# PHYS 170

Lecture 15 - Fluid Mechanics

## Introduction

Fluid could be a liquid (like water) or a gas (like air). In this chapter we will study both static fluid and the fluid flow.

The SI unit for volume is $m^3$. Unit conversion is a very basic skill for a science major. Make sure you know how to convert units to and from $cm^3$ and $L$ (liter) correctly or you will be penalized heavily. Here is a quick reminder: $$\begin{eqnarray} 1m &=& 100cm \tag{length}\\ \Rightarrow 1m^2 &=& (100cm)^2 = 10^4 cm^2 \tag{area}\\ \Rightarrow 1m^3 &=& (100cm)^3 = 10^6 cm^3 \tag{volume}\\ \end{eqnarray}$$ Using the above, we get: $$\begin{eqnarray} 1cm^3 &=& 10^{-6} m^3 \\ 1L &=& 1000cm^3 = 1000\times 10^{-6}m^3 = 10^{-3} m^3 \end{eqnarray}$$

### Density

Density $\rho$ is defined as the ratio of mass to volume:

$$\rho =\frac{m}{V}$$
The SI unit of density is $kg/m^3$.

Density of some common materials
Material Density ($kg/m^3$)
Air $1.2$
Water $1000$
Mercury $13600$
Gold $19300$

### Pressure

When a force $F$ is applied perpendicularly to a plance of area $A$, it exerts a pressure given by:

$$P = \frac{F}{A}$$
The SI unit is $Pa$ (Pascals), which is equivaldent to $N/m^2$.

A common non-SI unit for pressure is $atm$, or "atmospheric pressure". To convert to SI unit, use $1atm \approx 10^5Pa$.

Pressure on a person inside a fluid (such as a diver in deep sea, or you on Earth under the atmosphere) is given by the equation:

$$P = \rho g h$$
where $h$ is the depth of the person in the fluid, and $\rho$ is the density. The deeper you go inside a fluid, the higher is the pressure.

If you go diving, then the pressure acting on you comes from all the water above you as well as the atmospheric pressure above the surface of the water, in that case the pressure would be given by: $$P = P_{air} + P_{water} = P_0 + \rho_{water} g h$$ where $P_0$ is the atmospheric pressure.

## Buoyancy: Archimedes’ Principle

A fluid exerts an upward buoyant force on the body equal to the weight of the fluid displaced. $$B = m_{\text{water}}g$$

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Attribution - Tonyle [CC BY-SA 3.0], via Wikimedia Commons

Out of water the gold crown weights the same as a block of pure gold. If the gold crown contains impurities that is less dense than gold, then the gold crown would have a slightly larger volume compared to the pure gold. When immersed in water, the impure crown would then displace more water than the pure gold, generating a stronger buoyant force on the crown.
Attribution - MikeRun [CC BY-SA 4.0]
In the example shown in the figure, an object normally weights $4N$. When it is lowered into water, it displaces a volume of water weighing $3N$. Archimedes’ Principle tells us that the buoyant force must be $3N$ as well, pushing the mass upward. As a result the apparent weight of the immersed object is reduced from $4N$ to only $1N$.
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A ship can float in water because the bottom of the ship displaces a large volume of water. The weight of the water dispaced is $mg_{water}$, which by Archimedes’ Principle equals the buoyant force, therefore $B = mg_{water}$. When the amount of displaced water weighs the same as the ship itself, then the buoyant force will cancel the weight of the ship (i.e. $B=m_{ship}g$), keeping it afloat.
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The rock and the simple raft do not displace enough water, so the buoyant force is not strong enough to support the rock and it sinks. By modifying the shape of the raft, more water is displaced. This increases the weight of the water displaced ($mg_{water}$), therefore increasing the buoyant force $B=mg_{water}$, by Archimedes’ Principle). If the buoyant force is strong enough, then the rock and the raft could float.
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The Dead Sea has a high salt content. Since the density of salt water is higher than that of fresh water ($\rho_{salt\ water} > \rho_{fresh\ water}$)), salt water displaced by your body weighs more than an identical volume of fresh water. Since the weight of the fluid displaced is the same as the buoyant force, salt water will generate a stronger buoyant force, enough to float a person easily.
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Hot air balloon displaces a large volume of cold air, hence generating a buoyant force $B=m_{cold}g = \rho_{cold}Vg$. The weight of the air inside the ballon is hot, so it weighs $m_{hot}g = \rho_{hot}Vg$. Since the density of cold air is higher, we have $B = \rho_{cold}Vg > \rho_{hot}Vg$, in other words, the buoyant force is greater than the weight of the hot air in the balloon and creates a net upward force. When the net upward force is strong enough, it could float the balloon including the basket and the load it carries.
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The total volume of the submarine dispaces a fixed volume of water, so the total buoyant force on the submarine is fixed. By taking on water the submarine could increase its weight. When $m_{\text{submarine}}g > B$ the submarine sinks downward. By pumping out the water (using compressed air), the weight of the submarine decreases. When $m_{\text{submarine}}g < B$ the submarine surfaces.

## Fluid Flow - Bernoulli's Principle

In this course we will study incompressible fluid flow. "Incompressible fluid" is a (idealized) fluid that cannot be compressed, i.e. its volume always stays the same regardless of the pressure. When this assumption is satisified, the volume of fluid flowing into a pipe at point 1 must equal to the volume flowing out at point 2: $$A_1 v_1 = A_2 v_2$$ where $A$ and $v$ represent the cross-sectional area and the fluid velocity. This equation is called the "continuity equation".

Bernoulli's equation (along a flow line)):

$$P + \rho g h + \frac{1}{2} \rho v^2 = \text{constant}$$

Attribution - MannyMax (original) [CC BY-SA 3.0], via Wikimedia Commons
When apply to two points along the same flow line, it is usually written as: $$P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$$ For example, increasing $h$ decreases $P$ if velocity stays the same. In the otherwords, if the pipe's cross-sectional area stays the same (so that $v$ stays the same), the higher you are, the less water pressure you get. Symbollically:

Large $h \rightarrow$ Low $P$.

Attribution - User:HappyApple [Public domain], via Wikimedia Commons
A very interesting consequence of the Bernoulli's equation is how pressure decreases with increasing velocity (if $h$ stays the same). Symbollically:

Large $v \rightarrow$ Low $P$.

In the figure is a tube with uneven cross-section. Point 2 being narrower implies $v_2>v_1$ (because $A_1 v_1 = A_2 v_2$), as a result we have $P_2\lt P_1$.
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Attribution - Kraaiennest [CC BY-SA 3.0], via Wikimedia Commons
The wings of an airplane is shaped so that air flows at a higher velocity above the wings. By Bernoulli's Principle, $v_{\text{above}} > v_{\text{below}}$ implies $P_{\text{above}} \lt P_{\text{below}}$. The higher pressure below the wings therefore generates a lift pushing the plane upward.

Attribution - MatSouffNC858s [CC BY-SA 4.0], via Wikimedia Commons
The animation shows the principle behind a curve ball using a spinning cylinder. When a ball spins, it speeds up the air on one side but lowers the speed on the other. This difference in air velocity leads to a difference in pressure which curves the trajectory of the ball.

## Notations

Name Symbol Unit Meaning
Density $\rho$ $kg/m^3$ mass per unit volume
Pressure $P$ $Pa$ force per unit area
Buoyant force $B$ $N$ upward force from a fluid