§ 2.1  Module 2 — Energy: The One Equation

Bernoulli, Real Fluids, and Where the Energy Goes

Water keeps its books in metres, and it has never once posted a credit.

2.1.1Three ways to hold the same energy

A pump station has one equation. Everything else in this course — every friction factor, every K value, every pump curve and every surge calculation — is a method for putting a number on one of its terms. Learn to write it without hesitating and you can audit any station drawing you are handed, including the ones that are wrong.

Start where Daniel Bernoulli started: an idealised fluid with no viscosity, flowing steadily, with nothing adding energy and nothing taking it away. For such a fluid the quantity p + ½ρv² + ρgz is the same everywhere along a streamline. That form is energy per unit volume, and every term is in pascals. It is correct, and for our purposes the wrong form. Divide the whole statement through by ρg:

    p          v²
  ─────  +  ─────  +  z   =   constant        [metres]
    ρg         2g

  pressure     velocity     elevation
    head         head         head

Bernoulli for an ideal fluid, in the form hydraulics actually uses.

Every term is now a length, and that is not cosmetic. Energy per unit weight has units of joules per newton, which cancel to metres, so the whole energy balance of a pump station can be drawn on a section at the same scale as the concrete — which is exactly what a hydraulic profile is. The quantity has a name, head, and one property worth memorising first: the sum is conserved, not the individual terms. They trade against each other freely, and most confusion in this subject comes from watching one term move and concluding that energy has appeared or vanished.

Take the trade seriously, because it has a consequence you meet on every force main. Shut a pump down and hold the check valve closed: nothing flows, the velocity head is zero, and the sum is fixed at the elevation of the receiving water surface. Walk along the main and the two surviving terms swap continuously — every metre the pipe drops converts a metre of elevation head into a metre of pressure head. At the low point of the reference station used throughout this lesson the pipe sits 15.20 m below the discharge water surface, so the shut-down pressure there is 148.9 kPa (21.6 psi). Read that as the floor the low point sits at whenever the main is full, not as the ceiling: start the pump and the same tapping goes to 182.4 kPa (26.5 psi), higher by exactly the 3.41 m of friction still waiting downstream of it. What sets the pipe's pressure class is the highest sustained running pressure at that point plus an allowance for surge — and the surge allowance dwarfs both numbers, since stopping the column in a comparable 250 mm main is worth a 183 m Joukowsky rise (§1.3).

2.1.2What real water does to the statement

Real water has viscosity and a real pipe has a wall. Fluid shears against fluid, fluid shears against roughness, and at every fitting the flow separates and reattaches. The energy involved does not disappear — it ends as a vanishingly small rise in temperature — but it stops being available to move water. So the sum is no longer constant. It declines, monotonically, in the direction of flow, and only a machine can put any of it back.

Write the bookkeeping between two chosen points, 1 upstream and 2 downstream, with a pump somewhere between them. This is the extended energy equation, and it is the whole of Module 2:

   p₁      v₁²                     p₂      v₂²
  ──── + ───── + z₁  +  h_p  =  ──── + ───── + z₂  +  h_L
   ρg      2g                     ρg      2g

  ←──── total head at 1 ────→          ←──── total head at 2 ────→

  h_p  head added by the pump, m of water   (always on the upstream side)
  h_L  head lost between 1 and 2, m         (always on the downstream side)

The one conservation statement. Every later module sizes one of these terms.

1. Pick the two points before you write anything. They are almost always water surfaces, because there the pressure term is zero by definition and the velocity term is negligible. A wet well and a discharge chamber are the natural pair; picking flanges instead is the commonest way to lose a term.
2. h_p sits on the upstream side because the pump adds. It is the only term that can increase the sum, and only at one place — through the impeller.
3. h_L sits on the downstream side because loss subtracts. It is never negative. If your arithmetic returns a negative loss you have swapped 1 and 2, or you have signed a static lift backwards; you have not found a free lunch.
4. The equation does not care what is in between. Six bends, two check valves, a summit and a flow meter all land in h_L. That is what makes it a conservation statement rather than a model of a pipe — and why you can apply it before you can size anything.

Now put the reference station through it. Point 1 is the wet well surface, at elevation 4.00 m on the site datum: open to atmosphere and effectively at rest, so p₁/ρg and v₁²/2g both go. Point 2 is the pipe outlet, which discharges into the chamber at the receiving surface, elevation 16.00 m — atmospheric there too, so p₂/ρg goes as well, but the water is still moving at that instant, so v₂²/2g stays. Three of the eight terms are zero, and what is left is the design problem of a pump station in one line: h_p = 12.00 m + h_L + v₂²/2g. The 12.00 m is the static lift, off the drawing with a tape, and it is the surface-to-surface figure only because this outlet happens to sit at the receiving surface. The velocity head is arithmetic. Everything else in this course is finding h_L — and every one of those methods adds to the pump head. None subtract.

Taking point 2 at the outlet rather than in the standing water it discharges into is the one deliberate departure from rule 1, and it is worth seeing why the answer does not change. Move point 2 into the chamber and v₂²/2g vanishes with the other three — four terms zero, h_p = 12.00 m + h_L — but the jet still has to give its kinetic energy up to the standing water, and that appears inside h_L as an exit loss of one whole velocity head (Module 4). Same pump head, different ledger. What you may not do is count four zero terms and keep a velocity term outside h_L: that declares the same 0.076 m dead at the surface and alive in the equation on the same line, which is exactly the bookkeeping rule 1 exists to prevent.

2.1.3Why the currency is head and not pressure

A pressure gauge on a pump's discharge reads in kilopascals or psi; the manufacturer's curve for that same pump is drawn in metres or feet. The mismatch is not a convention chosen for convenience. There are three reasons head is the right unit, and the first is a property of the machine.

A centrifugal pump adds head, not pressure. Euler's pump equation gives the ideal head a rotating impeller imparts as H = u₂·cu2/g, where u₂ is the blade tip speed and c_u2 the tangential component of the fluid velocity leaving it. There is no density in that expression. A 300 mm impeller at 1750 rpm has a tip speed of 27.5 m/s, so its ideal ceiling — radial blades, no pre-rotation — is u₂²/g = 77 m, whatever it is spinning. Put the same pump on water and on a brine of specific gravity 1.20 at the same speed and flow and it produces the same metres; the discharge pressure and the shaft power both rise by 20%, because those are ρgH and ρgQH/η. Head is the quantity the machine actually controls, which is why the curve is drawn in it.

Everything else on the drawing is already a length. Invert elevations, water levels, freeboard, submergence, cover: all metres. Converting the one quantity that is not — pressure — puts the whole design on one scale, which is what makes a profile drawable at all (§2.2). And a gauge reads at its own elevation. A suction gauge and a discharge gauge on the same pump are rarely at the same height; convert each reading to head, add the elevation of its own tapping, and the two numbers become comparable. That is the calculation you will implement at the end of this lesson, and it is how a pump is field-tested.

  1 m of water      =  9.798 kPa   =  1.4211 psi
  1 psi             =  0.7037 m   =  2.309 ft        (handbook: 2.31 ft)
  1 bar (100 kPa)   =  10.21 m    =  33.5 ft
  1 atm (101.325)   =  10.34 m    =  33.9 ft

  1 m of head on 1 m³ of water  =  9.80 kJ  =  2.72 Wh
  the same, at 65% wire to water                =  4.19 Wh

Computed in src/core/hydraulics.js at 15 °C, where ρ = 999.13 kg/m³. The 2.31 ft/psi is the figure every water-works handbook carries, quoted at 60 °F.

One asymmetry in that table matters more than the conversions. Head has no upper bound worth worrying about, but pressure head has a hard floor: it cannot fall below the liquid's absolute vapour pressure, which at 15 °C is 1.71 kPa, or 0.17 m of water absolute. Since atmospheric pressure is only 10.34 m of water, no pump can lift water by suction more than about 10 m even in theory. That floor is the whole reason NPSH exists; it is Module 6's subject, and this lesson only asks you to notice it is there.

2.1.4Which term dominates, and which one never can

Put numbers on the reference station and the hierarchy is not close. At 60 L/s through a 250 mm main the mean velocity is 1.222 m/s and the velocity head is 0.076 m. The friction over 900 m of cement-lined ductile iron comes to 4.88 m — a hydraulic gradient of 0.542 m per 100 m, a number to treat here as measured, because producing it is §3.2's job. The static lift is 12.00 m, from the drawing. The pump must supply 16.95 m, and it divides 70.8% static, 28.8% friction, 0.45% velocity head.

The velocity head is not small by accident, and it is worth knowing why it cannot be anything else. A force main's velocity is boxed in from both sides. Below about 0.6 m/s (2 ft/s) — the minimum the Ten State Standards require for a force main — solids settle out, the main silts and then goes septic. Above roughly 2.4 to 3.0 m/s (8 to 10 ft/s), which is common practice rather than a standard, surge and wear on fittings stop being tolerable. So v²/2g is confined to a few centimetres up to under half a metre, permanently, in a system whose lift is tens of metres:

So the answer to "which term dominates" is never the velocity head. It is static lift or friction, and which one depends on the length of the main relative to the lift: static lift is fixed by the site and indifferent to flow, while friction is bought by the metre and grows as the square of flow. At the reference station's gradient, friction reaches the 12.00 m lift at about 2214 m of main — so a 900 m station is static-dominated and an otherwise identical station with 3 km of the same pipe is friction-dominated. That single ratio decides more downstream design than any other number in this course: it sets how much a variable-speed drive can ever save you (§6.6), how much a second pump in parallel actually buys (§7.3), and whether the money belongs in the pump or in the pipe (§7.6).

Then convert the winner into what it costs. Lifting 60 L/s through 16.95 m needs 9.97 kW delivered to the water; at a pump efficiency of 72% and a motor efficiency of 94% — both assumptions, both stated — the station spends 0.068 kWh for every cubic metre it passes. Of that, roughly 71% is buying elevation, which no design decision can avoid, and 29% is buying friction, which almost every design decision touches. That is the economic argument of this subject, and it rests entirely on the equation in §2.1.2.

2.1.5What this lesson has not accounted for

Four honest omissions, each of them somebody else's lesson. h_L has been a single symbol with a plausible magnitude: friction along the pipe wall is Module 3, and the fittings, valves and meters inside the station — which on a short main routinely exceed the pipe's own friction — are Module 4. The static lift has been one number, when the wet well level draws down as the pump runs and the receiving surface can rise, which turns one head into a range (§2.3). The two lines the sum and the sum-less-velocity-head trace along the main are not named until §2.2. And nothing here says how much flow arrives, or when.

What will not change is the equation. Every correction in the next six modules enters as a bigger h_L or a moved z, read off the same line of bookkeeping you have just written. That is why the module epigraph claims everything after it is bookkeeping: not because the rest is easy, but because it is arithmetic inside a structure that is now settled.