Laminar and turbulent pipe flow

The energy equation of Lesson 10 needs a head loss, and so far we have simply been given one. In this lesson we learn to calculate it. Most of the head a pump supplies in a long pipeline is spent overcoming friction at the pipe wall, and that friction depends on whether the flow is laminar or turbulent, on the pipe's diameter and roughness and on the velocity. Hana Lindqvist uses these methods to size new mains, to check whether an old cast-iron pipe has become badly tuberculated and to estimate how much energy a cleaning and lining program would save.

Laminar, transitional and turbulent flow

In 1883 Osborne Reynolds injected a thin stream of dye into water flowing through a glass tube. At low velocities the dye stayed in a neat straight thread: the flow was laminar, moving in smooth layers that slid past one another. As he raised the velocity, the thread began to waver, and beyond a certain point it broke up suddenly and spread through the whole tube: the flow had become turbulent, filled with irregular eddies of many sizes. Reynolds showed that the change depends not on velocity alone but on a dimensionless combination of velocity, diameter, density and viscosity, now called the Reynolds number.

The Reynolds number is the ratio of inertial forces to viscous forces. When it is small, viscosity damps out disturbances and the flow stays laminar; when it is large, disturbances grow and the flow becomes turbulent. In pipes, flow is generally laminar below \(Re \approx 2300\) and fully turbulent above about 4000, with a transitional range in between where the flow switches back and forth. Water mains run at Reynolds numbers in the hundreds of thousands to millions, so their flow is always turbulent. Laminar pipe flow occurs with viscous oils, in very small tubes and at very low velocities, such as in chemical dosing lines.

Entrance length and fully developed flow

Where a pipe leaves a tank, the velocity is nearly uniform across its entrance. The no-slip condition then slows the fluid next to the wall, and a boundary layer grows inward until it fills the pipe. Beyond that entrance length, the velocity profile stops changing along the pipe, and the flow is fully developed. For laminar flow the entrance length is about \(0.06\,Re\,D\), which can be very long; for turbulent flow it is roughly \(4.4\,Re^{1/6}D\), typically 20 to 40 diameters. The friction formulas in this lesson apply to fully developed flow. In long mains the entrance region is a small fraction of the length and is ignored, or folded into the entrance loss coefficient of Lesson 12.

The Darcy-Weisbach equation

For fully developed flow in a straight pipe of constant diameter, the head loss due to wall friction is given by the Darcy-Weisbach equation. It says that the loss is proportional to the pipe's length in diameters and to the velocity head, with a dimensionless friction factor \(f\) as the constant of proportionality. Dimensional analysis shows that \(f\) can depend only on the Reynolds number and the relative roughness \(\varepsilon/D\), the ratio of the average height of the wall's roughness elements to the diameter.

The Darcy-Weisbach equation is valid for any Newtonian fluid, laminar or turbulent, in a pipe of any material. All the difficulty lies in finding \(f\). Other formulas, notably the Hazen-Williams equation, remain popular in North American water practice because they are simple, but they are empirical, valid only for water near room temperature in turbulent flow, and less accurate for very smooth or very rough pipes. Hana uses Darcy-Weisbach for design and keeps Hazen-Williams for comparison with older records.

Laminar flow: an exact result

For fully developed laminar flow, the momentum equation and Newton's law of viscosity can be solved exactly. The result is the Hagen-Poiseuille flow: a parabolic velocity profile, with the centerline velocity twice the average, and a pressure drop proportional to the first power of the velocity. Substituting into Darcy-Weisbach gives the friction factor directly.

In laminar flow, wall roughness has no effect: the friction factor depends on the Reynolds number alone. Doubling the flow in a laminar dosing line doubles its pressure drop, and halving the tube diameter at the same flow multiplies the drop by sixteen. That steep dependence on diameter is why dosing tubes partly blocked by scale cause feed rates to fall sharply.

Turbulent flow and the Colebrook equation

Turbulent flow has no exact solution. Its friction factor comes from experiments, the most famous being Nikuradse's 1930s tests on pipes roughened with uniform sand grains, and from later tests on commercial pipes. In turbulent flow, a thin viscous sublayer lies next to the wall. If the roughness elements are buried inside it, the pipe behaves as hydraulically smooth and \(f\) depends only on \(Re\). If they poke through it, the pipe is rough and, at high enough Reynolds numbers, \(f\) depends only on \(\varepsilon/D\). Commercial pipes fall between these limits, and Colebrook combined the two into one implicit equation in 1939.

Because \(f\) appears on both sides, the Colebrook equation must be solved by iteration. Fortunately it converges quickly from almost any starting guess. Haaland's explicit formula, \(1/\sqrt{f} \approx -1.8\log_{10}[6.9/Re + (\varepsilon/D/3.7)^{1.11}]\), agrees within about 2 percent and makes an excellent first guess. The Moody chart, published in 1944, plots Colebrook's equation for a range of relative roughnesses and was the standard tool before computers; it remains the best way to see how friction behaves.

# Friction factor: laminar formula, Colebrook (solved by iteration) and Haaland
import math

def colebrook(Re, rel, tol=1e-12):
    if Re < 2300:
        return 64 / Re
    x = -1.8 * math.log10(6.9 / Re + (rel / 3.7) ** 1.11)   # Haaland as the starting guess, x = 1/sqrt(f)
    for _ in range(100):
        new = -2.0 * math.log10(rel / 3.7 + 2.51 * x / Re)
        if abs(new - x) < tol: break
        x = new
    return 1 / x**2

def haaland(Re, rel):
    return (-1.8 * math.log10(6.9 / Re + (rel / 3.7) ** 1.11)) ** -2

print("    Re      e/D      f Colebrook   f Haaland   diff %")
for Re in (1e3, 1e4, 1e5, 1e6, 1e7):
    for rel in (0.0, 1e-4, 1e-3, 1e-2):
        fc = colebrook(Re, rel)
        fh = haaland(Re, rel) if Re >= 2300 else fc
        print(f"{Re:8.0e}  {rel:7.4f}   {fc:10.5f}   {fh:9.5f}   {100*(fh-fc)/fc:6.2f}")

# Head loss in 1.8 km of 600 mm ductile iron main (cement lined, e = 0.25 mm assumed)
rho, mu, g = 998.0, 1.0e-3, 9.81
D, L, e, Q = 0.600, 1800.0, 0.25e-3, 0.45
V = Q / (math.pi * D**2 / 4)
Re = rho * V * D / mu
f = colebrook(Re, e / D)
hf = f * L / D * V**2 / (2 * g)
print()
print(f"Transmission main: V = {V:.2f} m/s, Re = {Re:,.0f}, f = {f:.4f}, hf = {hf:.2f} m ({rho*g*hf/1000:.0f} kPa)")

    Re      e/D      f Colebrook   f Haaland   diff %
   1e+03   0.0000      0.06400     0.06400     0.00
   1e+03   0.0001      0.06400     0.06400     0.00
   1e+03   0.0010      0.06400     0.06400     0.00
   1e+03   0.0100      0.06400     0.06400     0.00
   1e+04   0.0000      0.03088     0.03089     0.01
   1e+04   0.0001      0.03104     0.03099    -0.15
   1e+04   0.0010      0.03238     0.03217    -0.64
   1e+04   0.0100      0.04313     0.04304    -0.19
   1e+05   0.0000      0.01799     0.01782    -0.92
   1e+05   0.0001      0.01851     0.01827    -1.34
   1e+05   0.0010      0.02217     0.02197    -0.94
   1e+05   0.0100      0.03850     0.03854     0.09
   1e+06   0.0000      0.01165     0.01159    -0.50
   1e+06   0.0001      0.01344     0.01333    -0.86
   1e+06   0.0010      0.01994     0.01994    -0.01
   1e+06   0.0100      0.03796     0.03804     0.19
   1e+07   0.0000      0.00810     0.00813     0.35
   1e+07   0.0001      0.01217     0.01217    -0.00
   1e+07   0.0010      0.01967     0.01970     0.18
   1e+07   0.0100      0.03791     0.03799     0.20

Transmission main: V = 1.59 m/s, Re = 953,020, f = 0.0166, hf = 6.45 m (63 kPa)

Using the friction factor in design

Pipe problems come in three kinds. In the first, the flow and pipe are known and the head loss is wanted, which is direct, as in the example. In the second, the head available and the pipe are known and the flow is wanted; since the friction factor depends on the unknown flow through \(Re\), we guess \(f\), compute \(V\), update \(Re\) and \(f\), and repeat until it settles. In the third, the flow and allowable loss are known and the diameter is wanted, which also needs iteration, or a sweep of standard sizes. Hana's design spreadsheets do exactly this, sweeping pipe sizes and comparing the capital cost of each with the present value of the pumping energy it requires.

Head loss in turbulent flow rises roughly with the square of velocity, more precisely with about the 1.8th to 2nd power, so a pipe carrying double its design flow loses about four times the head. Diameter matters even more. At a fixed flow, velocity varies as \(1/D^2\), so the friction loss varies roughly as \(1/D^5\): going from a 500 mm to a 600 mm main cuts friction loss by about 60 percent. That steep dependence is why small savings in pipe diameter at construction often cost far more in energy over the pipe's life.