Fluid Statics – Comprehensive Study Notes

Introduction & Session Objectives

  • Lecture context: Fluid Mechanics 1, Lecture #5, dated 23-Feb-2020, instructors Ernest Adaze (MS, BSc, MASME, MSPE) & M. K. Osman (PhD)
  • Session goals
    • Define pressure & units
    • Distinguish between gauge, absolute, vacuum pressure
    • Explain hydraulic-machine principles & solve sample problems
    • Derive pressure-variation relations under different conditions (incompressible & compressible fluids)
    • Demonstrate pressure-measurement devices
    • Compute resultant forces on plane & curved surfaces immersed in fluids and locate their lines of action (centre of pressure)

Fluid Statics Fundamentals

  • Fluid statics studies fluids with no relative motion between particles ⇒ \text{velocity gradient}=0
    • Consequence: No shear stresses, only normal stresses (pressure forces) acting perpendicular to boundaries
  • Equilibrium requirements for a fluid element at rest
    • \sum F =0\quad \text{(all directions)}
    • \sum M =0\quad \text{(about any point)}

Pressure: Definition & Properties

  • Mathematical definition: p = \frac{F}{A} where
    • F = normal compressive force
    • A = area upon which force acts
  • Scalar quantity
    • Magnitude only, directionless in vector sense
    • Acts equally in all directions at a point (isotropic)

Stress-Equilibrium in an Element (Triangular Prism)

  • For point P inside a triangular prism (Fig. 1): pressures px,\;pz,\;p_n relate by equilibrium
    • pn = pz (from moment/force balance)
    • px = pn = p_z ⇒ pressure at a point is the same in all directions

Pascal’s Law & Pressure Transmission

  • Statement: A pressure change at any point in a confined incompressible fluid is transmitted undiminished to every point of the fluid and to the confining walls
    • Illustrated via closed system in Fig. 2
  • Applications (Fig. 3 & Fig. 4)
    • Hydraulic brakes, jacks, lifts
    • Compressor supplies air pressure → oil column → piston & car rise
  • Example (pressure amplifier)
    • Given small handle force F1=100\,\text{N} acting on small piston A1, determine supported load F2 on large piston A2 using \dfrac{F1}{A1}=\dfrac{F2}{A2} (neglect lift weight)

Absolute, Gauge & Vacuum Pressure

  • Absolute pressure referenced to perfect vacuum (p_{\text{abs}}=0)
  • Atmospheric pressure (p_a) is intermediate reference
  • Gauge pressure: p{\text{gauge}} = p{\text{abs}} - p_a
    • If p{\text{abs}}
  • Relation: p{\text{abs}} = pa + p_{\text{gauge}}
  • Units
    • SI: Pascal 1\,\text{Pa}=1\,\text{N/m}^2
    • Other: \text{psi},\;\text{kPa},\;\text{bar},\;\text{atm}
    • Annotation examples: 50\,\text{kPa gauge},\;150\,\text{kPa abs},\;8\,\text{psig},\;22\,\text{psia}

Pressure Variation with Elevation

Basic Hydrostatic Differential Equation

  • Consider cylindrical fluid element (Fig. 1) inclined at angle \alpha
  • Force equilibrium: p\,\Delta A - (p+\Delta p)\,\Delta A - \gamma\,\Delta A\,\Delta l\sin\alpha =0
    • Simplified: \dfrac{\Delta p}{\Delta l}= -\gamma\sin\alpha
  • As \Delta l \to 0,\; \sin\alpha = \dfrac{dz}{dl}
    • Result: \boxed{\dfrac{dp}{dz} = -\gamma}

Consequences

  • Pressure is constant on any horizontal plane (dz=0 \Rightarrow dp=0)
  • Pressure increases with depth (negative slope indicates rise when z decreases)

Uniform-Density (Incompressible) Fluid

  • \gamma = \text{const} ⇒ integrate:
    • p+\gamma z = C (piezometric pressure constant along same streamline)
    • Between two points 1 & 2: \dfrac{p1-p2}{\gamma}=z2-z1

Example 2.2

  • Tank: bottom 1\,\text{m} water + 0.5\,\text{m} kerosene open to atmosphere
    • Compute gauge pressure at bottom via layered integration:p=\gamma{\text{ker}}\,0.5+\gamma{\text{water}}\,1.0

Compressible Fluids (Ideal Gas)

  • p=\rho RT,\;\gamma=\rho g ⇒ \gamma =\dfrac{pg}{RT}
  • Substitute into hydrostatic equation:
    • \dfrac{dp}{dz}= -\dfrac{pg}{RT}
    • Rearranged: \boxed{\dfrac{dp}{p}= -\dfrac{g}{R T}dz} …… (*)
    • Requires T(z) relation for integration

Atmospheric Pressure Variation

  • Atmosphere divided into five layers; engineering focus on
    • Troposphere (sea level → 13.7\,\text{km})
    • Stratosphere (lower: 13.7\to16.8\,\text{km})

Troposphere (linear lapse rate)

  • T = T0 - \alpha(z-z0) with lapse rate \alpha
  • Integration of (*) gives
    \boxed{\left(\dfrac{p}{p0}\right)^{R\alpha/g}= \dfrac{T}{T0}}
    or common form
    \displaystyle p = p0\left[1 - \dfrac{\alpha (z - z0)}{T_0}\right]^{\frac{g}{R\alpha}}

Lower Stratosphere (isothermal)

  • T = \text{const} ⇒ integrate (*) directly:
    \boxed{p = p0\,e^{-\frac{g(z-z0)}{RT}}}

Pressure Measurement Devices

Barometer (Mercury)

  • Measures atmospheric pressure using Hg column of height h
    • pa = \gamma{Hg} h (vacuum above column at top)

Piezometer (Simple Manometer)

  • Transparent vertical tube attached to pipe
    • Gauge pressure at connection: p_1 = \gamma h
    • Suitable only for liquids, low pressures; impractical for gases or high p (requires tall column)

U-Tube Manometer

  • Contains manometric fluid (often mercury)
  • Pressure relation (Fig. ): from point 1 to atmosphere
    p1 = \gammam \Delta h - \gamma_l h
  • General pressure difference between two points (1 & 2):
    \boxed{p1 - p2 = \sum (\gamma \Delta z)_{\text{manometer path}}}

Differential Manometer

  • Connects two points in same/different pipes; measures p1 - p2
    • Horizontal configuration: p1-p2 = (\gamma_m-\gamma)\,\Delta h
    • Inclined configuration includes geometric offsets; general relation:
      p1-p2 = (\gammaB-\gammaA)\,\Delta h with correction terms if datum elevations differ

Bourdon-Tube Gauge

  • Curved metal tube straightens when pressurised
  • Pointer linkage converts motion → dial reading (gauge pressure = system – atmospheric)
  • Widely used for steam & compressed gas systems

Pressure Transducer

  • Converts pressure → electrical signal (strain-gage, piezoelectric, capacitive, etc.)
  • Output fed to oscillographs, digital indicators, or control circuits

Hydrostatic Forces on Plane Surfaces

Horizontal Plane Surface (Fig. 1)

  • Uniform pressure: p = \gamma zc where zc = depth of centroid
  • Resultant force: F = p A acting normal through centroid

Inclined or Vertical Plane Surface

  • Geometry (Fig. 3): plane of area A inclined at angle \alpha; vertical distance y = z/\sin\alpha
  • Differential force: dF = \gamma y \sin\alpha\,dA
  • Total force: F = \gamma\sin\alpha \intA y\,dA = \gamma A yc \sin\alpha
    • Since yc\sin\alpha = zc ⇒
      \boxed{F = \gamma A z_c} (pressure at centroid × area)

Centre of Pressure (Line of Action)

  • Moment equilibrium about surface trace o-o:
    y{cp}\,F = \gamma\sin\alpha \intA y^2 dA
  • Using second moment of area I0 and parallel-axis theorem (I0 = Ic + A yc^2): y{cp} = yc + \dfrac{Ic}{A yc}
    • Hence CP lies below centroid by \dfrac{Ic}{A yc}
  • Useful tabulated I_c values given in Appendix (Fig. A.1)

Summary Formulae (Plane Surface)

  • \boxed{F{hydro}=\gamma A zc}
  • \boxed{y{cp}=yc+\dfrac{Ic}{A yc}}

Hydrostatic Forces on Curved Surfaces

  • Pressure vectors normal to surface vary directionally ⇒ integrate via components
  • Procedure (Fig. 6)
    1. Envision fluid volume bounded by curved surface + vertical & horizontal projections
    2. Horizontal component: Fx = F{AC} (force on vertical projection AC) through its CP
    3. Vertical component: Fy = W + F{CB}
    • W weight of fluid directly above surface (acts at CG of that fluid)
    • F_{CB} force on horizontal projection CB (acts at its centroid)
    1. Resultant force on curved surface equals \sqrt{Fx^2+Fy^2}; direction from vector addition; line of action via moment summation

Illustrative Example Statements (no numeric solutions provided in transcript)

  • Example 1: Concrete form 2.44 m × 1.22 m; find force of fresh concrete (γ = 23.6 kN/m³)
  • Example 2: Elliptical gate 4 m dia, hinged top; water 8 m above top; determine opening force
  • Example 3: Gate on block with dimensions d = 12 m, h = 6 m, w = 6 m; compute gate force
  • Example 4: Circular arc AB (radius 2 m, width 1 m), EB = 4 m; atmospheric on both sides; find force magnitude & line of action
  • Pressure-layer Example: Oil (SG 0.80, depth 0.9 m) above water; total depth 3 m; compute gauge pressure at bottom (γw = 9810 N/m³, γoil = 7850 N/m³)
  • Multifluid manometer Example: Tank with air over oil & mercury columns distances l1 = 0.40 m, l2 = 1.00 m, l3 = 0.80 m; γoil = 7850 N/m³, γHg = 133 kN/m³; determine air pressure

Key Ethical & Practical Notes

  • Hydraulic devices exploit Pascal’s principle for mechanical advantage; importance of seal integrity (safety)
  • Pressure measurement accuracy depends on calibrations & fluid properties (density vs. temperature)
  • Centre-of-pressure calculations essential for dam, gate, and tank-wall design to prevent structural failure

Numerical & Unit Reminders

  • 1\,\text{Pa}=1\,\text{N/m}^2
  • 1\,\text{bar}=10^5\,\text{Pa}; 1\,\text{atm}\approx101.3\,\text{kPa}
  • Specific weight \gamma = \rho g; for water at 10^\circ\text{C},\;\gamma=9810\,\text{N/m}^3
  • Ideal gas constant (air): R=287\,\text{J/(kg·K)}

Study Tips

  • Memorise the hydrostatic equation dp/dz=-\gamma and its integrations for different density assumptions
  • Practice shifting between gauge & absolute pressures carefully; annotate units clearly (e.g., kPa g, kPa abs)
  • When using manometers, walk through the fluid columns sequentially, adding +\gamma\Delta z when moving down a column and -\gamma\Delta z when moving up
  • For plane-surface force problems, always locate centroid first, then use I_c values from tables
  • Curved-surface problems simplify by treating horizontal & vertical projections separately