4.1 Fluid Statics, Manometry, Hydrostatic Force & Buoyancy

4.1 Fluid Statics, Manometry, Hydrostatic Force & Buoyancy

Fluid statics is the branch of fluid mechanics that analyzes fluids at rest. In static fluids, no relative motion exists between adjacent fluid layers, meaning shear stresses are strictly zero ($\tau = 0$). Consequently, all forces exerted by static fluids on submerged surfaces act normal (perpendicular) to the boundary surface. Mastery of fluid statics, hydrostatic pressure distributions, manometry, and floating stability is vital for mechanical engineers designing hydraulic structures, pressure vessels, storage tanks, and naval vessels.

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1. Fundamental Fluid Properties

To analyze fluid behavior under static and dynamic conditions, key physical properties must be defined:

• Density ($\rho$): Mass per unit volume, measured in $\text{kg/m}^3$ (SI) or $\text{slugs/ft}^3$ (English). For pure water at standard condition ($4^\circ\text{C}$), $\rho_w = 1000\text{ kg/m}^3 = 1.94\text{ slugs/ft}^3$.
• Specific Weight ($\gamma$): Weight per unit volume, defined as $\gamma = \rho g$. For water at $4^\circ\text{C}$, $\gamma_w = 9810\text{ N/m}^3 = 9.81\text{ kN/m}^3 = 62.4\text{ lbf/ft}^3$.
• Specific Gravity ($SG$): Dimensionless ratio of fluid density to standard reference fluid density (water for liquids, air for gases): $SG = \frac{\rho}{\rho_{water}} = \frac{\gamma}{\gamma_{water}}$
• Dynamic Viscosity ($\mu$): Measure of fluid resistance to shear deformation, expressed in $\text{Pa}\cdot\text{s}$ ($\text{N}\cdot\text{s/m}^2$) or $\text{lbf}\cdot\text{s/ft}^2$.
• Kinematic Viscosity ($\nu$): Ratio of dynamic viscosity to density: $\nu = \frac{\mu}{\rho} \quad [\text{m}^2/\text{s} \text{ or } \text{ft}^2/\text{s}]$
• Bulk Modulus of Elasticity ($K$): Measure of fluid compressibility, defined as the pressure increase required to produce a unit relative volume decrease: $K = -V \frac{dP}{dV} = \rho \frac{dP}{d\rho}$
• Surface Tension ($\sigma$): Tensile force per unit length acting at liquid-gas or liquid-liquid interfaces due to molecular cohesion ($\text{N/m}$). Capillary rise or depression in a tube of diameter $d$ is given by: $h = \frac{4 \sigma \cos \theta}{\gamma d}$

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2. Pascal's Law & Hydrostatic Pressure Distribution

Pascal's Law states that pressure applied to a confined static fluid is transmitted undiminished in all directions throughout the fluid and acts with equal force on equal areas.

In a static fluid column subjected to gravity, the hydrostatic pressure variation is derived from force equilibrium on an infinitesimal fluid element $dAz$:

$$\frac{dP}{dz} = -\gamma = -\rho g$$

Integrating for an incompressible fluid (constant density $\rho$) from the free surface where pressure is $P_0$ at height $z_0$ down to depth $h = z_0 - z$ yields the Fundamental Hydrostatic Equation:

$$P = P_0 + \rho g h = P_0 + \gamma h$$

Gage Pressure vs. Absolute Pressure

• Absolute Pressure ($P_{abs}$): Measured relative to absolute vacuum ($0\text{ Pa abs}$).
• Gage Pressure ($P_{gage}$): Measured relative to local atmospheric pressure ($P_{atm} = 101.325\text{ kPa} = 14.7\text{ psi}$).

$$P_{abs} = P_{gage} + P_{atm}$$

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3. Manometry Principles

Manometers measure static fluid pressure differences using balanced columns of liquid with known specific weights.

Manometer Equation Rules:

1. Start at a known point of pressure (e.g., pipe center $P_A$ or open atmosphere $P_{atm}$).
2. Move along the fluid columns: add hydrostatic pressure increment $+\gamma h$ when moving vertically downward, and subtract $-\gamma h$ when moving vertically upward.
3. Continuous fluid layers of identical density across a continuous horizontal plane have equal pressure.
4. Equate the accumulated sum to the pressure at the destination point $P_B$.

For a U-tube differential manometer connected between pipe A and pipe B containing manometric fluid of specific weight $\gamma_m$:

$$P_A + \gamma_1 h_1 - \gamma_m h_m - \gamma_2 h_2 = P_B \implies P_A - P_B = \gamma_2 h_2 + \gamma_m h_m - \gamma_1 h_1$$

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4. Hydrostatic Force on Submerged Surfaces

Plane Submerged Surfaces

For a flat surface of total area $A$ inclined at angle $\theta$ to the free surface:

1. Resultant Force Magnitude ($F_R$): The total force equals the product of pressure at the centroid ($P_c$) and total surface area ($A$):

   $$F_R = P_c A = \gamma h_c A = \gamma (y_c \sin \theta) A$$

   where $h_c$ is vertical depth to centroid, and $y_c$ is inclined distance from free surface along the plane.

2. Center of Pressure ($y_{cp}$): The location where resultant force $F_R$ acts. Because hydrostatic pressure increases linearly with depth, the center of pressure always lies below the centroid ($y_{cp} > y_c$):

   $$y_{cp} = y_c + \frac{I_{\bar{x}}}{y_c A}$$ $$x_{cp} = x_c + \frac{I_{\bar{x}\bar{y}}}{y_c A}$$

   where $I_{\bar{x}}$ is area moment of inertia about the centroidal axis parallel to the free surface.

Common centroidal area moments of inertia:

• Rectangle ($b \times h$): $A = bh$, $y_c = h/2$, $I_{\bar{x}} = \frac{b h^3}{12}$
• Circle (diameter $d$): $A = \frac{\pi d^2}{4}$, $I_{\bar{x}} = \frac{\pi d^4}{64}$
• Triangle (base $b$, height $h$): $A = \frac{bh}{2}$, $y_c = \frac{2h}{3}$ from apex, $I_{\bar{x}} = \frac{b h^3}{36}$

Curved Submerged Surfaces

Hydrostatic force on a curved surface is calculated by resolving into horizontal and vertical components:

• Horizontal Component ($F_H$): Equal to the force on the projection of the curved surface onto a vertical plane:

  $$F_H = \gamma h_{c,proj} A_{proj}$$

  acting through the center of pressure of the projected vertical area.

• Vertical Component ($F_V$): Equal to the weight of the real or imaginary fluid column extending directly above the curved surface up to the free surface level:

  $$F_V = \gamma V_{volume}$$

  acting vertically through the center of gravity of the fluid volume.

• Resultant Hydrostatic Force ($F_R$):

  $$F_R = \sqrt{F_H^2 + F_V^2}, \quad \theta_{force} = \arctan\left(\frac{F_V}{F_H}\right)$$

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5. Archimedes' Principle & Floating Body Stability

Archimedes' Principle

Any body submerged in or floating on a fluid experiences an upward Buoyant Force ($F_B$) equal to the weight of the fluid displaced by the body:

$$F_B = \rho_f g V_{submerged} = \gamma_f V_{submerged}$$

• Submerged Body: $F_B$ acts vertically upward through the center of buoyancy ($B$), which is the centroid of the displaced fluid volume.
• Floating Equilibrium: A body of weight $W = \gamma_{body} V_{body}$ floats in equilibrium when $W = F_B$, yielding the submerged volume ratio: $$\frac{V_{submerged}}{V_{body}} = \frac{\rho_{body}}{\rho_{fluid}} = \frac{SG_{body}}{SG_{fluid}}$$

Stability of Floating Bodies

When a floating vessel heels through a small tilt angle $\theta$, the center of buoyancy shifts from $B$ to $B'$ due to the reconfigured displaced volume shape. Lines of action of original vertical buoyancy and tilted buoyancy intersect at the Metacenter ($M$).

• Metacentric Height ($GM$): The distance between center of gravity ($G$) and metacenter ($M$):

  $$GM = MB - GB$$

  where $GB$ is vertical distance between centroid of body ($G$) and initial center of buoyancy ($B$).

• Metacentric Radius ($MB$):

  $$MB = \frac{I_{waterline}}{V_{submerged}}$$

  where $I_{waterline}$ is minimum moment of inertia of the waterline cross-sectional area about the tilting axis.

• Stability Criteria:

  1. Stable ($GM > 0$): $M$ is above $G$. A positive restoring moment $T_R = W \cdot GM \sin \theta$ rights the vessel.
  2. Neutral ($GM = 0$): $M$ coincides with $G$. No restoring or overturning moment is produced.
  3. Unstable ($GM < 0$): $M$ is below $G$. An overturning moment causes the vessel to capsize.

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6. Worked Numerical Examples

Example 1: Hydrostatic Force on an Inclined Rectangular Gate

A rectangular gate $2.0\text{ m}$ wide and $3.0\text{ m}$ high is hinged along its top horizontal edge. The gate is submerged in water ($\gamma = 9.81\text{ kN/m}^3$) inclined at an angle of $60^\circ$ to the horizontal. The top edge of the gate is located at a vertical depth of $4.0\text{ m}$ below the water surface. Calculate:

1. The total hydrostatic force $F_R$ acting on the gate.
2. The location of center of pressure $y_{cp}$ along the gate plane from the water surface.

Solution Step-by-Step:

• Step 1: Calculate inclined distance to top edge ($y_1$) and centroid ($y_c$):

  $$y_1 = \frac{h_1}{\sin 60^\circ} = \frac{4.0}{\sin 60^\circ} = 4.619\text{ m}$$ $$y_c = y_1 + \frac{h_{gate}}{2} = 4.619 + \frac{3.0}{2} = 6.119\text{ m}$$ $$h_c = y_c \sin 60^\circ = 6.119 \times 0.8660 = 5.300\text{ m}$$

• Step 2: Calculate gate area and hydrostatic force:

  $$A = b \times h = 2.0\text{ m} \times 3.0\text{ m} = 6.0\text{ m}^2$$ $$F_R = \gamma h_c A = (9.81\text{ kN/m}^3)(5.300\text{ m})(6.0\text{ m}^2) = 311.94\text{ kN}$$

• Step 3: Calculate center of pressure $y_{cp}$:

  $$I_{\bar{x}} = \frac{b h^3}{12} = \frac{2.0 \times (3.0)^3}{12} = 4.50\text{ m}^4$$ $$y_{cp} = y_c + \frac{I_{\bar{x}}}{y_c A} = 6.119 + \frac{4.50}{6.119 \times 6.0} = 6.119 + 0.1226 = 6.2416\text{ m}$$

• Vertical depth to center of pressure $h_{cp} = y_{cp} \sin 60^\circ = 6.2416 \times 0.8660 = 5.405\text{ m}$.

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Example 2: U-Tube Differential Manometer

A differential U-tube manometer containing mercury ($SG_m = 13.6$, $\gamma_m = 133.4\text{ kN/m}^3$) is connected between two pipelines A and B. Pipeline A contains water ($\gamma_w = 9.81\text{ kN/m}^3$) under pressure, while pipeline B contains oil ($SG_o = 0.85$, $\gamma_o = 8.34\text{ kN/m}^3$). Center of pipe A is $1.5\text{ m}$ higher than center of pipe B. The mercury level in the leg connected to pipe A is $0.40\text{ m}$ below pipe A centerline, and the mercury deflection $h_m$ is $0.25\text{ m}$. Determine pressure difference $P_A - P_B$.

Solution Step-by-Step:

• Step 1: Express elevation levels and specific weights:

  $$\gamma_A = 9.81\text{ kN/m}^3$$ $$\gamma_m = 13.6 \times 9.81 = 133.416\text{ kN/m}^3$$ $$\gamma_B = 0.85 \times 9.81 = 8.3385\text{ kN/m}^3$$

• Step 2: Apply hydrostatic manometer equation starting at $P_A$:

  $$P_A + \gamma_A h_A - \gamma_m h_m - \gamma_B h_B = P_B$$

  where: $h_A = 0.40\text{ m}$ (water column down to mercury interface in left leg) $h_m = 0.25\text{ m}$ (mercury column deflection) $h_B = 0.40 + 1.50 - 0.25 = 1.65\text{ m}$ (oil column up to pipe B centerline)

• Step 3: Solve for $P_A - P_B$:

  $$P_A + (9.81 \times 0.40) - (133.416 \times 0.25) - (8.3385 \times 1.65) = P_B$$ $$P_A + 3.924 - 33.354 - 13.759 = P_B$$ $$P_A - 43.189\text{ kPa} = P_B \implies P_A - P_B = +43.19\text{ kPa}$$

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Example 3: Metacentric Height of a Rectangular Pontoon

A rectangular pontoon is $10.0\text{ m}$ long, $6.0\text{ m}$ wide, and $3.0\text{ m}$ deep. It has a total mass of $90,000\text{ kg}$ and floats in seawater ($\rho_{sw} = 1025\text{ kg/m}^3$). Its center of gravity $G$ is located $1.80\text{ m}$ above the flat bottom base. Calculate:

1. The draft $d$ of the pontoon.
2. The metacentric height $GM$.
3. The restoring moment for a heel tilt angle of $\theta = 5^\circ$.

Solution Step-by-Step:

• Step 1: Determine submerged draft ($d$):

  $$W = m \cdot g = (90,000\text{ kg})(9.81\text{ m/s}^2) = 882.9\text{ kN}$$ $$F_B = \rho_{sw} g V_{sub} = (1025)(9.81)(L \times B \times d) = W$$ $$1025 \times 10.0 \times 6.0 \times d = 90,000 \implies d = \frac{90,000}{61,500} = 1.4634\text{ m}$$

• Step 2: Find centers of buoyancy ($B$) and gravity ($G$):

  $$KB = \frac{d}{2} = \frac{1.4634}{2} = 0.7317\text{ m from bottom}$$ $$KG = 1.8000\text{ m from bottom}$$ $$GB = KG - KB = 1.8000 - 0.7317 = 1.0683\text{ m}$$

• Step 3: Calculate metacentric radius ($MB$) and metacentric height ($GM$):

  $$I_{waterline} = \frac{L \cdot B^3}{12} = \frac{10.0 \times (6.0)^3}{12} = 180.0\text{ m}^4$$ $$V_{sub} = L \times B \times d = 10.0 \times 6.0 \times 1.4634 = 87.804\text{ m}^3$$ $$MB = \frac{I_{waterline}}{V_{sub}} = \frac{180.0}{87.804} = 2.0500\text{ m}$$ $$GM = MB - GB = 2.0500\text{ m} - 1.0683\text{ m} = +0.9817\text{ m}$$

  Since $GM = +0.982\text{ m} > 0$, the pontoon is floating stably!

• Step 4: Calculate restoring moment at $\theta = 5^\circ$:

  $$T_R = W \cdot GM \sin\theta = (882.9\text{ kN})(0.9817\text{ m}) \sin(5^\circ)$$ $$T_R = 866.74 \times 0.087156 = 75.54\text{ kN}\cdot\text{m}$$