Fluid Mechanics solutions

This document provides solutions for Fluid Mechanics problems and assignments. It covers topics such as pressure calculation, force exerted by water, fluid velocity, and Reynolds number. The solutions are explained step by step with relevant formulas and calculations. This study material is suitable for engineering students studying Fluid Mechanics.

Engineering Assignment 1
Fluid Mechanics solutions
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Course Title:

Engineering Assignment 2
Q2
(a) Ignoring atmospheric pressure, calculate the pressure at a point in the Atlantic
Ocean, 8377 m below the surface. Assume a constant density of seawater of 1025 kg m-3.
Take g=9.81 m s-2. Give your answer in megapascals (MPa) to 3 significant figures.
Solution:
The expression to find the pressure below the sea level is given by:
P=hρg
Here P is the pressure, h is the depth of water, g is the force of gravity ¿ ρ is the density of water.
h=8377 m , ρ=1025 kg m-3 and g=9.81 m s-2.
P=8377 ×1025 × 9.81=84232829.25Pa
Converting to megapascals (MPa) we divide the answer by one million
84232829.25
1000000 =84.23282825
To make it to 3 significant figures we have 84.2 MPa.
(b) Calculate the force exerted by the water on the circular viewing window of a submersible
operating a depth of 3163m below the surface, if the radius of the window is 0.800 m.
Ignore the effect of atmospheric pressure and the air pressure inside the submersible and
assume a constant density of seawater of 1025 kg m-3.
Take π=3.14 and g=9.81 m s-2. Give your answer in meganewtons (MN) to 3 significant
figures.
Solution:
Force is calculated as, F=ma where a is the acceleration and m is the mass of the object
(Fujii, Y. 2006).
In this question we will treat gravity, g as the acceleration.
We need to get the mass first.
We know that Density=mass/volume.
Now to get mass=density×volume
Volume of the water =πr2 h= 3.14× 0.82 ×3163=6356.3648 m-3.
Mass of water = volume of water × density of water =1025×6356. 3648=6515273.92

Engineering Assignment 3
Finally, force will be given by;
F=mg=65155273.92×9.81=63914837.2N
To change it to meganewtons we divide it by one million
63914837,2 N
1000000 =63.9148372NM
To 3 significant figure we have 63.9NM
Q3
An incompressible fluid flows in a horizontal circular pipe of non-constant diameter. You may
assume the pressure of the fluid to be constant. The fluid has dynamic viscosity 1.002× 10−3
Pas, and density 998 kg m-3.
At point 1 the pipe diameter is 80.0 mm and the fluid velocity is 1.90 m s-1.
At point 2 the pipe diameter is 51.0 mm.
(a) Calculate the fluid velocity at point 2. Give your answer in m s−1 to 3 significant figures.
Solution:
Fluid velocity= 4. flow rate
π .( pipe diameter)2
We first need to get the flow rate by using the formula (De Nevers, N. 1997) below
Flow rate= 1
4 π ¿
Flow rate = 1
4 π ¿=0.0038793915
Thus fluid velocity= 4 × 0.0038793915
π .(0.051)2 =0.015517566
0.00816714 =1.90m s-1.

Engineering Assignment 4
(b) Calculate the Reynolds number at point 2. Give your answer to 3 significant figures.
Solution
Viscosity of the fluidμ=1.002×10−3 Pa s.
Density of the fluid ρ=998 kg m−3 .
Diameter of the fluid L at point 2= 5.1×10−2m
Velocity v=1.9m s-1.
The Reynold formula is given by; we denote Reynolds number by Re
Re = ρVL
μ = 998 kg m−3 ×1. 9 m s−×1 5.1× 10−2 m
1.002× 10−3 Pa s =96513.173653
To 3 significant figures we have 96500
Reynolds number is used to check whether the flow is turbulent or laminar.

Engineering Assignment 5
References
De Nevers, N. (1997). Fluid mechanics for chemical engineers.
Fujii, Y. (2006). Method for generating and measuring the micro-Newton level
forces. Mechanical Systems and Signal Processing, 20(6), 1362-1371.