12.1 Fluid Properties, Hydrostatic Pressure, and Manometry

12.1 Fluid Properties, Hydrostatic Pressure, and Manometry

Core FE Exam Principle: Fluid statics analyzes fluids in a state of rest or rigid-body motion where shear stresses are identically zero ((\tau = 0)). Pressure is an isotropic scalar quantity that acts normal to any solid surface or fluid boundary.

Fundamental Fluid Properties

To solve fluid mechanics problems on the NCEES FE exam, you must first master the physical properties that define fluid behavior under static and dynamic conditions.

Density, Specific Weight, and Specific Gravity

• Mass Density ((\rho)): Mass per unit volume. $\rho = \frac{m}{V} \quad [\text{kg/m}^3 \text{ or } \text{slug/ft}^3]$

• Specific Weight ((\gamma)): Weight force per unit volume. It directly incorporates gravitational acceleration (g): $\gamma = \rho g \quad [\text{N/m}^3, \text{ kN/m}^3, \text{ or } \text{lbf/ft}^3]$ Standard Water Values: (\rho_{w} = 1000 \text{ kg/m}^3) ((62.4 \text{ lbm/ft}^3)), (\gamma_{w} = 9.81 \text{ kN/m}^3) ((62.4 \text{ lbf/ft}^3)).

• Specific Gravity ((SG)): Dimensionless ratio comparing a fluid's density or specific weight to that of pure water at standard temperature ((4^\circ\text{C})): $SG = \frac{\rho}{\rho_{w}} = \frac{\gamma}{\gamma_{w}}$ Example: Mercury has (SG_{Hg} = 13.6), yielding (\rho_{Hg} = 13,600 \text{ kg/m}^3) and (\gamma_{Hg} = 133.4 \text{ kN/m}^3).

Viscosity and Newton's Law of Viscosity

Viscosity quantifies a fluid's internal resistance to shear deformation. Consider fluid contained between two parallel plates separated by distance (y), where the top plate moves at velocity (U):

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

where:

• (\tau) = Shear stress ((\text{N/m}^2) or (\text{Pa}), (\text{lbf/ft}^2))
• (\mu) = Dynamic (absolute) viscosity ((\text{Pa}\cdot\text{s} = \text{N}\cdot\text{s/m}^2) or (\text{lbf}\cdot\text{s/ft}^2))
• (\frac{du}{dy}) = Velocity gradient or shear rate ((\text{s}^{-1}))

Kinematic Viscosity ((\nu)): The ratio of dynamic viscosity to fluid mass density: $\nu = \frac{\mu}{\rho} \quad [\text{m}^2/\text{s} \text{ or } \text{ft}^2/\text{s}]$

Hydrostatic Pressure and Manometry

In a static fluid column, pressure increases linearly with increasing depth due to the weight of the fluid above.

Hydrostatic Pressure Equation

$\frac{dP}{dz} = -\gamma \implies P_2 - P_1 = \gamma (z_1 - z_2) = \gamma h$

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

• Gage Pressure ((P_{gage})): Pressure measured relative to local atmospheric pressure. (P_{gage} = 0) at standard atmosphere.
• Absolute Pressure ((P_{abs})): Total pressure measured relative to absolute zero pressure (vacuum). Must be used in all thermodynamic gas law calculations.
• Standard atmospheric pressure: (P_{atm} = 101.325 \text{ kPa} = 14.696 \text{ psia} = 760 \text{ mm Hg} = 29.92 \text{ in Hg}).

U-Tube and Differential Manometry

Manometers measure pressure differences by balancing fluid columns. The Golden Rule of Manometry is: Start at one end of the system, add (\gamma h) when moving downward through a continuous fluid column, subtract (\gamma h) when moving upward, and equate the resulting expression to the pressure at the other end.

$\text{Path Equation: } P_A + \sum (\gamma_i h_i)_{\text{downward}} - \sum (\gamma_j h_j)_{\text{upward}} = P_B$

Exam Tip: Points at the same elevation in a continuous, static body of the same fluid have identical pressures. Always use horizontal jump lines across identical fluid interfaces to simplify multi-fluid U-tube manometer calculations.

Hydrostatic Forces on Submerged Surfaces

Plane Surfaces

For a submerged flat plate inclined at angle (\theta) to the free surface:

1. Resultant Force Magnitude ((F_R)): Equal to the pressure at the area centroid (P_c) multiplied by the total plate surface area (A): $F_R = \gamma h_c A = P_c A$ where (h_c) is the vertical depth from the free surface to the centroid of the submerged plate.

2. Center of Pressure Location ((y_{cp})): The line of action of (F_R) intersects the plate at the center of pressure (cp): $y_{cp} = y_c + \frac{I_{xc}}{y_c A}$ where:

   • (y_c) = Inclined distance along the plate axis from the surface to the centroid
   • (I_{xc}) = Area moment of inertia about the centroidal horizontal axis (e.g., (I_{xc} = \frac{b h^3}{12}) for a rectangle, (\frac{\pi R^4}{4}) for a circle)
   • (y_{cp}) is always located deeper than the area centroid ((y_{cp} > y_c)).

Curved Surfaces

Hydrostatic forces on curved submerged boundaries are evaluated by resolving the resultant force into horizontal and vertical force components:

• Horizontal Component ((F_H)): Equal to the hydrostatic force acting on the vertical projection of the curved surface: $F_H = \gamma h_{c,vert} A_{projected}$
• Vertical Component ((F_V)): Equal to the weight of the fluid column extending vertically above the curved surface up to the real or imaginary free surface: $F_V = \gamma V_{\text{fluid block above curve}}$
• Resultant Force and Direction: $F_R = \sqrt{F_H^2 + F_V^2}, \quad \alpha = \arctan\left| \frac{F_V}{F_H} \right|$

Buoyancy and Metacentric Stability

Archimedes' Principle

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

$F_B = \gamma_{\text{fluid}} V_{\text{displaced}} = \rho_{\text{fluid}} g V_{\text{displaced}}$

• The buoyant force acts vertically upward through the Center of Buoyancy ((B)), which is the centroid of the displaced fluid volume.
• For a floating body in static equilibrium, the buoyant force equals the total weight of the body ((F_B = W_{\text{body}})): $\rho_{\text{fluid}} g V_{\text{submerged}} = \rho_{\text{body}} g V_{\text{total}} \implies \frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}} = \frac{SG_{\text{body}}}{SG_{\text{fluid}}}$

Metacentric Height and Floating Stability

For a floating vessel tilted by a small angle (\theta):

• Center of Gravity ((G)): Point where the total mass of the vessel acts.
• Metacenter ((M)): Intersection of the vertical buoyant force line of action with the tilted center line.
• Metacentric Height ((GM)): $GM = BM - BG = \frac{I_{\text{waterline}}}{V_{\text{displaced}}} - BG$ where (I_{\text{waterline}}) is the minimum area moment of inertia of the waterline cross-section.
• Stability Rule:
  • (GM > 0) (Point (M) above (G)): Stable equilibrium (restoring moment created).
  • (GM < 0) (Point (M) below (G)): Unstable equilibrium (overturning moment created).

Comprehensive Worked Engineering Example

Problem Statement

An inclined rectangular gate monitoring a reservoir discharge port is hinged at its top edge (A) as shown below. The gate is (2.5 \text{ m}) wide (into the page) and (3.0 \text{ m}) long, inclined at an angle (\theta = 60^\circ) to the horizontal. Water ((\gamma_w = 9.81 \text{ kN/m}^3)) covers the gate such that the top hinge (A) is located at a vertical depth of (h_A = 2.0 \text{ m}) below the water surface. A horizontal force (F_P) is applied at the bottom edge (B) to hold the gate closed.

Calculate:

1. The magnitude of the resultant hydrostatic force (F_R) acting on the gate.
2. The location of the center of pressure (y_{cp}) measured along the inclined gate axis from the free surface.
3. The magnitude of the required closing force (F_P) at bottom edge (B).

  Free Water Surface  ~~~~~~~~~~~~~~~~~~~~~~~
                               | h_A = 2.0 m
                               v
                          Hinge A
                           / \\  
                          /   \\  theta = 60 deg
                         /     \\  Gate Length L = 3.0 m
                        /       
                       /   F_R   
                      /----->o    
                     /             
                    /_______________
                    B <--- F_P

Step-by-Step Solution

Step 1: Compute Centroid Depth and Resultant Hydrostatic Force (F_R)

• Gate length (L = 3.0 \text{ m}), width (b = 2.5 \text{ m}), area (A = b \times L = 2.5 \times 3.0 = 7.5 \text{ m}^2).
• Distance along inclined axis from free surface to top hinge (A): $y_A = \frac{h_A}{\sin 60^\circ} = \frac{2.0}{\sin 60^\circ} = \frac{2.0}{0.86603} = 2.3094 \text{ m}$
• Distance along inclined axis from free surface to centroid (C): $y_c = y_A + \frac{L}{2} = 2.3094 + 1.50 = 3.8094 \text{ m}$
• Vertical depth of centroid (h_c): $h_c = y_c \sin 60^\circ = 3.8094 \times 0.86603 = 3.2990 \text{ m}$
• Hydrostatic force magnitude (F_R): $F_R = \gamma_w h_c A = 9.81 \text{ kN/m}^3 \times 3.2990 \text{ m} \times 7.5 \text{ m}^2 = 242.72 \text{ kN}$

Step 2: Compute Center of Pressure (y_{cp})

• Moment of inertia about gate centroidal axis: $I_{xc} = \frac{b L^3}{12} = \frac{2.5 \times (3.0)^3}{12} = \frac{67.5}{12} = 5.625 \text{ m}^4$
• Location of center of pressure along inclined axis (y_{cp}): $y_{cp} = y_c + \frac{I_{xc}}{y_c A} = 3.8094 + \frac{5.625}{3.8094 \times 7.5} = 3.8094 + \frac{5.625}{28.5705} = 3.8094 + 0.1969 = 4.0063 \text{ m}$
• Distance from hinge (A) to line of action of (F_R): $d_{cp} = y_{cp} - y_A = 4.0063 - 2.3094 = 1.6969 \text{ m}$

Step 3: Compute Required Closing Force (F_P) by Taking Moments About Hinge (A)

• Taking static moment equilibrium about hinge (A) ((\sum M_A = 0)): $F_R \cdot d_{cp} - (F_P \sin 60^\circ) \cdot L = 0$ $242.72 \text{ kN} \times 1.6969 \text{ m} - F_P \times (0.86603) \times 3.0 \text{ m} = 0$ $411.87 \text{ kN}\cdot\text{m} - 2.5981 \cdot F_P = 0 \implies F_P = \frac{411.87}{2.5981} = 158.53 \text{ kN}$

Final Answer: Resultant force (F_R = 242.7 \text{ kN}), (y_{cp} = 4.01 \text{ m}) from free surface along the inclined axis, and required closing force (F_P = 158.5 \text{ kN}).