6.1 Fluid Properties, Dimensionless Numbers, and Flow Regimes

Fluid Properties and the Continuum Hypothesis

In chemical engineering fluid mechanics, fluids (liquids and gases) are analyzed using the continuum hypothesis, which assumes that fluid properties vary continuously in space. This macromolecular treatment allows for the definition of point properties such as density ($\rho$), specific volume ($v$), specific weight ($\gamma$), and specific gravity ($SG$).

Density and Specific Volume: Density is defined as mass per unit volume ($\rho = m/V$). In SI units, it is expressed in $\text{kg/m}^3$, and in U.S. Customary System (USCS) units, it is expressed in $\text{lb/ft}^3$ or $\text{slug/ft}^3$. Specific volume is the reciprocal of density ($v = 1/\rho$), representing the volume occupied by a unit mass of fluid.

Specific Weight and Specific Gravity: Specific weight ($\gamma$) is the weight of a fluid per unit volume, related to density by $\gamma = \rho g$, where $g$ is the local acceleration due to gravity ($9.807\text{ m/s}^2$ or $32.174\text{ ft/s}^2$). Specific gravity ($SG$) is a dimensionless ratio of the fluid's density to the density of a reference fluid, typically pure water at $4^\circ\text{C}$ ($\rho_{\text{ref}} = 1000\text{ kg/m}^3$ or $62.43\text{ lb/ft}^3$) for liquids, and air at standard conditions for gases:

$SG = \frac{\rho}{\rho_{\text{ref}}}$

Viscosity: Viscosity is a fluid's resistance to gradual deformation by shear stress. According to Newton's law of viscosity, for a Newtonian fluid in laminar, one-dimensional flow, the shear stress ($\tau$) is directly proportional to the velocity gradient perpendicular to the plane of shear:

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

where $\mu$ is the dynamic (or absolute) viscosity, with SI units of $\text{Pa}\cdot\text{s}$ or $\text{kg}/(\text{m}\cdot\text{s})$ (commonly reported in centipoise, $\text{cP}$, where $1\text{ cP} = 10^{-3}\text{ Pa}\cdot\text{s}$), and USCS units of $\text{lb}\cdot\text{s/ft}^2$ or $\text{slug}/(\text{ft}\cdot\text{s})$. Kinematic viscosity ($\nu$) is the ratio of dynamic viscosity to density:

$\nu = \frac{\mu}{\rho}$

with SI units of $\text{m}^2\text{/s}$ (commonly reported in centistokes, $\text{cSt}$, where $1\text{ cSt} = 10^{-6}\text{ m}^2\text{/s}$) and USCS units of $\text{ft}^2\text{/s}$.

Surface Tension and Capillarity: Surface tension ($\sigma$) is the tensile force per unit length acting at the interface between two immiscible phases. This molecular cohesion leads to capillary action. The height ($h$) of capillary rise or depression in a circular tube of diameter $d$ is given by:

$h = \frac{4\sigma\cos\theta}{\rho g d}$

where $\theta$ is the contact angle. For water on clean glass, $\theta \approx 0^\circ$ (capillary rise), whereas for mercury on glass, $\theta \approx 140^\circ$ (capillary depression).

Hydrostatic Pressure: Pressure ($P$) is the compressive force per unit area. Absolute pressure is measured relative to a perfect vacuum, whereas gauge pressure is measured relative to local atmospheric pressure:

$P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}$

For an incompressible fluid at rest, the hydrostatic pressure difference between two elevations is:

$P_2 - P_1 = -\rho g (z_2 - z_1) = \rho g h$

where $h$ is the depth of the fluid.

Dimensionless Numbers in Transport Phenomena

Dimensionless groups characterize the relative importance of physical transport mechanisms. For the FE Chemical exam, three primary dimensionless numbers must be mastered:

1. Reynolds Number ($Re$): Represents the ratio of inertial forces to viscous forces. For flow in a circular pipe of diameter $D$ with average velocity $v$:

   $Re = \frac{D v \rho}{\mu} = \frac{D v}{\nu}$

   For non-circular ducts, the hydraulic diameter ($D_H$) is substituted for $D$:

   $D_H = \frac{4 A_c}{P_w}$

   where $A_c$ is the cross-sectional flow area and $P_w$ is the wetted perimeter.

2. Prandtl Number ($Pr$): The ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity ($\alpha = k / \rho C_p$):

   $Pr = \frac{C_p \mu}{k} = \frac{\nu}{\alpha}$

   It depends solely on fluid properties and governs the relative thickness of velocity and thermal boundary layers.

3. Schmidt Number ($Sc$): The ratio of momentum diffusivity to mass diffusivity ($D_{AB}$):

   $Sc = \frac{\mu}{\rho D_{AB}} = \frac{\nu}{D_{AB}}$

   It characterizes the relative thickness of velocity and concentration boundary layers.

4. Péclet Number ($Pe$): Characterizes the ratio of advective transport to diffusive transport. For heat transfer, $Pe = Re \cdot Pr$; for mass transfer, $Pe_M = Re \cdot Sc$.

Flow Regimes: Laminar vs. Turbulent Flow

The Reynolds number is the primary criterion for determining the flow regime. For internal flow in pipes:

• Laminar Flow ($Re < 2100$): Viscous forces dominate. Fluid flows in parallel, concentric streamlines with no macroscopic mixing. The velocity profile is parabolic:

  $v(r) = v_{\text{max}} \left[ 1 - \left(\frac{r}{R}\right)^2 \right]$

  where the average velocity is exactly half the maximum velocity ($v_{\text{avg}} = 0.5 v_{\text{max}}$).

• Transition Flow ($2100 \le Re \le 4000$): The flow is unstable and can oscillate between laminar and turbulent regimes.

• Turbulent Flow ($Re > 4000$): Inertial forces dominate. The flow is characterized by chaotic, three-dimensional eddy motions that enhance heat, mass, and momentum transfer. The velocity profile is flatter due to turbulent mixing, often modeled by the empirical one-seventh power-law:

  $v(r) = v_{\text{max}} \left( 1 - \frac{r}{R} \right)^{1/7}$

  where $v_{\text{avg}} \approx 0.817 v_{\text{max}}$.

Summary of Key Fluid Properties

Worked Example 1: Reynolds Number and Flow Regime Determination

Problem: A liquid hydrocarbon mixture ($SG = 0.82$, $\mu = 1.8\text{ cP}$) flows through a rectangular duct of cross-section $15\text{ cm} \times 30\text{ cm}$ at a volumetric flow rate of $0.045\text{ m}^3\text{/s}$. Determine the hydraulic diameter, the average velocity, and the flow regime.

Solution:

First, calculate the fluid density:

$\rho = SG \cdot \rho_{\text{water}} = 0.82 \cdot 1000\text{ kg/m}^3 = 820\text{ kg/m}^3$

Convert viscosity to SI units:

$\mu = 1.8\text{ cP} = 1.8 \times 10^{-3}\text{ Pa}\cdot\text{s} = 1.8 \times 10^{-3}\text{ kg}/(\text{m}\cdot\text{s})$

Calculate the cross-sectional area ($A_c$) and wetted perimeter ($P_w$):

$A_c = 0.15\text{ m} \times 0.30\text{ m} = 0.045\text{ m}^2$

$P_w = 2 \times (0.15\text{ m} + 0.30\text{ m}) = 0.90\text{ m}$

Compute the hydraulic diameter ($D_H$):

$D_H = \frac{4 A_c}{P_w} = \frac{4 \cdot 0.045\text{ m}^2}{0.90\text{ m}} = 0.20\text{ m}$

Determine the average velocity ($v_{\text{avg}}$):

$v_{\text{avg}} = \frac{Q}{A_c} = \frac{0.045\text{ m}^3\text{/s}}{0.045\text{ m}^2} = 1.0\text{ m/s}$

Compute the Reynolds number ($Re$):

$Re = \frac{D_H v_{\text{avg}} \rho}{\mu} = \frac{0.20\text{ m} \cdot 1.0\text{ m/s} \cdot 820\text{ kg/m}^3}{1.8 \times 10^{-3}\text{ kg}/(\text{m}\cdot\text{s})} = 91,111$

Since $Re = 91,111 > 4000$, the flow is highly turbulent.

Worked Example 2: Capillary Rise Calculation

Problem: Estimate the capillary rise of pure water at $20^\circ\text{C}$ ($\sigma = 0.0728\text{ N/m}$, $\theta \approx 0^\circ$) in a glass tube with an internal diameter of $1.5\text{ mm}$.

Solution:

Using the capillary rise formula:

$h = \frac{4\sigma\cos\theta}{\rho g d}$

Substitute the given values ($\rho = 1000\text{ kg/m}^3$, $g = 9.807\text{ m/s}^2$, $d = 0.0015\text{ m}$):

$h = \frac{4 \cdot (0.0728\text{ N/m}) \cdot \cos(0^\circ)}{(1000\text{ kg/m}^3) \cdot (9.807\text{ m/s}^2) \cdot (0.0015\text{ m})} = \frac{0.2912}{14.71} \approx 0.0198\text{ m} = 19.8\text{ mm}$