Fluid Mechanics solutions

Engineering Assignment 1
Fluid Mechanics solutions
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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