Fluid Mechanics: Fluid Properties, Pressure, Density, and Viscosity

Fundamental Concepts and Fluid Definitions

Definition of a Fluid:
- A fluid is defined as matter that deforms continuously to take the shape of its container.
- A fluid is defined as matter that cannot sustain a shear stress while remaining at rest.
States of Matter and Phase Transitions:
- Solid to Liquid: Melting / Fusion
- Liquid to Solid: Freezing / Solidification
- Liquid to Gas: Evaporation
- Gas to Liquid: Condensation
- Solid to Gas: Sublimation
- Gas to Solid: Deposition

Phase Transitions Between Solid, Liquid, and Gas States

Scope and Branches of Fluid Mechanics:
- Fluid Mechanics is a fundamental branch of general Continuum Mechanics with extensive engineering applications.
- Fluid Statics: The study of fluids at rest.
- Fluid Dynamics: The study of fluids in motion.
- Core Engineering Goals:
  - Accurate flow measurement.
  - Predicting forces acting on bodies submerged in fluids or containing moving fluid masses.
  - Predicting flow regimes, sizing pipes, and designing pumps.
  - Calculating total energy consumption and system losses.

Units and Dimensional Analysis in Fluid Mechanics

Fundamental SI Units:
- Fundamental units cannot be subdivided into other units and form the basis of the SI system:
  - Mass: kilogram ( $\text{kg}$ ), dimension $[\text{M}]$
  - Length: metre ( $\text{m}$ ), dimension $[\text{L}]$
  - Time: second ( $\text{s}$ ), dimension $[\text{T}]$
  - Temperature: kelvin ( $\text{K}$ ), dimension $[\Theta]$
  - Amount of Substance: mole ( $\text{mol}$ ), dimension $[\text{N}]$
  - Electric Current: ampere ( $\text{A}$ ), dimension $[\text{I}]$
  - Luminous Intensity: candela ( $\text{cd}$ ), dimension $[\text{J}]$
Derived SI Units Named After Scientists:
- Derived units represent products or quotients of fundamental units raised to powers:
  - Force: newton ( $\text{N}$ ), fundamental unit equivalent: $\text{kg}\cdot\text{m/s}^2$
  - Pressure: pascal ( $\text{Pa}$ or $\text{N/m}^2$ ), fundamental unit equivalent: $\text{kg}/(\text{m}\cdot\text{s}^2)$
  - Energy: joule ( $\text{J}$ or $\text{N}\cdot\text{m}$ ), fundamental unit equivalent: $\text{kg}\cdot\text{m}^2/\text{s}^2$
  - Power: watt ( $\text{W}$ or $\text{J/s}$ ), fundamental unit equivalent: $\text{kg}\cdot\text{m}^2/\text{s}^3$
  - Frequency: hertz ( $\text{Hz}$ ), fundamental unit equivalent: $\text{s}^{-1}$
  - Angle: radian ( $\text{rad}$ ), dimensionless
Key Fluid Mechanics Quantities and Symbols:
- Area ( $A$ ): $\text{m}^2$
- Volume ( $V$ ): $\text{m}^3$
- Density ( $\rho$ ): $\text{kg/m}^3$
- Specific Volume ( $v$ ): $\text{m}^3/\text{kg}$
- Average Velocity ( $u$ ): $\text{m/s}$
- Point Velocity ( $v$ ): $\text{m/s}$
- Mass Flow Rate ( $\dot{m}$ ): $\text{kg/s}$
- Volumetric Flow Rate ( $Q$ ): $\text{m}^3/\text{s}$
- Pressure / Normal Stress ( $P$ or $\sigma$ ): $\text{Pa}$
- Shear Stress ( $\tau$ ): $\text{Pa}$
- Shear Rate ( $\dot{\gamma}$ ): $\text{s}^{-1}$
- Dynamic Viscosity ( $\mu$ ): $\text{Pa}\cdot\text{s}$
- Surface Tension ( $\sigma$ ): $\text{N/m}$
- Compressibility ( $c$ ): $\text{Pa}^{-1}$

Density, Specific Properties, and Equations of State

Behavioral Distinction Between Liquids and Gases:
- Liquids: Occupy an enclosing container only to the extent of their own volume and are treated as incompressible.
- Gases: Expand completely to occupy the total enclosure volume and are treated as compressible.
Specific Fluid Properties:
- Specific properties are intensive properties, meaning they are independent of the total size or mass of the system.
- Density ( $\rho$ ): Mass per unit volume: $\rho = \frac{m}{V} \quad [\text{kg/m}^3]$
- Specific Volume ( $v$ ): Volume per unit mass, defined as the reciprocal of density: $v = \frac{1}{\rho} \quad [\text{m}^3/\text{kg}]$
- Specific Weight ( $w$ ): Weight per unit volume: $w = \rho g \quad [\text{N/m}^3]$
- Specific Gravity ( $\gamma$ ): Ratio of the density of a given fluid to the density of standard water: $\gamma = \frac{\rho_f}{\rho_w}$
Density Variations in Liquids vs. Gases:
- Liquids: Density depends primarily on temperature and varies minimally with pressure. For example, water density varies with temperature from $\sim 1000\,\text{kg/m}^3$ at $20\,^{\circ}\text{C}$ to $\sim 958\,\text{kg/m}^3$ at $100\,^{\circ}\text{C}$ . Empirical density variations can be expressed as: $\frac{1}{\rho} = A + B T$
- Gases: Density depends strongly on both temperature and pressure, governed by an Equation of State (EoS).
The Ideal Gas Law:
- An Equation of State relates the state variables of pressure, temperature, and density.
- Molar Form: $P V = n R_0 T$
- Mass Form: $P V = m R_{\text{spec}} T$
- Density Form: $\rho = \frac{P}{R_{\text{spec}} T}$
- Universal Gas Constant ( $R_0$ ): $R_0 = 8.314\,\text{J}/(\text{mol}\cdot\text{K})$
- Specific Gas Constant ( $R_{\text{spec}}$ ): $R_{\text{spec}} = 1000 \times \frac{R_0}{M} \quad [\text{J}/(\text{kg}\cdot\text{K})]$ where $M$ is the molar mass in $\text{g/mol}$ .
- Specific Gas Constant Values:
  - Air: $287\,\text{J}/(\text{kg}\cdot\text{K})$
  - Hydrogen ( $\text{H}_2$ ): $4160\,\text{J}/(\text{kg}\cdot\text{K})$
  - Carbon Dioxide ( $\text{CO}_2$ ): $190\,\text{J}/(\text{kg}\cdot\text{K})$ (or $189\,\text{J}/(\text{kg}\cdot\text{K})$ derived from $M = 44\,\text{g/mol}$ )
- Temperature Units: Temperature $T$ must always be converted to absolute temperature in kelvin ( $\text{K}$ ) where $T = t [^{\circ}\text{C}] + 273.15$ .

Stress, Tensor Representation, and Hydrostatic Pressure

Distinction Between Force and Stress:
- Stress represents force applied over a specific area ( $\text{Stress} = \frac{\text{Force}}{\text{Area}}$ ).
- Normal Stress ( $\sigma$ ): Applied perpendicular to a surface, producing compression or tension.
- Shear Stress ( $\tau$ ): Applied parallel to a surface, causing relative sliding or flow.
- The surface areas over which normal stress and shear stress act are physically distinct.
The Stress Tensor:
- The complete state of stress at any point within a fluid element is described by a second-order Stress Tensor ( $\boldsymbol{\sigma}$ ): $\boldsymbol{\sigma} = \begin{pmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_{zz} \end{pmatrix}$
- Each component $\sigma_{ij}$ or $\tau_{ij}$ is a vector operating on a face perpendicular to axis $i$ in the direction of axis $j$ .

Normal Stress Components on a 3D Fluid Element

Shear Stress Components on a 3D Fluid Element

Hydrostatic Pressure:
- Hydrostatic pressure ( $P$ ) is an overall scalar measure of uniform, isotropic compression.
- Mathematically, hydrostatic pressure is defined as one-third of the trace of the stress tensor: $P = \frac{1}{3}\text{tr}(\boldsymbol{\sigma}) = \frac{\sigma_{xx} + \sigma_{yy} + \sigma_{zz}}{3}$
Physical Origin and Units of Pressure:
- Pressure is a physical consequence of continuous molecular activity and collisions against a bounding surface.
- SI Unit: Pascal ( $1\,\text{Pa} = 1\,\text{N/m}^2$ ).
- Standard Unit Conversions:
  - $1\,\text{bar} = 10^5\,\text{N/m}^2 = 10^5\,\text{Pa}$
  - $1\,\text{atm} = 101325\,\text{N/m}^2 = 101325\,\text{Pa}$

Absolute Pressure vs Gauge Pressure

Reference Baselines:
- Absolute Zero Reference: Perfect vacuum ( $P = 0\,\text{Pa}$ ).
- Atmospheric Pressure Reference ( $P_{\text{atm}}$ ): Standard atmospheric pressure ( $P_{\text{atm}} = 1\,\text{atm} = 101325\,\text{Pa}$ ), as measured by a barometer.
Pressure Definitions and Mathematical Relationships:
- Absolute Pressure ( $P_{\text{abs}}$ ): Total pressure measured relative to a perfect vacuum ( $P = 0\,\text{Pa}$ ).
- Gauge Pressure ( $P_{\text{gauge}}$ ): Pressure measured relative to local atmospheric pressure ( $P_{\text{atm}}$ ). $P_{\text{gauge}} = P_{\text{abs}} - P_{\text{atm}}$
Operating Regimes:
- Positive Gauge Pressure ( $P_{\text{gauge}} > 0\,\text{Pa}$ ): Absolute pressure exceeds atmospheric pressure ( $P_{\text{abs}} > P_{\text{atm}}$ ).
- Negative Gauge Pressure / Vacuum ( $P_{\text{gauge}} < 0\,\text{Pa}$ ): Absolute pressure is below atmospheric pressure ( $P_{\text{abs}} < P_{\text{atm}}$ ).

Shear Stress, Viscosity, and Newton's Law

Response to Shear Stress: Solids vs. Fluids:
- Elastic Solid:
  - Resists an applied shear stress $\tau = \frac{F}{A}$ .
  - Deforms in a reversible way (or breaks permanently).
  - The deformation or shear strain ( $\gamma$ ) is directly proportional to the applied force.
  - Solids remain deformed for as long as the force is applied. Solids do NOT flow, they DEFORM.

Deformation of an Elastic Solid under Shear Stress

*   **Fluid**:
    *   Cannot resist a shear stress while at rest.
    *   Upon application of  $\tau$ , the top layer slides over adjacent lower layers, dragging them along via fluid friction.
    *   Continuous slipping occurs between layers as long as  $\tau$  is applied. Fluids FLOW, and do NOT deform statically.

Layer Sliding and Flow of a Fluid under Shear Stress

Definition and Geometric Derivation of Shear Rate:
- Consider a fluid element deforming between time $t = t_0$ and $t = t_0 + \Delta t$ .
- The top face moves at velocity $u + \Delta u$ while the bottom face moves at velocity $u$ .
- The angular deformation (shear strain $\Delta\gamma$ ) is given by: $\tan(\Delta\gamma) = \frac{[x + (u + \Delta u)\Delta t] - [x + u\Delta t]}{(y + \Delta y) - y} = \frac{\Delta u \Delta t}{\Delta y}$
- Taking the differential limit yields the shear rate ( $\dot{\gamma}$ ): $\dot{\gamma} = \frac{du}{dy}$
- Shear rate represents the velocity gradient perpendicular to the direction of flow/force application.

Newton's Law of Viscosity:
- States that the shear rate generated in a fluid is directly proportional to the applied shear stress: $\tau = \mu \dot{\gamma} = \mu \frac{du}{dy}$
- Fluids that obey this linear relation are known as Newtonian Fluids.
- Crucial Rule: In fluids, shear stress $\tau$ is proportional to the shear rate $\dot{\gamma}$ , NOT the shear strain $\gamma$
Viscosity Definitions and Units:
- Dynamic Viscosity ( $\mu$ ): Fluid shear viscosity representing resistance to sliding friction between fluid layers.
- SI Units of Dynamic Viscosity: $\text{kg}/(\text{m}\cdot\text{s}) = \text{N}\cdot\text{s}/\text{m}^2 = \text{Pa}\cdot\text{s}$
- Traditional Units (Poise / Centipoise): $1\,\text{P} = 100\,\text{cP} = 0.1\,\text{Pa}\cdot\text{s}$
- Kinematic Viscosity ( $\nu$ ): Ratio of viscous forces to inertial forces: $\nu = \frac{\mu}{\rho} \quad [\text{m}^2/\text{s}]$

Worked Examples and Engineering Calculations

Problem 1: Force Calculation for Sled Sliding on Thin Water Layer over Ice:
- Problem Statement: A skier pushes a sled over ice. Friction melts a thin layer of water of thickness $t = 1\,\mu\text{m} = 10^{-6}\,\text{m}$ . Contact area is a rectangle of length $1\,\text{m}$ and width $20\,\text{cm} = 0.2\,\text{m}$ . Water viscosity $\mu = 10^{-3}\,\text{Pa}\cdot\text{s}$ . Find the required force when moving at $u = 60\,\text{cm/s} = 0.60\,\text{m/s}$ .

Sled Moving on Thin Water Layer over Ice

*   **Contact Area**:

$A = 1 \times 0.2 = 0.2\,\text{m}^2$ * Shear Rate: $\dot{\gamma} = \frac{du}{dy} = \frac{\Delta u}{\Delta y} = \frac{0.60 - 0}{10^{-6}} = 6 \times 10^5\,\text{s}^{-1}$ * Shear Stress: $\tau = \mu \dot{\gamma} = 0.001 \times (6 \times 10^5) = 600\,\text{Pa}$ * Required Force: $F = \tau A = 600 \times 0.2 = 120\,\text{N}$

Problem 2: Force Calculation for Shaft Pulled Linearly Through Bearing:
- Problem Statement: A shaft of diameter $D = 25\,\text{mm} = 0.025\,\text{m}$ and length $L = 50\,\text{mm} = 0.05\,\text{m}$ passes through a bearing with uniform radial clearance $\Delta y = 0.1\,\text{mm} = 0.1 \times 10^{-3}\,\text{m}$ filled with oil ( $\mu = 0.25\,\text{Pa}\cdot\text{s}$ ). Calculate force required to pull shaft at $u = 10\,\text{m/s}$ .
- Shear Rate: $\dot{\gamma} = \frac{du}{dy} = \frac{\Delta u}{\Delta y} = \frac{10 - 0}{0.1 \times 10^{-3}} = 10^5\,\text{s}^{-1}$
- Shear Stress: $\tau = \mu \dot{\gamma} = 0.25 \times 10^5 = 2.5 \times 10^4\,\text{Pa}$
- Surface Area: $A = \pi D L = \pi \times 0.025 \times 0.05 = 3.927 \times 10^{-3}\,\text{m}^2$
- Pulling Force: $F = \tau A = (2.5 \times 10^4) \times (\pi \times 0.025 \times 0.05) = 98.2\,\text{N}$
Problem 3: Torque Calculation for Shaft Rotated in Bearing:
- Problem Statement: For the same shaft system ( $D = 25\,\text{mm}$ , $L = 50\,\text{mm}$ , clearance $\Delta y = 0.1\,\text{mm}$ , $\mu = 0.25\,\text{Pa}\cdot\text{s}$ ), calculate the torque needed to rotate the shaft at $1000\,\text{rev/min}$ .
- Surface Velocity: $u_x = r \omega = \left(\frac{D}{2}\right) \left(\frac{1000 \times 2\pi}{60}\right) = \frac{0.025}{2} \times 104.72 = 1.31\,\text{m/s}$
- Shear Rate: $\dot{\gamma} = \frac{\Delta u}{\Delta y} = \frac{1.31 - 0}{0.1 \times 10^{-3}} = 1.31 \times 10^4\,\text{s}^{-1}$
- Shear Stress: $\tau = \mu \dot{\gamma} = 0.25 \times (1.31 \times 10^4) = 3.27 \times 10^3\,\text{Pa}$
- Tangential Force: $F = \tau A = (3.27 \times 10^3) \times (\pi \times 0.025 \times 0.05) = 12.9\,\text{N}$
- Required Torque: $T = r F = (12.5 \times 10^{-3}) \times 12.9 = 0.161\,\text{N}\cdot\text{m}$
Problem 4: Air Mass Calculation in Elevated Spherical Balloon:
- Problem Statement: Calculate mass of air contained in a spherical balloon of diameter $D = 10\,\text{m}$ ( $r = 5\,\text{m}$ ) at altitude $2000\,\text{m}$ where $T = -10\,^{\circ}\text{C}$ and $P = 80000\,\text{Pa}$ . Given $R_{\text{spec,air}} = 287\,\text{J}/(\text{kg}\cdot\text{K})$ .
- Absolute Temperature: $T = 273.15 - 10 = 263.15\,\text{K}$
- Sphere Volume: $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (5)^3 = 523.6\,\text{m}^3$
- Mass Evaluation: $m = \frac{P V}{R_{\text{spec}} T} = \frac{80000 \times 523.6}{287 \times 263.15} = 554.6\,\text{kg}$
Problem 5: CO2 Production Rate from Thermal Plant:
- Problem Statement: A thermal plant produces $100\,\text{tons/h} = 100000\,\text{kg/h}$ of $\text{CO}_2$ exiting at $T = 600\,\text{K}$ and $P = 1\,\text{atm} = 101325\,\text{Pa}$ . Calculate volumetric flow rate in $\text{m}^3/\text{h}$ .
- Specific Gas Constant for CO2: $R_{\text{spec,CO2}} = \frac{8314\,\text{J}/(\text{kmol}\cdot\text{K})}{44\,\text{kg/kmol}} = 189\,\text{J}/(\text{kg}\cdot\text{K})$
- Gas Density: $\rho_{\text{CO2}} = \frac{P}{R_{\text{spec}} T} = \frac{101325}{189 \times 600} = 0.89\,\text{kg/m}^3$
- Volumetric Flow Rate: $Q_{\text{flow}} = \frac{m_{\text{flow}}}{\rho} = \frac{100000\,\text{kg/h}}{0.89\,\text{kg/m}^3} = 112360\,\text{m}^3/\text{h}$
Problem 6: Cryogenic Hydrogen Storage Temperature:
- Problem Statement: Calculate temperature of $2\,\text{kg}$ of hydrogen ( $\text{H}_2$ ) stored in a cubic deposit of side length $l = 0.6\,\text{m}$ at $P = 80\,\text{bar} = 80 \times 10^5\,\text{Pa}$ . Given $M_{\text{H2}} = 2\,\text{g/mol}$ .
- Deposit Volume: $V = l^3 = (0.6)^3 = 0.216\,\text{m}^3$
- Specific Gas Constant for H2: $R_{\text{spec,H2}} = \frac{8.314\,\text{J}/(\text{mol}\cdot\text{K})}{2\,\text{g/mol}} = 4.16\,\text{J}/(\text{g}\cdot\text{K})$ $R_{\text{spec,H2}} = 4.16 \times 1000 = 4160\,\text{J}/(\text{kg}\cdot\text{K})$
- Temperature Evaluation: $T = \frac{P V}{m R_{\text{spec}}} = \frac{8000000 \times 0.216}{2 \times 4160} = 207.7\,\text{K}$