12.5 Flow in Pipes, Friction Factors (Moody Diagram), and Head Losses

12.5 Flow in Pipes, Friction Factors (Moody Diagram), and Head Losses

Core FE Exam Principle: Viscous friction along internal conduit walls degrades mechanical fluid energy into thermal energy. Pipe flow performance depends critically on whether the boundary layer flow regime is laminar, transitional, or turbulent.

Characterizing Flow Regimes: The Reynolds Number

The Reynolds Number ((Re)) represents the ratio of inertial forces to viscous forces within a fluid element:

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

where:

• (V) = Average flow velocity ((\text{m/s}) or (\text{ft/s}))
• (D) = Pipe internal diameter ((\text{m}) or (\text{ft}))
• (\mu) = Dynamic viscosity ((\text{Pa}\cdot\text{s}))
• (\nu) = Kinematic viscosity ((\text{m}^2/\text{s}))

Critical Flow Boundaries for Circular Pipes

• Laminar Flow ((Re < 2300)): Fluid moves in smooth parallel layers (laminae) without lateral mixing. Viscous forces dominate.
• Transitional Flow ((2300 \le Re \le 4000)): Unstable flow fluctuating unpredictably between laminar and turbulent behavior.
• Turbulent Flow ((Re > 4000)): Chaotic, highly agitated flow characterized by random three-dimensional velocity fluctuations and rapid scalar mixing.

Non-Circular Conduits and Hydraulic Diameter

For non-circular cross-sections (e.g., rectangular HVAC ducts or concentric annular pipes), define the Hydraulic Diameter ((D_h)):

$D_h = \frac{4 A}{P_w}$

where (A) is the fluid cross-sectional area and (P_w) is the wetted perimeter in contact with the fluid.

• Rectangular Duct ((a \times b)): (D_h = \frac{4(ab)}{2(a+b)} = \frac{2ab}{a+b}).
• Concentric Annulus (Inner (D_i), Outer (D_o)): (D_h = \frac{4 \cdot \frac{\pi}{4}(D_o^2 - D_i^2)}{\pi(D_o + D_i)} = D_o - D_i).

Laminar Pipe Flow (Hagen-Poiseuille Flow)

In fully developed laminar pipe flow, exact analytical solutions describe the velocity field.

Laminar Velocity Profile

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

• The velocity distribution is parabolic, reaching maximum velocity (V_{\text{max}}) along the pipe centerline ((r=0)).
• Average Velocity Relation: $V_{\text{avg}} = \frac{1}{2} V_{\text{max}}$
• Wall Shear Stress ((\tau_w)): $\tau_w = -\mu \left. \frac{du}{dr} \right|_{r=R} = \frac{4 \mu V_{\text{avg}}}{R} = \frac{8 \mu V_{\text{avg}}}{D}$

Laminar Friction Factor

By equating wall shear stress to pressure drop, the friction factor for laminar pipe flow simplifies to an inverse function of Reynolds number, completely independent of pipe wall roughness ((\epsilon)):

$f = \frac{64}{Re}$

Substituting (f = \frac{64}{Re}) into the Darcy-Weisbach equation yields the Hagen-Poiseuille Equation for laminar pressure drop:

$\Delta P = \frac{32 \mu L V_{\text{avg}}}{D^2}$

Turbulent Pipe Flow and the Moody Diagram

In turbulent flow, random eddy motion increases wall momentum transfer, making the Darcy friction factor (f) a function of both Reynolds number ((Re)) and relative roughness ((\epsilon/D)).

Equivalent Surface Roughness ((\epsilon))

Friction Factor Equations

1. Colebrook Equation (Implicit): Standard benchmark for turbulent pipe flow across all roughness ranges: $\frac{1}{\sqrt{f}} = -2.0 \log_{10} \left( \frac{\epsilon/D}{3.7} + \frac{2.51}{Re \sqrt{f}} \right)$

2. Swamee-Jain Equation (Explicit): Approximates Colebrook within 1% for (5000 \le Re \le 10^8) and (10^{-6} \le \epsilon/D \le 10^{-2}): $f = \frac{0.25}{\left[ \log_{10} \left( \frac{\epsilon/D}{3.7} + \frac{5.74}{Re^{0.9}} \right) \right]^2}$

3. Wholly Turbulent (Fully Rough) Zone: At extremely high Reynolds numbers, viscous forces become negligible relative to roughness drag. The friction factor becomes independent of (Re): $\frac{1}{\sqrt{f}} = -2.0 \log_{10} \left( \frac{\epsilon/D}{3.7} \right)$

Major and Minor Head Loss Calculations

Total system head loss combines continuous pipe skin friction (major loss) with localized flow separation losses caused by fittings, valves, and geometry changes (minor loss).

$h_L = h_{\text{major}} + h_{\text{minor}} = \sum f \left(\frac{L}{D}\right) \frac{V^2}{2g} + \sum K \left(\frac{V^2}{2g}\right)$

Major Friction Head Loss (Darcy-Weisbach Equation)

$h_f = f \left( \frac{L}{D} \right) \frac{V^2}{2g} \quad \text{and} \quad \Delta P_f = \gamma h_f = f \left( \frac{L}{D} \right) \frac{\rho V^2}{2}$

Minor Loss Coefficient Method ((K))

$h_m = K \left( \frac{V^2}{2g} \right)$

• Entrance Losses:
  • Sharp-Edged Entrance: (K = 0.5)
  • Well-Rounded Entrance ((r/D \ge 0.15)): (K = 0.04)
  • Re-entrant Entrance (pipe projecting inward): (K = 0.8)
• Exit Loss (Submerged Pipe Discharge into Reservoir): (K = 1.0) (all kinetic energy is dissipated as heat).
• Valves and Fittings: Gate valve (fully open (K=0.2)), Globe valve (fully open (K=10.0)), (90^\circ) standard elbow ((K=0.9)).

  Entrance Losses K:              Exit Loss K = 1.0:
  ------------------              ------------------
  Re-entrant:  K = 0.8            Submerged Discharge:
  Sharp-edge:  K = 0.5             ===========\\  
  Rounded:     K = 0.04           ===========/ ===> Discharges into open tank
                                                   Entire kinetic energy lost!

Pipe Networks: Series and Parallel Systems

Series Pipe Systems

Pipes connected end-to-end carrying a single flow stream.

• Flow Rate: (Q_1 = Q_2 = Q_3 = Q_{\text{total}})
• Total Head Loss: (h_{L,\text{total}} = h_{L,1} + h_{L,2} + h_{L,3})

Parallel Pipe Systems

Flow splits into two or more parallel branches connecting common junction nodes (A) and (B).

• Flow Rate: (Q_{\text{total}} = Q_1 + Q_2 + Q_3)
• Head Loss Equality: Head loss across all parallel branches must be equal: $h_{L,1} = h_{L,2} = h_{L,3} = h_{L, AB}$

$\text{Since } h_f = \frac{8 f L Q^2}{\pi^2 g D^5}, \quad \frac{f_1 L_1 Q_1^2}{D_1^5} = \frac{f_2 L_2 Q_2^2}{D_2^5} \implies \frac{Q_1}{Q_2} = \sqrt{\frac{f_2 L_2 D_1^5}{f_1 L_1 D_2^5}}$

Comprehensive Worked Engineering Example

Problem Statement

A commercial steel pipeline ((\epsilon = 0.045 \text{ mm})) with internal diameter (D = 200 \text{ mm}) and total length (L = 250 \text{ m}) conveys water ((\rho = 1000 \text{ kg/m}^3), (\nu = 1.0 \times 10^{-6} \text{ m}^2/\text{s}), (\gamma = 9.81 \text{ kN/m}^3)) at a volumetric flow rate (Q = 0.06283 \text{ m}^3/\text{s}).

The pipeline contains the following fittings:

• 1 sharp-edged entrance ((K_1 = 0.5))
• 4 standard (90^\circ) threaded elbows ((K_2 = 0.9) each)
• 1 fully open globe valve ((K_3 = 10.0))
• 1 submerged exit ((K_4 = 1.0))

Calculate:

1. The Reynolds number (Re) and flow regime.
2. The friction factor (f) using the Swamee-Jain equation.
3. The total head loss (h_L) across the pipeline.
4. The required pressure drop (\Delta P) along a horizontal installation.

Step-by-Step Solution

Step 1: Compute Flow Velocity and Reynolds Number

• Pipe cross-sectional area: $A = \frac{\pi D^2}{4} = \frac{\pi (0.200)^2}{4} = 0.031416 \text{ m}^2$
• Average flow velocity: $V = \frac{Q}{A} = \frac{0.06283 \text{ m}^3/\text{s}}{0.031416 \text{ m}^2} = 2.00 \text{ m/s}$
• Velocity head: $\frac{V^2}{2g} = \frac{(2.00)^2}{2 \times 9.81} = \frac{4.00}{19.62} = 0.20387 \text{ m}$
• Reynolds number: $Re = \frac{V D}{\nu} = \frac{2.00 \times 0.200}{1.0 \times 10^{-6}} = 400,000 = 4.0 \times 10^5$ Since (Re = 4.0 \times 10^5 > 4000), the flow is fully turbulent.

Step 2: Compute Relative Roughness and Friction Factor (f)

• Relative roughness: $\frac{\epsilon}{D} = \frac{0.045 \text{ mm}}{200 \text{ mm}} = 0.000225$
• Swamee-Jain explicit equation: $f = \frac{0.25}{\left[ \log_{10} \left( \frac{0.000225}{3.7} + \frac{5.74}{(400,000)^{0.9}} \right) \right]^2}$ $\frac{0.000225}{3.7} = 6.081 \times 10^{-5}, \quad (400,000)^{0.9} = 115,974 \implies \frac{5.74}{115,974} = 4.949 \times 10^{-5}$ $\text{Sum} = 6.081 \times 10^{-5} + 4.949 \times 10^{-5} = 1.103 \times 10^{-4}$ $\log_{10}(1.103 \times 10^{-4}) = -3.9574 \implies (-3.9574)^2 = 15.661$ $f = \frac{0.25}{15.661} = 0.01596 \approx 0.0160$

Step 3: Compute Major and Minor Head Losses

• Major friction head loss: $h_f = f \left(\frac{L}{D}\right) \frac{V^2}{2g} = 0.01596 \times \left(\frac{250}{0.200}\right) \times 0.20387 = 0.01596 \times 1250 \times 0.20387 = 4.067 \text{ m}$
• Minor loss coefficient sum: $\sum K = K_{\text{entrance}} + 4(K_{\text{elbow}}) + K_{\text{valve}} + K_{\text{exit}} = 0.5 + 4(0.9) + 10.0 + 1.0 = 0.5 + 3.6 + 10.0 + 1.0 = 15.1$
• Minor head loss: $h_m = \sum K \left(\frac{V^2}{2g}\right) = 15.1 \times 0.20387 = 3.078 \text{ m}$
• Total head loss: $h_L = h_f + h_m = 4.067 + 3.078 = 7.145 \text{ m}$

Step 4: Compute Required Pressure Drop

• For a horizontal pipeline ((z_1 = z_2)) of constant diameter ((V_1 = V_2)): $\Delta P = \gamma h_L = 9.81 \text{ kN/m}^3 \times 7.145 \text{ m} = 70.09 \text{ kPa}$

Final Answer: (Re = 4.0 \times 10^5) (turbulent), friction factor (f = 0.0160), total head loss (h_L = 7.15 \text{ m}), and pressure drop (\Delta P = 70.1 \text{ kPa}).