Skip to content

Pressure

Intuition

Dive to the bottom of a pool and your ears hurt — equally, whichever way you tilt your head. That's the two essential facts about pressure in one experience: it grows with the weight of fluid above you, and at any single point it pushes the same in every direction. Pressure is not a vector pointing somewhere; it is a scalar intensity of squeezing.

Formal definition

Pressure is normal force per unit area, in the limit of a vanishing area element:

\[P = \lim_{\Delta A \to 0} \frac{\Delta F}{\Delta A}\]

with SI unit the pascal (\(\text{Pa} \equiv \text{N m}^{-2}\)). The force due to pressure is always normal to a surface, and pressure is isotropic — the same in all directions at a point.

Two conventions:

Quantity Reference Sign
Absolute pressure \(P_\text{abs}\) perfect vacuum always \(\geq 0\)
Gauge pressure \(P_\text{gauge}\) local atmosphere either sign
\[P_\text{abs} = P_\text{gauge} + P_\text{atm}, \qquad P_\text{atm} \approx 101.3\ \text{kPa at sea level}.\]

Derivation: why pressure is isotropic

Take a small triangular wedge of fluid at rest (base \(\Delta x\), height \(\Delta z\), hypotenuse at angle \(\theta\)), with pressures \(P_1, P_2, P_3\) on its three faces. Force balance:

\[\sum F_x = 0:\quad P_1 \Delta y\,\Delta z - P_3 \Delta y\, l \sin\theta = 0\]
\[\sum F_z = 0:\quad P_2 \Delta x\,\Delta y - P_3 \Delta y\, l\cos\theta - \tfrac{1}{2}\rho g\,\Delta x\,\Delta y\,\Delta z = 0\]

With \(\Delta x = l\cos\theta\) and \(\Delta z = l\sin\theta\), the first gives \(P_1 = P_3\); the second gives \(P_2 - P_3 = \tfrac{1}{2}\rho g \Delta z \to 0\) as the wedge shrinks. Hence

\[P_1 = P_2 = P_3\]

— pressure at a point is independent of the orientation of the surface you measure it on. This is why it is a scalar.

Worked example: the Venturi effect

A fluid of density \(\rho\) flows through a pipe that narrows from area \(A_1\) to \(A_2\). Manometer columns show a height difference \(h\). Find \(h\) in terms of \(v_1, A_1, A_2, g\).

From the continuity equation, \(A_1 v_1 = A_2 v_2\). From Bernoulli at equal elevation:

\[p_1 - p_2 = \tfrac{1}{2}\rho\left(v_2^2 - v_1^2\right) = \rho g h\]
\[\boxed{\,h = \frac{v_1^2}{2g}\left(\frac{A_1^2}{A_2^2} - 1\right)}\]

Faster flow ⇒ lower pressure: the fluid level drops over the constriction.

Common mistakes

  • Treating pressure as a vector. The force \(P\,d\mathbf{A}\) has direction; \(P\) itself does not.
  • Mixing gauge and absolute pressure in the ideal-gas law or in barometric formulas — those need absolute pressure.
  • Assuming pressure pushes only downward. Water pushes up on the bottom of a boat, sideways on a dam. Direction comes from the surface, not from \(P\).

Knowledge graph position

Prerequisites: What is a fluid? Leads to: Hydrostatic equilibrium, Buoyancy, Bernoulli's principle.

Quiz

Q1 (computational). A scuba diver is at 20 m depth in seawater (\(\rho = 1025\ \text{kg m}^{-3}\)). What is the gauge pressure and the absolute pressure?

Answer

\(P_\text{gauge} = \rho g h = 1025 \times 9.8 \times 20 \approx 201\ \text{kPa}\). \(P_\text{abs} = 201 + 101.3 \approx 302\ \text{kPa}\) — about 3 atm.

Q2 (conceptual). The wedge derivation assumed the fluid was at rest. Where exactly did that assumption enter?

Answer

We set the net force to zero (equilibrium) and included no shear stresses on the faces — only a fluid at rest exerts purely normal stresses. In a moving viscous fluid the stress tensor gains off-diagonal (shear) terms.

Q3 (multiple choice). A vertical dam wall holds back water. The pressure force on the wall is:

  • (a) vertical, equal to the weight of the water
  • (b) horizontal, increasing linearly with depth
  • (c) uniform over the wall
  • (d) zero, since the water is static
Answer

(b). Pressure acts normal to the wall (horizontally) and grows as \(\rho g h\) with depth — see forces on submerged surfaces.