Fluid Dynamics Assignment: Problems and Detailed Solutions

This assignment provides detailed solutions to several fluid dynamics problems. It covers topics such as calculating angular velocity and linear speed of a rotating shaft, determining angular and linear acceleration, and analyzing tension in cables during lifting operations. It also addresses problems related to pile drivers, including velocity at impact and average resisting force, as well as frequency of oscillation, maximum acceleration, and velocity in spring-mass systems. Numerical answers are calculated and explained in detail. Desklib offers more solved assignments for students.

Fluid Dynamics 1
FLUID DYNAMICS
by [NAME]
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Fluid Dynamics 2
Fluid Dynamics
Question 1
Shaft diameter = 400 mm
RPM = 2550
a. Angular Velocity
Each revolution generates an angle of 2 π radians
Therefore the angular velocity ¿ number of rovolutions per minute × 2 π radians
1revolution
¿ 2550
1minute × 2 π radians
1 revolution
¿ 5100 π radians per minute
¿ 85 π rad /sec
¿ 267.035 rad /s
b. Linear speed of a point on the circumference of the shaft
The linear speed = radius × angular velocity
¿ 0.2 m× 267.035
¿ 53.407 m/s
c. Angular acceleration
angular acceleretaion α = ∆ ω
∆ t

Fluid Dynamics 3
RPM=4500,
Therefore angular velocity ¿ 4500
1minute × 2 π radians
1 revolution
¿ 9000 π radians
minute
¿ 150 πradians /sec
¿ 471.239 rad / s
Therefore α = ∆ ω
∆ t = 471.239−267.035
30
¿ 6.8068 rad / s2
d. Linear acceleration
at=r ∆ ω
∆ t
But α = ∆ ω
∆ t
Therefore at=r α
¿ 0.2 ×6.8068
¿ 1.36136 m/s2
Question 2
a. Diagram of the System

Fluid Dynamics 4
b. Tension above the balancing mass
The machine moves for 3m
Therefore the diameter ¿ 3
π =0.9549m
Thus r =0.4775 m
ω= v
r = 2.4
0.4775
¿ 5.0265 rad / s
tension=mr ω2−mg
¿ 300 ×3 ×5.02652 −300× 9.81

Fluid Dynamics 5
¿ 19789.132 N
c. Tension above the machine
t ension=mr ω2−mg
¿ 150 ×3 ×5.02652−150× 9.81
¿ 9898.066 N
Question 3
a. Velocity of the Hammer
v2=u2 +2 as
But u=0
Therefore v2=2 as
¿ 2 ×9.81 ×2.25
¿ 44.145
v=6.644 m/s
b. Common Velocity
The potential energy of the pile driver¿ mgh=1000 × 9.81× 0.6
¿ 5886 N
The force of the hammer on the pile driver = the potential energy of the pile driver
Force of the hammer 5886=500 a

Fluid Dynamics 6
Therefore a=11.772 m/s2
v2=u2 +2 as
v2=2× 11.772× 2.25
v2=52.974
v=7.278 m/s
c. Average Resisting Force
Resisting force of the ground = the force exerted by the harmer and the pile
¿ 5886+5886
¿ 11772 N
Question 4
a. Frequency of Oscillation
F=18 N
x=25 mm
But F=kxwhere k is the spring constant
Therefore 18=0.025 k
k =720 N /m
ω= √ k
m= √ 720
20 =36 rad /s

Fluid Dynamics 7
Frequency f = ω
2 π = 36
2 π
¿ 5.73 Hz
b. Maximum Acceleration
amax=xmax × ω2 where xmax =maximum displacement
Therefore amax=0.025× 362
¿ 32.4 m/s2
c. Velocity when at 10 mm away
v=ω × A
¿ 36 ×0.01
¿ 0.36 m/s

Fluid Dynamics 8
References
Johnson, R.W., 2016. Handbook of fluid dynamics. Crc Press.
Munson, B.R., Okiishi, T.H., Huebsch, W.W. and Rothmayer, A.P., 2013. Fluid mechanics.
Singapore: Wiley.
Pain, H.J., 2017. THE PHYSICS OF VIBRATIONS AND WAVES Sixth Edition. John Wiley &
Sons Ltd.
Sone, Y., 2012. Kinetic theory and fluid dynamics. Springer Science & Business Media.