Physics - Classical Mechanics - Exercises on Fluid Statics (part 1)

avatar

[Image1]

Introduction

Hey it's a me again @drifter1!

Today we continue with Physics, and more specifically the branch of "Classical Mechanics", in order to get into Exercises on Fluid Statics. This is part 1, where we will only cover density and pressure.

So, without further ado, let's get straight into it!


Density Examples

Let's fill out the following table.

Applying the definition of density ρ = m / V and knowing 3D Geometry it's a piece of cake.

Cuboid

First of all, the volume of the cuboid is 2000 cm3 or 2 x 10-3 m3, because 1 cm3 equals 1 x 10-6 m3.

Therefore, the density is:

Cube

The volume of the cube is:

and thus the length of each edge is approximately:

Sphere

The volume of the sphere can be calculated from its radius as follows:

and so the mass is:


Pressure Examples

Next up, let's calculate the pressure in fluids for various cases.

Below the Ocean

Calculate the Absolute pressure (p) and Gauge pressure (pg) at 4 km below the ocean. The density of seawater is 1.03 x 103 Kg/m3. Suppose an atmospheric pressure (patm) of 1.013 x 105 Pa and acceleration of gravity (g) of 9.8 m/s2.

All quantities are known, and so the absolute pressure is a simple substitution:

For gauge pressure we need to subtract the atmospheric pressure from that result, yielding:

Water Barrel with Oil Layer

Consider a water barrel with a layer of oil on top. The layer of oil is 10 cm thick, whilst the barrel has an overall depth of 60 cm. The density of water is about 1 x 103 Kg/m3. For the oil suppose a density of 0.7 x 103 Kg/m3.

Calculate the gauge pressure at:

  • the oil-water interface, and
  • the bottom of the barrel

At the oil-water interface only the weight of the oil adds to the pressure. In other words, the gauge pressure is:

At the bottom of the barrel, both the oil and the water need to be considered. The pressure at the top of the water is equal to that at the bottom of the oil, and so 686 Pa. The height of the water "layer" is 60 - 10 = 50 cm. And so the gauge pressure at the bottom is:

Manometer

Consider a manometer in the form of U-tube. One side is open to atmosphere, whilst pressure is applied to the other. The manometer is filled with mercury, which has a density of about 13.6 g/cm3.

For a difference in height between the two columns of 10 cm, calculate the applied pressure.

The difference in height indicates the pressure. All the quantities are known, and so the pressure can be calculated directly from ρgh.

Let's calculate only the gauge pressure:


RESOURCES:

Images

  1. https://pxhere.com/en/photo/1045542

Mathematical equations used in this article, where made using quicklatex.

Visualizations were made using draw.io.


Previous articles of the series

Rectlinear motion

Plane motion

Newton's laws and Applications

Work and Energy

Momentum and Impulse

Angular Motion

Equilibrium and Elasticity

Gravity

Periodic Motion

Fluid Mechanics


Final words | Next up

And this is actually it for today's post!

Next time we will get into examples around the remaining Fluid Static topics...

See ya!

Keep on drifting!

Posted with STEMGeeks



0
0
0.000
0 comments