4.1 Fluid Statics, Manometry, Submerged Surfaces & Buoyancy

Fluid Statics, Manometry, Submerged Surfaces & Buoyancy

Fluid statics investigates fluids at rest where shear stresses are identically zero ($\tau = 0$), leaving normal compressive stress—pressure ($p$)—as the sole active surface force. Mastery of fluid properties, hydrostatic pressure distribution, multi-fluid manometer balancing, submerged plane and curved surface loading, and rotational buoyancy stability constitutes the cornerstone of thermal-fluid engineering on the NCEES PE Mechanical exam.

+---------------------------------------------------------------------------------------------------+
|                                 FLUID STATICS CORE ARCHITECTURE                                   |
|                                                                                                   |
|   [FLUID PROPERTIES]           [PRESSURE FIELDS]             [SURFACE RESULTANTS]                 |
|   - Density (\rho), \gamma     - p = \rho*g*h = \gamma*h     - Plane: F_R = \gamma * h_c * A      |
|   - Specific Gravity (SG)      - Multi-fluid Manometry       - Center of Pressure: y_cp           |
|   - Viscosity (\mu, \nu)       - Absolute vs Gauge           - Curved: F_H (proj) & F_V (weight)  |
|                                                                       |                           |
|                                                                       v                           |
|                                                              [BUOYANCY & STABILITY]               |
|                                                              - F_b = \gamma * V_disp              |
|                                                              - Metacenter: GM = BM - BG           |
|                                                              - BM = I_waterline / V_disp          |
+---------------------------------------------------------------------------------------------------+

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1. Fluid Properties and Continuum Fundamentals

A fluid is a substance that deforms continuously under the application of a shear stress, no matter how small that shear stress may be.

Density, Specific Weight, and Specific Gravity

• Mass Density ($\rho$): Mass per unit volume ($\text{kg/m}^3$ in SI; $\text{slug/ft}^3$ or $\text{lbm/ft}^3$ in US Customary). For standard liquid water at $4^\circ\text{C}$ ($39.2^\circ\text{F}$): $$\rho_{\text{water}} = 1000\text{ kg/m}^3 = 1.940\text{ slugs/ft}^3 = 62.43\text{ lbm/ft}^3$$
• Specific Weight ($\gamma$): Gravitational weight per unit volume: $$\gamma = \rho g$$ For standard water at $68^\circ\text{F}$ ($20^\circ\text{C}$), $\gamma_{\text{water}} = 9.807\text{ kN/m}^3 = 62.4\text{ lbf/ft}^3$.
• Specific Gravity ($SG$): Dimensionless ratio of fluid density to standard water density: $$SG = \frac{\rho}{\rho_{\text{water, } 4^\circ\text{C}}} = \frac{\gamma}{\gamma_{\text{water, } 4^\circ\text{C}}}$$ $$\implies \rho = SG \times \rho_{\text{water}}, \quad \gamma = SG \times \gamma_{\text{water}}$$

Viscosity: Dynamic and Kinematic

Shear stress in a Newtonian fluid in laminar shear flow is governed by Newton's Law of Viscosity:

$$\tau = \mu \frac{du}{dy}$$

Where:

• $\tau = \text{Shear stress (Pa or } \text{lbf/ft}^2\text{)}$
• $\mu = \text{Dynamic (absolute) viscosity } (\text{Pa}\cdot\text{s} = \text{N}\cdot\text{s/m}^2 = \text{kg/(m}\cdot\text{s)} \text{ in SI; } \text{lbf}\cdot\text{s/ft}^2 = \text{slug/(ft}\cdot\text{s)} \text{ in US Customary})$
• $\frac{du}{dy} = \text{Velocity gradient or shear rate } (\text{s}^{-1})$

+---------------------------------------------------------------------------------------------------+
|                             VISCOSITY CONVERSIONS & DEFINITIONS                                   |
|                                                                                                   |
|   Kinematic Viscosity (\nu):         \nu = \frac{\mu}{\rho}  [\text{m}^2/\text{s} \text{ or } \text{ft}^2/\text{s}]    |
|                                                                                                   |
|   Common Metric Units:               1 Poise (P) = 0.1 Pa·s = 100 Centipoise (cP)                 |
|                                      1 Stoke (St) = 10^-4 m^2/s = 100 Centistokes (cSt)           |
|                                      Water at 20°C: \mu \approx 1.0 cP = 1.002 \times 10^-3 Pa·s   |
|                                                     \nu \approx 1.0 cSt = 1.004 \times 10^-6 m^2/s |
+---------------------------------------------------------------------------------------------------+

Bulk Modulus of Elasticity and Surface Tension

• Bulk Modulus ($E_v$ or $K$): Quantifies fluid compressibility under isotropic hydrostatic pressure: $$E_v = -v \frac{dP}{dv} = \rho \frac{dP}{d\rho}$$ The speed of sound $c$ in a liquid medium is directly related to bulk modulus: $c = \sqrt{E_v / \rho}$. For water, $E_v \approx 2.2\text{ GPa}$ ($3.19 \times 10^5\text{ psi}$), confirming liquids are virtually incompressible under conventional engineering pressures.
• Surface Tension ($\sigma$): Tensile force per unit length acting at liquid-gas or liquid-liquid interfaces ($\text{N/m}$ or $\text{lbf/ft}$). Capillary height $h$ in a circular tube of inner diameter $d$ with contact angle $\theta$ is: $$h = \frac{4 \sigma \cos \theta}{\gamma d}$$

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2. Hydrostatic Pressure Field and Absolute vs. Gauge Pressure

In a static fluid, the pressure gradient depends solely on the local body force of gravity:

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

Integrating for an incompressible fluid ($\rho = \text{const}$) from free surface elevation $z_0$ where pressure is $p_0$ down to elevation $z$ at depth $h = z_0 - z$:

$$p(h) = p_0 + \rho g h = p_0 + \gamma h$$

+---------------------------------------------------------------------------------------------------+
|                             PRESSURE DATUM RELATIONSHIPS                                          |
|                                                                                                   |
|   [Absolute Pressure (p_abs)] = [Gauge Pressure (p_gauge)] + [Atmospheric Pressure (p_atm)]       |
|                                                                                                   |
|   [Vacuum Pressure (p_vac)]   = [Atmospheric Pressure (p_atm)] - [Absolute Pressure (p_abs)]     |
|                                                                                                   |
|   Standard Atmospheric Constants:                                                                 |
|   1 atm = 101.325 kPa = 1.01325 bar = 14.696 psia = 2116.2 lbf/ft^2                               |
|         = 29.92 inHg = 760 mmHg = 33.91 ft of water = 10.33 m of water                            |
+---------------------------------------------------------------------------------------------------+

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3. Multi-Fluid Manometers and Differential Pressure Balancing

A manometer measures pressure differences across fluid columns using hydrostatic principles. The golden rule of manometer analysis is:

$$\text{Start at point 1} \longrightarrow \text{Add } (+\gamma_i h_i) \text{ when moving DOWN} \longrightarrow \text{Subtract } (-\gamma_i h_i) \text{ when moving UP} \longrightarrow \text{Equate to point 2}$$

+---------------------------------------------------------------------------------------------------+
|                         DIFFERENTIAL U-TUBE MANOMETER SCHEMATIC                                   |
|                                                                                                   |
|       Pipe A (Fluid A, \gamma_A)                       Pipe B (Fluid B, \gamma_B)                 |
|             (•) p_A                                          (•) p_B                              |
|              |                                                |                                   |
|              | h_1                                            | h_3                               |
|              |                                                |                                   |
|              +-----[Interface 1]                              |                                   |
|                     \                                         |                                   |
|                      \  Manometer Fluid (\gamma_m)     +-----[Interface 2]                        |
|                       \                               /                                           |
|                        \                             / h_2 (Deflection)                           |
|                         +---------------------------+                                             |
|                               (Datum Level)                                                       |
+---------------------------------------------------------------------------------------------------+

Equation Setup for Differential Manometer

Tracing the path from Pipe A to Pipe B:

$$p_A + \gamma_A h_1 - \gamma_m h_2 - \gamma_B h_3 = p_B$$ $$\implies p_A - p_B = \gamma_m h_2 + \gamma_B h_3 - \gamma_A h_1$$

Inclined Manometers for High Sensitivity

To measure tiny gas pressure differentials $\Delta p$, the manometer limb is inclined at an angle $\theta$ to the horizontal. The scale reading $L$ along the incline magnifies the vertical deflection $h = L \sin \theta$:

$$\Delta p = \gamma_m L \sin \theta \implies L = \frac{\Delta p}{\gamma_m \sin \theta}$$

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4. Hydrostatic Forces on Submerged Plane Surfaces

When a planar surface of arbitrary geometry and area $A$ is submerged in a liquid at an angle $\theta$ relative to the free surface:

+---------------------------------------------------------------------------------------------------+
|                        SUBMERGED PLANE SURFACE GEOMETRY & LOADING                                 |
|                                                                                                   |
|   Free Surface //////////////////////////////////////////////////////                             |
|               \                                                                                   |
|                \  Angle \theta                                                                    |
|                 \                                                                                 |
|                  \ y_c (Distance along incline to Centroid C)                                     |
|                   \-----------------------> [ CENTROID (C) ]  Depth h_c = y_c * sin(\theta)       |
|                    \                        Pressure p_c = \gamma * h_c                           |
|                     \ y_cp (To Center of P)                                                       |
|                      \--------------------> [ CENTER OF PRESSURE (CP) ]                           |
|                                             Line of Action for Total Force F_R                    |
+---------------------------------------------------------------------------------------------------+

Resultant Hydrostatic Force Magnitude

The total resultant force $F_R$ equals the pressure at the centroid multiplied by the total plate area:

$$F_R = \int_A p \, dA = \int_A (\gamma y \sin \theta) \, dA = \gamma \sin \theta \int_A y \, dA = \gamma \sin \theta (y_c A) = p_c A$$ $$F_R = \gamma h_c A$$

Where:

• $h_c = y_c \sin \theta = \text{Vertical depth from free surface to the centroid of the area}$
• $A = \text{Total submerged surface area}$

Center of Pressure Location ($y_{cp}, x_{cp}$)

Because hydrostatic pressure increases linearly with depth, the pressure distribution is trapezoidal (or triangular). The resultant force acts below the centroid at the center of pressure $(x_{cp}, y_{cp})$:

$$y_{cp} = y_c + \frac{I_{xc}}{y_c A}$$ $$x_{cp} = x_c + \frac{I_{xyc}}{y_c A}$$

Where $I_{xc}$ is the area moment of inertia of the surface about its horizontal centroidal axis. For symmetric shapes, $I_{xyc} = 0$, so $x_{cp} = x_c$.

Standard Geometric Section Properties

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5. Hydrostatic Forces on Submerged Curved Surfaces

For curved surfaces, the pressure vectors vary continuously in direction across every differential element $dA$. Directly integrating vectors is avoided by resolving the total force into horizontal ($F_H$) and vertical ($F_V$) components.

+---------------------------------------------------------------------------------------------------+
|                         FORCES ON SUBMERGED CURVED SURFACES                                       |
|                                                                                                   |
|   Free Surface ///////////////////////////////////////                                            |
|               |                     |                                                             |
|               |  Fluid Volume (V)   |                                                             |
|               |  Directly Above     |                                                             |
|               |  Curved Surface AB  |                                                             |
|               |                     v                                                             |
|               +=================( Surface AB )                                                    |
|               | <--- F_H         /                                                                |
|               |                 /  F_R = \sqrt{F_H^2 + F_V^2}                                     |
|               v Projection A_v /                                                                  |
|                                                                                                   |
|   1. Horizontal Component F_H = Hydrostatic force on vertical projection A_v                      |
|   2. Vertical Component   F_V = Total weight of fluid column situated above surface AB            |
+---------------------------------------------------------------------------------------------------+

Formulation Matrix

1. Horizontal Force Component ($F_H$): $$F_H = \gamma h_{c,v} A_v$$ Where $A_v$ is the area of the vertical projection of the curved surface, and $h_{c,v}$ is the depth to the centroid of this projected vertical plane. $F_H$ acts through the center of pressure of the vertical projection ($y_{cp,v} = y_{c,v} + I_{xc,v}/(y_{c,v} A_v)$).
2. Vertical Force Component ($F_V$): $$F_V = \gamma V$$ Where $V$ is the volume of the real (or imaginary) fluid column extending vertically from the curved surface up to the free surface plane. $F_V$ acts vertically downward (if fluid is above the surface) or upward (if fluid is below) through the centroid of the fluid volume $V$.
3. Resultant Force Magnitude and Direction: $$F_R = \sqrt{F_H^2 + F_V^2}, \quad \theta = \arctan\left(\frac{F_V}{F_H}\right)$$ For circular-arc surfaces (e.g., Tainter radial gates), the local pressure vector on every surface element is normal to the boundary and thus passes through the center of curvature $O$. Therefore, the resultant force $F_R$ must pass directly through the arc's center of curvature.

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6. Archimedes' Principle, Buoyancy and Rotational Stability

Archimedes' Principle

Any body completely or partially submerged in a static fluid experiences an upward buoyant force $F_b$ equal to the weight of the fluid displaced by the body:

$$F_b = \rho_f g V_{\text{disp}} = \gamma_f V_{\text{disp}}$$

• The buoyant force acts vertically upward through the Center of Buoyancy ($B$), which is the geometric centroid of the displaced fluid volume $V_{\text{disp}}$.
• The gravitational force $W = m g = \rho_{\text{body}} g V_{\text{body}}$ acts vertically downward through the Center of Gravity ($G$) of the body.

+---------------------------------------------------------------------------------------------------+
|                       STABILITY OF SUBMERGED VS. FLOATING BODIES                                  |
|                                                                                                   |
|   SUBMERGED BODY STABILITY:                    FLOATING BODY ROTATIONAL STABILITY:                |
|                                                                                                   |
|   - Stable:   G is BELOW B                     - Metacenter M: Intersection of buoyant line       |
|               (Weight creates restoring couple)                with centerline during tilt        |
|   - Unstable: G is ABOVE B                     - Metacentric Height: GM = BM - BG                 |
|               (Weight creates overturning)     - Metacentric Radius: BM = I_w / V_disp            |
|   - Neutral:  G coincides with B               - Stable:   GM > 0 (M is ABOVE G)                  |
|                                                - Neutral:  GM = 0 (M coincides with G)            |
|                                                - Unstable: GM < 0 (M is BELOW G)                  |
+---------------------------------------------------------------------------------------------------+

                     Tilted Floating Hull Cross-Section:
                                       CL
                                       | 
                                      (M) Metacenter
                                     / | 
                                    /  |  <--- Restoring Moment Arm: GZ = GM * sin(\theta)
                                   /  (G) Center of Gravity (Hull weight W)
                                  /    |
                        (B') ----+----(B) Original Center of Buoyancy
                         ^             |
                         |             v
                    Buoyant Force F_b

Metacentric Height Calculation Formulation

1. Distance $BG$: Vertical distance from original center of buoyancy $B$ to center of gravity $G$: $$BG = z_G - z_B$$
2. Metacentric Radius ($BM$): $$BM = \frac{I_w}{V_{\text{disp}}}$$ Where $I_w$ is the second moment of area of the horizontal waterline cross-section about the longitudinal tilting axis (for a rectangular hull of length $L$ and beam width $b$, $I_w = \frac{L b^3}{12}$).
3. Metacentric Height ($GM$): $$GM = BM - BG = \frac{I_w}{V_{\text{disp}}} - (z_G - z_B)$$
4. Righting Moment ($M_R$): For small heel angles $\theta < 10^\circ$: $$M_R = W \cdot GM \sin \theta = (\gamma_f V_{\text{disp}}) \cdot GM \sin \theta$$

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7. Step-by-Step Worked Engineering Problem

Problem Statement

A rectangular sluice gate of width $w = 2.5\text{ m}$ (into the page) and inclined length $L = 4.0\text{ m}$ is hinged at its top edge $A$, located at a vertical depth of $3.0\text{ m}$ below the free water surface. The gate is inclined at an angle of $\theta = 60^\circ$ to the horizontal. Water density is $\rho = 1000\text{ kg/m}^3$ ($\gamma = 9.81\text{ kN/m}^3$).

Determine:

1. The total resultant hydrostatic force $F_R$ acting on the gate.
2. The location of the center of pressure $y_{cp}$ along the incline from the free surface.
3. The horizontal clamping force $F_{\text{stop}}$ required at the bottom edge $B$ to keep the gate closed.

   Free Surface //////////////////////////////////////////////
                 \  
                  \ 3.0 m (Vertical to Hinge A)
                   \  y_A = 3.0 / sin(60°) = 3.464 m
                    (A) HINGE
                     \ 
                      \  L = 4.0 m, w = 2.5 m, \theta = 60°
                       \   
                        \---- [ Centroid C: y_c = y_A + 2.0 m ]
                         \   
                          \-- [ Center of Pressure CP: y_cp ]
                           \ 
                            (B) BOTTOM STOP (Horizontal Force F_stop)

Step-by-Step Solution

Step 1: Determine Coordinate Locations Along the Incline

• Incline distance from surface datum to hinge $A$: $$y_A = \frac{h_A}{\sin \theta} = \frac{3.0\text{ m}}{\sin 60^\circ} = \frac{3.0}{0.8660} = 3.4641\text{ m}$$
• Distance along incline to centroid $C$ (midpoint of gate $L/2 = 2.0\text{ m}$): $$y_c = y_A + \frac{L}{2} = 3.4641 + 2.0000 = 5.4641\text{ m}$$
• Vertical depth of centroid $h_c$: $$h_c = y_c \sin 60^\circ = 5.4641 \times 0.8660 = 4.732\text{ m}$$

Step 2: Calculate Resultant Hydrostatic Force ($F_R$)

• Gate surface area $A = w \times L = 2.5\text{ m} \times 4.0\text{ m} = 10.0\text{ m}^2$.
• Total force: $$F_R = \gamma h_c A = (9810\text{ N/m}^3) \times (4.732\text{ m}) \times (10.0\text{ m}^2) = 464,209\text{ N} = 464.2\text{ kN}$$

Step 3: Calculate Center of Pressure ($y_{cp}$)

• Centroidal moment of inertia for rectangle: $$I_{xc} = \frac{w L^3}{12} = \frac{2.5 \times (4.0)^3}{12} = \frac{160}{12} = 13.333\text{ m}^4$$
• Location along incline: $$y_{cp} = y_c + \frac{I_{xc}}{y_c A} = 5.4641 + \frac{13.333}{5.4641 \times 10.0} = 5.4641 + 0.2440 = 5.7081\text{ m}$$
• Distance from hinge $A$ to center of pressure: $$L_{cp} = y_{cp} - y_A = 5.7081 - 3.4641 = 2.244\text{ m}$$

Step 4: Moment Equilibrium About Hinge $A$

• Taking moments about hinge $A$ ($\sum M_A = 0$):
  • Hydrostatic force acts perpendicularly to the gate at distance $L_{cp} = 2.244\text{ m}$ from hinge $A$.
  • Horizontal force $F_{\text{stop}}$ acts at bottom edge $B$ (distance along gate is $4.0\text{ m}$, vertical moment arm is $4.0 \sin 60^\circ = 3.464\text{ m}$):
  $$\sum M_A = (F_R \times L_{cp}) - (F_{\text{stop}} \times L \sin 60^\circ) = 0$$ $$(464.21\text{ kN} \times 2.244\text{ m}) - (F_{\text{stop}} \times 4.0\text{ m} \times 0.8660) = 0$$ $$1041.69\text{ kN}\cdot\text{m} = F_{\text{stop}} \times 3.4641\text{ m}$$ $$F_{\text{stop}} = \frac{1041.69}{3.4641} = 300.7\text{ kN}$$

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8. Common Exam Traps & PE Pro-Tips