachute-design-and-drop/

2. Nguyen, Y. V. (1989). The parachute. Place of publication not identified: publisher not
identified.

3. H.M.S.O. (1951). Parachute design.

4. Parker, D., & Ottley, M. (2016). Parachute. Grand Rapids, MI: Eerdmans Books for
Young Readers.

5. Rysev, O. (1993). Mathematical simulation in parachute design. Aerospace Design

Conference. doi: 10.2514/6.1993-1202