Fluid Mechanics · Chapter 8
Fluid Mechanics · Chapter 8

Flow in Pipes

Pipe flow is the bread and butter of applied fluid mechanics: how much pressure drop, and how much pumping power, does it take to move a fluid through a system? This chapter builds the toolkit — the Reynolds number and the laminar/turbulent divide, the Darcy–Weisbach equation and the Moody chart, minor losses, piping networks, pump matching, and the meters that measure it all.

Learning objectives

  • Distinguish laminar from turbulent flow and analyze fully developed flow in pipes.
  • Calculate major and minor losses, and the pumping power they demand.
  • Use the Moody chart and stated correlations only within their documented ranges.
  • Define energy-equation boundaries, coordinates, pressure references, and pump signs.
  • Iterate coupled flow, friction, and pump operating-point calculations.
  • Check Reynolds and Mach regimes, NPSH margin, units, and uncertainty.
Section 8–1

Introduction

Liquid or gas flow through pipes and ducts is everywhere — household plumbing, city water mains, oil pipelines, even blood vessels. The central practical questions are how much pressure (and pumping power) a flow demands, and how that depends on flow rate, pipe size, and roughness.

Average Velocity

Because velocity varies across a pipe cross-section (zero at the wall, maximum at the center), analysis uses an average velocity \(V_{\text{avg}}\) defined so that it carries the correct mass flow rate. For incompressible flow in a circular pipe, \(\dot m = \rho V_{\text{avg}} A_c\).

Physical meaning

Average velocity preserves mass flow through a section; it is not generally the centerline velocity. A pipe-system control volume should cross sections where pressure, elevation, and mean velocity are known or can be modeled.

Round vs. Noncircular Pipes

Most pipes are round because a circular cross-section withstands large pressure differences with the least material. Ducts in air-handling systems are often rectangular and are handled using the hydraulic diameter.

Section 8–2

Laminar & Turbulent Flows

Two Flow Regimes

Laminar flow is smooth and orderly, moving in coherent layers; turbulent flow is chaotic, marked by rapid velocity fluctuations and mixing. Osborne Reynolds showed with a dye-streak experiment that which regime occurs is governed by a single dimensionless ratio.

Reynolds Number

The ratio of inertial forces to viscous forces. Large values mean inertia dominates and viscous damping cannot suppress fluctuations (turbulent); small values mean viscosity keeps the flow orderly (laminar).

Reynolds Number for Pipe Flow
\[ \mathrm{Re} = \frac{\rho V_{\text{avg}} D}{\mu} = \frac{V_{\text{avg}} D}{\nu} \]
Critical values for a circular pipe: flow is laminar for \(\mathrm{Re} \lesssim 2300\), transitional for \(2300 \lesssim \mathrm{Re} \lesssim 4000\), and fully turbulent for \(\mathrm{Re} \gtrsim 4000\). The common design threshold is \(\mathrm{Re}_{cr} \approx 2300\).

These thresholds are approximate and disturbance-sensitive. For gas flow, also check Mach number; an incompressible density model is commonly unsuitable once density variation becomes material. Froude number is usually irrelevant in a completely filled pressurized line but matters when a free surface appears.

Hydraulic Diameter

For noncircular passages, the Reynolds number and loss correlations use the hydraulic diameter, defined so it reduces to the ordinary diameter for a round pipe. Here \(A_c\) is the cross-sectional area and \(p\) the wetted perimeter.

Hydraulic Diameter
\[ D_h = \frac{4 A_c}{p} \]
Section 8–3

The Entrance Region

Developing vs. Fully Developed Flow

Near a pipe inlet a boundary layer grows from the wall until it fills the pipe. The region where the velocity profile is still changing is the hydrodynamic entrance region; beyond it the profile no longer changes with distance and the flow is fully developed.

Entrance Length (approximate)
\[ \frac{L_{h,\text{lam}}}{D} \approx 0.05\,\mathrm{Re}, \qquad \frac{L_{h,\text{turb}}}{D} \approx 1.359\,\mathrm{Re}^{1/4} \]

In fully developed flow the wall shear stress — and hence the friction factor — stays constant along the pipe, which is what makes the pressure drop predictable.

Section 8–4

Laminar Flow in Pipes

For steady, fully developed laminar flow in a round pipe, the momentum balance can be solved exactly, giving a parabolic velocity profile and clean closed-form results.

Velocity Profile (parabolic)
\[ u(r) = 2V_{\text{avg}}\left(1 - \frac{r^2}{R^2}\right) \]

The maximum (centerline) velocity is exactly twice the average. The volume flow rate that follows is the Hagen–Poiseuille result.

Hagen–Poiseuille Flow Rate
\[ \dot{V} = \frac{\pi D^4 \,\Delta P}{128\,\mu L} \]
Friction Factor (laminar)
\[ f = \frac{64}{\mathrm{Re}} \]
pipe wall (u = 0, no slip) u_max = 2 V_avg parabolic profile
Figure 8–1. Fully developed laminar pipe flow has a parabolic velocity profile; the centerline velocity is twice the cross-sectional average. (Original schematic.)
EXAMPLE 1Viscous calibration-line pressure drop
A supplied calibration liquid flows at \(1.20\times10^{-6}\ \text{m}^3/\text{s}\) through a horizontal \(4.00\ \text{mm}\)-diameter, \(1.50\ \text{m}\)-long tube. Find the pressure drop.

Recognition strategy

A straight small tube with supplied viscosity suggests fully developed laminar analysis, but Reynolds number must confirm the regime.

Given information

  • At \(20^\circ\text{C}\), supplied \(\rho=920\ \text{kg/m}^3\), \(\mu=0.120\ \text{Pa}\!\cdot\!\text{s}\); \(Q=1.20\times10^{-6}\ \text{m}^3/\text{s}\).
  • \(D=0.00400\ \text{m}\), \(L=1.50\ \text{m}\); \(x\) is downstream and \(\Delta p=p_{\rm in}-p_{\rm out}>0\), using gauge pressure.

Model and interpretation

A steady incompressible tube control volume is horizontal, Newtonian, and fully developed. Hagen–Poiseuille is exact under those idealizations and approximate for the real entrance.

Step-by-step solution
1. Find area and speed. \(A=\pi D^2/4=1.257\times10^{-5}\ \text{m}^2\), so \(V=Q/A=0.0955\ \text{m/s}\).
2. Verify regime. \(\mathrm{Re}=\rho VD/\mu=2.93\), safely laminar rather than transitional.
3. Apply the exact model. \(\Delta p=128\mu LQ/(\pi D^4)\) for fully developed circular flow.
4. Substitute. \(\Delta p=128(0.120)(1.50)(1.20\times10^{-6})/[\pi(0.00400)^4]=34.4\ \text{kPa}\).

Verification

Units: \(\mu LQ/D^4\) reduces from \(\text{Pa}\!\cdot\!\text{s}\!\cdot\!\text{m}\!\cdot\!\text{m}^3/\text{s}/\text{m}^4\) to pascals.

Signs and direction: Positive downstream flow requires pressure to decrease, so \(p_{\rm in}-p_{\rm out}\) is positive.

Plausibility: The very low Reynolds number supports laminar flow; the large loss is credible because resistance varies as \(D^{-4}\), though entrance length and property uncertainty should be checked.

Final result: \(\Delta p=34.4\ \text{kPa}\) as a positive inlet-to-outlet gauge pressure drop.

Engineering interpretation

The small bore makes pressure highly sensitive to measured diameter. Temperature control is also important because the supplied viscosity basis is \(20^\circ\text{C}\).

Section 8–5

Turbulent Flow in Pipes

Head Loss & the Darcy Friction Factor

The pressure drop in fully developed flow is expressed as a head loss, proportional to pipe length, the square of velocity, and a dimensionless Darcy friction factor \(f\). This single relation — the Darcy–Weisbach equation — is valid for both laminar and turbulent flow.

Darcy–Weisbach Head Loss
\[ h_L = f\,\frac{L}{D}\,\frac{V_{\text{avg}}^2}{2g} \]

The Colebrook Equation & Moody Chart

In turbulent flow the friction factor depends on both Reynolds number and the relative roughness \(\varepsilon/D\). The implicit Colebrook equation captures this dependence, and the Moody chart is its graphical solution — the everyday tool for finding \(f\).

Colebrook Equation (turbulent, implicit in f)
\[ \frac{1}{\sqrt{f}} = -2.0\,\log\!\left(\frac{\varepsilon/D}{3.7} + \frac{2.51}{\mathrm{Re}\,\sqrt{f}}\right) \]
The three problem types: (1) find head loss given flow rate — direct; (2) find flow rate given head loss — iterative; (3) find pipe diameter given both — iterative. All three use Darcy–Weisbach plus the Moody chart.

Colebrook is an empirical turbulent correlation, not a transitional-flow model. Iterate until both friction factor and energy balance meet a stated tolerance; compare with the Moody chart or an independently valid explicit correlation.

Problem-solving strategy

  1. Draw the system and choose inlet and outlet control surfaces.
  2. Declare \(+x\), elevations, gauge or absolute pressures, and pump-head sign.
  3. Use continuity to determine every section velocity.
  4. Compute Reynolds number and relative roughness before selecting friction relations.
  5. Apply the energy balance with major and supplied minor-loss coefficients.
  6. Iterate coupled unknowns and check residual, regime, power, Mach number, and cavitation risk.
Supplied Haaland approximation
\[ \frac{1}{\sqrt{f}}=-1.8\log_{10}\!\left[\left(\frac{(\varepsilon/D)}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right] \]

This is an approximate turbulent-flow correlation for the Darcy friction factor. Use it only after confirming turbulent, fully developed internal flow and that the Reynolds number and relative roughness lie within the range specified by the supplied correlation source or design standard.

EXAMPLE 2Turbulent transfer-line head loss
Water flows at \(0.0120\ \text{m}^3/\text{s}\) through \(80.0\ \text{m}\) of \(0.100\ \text{m}\) pipe. Compute major and supplied minor losses.

Recognition strategy

Known flow makes this a direct loss calculation: establish regime, use the supplied roughness correlation, then add fitting losses.

Given information

  • At \(20^\circ\text{C}\), supplied \(\rho=998\ \text{kg/m}^3\), \(\mu=1.002\times10^{-3}\ \text{Pa}\!\cdot\!\text{s}\), \(\varepsilon=0.045\ \text{mm}\), \(\sum K=5.0\).
  • Gauge pressure is used; \(x\) and flow are positive downstream, \(g=9.81\ \text{m/s}^2\), and the line is steady and completely filled.

Model and interpretation

The control volume covers pipe and fittings. Incompressibility and fully developed major loss are approximations; Haaland is supplied as valid here and uses the Darcy factor.

Step-by-step solution
1. Determine velocity. \(V=Q/(\pi D^2/4)=1.528\ \text{m/s}\).
2. Establish regime. \(\mathrm{Re}=\rho VD/\mu=1.52\times10^5\) and \(\varepsilon/D=4.50\times10^{-4}\).
3. Find friction factor. The supplied Haaland relation gives the approximate Darcy value \(f=0.01888\), or \(0.0189\) when rounded.
4. Sum losses. \(h_f=f(L/D)V^2/(2g)=1.797\ \text{m}\), approximately \(1.80\ \text{m}\); \(h_m=(5.0)V^2/(2g)=0.5949\ \text{m}\), approximately \(0.595\ \text{m}\); thus \(h_L=2.392\ \text{m}\), approximately \(2.39\ \text{m}\).

Verification

Units: Both \(fL/D\) and \(K\) are dimensionless, while \(V^2/(2g)\) has metres of fluid head.

Signs and direction: The positive total loss \(h_L=2.392\ \text{m}\) produces a downstream decrease in total head for the declared positive flow direction.

Plausibility: Turbulent Reynolds number lies in the supplied correlation range; the \(0.5949\ \text{m}\) minor loss is about one quarter of the \(2.392\ \text{m}\) total and cannot be discarded. Froude is not governing and liquid Mach number is negligible.

Final result: \(f=0.01888\), \(h_f=1.797\ \text{m}\) (approximately \(1.80\ \text{m}\)), \(h_m=0.5949\ \text{m}\) (approximately \(0.595\ \text{m}\)), and \(h_L=2.392\ \text{m}\) (approximately \(2.39\ \text{m}\)).

Engineering interpretation

The \(0.5949\ \text{m}\) fitting loss contributes about \(25\%\) of the \(2.392\ \text{m}\) system loss. Roughness and \(K\) values are supplied data, so procurement or as-built changes require recalculation.

Section 8–6

Minor Losses

Minor Losses & Loss Coefficient

Valves, bends, tees, inlets, exits, and area changes add their own losses, accounted for with a dimensionless loss coefficient \(K_L\). Despite the name, in a system with many fittings these "minor" losses can dominate.

Minor Loss
\[ h_{L,\text{minor}} = K_L\,\frac{V^2}{2g} \]
Total Head Loss
\[ h_{L,\text{total}} = \sum_i f_i\frac{L_i}{D_i}\frac{V_i^2}{2g} + \sum_j K_{L,j}\frac{V_j^2}{2g} \]

An alternative is the equivalent length — the length of straight pipe that would give the same loss as a fitting, \(L_{\text{equiv}} = (K_L/f)D\).

Section 8–7

Piping Networks & Pump Selection

Series & Parallel Pipes

Pipes in series carry the same flow rate while head losses add. Pipes in parallel share the same head loss across them while flow rates add — the same rules as electrical resistors, with head loss playing the role of voltage drop.

Operating Point

A pump's supplied head falls as flow rate rises (its characteristic curve), while the system's required head rises with flow rate (the system curve). The flow that actually occurs is the operating point where the two curves intersect.

Useful Pumping Power
\[ \dot{W}_{\text{pump,u}} = \rho g \dot{V}\, h_{\text{pump}}, \qquad \dot{W}_{\text{shaft}} = \frac{\dot{W}_{\text{pump,u}}}{\eta_{\text{pump}}} \]
flow rate V̇ head H pump curve system curve operating point
Figure 8–2. The operating point sits where the falling pump curve meets the rising system curve — the only flow rate at which supplied and required head agree. (Original schematic.)
EXAMPLE 3Pump operating point and suction audit
A supplied pump curve is \(H_p=28.0-8000Q^2\) and a system curve is \(H_s=10.0+4000Q^2\), with head in metres and \(Q\) in \(\text{m}^3/\text{s}\). Find the operating point and available NPSH.

Recognition strategy

The operating point is the positive curve intersection; a separate absolute-pressure suction balance is required for cavitation screening.

Given information

  • Supplied curves are valid over this range; supplied efficiency is \(0.72\), \(\rho=998\ \text{kg/m}^3\), and \(g=9.81\ \text{m/s}^2\).
  • At \(20^\circ\text{C}\), supplied \(p_{\rm atm}=101.3\ \text{kPa absolute}\), \(p_v=2.34\ \text{kPa absolute}\), suction lift \(z=3.0\ \text{m}\), suction loss \(h_{Ls}=1.2\ \text{m}\); upward lift is positive.

Model and interpretation

Pump head is positive energy added to a steady incompressible control volume. Curve equality is approximate empirical modeling; \(\mathrm{NPSH}_A=p_{\rm atm}/(\rho g)-p_v/(\rho g)-z-h_{Ls}\).

Step-by-step solution
1. Intersect curves. \(28-8000Q^2=10+4000Q^2\), hence \(Q^2=18/12000\).
2. Select physical root. Positive flow gives \(Q=0.0387\ \text{m}^3/\text{s}\) and \(H=16.0\ \text{m}\).
3. Compute shaft power. \(\dot W_s=\rho gQH/\eta=8.43\ \text{kW}\), positive power input.
4. Audit suction. \(\mathrm{NPSH}_A=(101.3-2.34)10^3/(998(9.81))-3.0-1.2=5.91\ \text{m}\).

Verification

Units: Curve coefficients produce metres, \(\rho gQH\) produces watts, and each NPSH term is metres of liquid.

Signs and direction: Pump head and shaft input are positive additions; suction lift and loss reduce available NPSH as expected.

Plausibility: Substitution gives both curves \(16.0\ \text{m}\). Cavitation acceptability cannot be claimed without the manufacturer-supplied required NPSH plus project margin.

Final result: \(Q=0.0387\ \text{m}^3/\text{s}\), \(H=16.0\ \text{m}\), \(\dot W_s=8.43\ \text{kW}\), and \(\mathrm{NPSH}_A=5.91\ \text{m}\).

Engineering interpretation

The hydraulic solution is not pump approval. Confirm vendor curve tolerance, operating envelope, required NPSH, motor rating, transient conditions, and fluid temperature.

Section 8–8

Flow Rate & Velocity Measurement

Obstruction Meters (Orifice, Nozzle, Venturi)

These place a constriction in the pipe and infer flow rate from the resulting pressure drop, via the Bernoulli relation with a discharge coefficient. The Venturi recovers the most pressure (lowest permanent loss); the orifice plate is cheapest but loses the most.

Obstruction-Meter Flow Rate
\[ \dot{V} = C_d A_0\sqrt{\frac{2(P_1 - P_2)}{\rho\,(1-\beta^4)}}, \qquad \beta = \frac{d}{D} \]

Other Methods

Pitot-static probes give local velocity from a stagnation-pressure difference; rotameters read flow from the height of a float; turbine, ultrasonic, and electromagnetic meters give electronic readouts without a large pressure penalty.

Instructor Note

This is the most exam-heavy chapter of the course. Build the habit: write the energy equation between two points, then expand \(h_L\) into major (\(f\,L/D\)) plus minor (\(\sum K_L\)) terms. Get the Reynolds number first — it decides whether you use \(f = 64/\mathrm{Re}\) or the Moody chart. Almost every pipe problem is this one template.

Chapter recap

  • Continuity fixes average velocities and Reynolds number selects the flow model.
  • Major and minor losses enter one consistent mechanical-energy balance.
  • Unknown flow, diameter, or operating point generally requires iteration and residual checks.
  • Before moving on, be able to define pressure and pump signs, choose a valid friction relation, and audit power, Mach regime, and NPSH.