Fluid Mechanics · Chapter 3
Fluid Mechanics · Chapter 3

Pressure & Fluid Statics

When a fluid is at rest, the only stress it carries is pressure, and pressure variation is governed entirely by weight. This chapter builds from the definition of pressure to the tools of hydrostatics — manometers and barometers, hydrostatic forces on submerged surfaces, buoyancy and stability, and the pressure fields of fluids accelerating or rotating as a rigid body.

Chapter Objectives

  • Determine how pressure varies in a fluid at rest.
  • Calculate the forces a static fluid exerts on plane or curved submerged surfaces.
  • Analyze the rigid-body motion of fluids during linear acceleration or rotation.
  • Verify pressure references, force directions, displaced volumes, and limiting behavior.
Section 3–1

Pressure

Pressure

The normal force exerted by a fluid per unit area. Its SI unit is the pascal, \(1\ \text{Pa} = 1\ \text{N/m}^2\). The pascal is small, so kPa and MPa are common; the bar, standard atmosphere, and psi are also used (\(1\ \text{atm} = 101.325\ \text{kPa} = 14.696\ \text{psi}\)).

Absolute, Gage & Vacuum Pressure

Absolute pressure is measured from a perfect vacuum. Most instruments read gage pressure — the amount above local atmospheric. Pressures below atmospheric are vacuum pressures.

Pressure Relations
\[ P_{\text{gage}} = P_{\text{abs}} - P_{\text{atm}}, \qquad P_{\text{vac}} = P_{\text{atm}} - P_{\text{abs}} \]

Pressure at a Point

Pressure at any point in a fluid is the same in all directions — it is a scalar. A force balance on a wedge-shaped element confirms \(P_1 = P_2 = P_3\) regardless of orientation.

Variation of Pressure with Depth

Pressure does not change horizontally in a static fluid, but it increases with depth because deeper layers support the weight above. For constant density, the increase is linear with depth.

Pressure with Depth
\[ \Delta P = \rho g\,\Delta z = \gamma_s\,\Delta z, \qquad P = P_{\text{atm}} + \rho g h \]
General Variation (variable density)
\[ \frac{dP}{dz} = -\rho g \]
Pascal's Law

Pressure applied to a confined fluid increases the pressure throughout by the same amount. Because force equals pressure times area, two connected cylinders of different area form a hydraulic lever: a small force on the small piston balances a large force on the large piston.

\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \;\Rightarrow\; \frac{F_2}{F_1} = \frac{A_2}{A_1} \]
Section 3–2

The Manometer

Manometer Principle

A manometer measures pressure differences using a column of liquid. Three rules make any manometer solvable: (1) pressure changes by \(\rho g h\) across a fluid column of height \(h\); (2) pressure increases downward and decreases upward; (3) two points at the same elevation in the same continuous fluid are at equal pressure — so you may "jump" across at that level.

Basic Manometer Reading
\[ P = P_{\text{atm}} + \rho g h \]
Differential Manometer (across a device)
\[ P_1 - P_2 = (\rho_2 - \rho_1)\,g\,h \]
Gas, P h open to atm
Figure 3–1. A simple U-tube manometer. The pressure of the trapped gas equals atmospheric pressure plus the \(\rho g h\) of the manometer-fluid column. (Original schematic.)
Section 3–3

The Barometer & Atmospheric Pressure

Barometer

Atmospheric pressure is measured by a barometer: an inverted tube in a mercury reservoir, where the height of the mercury column balances the weight of the atmosphere. Pressure is read directly as a column height.

Barometric Pressure
\[ P_{\text{atm}} = \rho g h \]

Atmospheric pressure declines with altitude, with everyday consequences: water boils at lower temperature, naturally aspirated engines lose power, and aircraft need longer runways in thin air.

Section 3–4

Introduction to Fluid Statics

Fluid Statics

The study of fluids at rest (or in rigid-body motion). With no relative motion between fluid layers, there is no shear stress; the only stress is pressure, and pressure variation comes solely from the fluid's weight. It is called hydrostatics for liquids and aerostatics for gases.

Section 3–5

Hydrostatic Forces on Submerged Plane Surfaces

Resultant Hydrostatic Force

The total force a static fluid exerts on a plane surface equals the pressure at the surface's centroid times the area. Because pressure rises with depth, the line of action lies below the centroid, at the center of pressure.

Resultant Force on a Plane Surface
\[ F_R = P_C\,A = \big(P_0 + \rho g h_C\big)A \]
Center of Pressure (location below centroid)
\[ y_P = y_C + \frac{I_{xx,C}}{\,y_C\,A\,} \]

Here \(h_C\) is the centroid depth, \(y_C\) its slant distance from the free surface, and \(I_{xx,C}\) the second moment of area about the centroidal axis. For a surface symmetric about the vertical axis, the center of pressure lies directly below the centroid.

free surface gate P increases with depth center of pressure centroid
Figure 3–2. Pressure on a submerged vertical gate grows linearly with depth, so the resultant force acts at the center of pressure, located below the centroid. (Original schematic.)
Section 3–6

Buoyancy & Stability

Archimedes' Principle

The buoyant force on a body immersed in a fluid equals the weight of the fluid the body displaces, and it acts upward through the centroid of the displaced volume.

\[ F_B = \rho_f\, g\, V_{\text{sub}} \]

A body floats when the buoyant force balances its weight, so a floating body displaces a fluid volume whose weight equals its own. The apparent weight of a submerged object is its true weight minus the buoyant force.

Stability of Immersed & Floating Bodies

An immersed body is rotationally stable when its center of gravity \(G\) lies below the center of buoyancy \(B\); a disturbance then creates a restoring moment. A floating body can be stable even with \(G\) above \(B\), provided the shifting shape of the displaced volume moves \(B\) enough to restore it (a positive metacentric height).

Section 3–7

Fluids in Rigid-Body Motion

When a fluid container accelerates steadily or rotates at constant speed, the fluid eventually moves as a rigid body — no relative motion, hence no shear. Newton's second law then gives the pressure field.

Linear Acceleration

A liquid in a tank under constant horizontal acceleration \(a_x\) tilts its free surface; surfaces of constant pressure (including the free surface) become parallel inclined planes whose slope depends on the acceleration.

Free-Surface Slope under Linear Acceleration
\[ \frac{dz_{\text{isobar}}}{dx} = -\frac{a_x}{\,g + a_z\,} \]

Rotation in a Cylinder

A liquid rotating at constant angular velocity \(\omega\) is flung outward, and its free surface becomes a paraboloid of revolution. Surfaces of constant pressure are also paraboloids.

Free-Surface Shape under Rotation
\[ z_s(r) = h_c + \frac{\omega^2 r^2}{2g} \]
Instructor Note

For hydrostatic-force problems, draw the pressure triangle first — it tells you both the magnitude (area × centroidal pressure) and roughly where the center of pressure sits. And never forget atmospheric pressure cancels on surfaces exposed to it on both sides; work in gage pressure whenever you can.

Physical meaning

Static pressure is the isotropic normal stress required to balance body force. With \(+z\) upward, \(dP/dz=-\rho g\): pressure rises downward. Integrating that field gives instrument readings, resultant surface forces, buoyancy, and the free-surface shape in an accelerating frame.

Problem-solving strategy

  1. Confirm static or rigid-body motion; otherwise shear and flow balances may be required.
  2. Draw the fluid boundary and interfaces, declare \(+z\), and label every pressure as gauge or absolute.
  3. March through manometer columns or integrate the pressure field without changing sign conventions.
  4. For surfaces, integrate pressure for force and moment; for bodies, balance weight and buoyancy.
  5. Check units, pressure reference, force direction, center-of-pressure location, displaced volume, and limiting cases.
EXAMPLE 1Absolute pressure at a submerged sensor
A sensor lies \(2.40\ \text{m}\) below an open liquid surface. Given \(\rho=875\ \text{kg/m}^3\) at \(22^\circ\text{C}\), \(P_{\rm atm}=94.0\ \text{kPa}\) absolute, and \(g=9.81\ \text{m/s}^2\), find gauge and absolute pressure.

Recognition strategy

An open, constant-density liquid at rest is a direct hydrostatic pressure-rise problem; no Bernoulli or loss relation is appropriate.

Given information

  • Liquid property supplied at \(22^\circ\text{C}\): \(\rho=875\ \text{kg/m}^3\); depth \(h=2.40\ \text{m}\), \(P_{\rm atm,abs}=94.0\ \text{kPa}\).
  • Choose \(+z\) upward; the sensor is below the surface, so moving from surface to sensor is downward.

Model and interpretation

Use a static, incompressible, uniform-density model. \(dP/dz=-\rho g\) is exact for static equilibrium; constant \(\rho\) makes the finite-depth result approximate.

Step-by-step solution
1. Set the reference. The free surface has zero gauge pressure and \(94.0\ \text{kPa}\) absolute pressure.
2. Integrate downward. Constant density gives \(P_{\rm g}=\rho gh\), positive because pressure rises as \(z\) decreases.
3. Evaluate gauge pressure. \(P_{\rm g}=(875)(9.81)(2.40)=20.6\ \text{kPa}\).
4. Convert reference. \(P_{\rm abs}=P_{\rm atm}+P_{\rm g}=94.0+20.6=114.6\ \text{kPa}\).

Verification

Units: \(\text{kg/m}^3\) times \(\text{m/s}^2\) times metres gives \(\text{N/m}^2=\text{Pa}\).

Signs and direction: The sensor is lower than the free surface, so its pressure must exceed atmospheric; both computed increments have the correct positive sense.

Plausibility: This liquid gives about \(8.58\ \text{kPa}\) per metre, so \(2.4\ \text{m}\) should add about \(20.6\ \text{kPa}\).

Final result: \(P_{\rm g}=20.6\ \text{kPa}\); \(P_{\rm abs}=114.6\ \text{kPa}\).

Engineering interpretation

The sensor specification must use the same pressure reference. Density variation, surface acceleration, atmospheric uncertainty, and calibration would enter a field uncertainty budget.

EXAMPLE 2Force and line of action on a service gate
A vertical rectangular gate is \(1.20\ \text{m}\) wide and \(1.80\ \text{m}\) high, with its top \(0.60\ \text{m}\) below an open water surface. Given \(\rho=998\ \text{kg/m}^3\) at \(20^\circ\text{C}\), find resultant force and center-of-pressure depth.

Recognition strategy

A submerged plane under a linear hydrostatic distribution requires both the pressure integral and its moment, not pressure at a single edge.

Given information

  • \(b=1.20\ \text{m}\), \(H=1.80\ \text{m}\), top depth \(0.60\ \text{m}\), supplied \(\rho=998\ \text{kg/m}^3\) at \(20^\circ\text{C}\), \(g=9.81\ \text{m/s}^2\).
  • Atmosphere acts on both sides and cancels, so pressures are gauge; \(+z\) is upward and depth \(h\) is positive downward.

Model and interpretation

Choose the gate as the surface system. Water is static and approximately incompressible; \(F_R=\rho g h_C A\) and \(h_P=h_C+I_C/(h_CA)\) follow exactly from the linear gauge-pressure field.

Step-by-step solution
1. Locate the centroid. \(h_C=0.60+1.80/2=1.50\ \text{m}\), and \(A=bH=2.16\ \text{m}^2\).
2. Integrate pressure magnitude. \(F_R=\rho g h_C A=(998)(9.81)(1.50)(2.16)=31.7\ \text{kN}\).
3. Form the area moment. For the supplied rectangle geometry, \(I_C=bH^3/12=0.5832\ \text{m}^4\).
4. Locate the resultant. \(h_P=1.50+0.5832/[(1.50)(2.16)]=1.68\ \text{m}\) below the surface.

Verification

Units: Pressure times area yields newtons, while \(I_C/(h_CA)\) has metres and can be added to centroid depth.

Signs and direction: Fluid force acts normal to the vertical gate; increasing downward pressure places the center of pressure below the centroid.

Plausibility: The line of action at \(1.68\ \text{m}\) lies within the gate's \(0.60\) to \(2.40\ \text{m}\) depth span and below \(h_C=1.50\ \text{m}\).

Final result: \(F_R\approx31.7\ \text{kN}\); \(h_P\approx1.68\ \text{m}\) below the free surface.

Engineering interpretation

The hinge and frame must resist both force and moment. Structural design still needs dynamic water loads, corrosion, seals, tolerances, codes, and safety factors.

EXAMPLE 3Rotating wash-tank free-surface clearance
A cylindrical wash tank of radius \(0.450\ \text{m}\) rotates at \(90.0\ \text{rpm}\). Initially still, its liquid surface is level and \(0.500\ \text{m}\) below the rim. After rigid-body motion is established, determine the axis-to-wall difference and the remaining wall freeboard.

Recognition strategy

Constant rotation after transients produces rigid-body motion, so the free surface is an isobaric paraboloid rather than a flat hydrostatic surface.

Given information

  • Tank radius \(R=0.450\ \text{m}\), speed \(N=90.0\ \text{rpm}\), and initial still-surface-to-rim freeboard \(0.500\ \text{m}\); \(g=9.81\ \text{m/s}^2\).
  • Use \(+z\) upward from the initial still surface and \(r\) outward; the free surface is at atmospheric pressure, so gauge pressure there is zero. Temperature and density do not affect the shape.

Model and interpretation

Assume steady axisymmetric rigid-body rotation, negligible waves, no spill, and no slip after spin-up. Conserved liquid volume holds the mean surface elevation at its initial datum, so the axis falls and wall rises by half the axis-to-wall difference. The paraboloid relation is exact for this idealization; clearance is approximate because transients are omitted.

Step-by-step solution
1. Convert angular speed. \(\omega=2\pi N/60=2\pi(90.0)/60=9.425\ \text{rad/s}\).
2. Use the isobar condition. Integration gives \(z(r)-z(0)=\omega^2r^2/(2g)\).
3. Evaluate the axis-to-wall difference. \(\Delta z=(9.425)^2(0.450)^2/[2(9.81)]=0.917\ \text{m}\).
4. Enforce volume and compare clearance. The conserved-volume mean remains at the initial datum, so wall rise is \(\Delta z/2=0.4585\ \text{m}\) and wall freeboard is \(0.500-0.4585=0.0415\ \text{m}\).

Verification

Units: \(\omega^2R^2/g\) reduces from \(\text{s}^{-2}\text{m}^2/(\text{m/s}^2)\) to metres, and halving \(\Delta z\) preserves the elevation units used for freeboard.

Signs and direction: Elevation increases with outward radius because centrifugal acceleration is outward; relative to the initial datum, the wall is \(+0.4585\ \text{m}\) and the axis is \(-0.4585\ \text{m}\).

Plausibility: Conserved volume requires the paraboloid's area-average elevation to remain at the original level, giving equal-magnitude center fall and wall rise; the result tends to zero with either radius or speed.

Final result: Axis-to-wall difference \(\Delta z\approx0.917\ \text{m}\); wall rise \(0.4585\ \text{m}\); remaining wall freeboard \(0.0415\ \text{m}\).

Engineering interpretation

The no-spill idealization leaves only about \(42\ \text{mm}\) at the wall, a small margin rather than a \(0.417\ \text{m}\) deficit. Fill volume, wall datum, startup waves, vortex behavior, guards, and operating tolerances still require review.

Chapter recap

  • Integrate \(dP/dz=-\rho g\) with an explicit coordinate and reference.
  • Use pressure and pressure moment to find surface resultants.
  • Use displaced-fluid weight for buoyancy and effective body force for rigid motion.

Before moving on: you should be able to solve pressure columns, distinguish gauge from absolute pressure, locate a hydrostatic resultant, balance buoyancy, and assess an accelerating free surface.