The Bernoulli equation and its limits

During a power failure two summers ago, the Northside clearwell had to be partly drained without pumps, and Ray Mendoza's crew rigged a 100 mm siphon hose over the clearwell wall. It worked, but only after they lowered the hose crest: with the crest too high above the water, the flow stopped and the hose filled with bubbles. Jun Park wanted to know why. The answer lies in the most famous equation in fluid mechanics, the Bernoulli equation, which relates pressure, velocity and elevation along a streamline. It is powerful, simple and frequently misused, so this lesson pays as much attention to its limits as to its uses.

Newton's second law along a streamline

Consider a small fluid particle moving along a streamline in steady flow. Measure distance \(s\) along the streamline. The forces acting along the streamline are the pressure difference across the particle's ends and the component of its weight; we neglect viscous forces. Newton's second law, with the particle's acceleration along the streamline equal to \(V\,dV/ds\) in steady flow (the convective acceleration of Lesson 5), gives the Euler equation along a streamline.

For an incompressible fluid, every term can be integrated along the streamline. The result is the Bernoulli equation: along a streamline in steady, incompressible, frictionless flow, the sum of pressure, kinetic energy per unit volume and potential energy per unit volume stays constant.

Dividing by \(\gamma = \rho g\) gives the head form, in which every term is a length. The pressure head \(p/\gamma\), the velocity head \(V^2/(2g)\) and the elevation head \(z\) add up to the total head \(H\), which stays constant along the streamline. Engineers working with pumps and pipelines almost always use the head form, because heads can be drawn directly on a profile of the pipeline.

Static, dynamic and stagnation pressure

In the pressure form, \(p\) is the static pressure, the pressure a gauge moving with the fluid would read, and the one measured at a tap flush with the pipe wall. The term \(\tfrac{1}{2}\rho V^2\) is the dynamic pressure. Their sum, plus the hydrostatic term, is the total pressure. At a stagnation point, where the fluid comes to rest on the nose of a body, the velocity term converts entirely into pressure, and the local stagnation pressure equals static plus dynamic pressure.

This is the principle of the pitot tube, an open tube facing upstream, which reads the stagnation pressure, combined with a static tap, which reads the static pressure. The difference gives the velocity. Pitot-static probes measure aircraft airspeed, and in water systems a pitot rod inserted through a corporation stop measures the velocity at points across a large main, from which the flow rate can be found. In the 600 mm main, a differential of 1.28 kPa corresponds to 1.60 m/s.

Applying the Bernoulli equation

The equation is used between two points on the same streamline, often a free surface and a jet or outlet. The standard steps are: draw the streamline, choose points where as much as possible is known, pick an elevation datum, write the equation, and simplify. At a large reservoir's free surface, velocity is negligible and pressure is atmospheric. At a free jet discharging to the atmosphere, the pressure across the jet is atmospheric. Using gauge pressure makes those atmospheric terms zero.

One famous result follows immediately. For a tank draining through a small hole a depth \(h\) below the surface, apply Bernoulli from the surface, where \(p = 0\) gauge and \(V \approx 0\), to the jet, where \(p = 0\) gauge: the jet velocity is \(\sqrt{2gh}\), the speed a stone gains falling through height \(h\). This is Torricelli's result, used in Lesson 6. Real openings deliver somewhat less flow, because friction slows the jet slightly and the jet contracts after it leaves a sharp-edged hole; the discharge coefficient of about 0.6 covers both effects.

# Bernoulli along a streamline: a siphon over the clearwell wall and a pitot probe
import math
g, rho = 9.81, 998.0
patm, pv = 97.4e3, 2.34e3       # Pa: barometer at the station, pressure at which water boils at 20 C

# Siphon: free surface at z = 0, crest at z = +zc, discharge at z = -zd
def siphon(zc, zd):
    V = math.sqrt(2 * g * zd)                 # Bernoulli from surface to outlet
    p_crest = patm - rho * g * zc - rho * V**2 / 2
    return V, p_crest

d = 0.10
for zc in (2.0, 4.0, 6.0):
    V, pc = siphon(zc, 3.0)
    Q = V * math.pi * d**2 / 4
    flag = "close to boiling pressure" if pc < 4 * pv else "ok"
    print(f"crest {zc:3.1f} m: V = {V:4.2f} m/s, Q = {Q*1000:5.1f} L/s, p_crest = {pc/1000:6.1f} kPa abs  {flag}")

# Highest crest before the water starts to boil (outlet 3 m below)
V = math.sqrt(2 * g * 3.0)
zmax = (patm - pv - rho * V**2 / 2) / (rho * g)
print(f"Maximum crest height with a 3.0 m drop: {zmax:.2f} m")

# Pitot-static probe in the 600 mm main: dynamic pressure from a differential gauge
for dp in (0.5e3, 1.28e3, 2.0e3):
    V = math.sqrt(2 * dp / rho)
    print(f"pitot dp = {dp/1000:4.2f} kPa -> V = {V:.2f} m/s")

crest 2.0 m: V = 7.67 m/s, Q =  60.3 L/s, p_crest =   48.4 kPa abs  ok
crest 4.0 m: V = 7.67 m/s, Q =  60.3 L/s, p_crest =   28.9 kPa abs  ok
crest 6.0 m: V = 7.67 m/s, Q =  60.3 L/s, p_crest =    9.3 kPa abs  close to boiling pressure
Maximum crest height with a 3.0 m drop: 6.71 m
pitot dp = 0.50 kPa -> V = 1.00 m/s
pitot dp = 1.28 kPa -> V = 1.60 m/s
pitot dp = 2.00 kPa -> V = 2.00 m/s

The program shows that the flow rate does not depend on the crest height at all, in this frictionless model, while the crest pressure falls by about 9.8 kPa for every meter of extra height. It also computes the highest workable crest, about 6.7 m above the water surface with a 3 m drop at the station's elevation. In practice the limit is lower, because dissolved air comes out of solution well above the vapor pressure and because friction in the hose lowers the pressure further. Field crews keep siphon crests below about 5 m.

The energy grade line and hydraulic grade line

Plotting heads along a pipeline makes Bernoulli visual. The energy grade line (EGL) is a line at height \(H = p/\gamma + V^2/(2g) + z\) above the datum. The hydraulic grade line (HGL) lies one velocity head below it, at \(p/\gamma + z\). The HGL is the level to which water would rise in a piezometer tube tapped into the pipe at that point. Wherever the pipe rises above the HGL, the pressure inside is below atmospheric, a warning sign for air entry, collapse of thin-walled pipe or cavitation.

In the frictionless flow assumed by Bernoulli, the EGL is horizontal. In a real pipe it slopes downward in the direction of flow because friction converts mechanical energy into heat, and it jumps up at a pump and drops at a turbine. Lessons 10 to 12 add those effects. Experienced engineers sketch the EGL and HGL for every pipeline, because the sketch shows at a glance where pressures are high, where they are dangerously low, and where an air-release valve is needed.

When Bernoulli applies, and when it does not

The Bernoulli equation rests on four assumptions, and each can fail. The flow must be steady; during a valve closure or pump start, an extra unsteady term appears. The fluid must be incompressible, which is fine for liquids and for gases below a Mach number of about 0.3. The flow must be frictionless, which is reasonable over short distances in smooth, accelerating flows, such as through a nozzle or contraction, but poor over long pipes or through separated flows. And the equation holds along a single streamline, unless the flow is irrotational, in which case the constant is the same throughout.

Bernoulli also applies only where no shaft work or heat transfer occurs between the two points. It cannot be applied across a pump, a turbine or a fan. And it fails badly in regions of flow separation: downstream of a sudden enlargement, behind a partly closed valve, or in the wake of a bluff body. There, large eddies dissipate mechanical energy, and the pressure does not recover as the equation predicts. The energy equation of Lesson 10, with its head-loss term, handles these cases.

A common misconception deserves mention. It is often said that an aircraft wing generates lift because air over the curved top must travel farther in the same time and therefore goes faster. The conclusion that the top flow is faster is true, but the reasoning is false: the air over the top arrives at the trailing edge well before the air beneath it. Lift comes from the wing turning the flow downward, and Bernoulli correctly relates the resulting velocity and pressure differences, but it does not by itself explain why the velocities differ.