Fluids

Pressure

$P = \frac{F}{A}$ Unit: pascal ($1\ \text{Pa} = 1\ \text{N/m}^2$).

Pressure at depth in a fluid: $P = P_0 + \rho g h$ where $P_0$ is atmospheric pressure, $\rho$ is fluid density, and $h$ is depth.

Pascal's Law

Pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid: $\frac{F_1}{A_1} = \frac{F_2}{A_2}$

Buoyancy and Archimedes' Principle

The buoyant force equals the weight of the displaced fluid: $F_b = \rho_{\text{fluid}} V_{\text{displaced}} g$

• Object floats if $\rho_{\text{object}} < \rho_{\text{fluid}}$.
• Object sinks if $\rho_{\text{object}} > \rho_{\text{fluid}}$.

Continuity Equation (Conservation of Mass)

For an incompressible fluid: $A_1 v_1 = A_2 v_2$

Bernoulli's Equation (Conservation of Energy)

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g y_2$

Special case — Torricelli's theorem (draining tank): $v = \sqrt{2gh}$

Problem: A wooden block ($\rho = 600\ \text{kg/m}^3$) with volume $0.01\ \text{m}^3$ is placed in water ($\rho = 1000\ \text{kg/m}^3$). What fraction is submerged?

Solution:

At equilibrium, buoyant force = weight: $\rho_w V_{\text{sub}} g = \rho_{\text{block}} V_{\text{block}} g$ $\frac{V_{\text{sub}}}{V_{\text{block}}} = \frac{\rho_{\text{block}}}{\rho_w} = \frac{600}{1000} = 0.6$

60% of the block is submerged.

• Pressure increases with depth: $P = P_0 + \rho g h$.
• Buoyant force = weight of displaced fluid.
• Continuity: $A_1 v_1 = A_2 v_2$ for incompressible flow.
• Bernoulli's equation relates pressure, speed, and height along a streamline.