3.2 Incompressible Flow, Friction Factors, and the Moody Diagram

3.2 Incompressible Flow, Friction Factors, and the Moody Diagram

In chemical process plants, fluids are transported between unit operations—reactors, distillation columns, absorbers, and heat exchangers—through extensive piping networks. Designing piping, specifying pipe schedules, and selecting pump duty points require calculating frictional head loss. For liquids and gases experiencing small fractional pressure variations ($\Delta P / P_1 < 0.10$), the fluid is modeled as incompressible (constant density $\rho$).

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1. The Mechanical Energy Balance (MEB)

Integrating the Navier-Stokes equations along a streamline for a steady-state, one-dimensional, incompressible fluid flow between upstream station 1 and downstream station 2 yields the Mechanical Energy Balance (MEB):

$\frac{\Delta P}{\rho} + \frac{\Delta(v^2)}{2\alpha} + g\Delta z + \hat{w}_s + \hat{h}_f = 0$

Expressed in the traditional head form (dividing through by $g$, where each term has units of length, such as meters or feet of flowing fluid):

$\frac{P_1}{\rho g} + \frac{\alpha_1 v_1^2}{2g} + z_1 + H_{pump} = \frac{P_2}{\rho g} + \frac{\alpha_2 v_2^2}{2g} + z_2 + H_{turbine} + h_L$

where:

• $P / (\rho g)$ is the static pressure head ($\text{m}$ or $\text{ft}$),
• $\alpha v^2 / (2g)$ is the velocity head corrected for cross-sectional velocity profile,
• $z$ is the elevation head relative to a fixed datum,
• $H_{pump} = -\hat{w}_s / g$ is the total dynamic head added to the fluid by a pump,
• $H_{turbine} = \hat{w}_{s,out} / g$ is shaft work extracted by a hydraulic turbine,
• $h_L = \hat{h}_f / g$ is the total head loss due to skin friction and minor fitting losses.

The Kinetic Energy Correction Factor ($\alpha$)

Fluid velocity is not uniform across a pipe cross-section; it is zero at the wall (no-slip condition) and reaches a maximum at the pipe centerline. The kinetic energy correction factor $\alpha$ reconciles the average velocity $v = Q/A$ with the true kinetic energy flux $\int \frac{1}{2}\rho u(r)^3 dA$:

• Laminar Flow (parabolic profile): $\alpha = 0.50$ (or $1/2$).
• Turbulent Flow (blunted $1/7$-th power profile): $\alpha \approx 1.04 - 1.08$, universally taken as $\alpha = 1.0$ in engineering calculations.

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2. Dimensionless Flow Regimes and the Reynolds Number

The flow regime in a circular conduit of inside diameter $D$ is governed by the Reynolds number ($Re$), representing the ratio of inertial forces to viscous forces:

$Re = \frac{\rho v D}{\mu} = \frac{v D}{\nu} = \frac{4 \dot{m}}{\pi D \mu} = \frac{4 \rho Q}{\pi D \mu}$

where $\mu$ is dynamic shear viscosity ($\text{Pa}\cdot\text{s}$ or $\text{lb}_m/(\text{ft}\cdot\text{s})$), $\nu = \mu / \rho$ is kinematic viscosity ($\text{m}^2/\text{s}$ or $\text{ft}^2/\text{s}$), and $\dot{m} = \rho Q$ is mass flow rate.

Regimes in Circular Pipes

1. Laminar Regime ($Re < 2,100$): Viscous forces dominate. Fluid particles move in parallel concentric laminae without macroscopic mixing or radial velocity fluctuations.
2. Transition Regime ($2,100 \le Re \le 4,000$): Unstable flow. Intermittent turbulent bursts occur; flow fluctuates unpredictably between laminar and turbulent states.
3. Turbulent Regime ($Re > 4,000$): Inertial forces dominate. Chaotic, three-dimensional turbulent eddies enhance radial momentum transfer, producing a flattened core velocity profile bounded by a thin laminar/viscous sublayer at the pipe wall.

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3. The Great Friction Factor Divide: Darcy-Weisbach vs. Fanning

A paramount source of confusion on the NCEES PE Chemical exam is the distinction between the Darcy-Weisbach friction factor ($f_D$, also designated as the Moody friction factor $f$) and the Fanning friction factor ($f_F$). Both are dimensionless, but their fundamental definitions differ by a factor of 4.

[!IMPORTANT] Always verify the friction factor identity before calculating! Examine the laminar formula on the chart or equation sheet. If laminar flow is represented by $64/Re$, it is Darcy-Weisbach ($f_D$). If it is $16/Re$, it is Fanning ($f_F$).

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4. Laminar Flow: The Hagen-Poiseuille Equation

For laminar flow in a horizontal cylindrical pipe, momentum conservation yields a parabolic velocity profile:

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

Notice that the centerline velocity is exactly twice the cross-sectional average velocity: $u_{max} = 2 v$.

Integrating the shear stress across the radius leads to the analytical Hagen-Poiseuille equation:

$\Delta P = \frac{128 \mu L Q}{\pi D^4} = \frac{32 \mu v L}{D^2}$

Expressed in head loss form:

$h_f = \frac{32 \mu v L}{\rho g D^2} = \left(\frac{64}{Re}\right) \frac{L}{D} \frac{v^2}{2g}$

Key physical takeaway: In laminar flow, frictional pressure drop is directly proportional to velocity ($\Delta P \propto v$) and viscosity ($\Delta P \propto \mu$), and inversely proportional to diameter to the fourth power ($\Delta P \propto D^{-4}$). Wall roughness has zero effect on laminar friction because surface asperities remain buried within the streamlined laminae.

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5. Turbulent Flow Correlations and the Moody Diagram

In turbulent flow, analytical integration is precluded by Reynolds shear stresses ($-\rho \overline{u'v'}$). Friction factors are determined semi-empirically based on $Re$ and relative roughness ($\varepsilon / D$), where $\varepsilon$ is the absolute equivalent sand-grain roughness of the pipe material.

Typical Absolute Roughness Values ($\varepsilon$)

• Drawn copper, glass, plastic (PVC, HDPE): $\varepsilon = 0.0015\text{ mm} = 0.000005\text{ ft}$
• Commercial carbon steel, welded steel: $\varepsilon = 0.045 - 0.046\text{ mm} = 0.00015\text{ ft}$
• Galvanized iron: $\varepsilon = 0.15\text{ mm} = 0.0005\text{ ft}$
• Cast iron: $\varepsilon = 0.26\text{ mm} = 0.00085\text{ ft}$

Governing Turbulent Correlations (Darcy Form)

1. The Implicit Colebrook-White Equation (Moody Chart Basis): $\frac{1}{\sqrt{f_D}} = -2.0 \log_{10} \left( \frac{\varepsilon / D}{3.7} + \frac{2.51}{Re \sqrt{f_D}} \right)$

2. The Explicit Haaland Equation (Recommended for Rapid PE Solving): Developed by S. E. Haaland in 1983, this non-iterative formula provides accuracy within $\pm 1.5%$ of Colebrook-White for $4,000 \le Re \le 10^8$ and $10^{-6} \le \varepsilon/D \le 0.05$: $\frac{1}{\sqrt{f_D}} = -1.8 \log_{10} \left[ \left(\frac{\varepsilon / D}{3.7}\right)^{1.11} + \frac{6.9}{Re} \right]$

3. The Blasius Correlation (Smooth Pipes, Low $Re$): Applicable for hydraulically smooth pipes ($\varepsilon / D \approx 0$) in the range $4,000 \le Re \le 100,000$: $f_D = 0.3164 Re^{-0.25} \quad \Longleftrightarrow \quad f_F = 0.0791 Re^{-0.25}$

4. The Fully Rough (Wholly Turbulent) Regime: At sufficiently high Reynolds numbers (the right side of the Moody diagram), the laminar sublayer becomes thinner than the roughness asperities. Form drag over the asperities completely overwhelms viscous shear, making the friction factor independent of $Re$ (the von Kármán equation): $\frac{1}{\sqrt{f_D}} = -2.0 \log_{10} \left(\frac{\varepsilon / D}{3.7}\right) = 1.14 - 2 \log_{10} \left(\frac{\varepsilon}{D}\right)$ In this regime, $\Delta P$ is strictly proportional to $v^2$ (or $Q^2$).

5. The Universal Churchill Correlation: Valid across all regimes (laminar, transition, turbulent) without piecewise discontinuity: $f_D = 8 \left[ \left(\frac{8}{Re}\right)^{12} + \frac{1}{(A + B)^{1.5}} \right]^{1/12}$ where $A = \left[-2.457 \ln\left( \left(\frac{7}{Re}\right)^{0.9} + 0.27 \frac{\varepsilon}{D} \right)\right]^{16}$ and $B = \left(\frac{37,530}{Re}\right)^{16}$.

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6. Hydraulic Diameter for Non-Circular Conduits

To apply circular pipe friction charts and equations to non-circular passages (e.g., heat exchanger annuli, rectangular ducts, jacketed vessels), define the hydraulic diameter ($D_H$):

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

where $A_c$ is cross-sectional flow area and $P_w$ is the wetted perimeter (the boundary perimeter in direct contact with the flowing fluid).

• Concentric Annulus (Outer ID = $D_o$, Inner OD = $D_i$): $A_c = \frac{\pi}{4}(D_o^2 - D_i^2), \quad P_w = \pi(D_o + D_i) \implies D_H = \frac{4 \frac{\pi}{4}(D_o^2 - D_i^2)}{\pi(D_o + D_i)} = D_o - D_i$
• Rectangular Duct (width $a$, height $b$): $A_c = a b, \quad P_w = 2(a + b) \implies D_H = \frac{4 a b}{2(a + b)} = \frac{2 a b}{a + b}$

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7. Step-by-Step Worked Numerical Example

Problem Statement

Liquid toluene at $25^\circ\text{C}$ (density $\rho = 862\text{ kg/m}^3$, absolute viscosity $\mu = 0.560\text{ cP} = 0.560 \times 10^{-3}\text{ Pa}\cdot\text{s}$) is pumped through a horizontal $150\text{ m}$ run of commercial carbon steel pipe ($\varepsilon = 0.046\text{ mm}$) at a volumetric flow rate of $Q = 38.0\text{ m}^3/\text{h}$. The piping is nominal 3-inch Schedule 40.

From standard pipe schedule tables:

• Nominal size: $3\text{ in}$
• Actual inside diameter: $D = 3.068\text{ in} = 0.07793\text{ m}$
• Internal cross-sectional area: $A_c = 0.004769\text{ m}^2$

Calculate:

1. The mean flow velocity $v$ and Reynolds number $Re$.
2. The Darcy-Weisbach friction factor $f_D$ using the Haaland equation.
3. The total frictional head loss $h_f$ in meters of toluene.
4. The pressure drop $\Delta P_f$ in $\text{kPa}$ and $\text{psi}$.

Solution

Step 1: Calculate velocity and Reynolds number. Convert flow rate to SI base units:

$Q = \frac{38.0\text{ m}^3/\text{h}}{3,600\text{ s/h}} = 0.010556\text{ m}^3/\text{s}$

Mean velocity:

$v = \frac{Q}{A_c} = \frac{0.010556\text{ m}^3/\text{s}}{0.004769\text{ m}^2} = 2.2135\text{ m/s}$

Reynolds number:

$Re = \frac{\rho v D}{\mu} = \frac{862\text{ kg/m}^3 \times 2.2135\text{ m/s} \times 0.07793\text{ m}}{0.560 \times 10^{-3}\text{ Pa}\cdot\text{s}} = \frac{148.69}{0.000560} = 265,520$

Since $Re = 2.66 \times 10^5 > 4,000$, the flow is fully turbulent.

Step 2: Determine the Darcy friction factor using Haaland's formula. Relative roughness:

$\frac{\varepsilon}{D} = \frac{0.046\text{ mm}}{77.93\text{ mm}} = 0.0005903$

Applying the Haaland equation:

$\frac{1}{\sqrt{f_D}} = -1.8 \log_{10} \left[ \left(\frac{0.0005903}{3.7}\right)^{1.11} + \frac{6.9}{265,520} \right]$

Evaluate intermediate terms:

$\frac{\varepsilon / D}{3.7} = \frac{0.0005903}{3.7} = 1.5954 \times 10^{-4}$ $\left(1.5954 \times 10^{-4}\right)^{1.11} = 6.307 \times 10^{-5}$ $\frac{6.9}{Re} = \frac{6.9}{265,520} = 2.5986 \times 10^{-5}$ $\text{Bracket Sum} = 6.307 \times 10^{-5} + 2.599 \times 10^{-5} = 8.906 \times 10^{-5}$ $\log_{10}\left(8.906 \times 10^{-5}\right) = -4.0503$ $\frac{1}{\sqrt{f_D}} = -1.8 \times (-4.0503) = 7.2906$ $\sqrt{f_D} = \frac{1}{7.2906} = 0.13716$ $f_D = (0.13716)^2 = 0.01881$

(Note: If using the Fanning convention, $f_F = f_D / 4 = 0.00470$.)

Step 3: Calculate frictional head loss ($h_f$). Using the Darcy-Weisbach head loss equation:

$h_f = f_D \frac{L}{D} \frac{v^2}{2g} = 0.01881 \times \left(\frac{150\text{ m}}{0.07793\text{ m}}\right) \times \frac{(2.2135\text{ m/s})^2}{2 \times 9.807\text{ m/s}^2}$ $h_f = 0.01881 \times 1,924.8 \times \frac{4.8996}{19.614} = 36.205 \times 0.2498 = 9.044\text{ m of toluene}$

Step 4: Calculate frictional pressure drop ($\Delta P_f$).

$\Delta P_f = \rho g h_f = 862\text{ kg/m}^3 \times 9.807\text{ m/s}^2 \times 9.044\text{ m} = 76,450\text{ Pa} = 76.45\text{ kPa}$

Convert to psi:

$\Delta P_f = 76.45\text{ kPa} \times \left(\frac{1\text{ psi}}{6.89476\text{ kPa}}\right) = 11.09\text{ psi}$

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8. Common PE Exam Traps in Incompressible Flow

1. The Factor-of-Four Blunder: Using Fanning friction factor $f_F$ in Darcy's equation $\Delta P = f (L/D) (\rho v^2 / 2)$ without the factor of 4, producing an answer that is exactly $25%$ of the true value.
2. Nominal vs. Actual Pipe Diameter: Using $3.00\text{ in}$ instead of $3.068\text{ in}$ for Schedule 40 pipe. Because pressure drop scales as $D^{-5}$, an error of just $2.3%$ in diameter creates a $(3.00/3.068)^{-5} - 1 = +11.8%$ error in computed pressure drop!
3. Omitting the Kinetic Energy Correction Factor in Laminar Balance: When solving laminar energy balances where upstream and downstream velocities differ, using $\alpha = 1.0$ instead of $\alpha = 0.50$.
4. Assuming Roughness Always Matters: In laminar flow ($Re < 2,100$), wall roughness $\varepsilon$ is completely irrelevant. Do not waste time looking up $\varepsilon / D$ if $Re < 2,100$.