§ 4.3  Module 4 — Head Loss: Everything That Is Not Straight Pipe

Entrances, Exits, and the Velocity Head You Forgot

A pipe has two ends, and at both of them the water is doing something it cannot be talked out of.

§4.1 gave you a table of K values and §4.2 gave you two ways to write them down. Both ends of the pipe are still missing, and they are the two entries in that table which are not fittings at all. An entrance is where the water has to decide how to get into a hole; an exit is where it stops being your problem. Both cost head, and one of them — the exit — you pay whatever you bolt on the end. Getting that term into a total dynamic head exactly once is the bookkeeping this lesson is named for.

4.3.1Both ends of a pipe are the same equation

Where a stream meets an abrupt increase in area it cannot follow the wall. It separates, carries on as a jet into slower fluid, and mixes. The mixing is where the energy goes, and the price is set by momentum and energy together rather than by a coefficient anyone had to measure:

  momentum, across the step, on the larger area A₂:
      (p₂ − p₁)/ρg  =  V₂(V₁ − V₂)/g              ← the pressure RISES

  energy, from just upstream to just downstream:
      h  =  (p₁ − p₂)/ρg  +  (V₁² − V₂²)/2g

  together:
      h  =  (V₁ − V₂)² / 2g

The Borda-Carnot loss at a sudden enlargement. V₁ is the velocity in the smaller section, V₂ in the larger. Nothing here is empirical.

Read the two boundary conditions off that one line and you have both ends of the pipe.

An entrance is that equation applied to the flow's own contraction. At a square-edged inlet the streamlines cannot turn the corner, so they separate at the lip and the jet narrows to a vena contracta of area C_c·A before re-expanding to fill the bore. The velocity at the throat is V/C_c, so putting V₁ = V/C_c and V₂ = V into Borda-Carnot gives the coefficient directly:

  K  =  (1/C_c  −  1)²          referenced to the PIPE velocity head, V²/2g

      C_c = 0.586   →   K = 0.499        the tabulated flush square entry
      C_c = 0.611   →   K = 0.405        the classic sharp-orifice value
      C_c = 0.500   →   K = 1.000        an ideal thin re-entrant tube

The entrance coefficient is a contraction coefficient in disguise.

That last row is worth dwelling on. An ideal thin tube projecting into a large vessel contracts to exactly half its own area — a momentum result on the vessel, not a measurement — so its coefficient is exactly 1.0. Real projecting pipe has wall thickness, which guides the flow a little, and the tabulated 0.8 is that fitting with its edge blunted. Everything in the entrance column of a K table is one geometry question: how much does the jet neck down before it fills the pipe?

Be honest about where that third column stops explaining anything. For the square and projecting rows it is the physics: there really is a separated jet, and 0.586 is close to the contraction measured at a sharp orifice. For the bellmouth it is not. A well-formed bell produces almost no separation, and its residual 0.05 is skin friction and a little unsteadiness — so inverting the formula yields an arithmetically valid C_c = 0.82 that describes nothing. The bellmouth's 0.05 is a measured number, and the Hydraulic Institute's ANSI/HI 9.8 Pump Intake Design is where the geometry that earns it is specified.

4.3.2The exit: one velocity head, and the only choice you have

Now take the other boundary condition. A pipe discharging into a tank, a wet well, a receiving manhole or a reservoir is an enlargement into an area so large that the receiving water is at rest. Put V₂ = 0 into Borda-Carnot and there is no coefficient left to choose:

  h  =  (V₁ − 0)² / 2g  =  V₁²/2g          so  K = 1.0,  exactly

  The pipe's kinetic energy is dissipated in the receiving body. Nothing
  downstream of the release point recovers it, because there is no
  downstream — the water has arrived.

The exit loss is not a fitting's property. It is what "into still water" means.

This is where the lesson's title comes from, and it is worth being exact about what is unavoidable. It is not "one velocity head" in the abstract. It is one velocity head at the velocity you release the water. No detail on the end of the pipe can make the coefficient less than 1.0 — but a detail can change the velocity, and V²/2g is quadratic, so that is a real lever. It is also a lever with a trap in it.

Suppose you fit a short stepped enlargement on the outlet of the worked station's 250 mm main. The water now pays twice: (V₁−V₂)²/2g at the step, and V₂²/2g at the mouth. Enlarging the outlet makes the second term smaller and the first term larger, and in the limit of an enormous outlet the step is an exit — the whole velocity head is dumped a metre earlier and you have saved nothing. Differentiate the sum and the optimum falls out: V₂ = V₁/2, which is twice the area and therefore √2 times the diameter, and it costs exactly half a velocity head.

Both ends of that table are one full velocity head and the curve between them turns once. Doubling the diameter instead of the area — 500 mm rather than 353.55 mm — costs 25% more than the optimum, which is what remembering a result without its exponent is worth. A real conical flare beats a step, and the usual convention is to charge a fraction k_d of the Borda-Carnot loss: k_d = 1 for a stepped fitting, falling toward about 0.2 for a long well-proportioned cone. That is a fitting-handbook number, not physics, so name the source. The optimum then moves to D₁·√((1+k_d)/k_d) and the best achievable loss to k_d/(1+k_d) velocity heads: at k_d = 0.2, 612 mm and 0.0127 m — exactly one third of the best a step can do.

4.3.3Counting it once: a bookkeeping convention for TDH

The exit term is the single most commonly mis-counted line in a total dynamic head schedule, and the reason is that there are two correct ways to write it and they do not look alike. Both are just the energy equation with the control volume closed in a different place.

1. Surface to surface. Take the static term between the suction water surface and the discharge water surface. Both surfaces are still water, so neither carries velocity head, and the loss schedule then includes an exit loss of 1.0 × V²/2g at the release velocity. No separate velocity-head term exists. This is the convention for any submerged outlet, and it is the one a hydraulic profile drawing supports directly.
2. Free discharge. Take the static term to the outlet centreline and close the control volume at the pipe mouth, where the pressure is atmospheric. The water leaves carrying V²/2g which is never recovered, so you add it as a residual velocity-head term instead of an exit loss. This is the convention for a pipe that discharges to air — over a weir, into a channel, above the receiving level.

They are the same station. Assembled on the worked station — 24.55 m of static lift between §1.5's water surfaces, 0.2031 m on the suction side, 0.4154 m in the station's own discharge pipework, 6.5041 m of friction in 1 200 m of 250 mm main, and 0.0762 m at the outlet — the surface-to-surface total is 31.749 m and the free-discharge total is 30.349 m. They differ by 1.400 m, which is exactly the submergence of the outlet: 27.30 m of receiving water surface less 25.90 m of pipe centreline. That is a subtraction off the drawing, not a hydraulic disagreement. Move the datum and the number moves with it; use the same datum twice and the two conventions agree to the last digit.

The audit rule is short. Write down which water surface or pipe elevation each end of the static term is measured to. Then check that the outlet velocity head appears exactly once in the whole schedule — under either name. If you are using this course's totalDynamicHead in src/core/hydraulics.js, that means one of these and not both: either put the exit loss in dischargeLossM and leave residualVelocityHeadM at zero, or keep it out of the loss list and pass it as residualVelocityHeadM. The function will happily add both.

4.3.4What it is worth, and where "minor" stops being true

There is a clean way to price one velocity head against pipe. Setting K·V²/2g equal to f·(L/D)·V²/2g gives §4.2's equivalent length, L_e = K·D/f, and for K = 1 that is simply D/f. At municipal duty f sits near 0.018, so:

  one velocity head  =  D/f  ≈  56 pipe diameters of straight pipe

      150 mm:  1/f = 53.4        250 mm:  1/f = 56.2   →   14.05 m of main
      200 mm:  1/f = 55.3        300 mm:  1/f = 56.5

Computed with equivalentLength() and headLossDarcy(), 60 L/s, ε = 0.10 mm.

Fifty-six diameters is a rule worth carrying. It says an exit loss is worth 14 m of 250 mm main, and therefore that whether it is "minor" is entirely a question about the length of the pipeline it is attached to — which is exactly what the module epigraph claims:

Both halves of the folklore are right, about different pipes. On a 3 km transmission main the exit loss is genuinely rounding, and that is the pipeline "minor" was coined on; on the short runs between structures it is a term you can see. The entrance detail is not measured against the force main at all — it lands on the suction side, where the entire loss budget is a fraction of a metre and where Module 6 will show that a fraction of a metre is the currency. That is the asymmetry worth carrying: on the discharge side these are one line among many; on the suction side they are a large share of everything there is.

What this lesson has not accounted for. The friction of the flare itself, a few diameters long and neglected throughout — defensible here, not if you flare over twenty metres. The submergence an outlet needs to stay submerged, and the vortexing a badly submerged inlet does: Module 8. Anything dynamic — a check valve is K = 2.0 here and nothing else, and what it does while closing is Module 8 too. Drawing this as a profile is Module 5. And the pump is still what it has been since Module 1: a black box producing whatever head the schedule demands. Whether one exists that will produce 31.749 m at 60 L/s, and what the 0.1395 m a bellmouth saved is worth to its suction, is Module 6.