💼 Case: Consider a column of liquid of constant density $\rho$ (incompressible)

Force on top face = $P_0 A$

Weight of liquid above $h_1=A(h_0-h_1)\rho\times g$

Force at $h_1=P_0A+A\rho (h_0-h_1)g$

Pressure at $h_1=P(h_1)=P_0+\rho g (h_0-h_1)$

$P_0, h_0$ are constant, so $P_c=P_0+\rho g h_0$ is constant

$$ \begin{aligned} \text{thus:} \qquad P(h_1)&=P_c-\rho gh_1 \\ \text{or: } \; P(h)+\rho gh&= \text{constant}

\end{aligned} $$

✨ Pressure increases with depth $d$ at below the surface

$$ P(d)=P_0+\rho gd $$

Buoyancy and Archimedes’ principle

💼 Case: Consider a submerged shape of volume $V$ of density $\rho_L$ and contains liquid of density $\rho_L$

Forces:

$$ \begin{aligned} &1) \quad W=\rho_L Vg \\ &2) \quad \text{Complicated set of forces across} \\ &\quad\;\;\,\text{ the surface} \\ & \qquad \hookrightarrow \text{net buyonancy force}\\&\qquad\quad\;\;\, \text{(upthrust)}

\end{aligned} $$

In equilibrium 1) and 2) balance

$$ F_b=W=\rho_L Vg $$

💼 Case: Same situation as above except now the shape has density $\rho_s$

Again 2 Forces:

• Same buoyancy force Archimedes’ principle

$$ F_b=\rho _L Vg $$

• Weight is now

$$ W=\rho_sVg $$

Conservation of mass (the continuity equation)

For an incompressible liquid ($\rho$=constant), the rate of mass flow through a surface element $\text d\vec S$ is $\rho(\vec v\cdot\text d\vec S)$. Thus the rate of mass flow into a volume bound by surface S is:

$$ \rho\int_S\vec v\cdot \text d \vec S=0 $$

🗒️ Must be zero otherwise $\rho$ would not be constant

• Using divergence theorem we get $\vec \nabla \cdot \vec v=0$

🗞️ Example: Flow through a horizontal pipe

$$ \begin{aligned} \int_S \vec v\cdot \text d \vec S&= -v_1A_1+v_2A_2 \\&=0 \\ \Rightarrow v_1A_1&=v_2A_2

\end{aligned} $$

Bernoulli's principle

👷 Conditions: assume no viscosity or other surface forces and assume steady flow ( $\vec v$ only depends on $\vec r$ and not $t$ ). Particles in the liquid will follow lines called ‘streamlines’ with $\vec v (\vec r)$ always tangential to the lines.

💼 Case: Consider a tube of adjacent streamlines:

Area $A_1$ and pressure $P_1$ at $\text{\textcircled{\small 1}}$ ; $A_2$ and $P_2$ at $\text{\textcircled{\small 2}}$

• Continuity equation:

$$ |\vec v_1 |A_1=|\vec v_2| A_2 $$

• Work

$$ \begin{aligned} \text dW_1&=\overbrace{P_1A_1}^\text{Force}\times \overbrace{|\vec v_1|\text dt}^\text{Distance} \\ \text dW_2&=P_2A_2|\vec v_2|\text dt \end{aligned} $$