§ 2.5 Module 2 — Energy: The One Equation
Four terms, and only one of them is a property of the ground. The other three are a property of how hard you are pushing.
By the end of this lesson
A pump does not know where it is. It knows one thing: how much head it is being asked to produce at the flow it happens to be passing. Everything the station is — its lift, its pipe, its valves, the level in its wet well this minute — reaches the impeller as one number, and that number is the total dynamic head. Assembling it is the whole of this module's bookkeeping, and getting it wrong is how a station ends up unable to deliver what it was built for.
The extended energy equation of §2.1 already contains the answer. Write it between the two water surfaces the station works between and almost everything cancels: both are at atmospheric pressure, so the pressure terms go, and both are near enough still, so the velocity terms go. Cancelling those two is not the same as saying no kinetic energy is spent: the water is moving in between, and whatever brings it to rest again has to be paid for. What is left is the elevation difference, the losses in between — including that last one — and the pump term that pays for all of it:
TDH = static lift + friction + minor losses + velocity head
static lift = discharge water surface − suction water surface
friction = loss along the pipe walls (Module 3)
minor losses = loss at every fitting and valve (Module 4)
velocity head = v²/2g, spent once, at the far end
The bookkeeping form. Every term is in metres of water (or feet). One velocity head is always spent at the far end; §2.5.4 is which of the last two columns it belongs in.
Two habits of language prevent real errors. First, TDH is dynamic: it is the head required at a stated flow. "The TDH of this station is 33 m" is not a complete sentence; "33 m at 60 L/s with the well drawn down" is. Second, static lift is measured between water surfaces, never between flanges or pump centrelines (§2.3). Measure to the discharge flange and you have deleted the depth of water standing above it, and specified a pump that cannot fill the receiving structure.
Sort the four terms by what they respond to, and the shape of the whole subject appears.
| Term | Set by | With flow | Moves when | Failure if you get it wrong |
|---|---|---|---|---|
| Static lift | Two water surfaces | Unchanged | The wet well draws down; the receiving level rises | Measured to a flange: pump too small, and it never reaches the far end |
| Friction | Pipe length, bore, roughness | Rises as about Q1.9 | The pipe ages; you re-rate the station upward | Under-estimated: the station makes less flow than the drawing promised |
| Minor losses | ΣK of every fitting | Rises as Q² | A valve is throttled; a check valve wears; a strainer blinds | Omitted inside the station, where the fittings are dense and the pipe is short |
| Velocity head | Flow and bore alone | Rises as Q² | Never, at fixed flow and bore | Counted twice, or dropped on a low-head station where it is a real fraction |
The static term does not care how hard you push. It is elevation, and elevation is indifferent to flow: a station lifting 25.95 m (85.14 ft) lifts 25.95 m at 10 L/s and at 100 L/s. It does move — with level, not with flow — which is why it produces a band rather than an error, and why the wet well's working depth ends up in the pump specification.
The other three grow with flow, and grow fast. Minor losses and velocity head go exactly as Q², because both are multiples of v²/2g and velocity is proportional to flow in a fixed bore. Friction is nearly so: over this station's working band the exponent is about 1.9, not 2, because the friction factor drifts slowly with flow. Module 3 owns that drift. What matters here is structural — the loss terms scale together and the static term does not scale at all — and that asymmetry generates the system curve, the reason parallel pumps disappoint, and the reason a variable-speed drive helps a friction-dominated station and barely touches a static one.
Why head became bookkeeping
Bookkeeping arrives in engineering when the thing being counted gets too big to hold in the head. In 1936 the Engineering Experiment Station of the University of Illinois published Bulletin No. 286, Analysis of Flow in Networks of Conduits or Conductors, by Hardy Cross (1885–1959) — a professor of structural engineering who had already given structural analysis its moment-distribution method. His method is iterative: assume flows, add up the head lost around each closed loop, and correct until every loop closes. It works because of the statement this module is built on — around any loop the head changes must sum to zero — and it is usable because head is written in one column, where an omission is visible.
The other half is commercial: a buyer who cannot measure head cannot reject a pump. The Hydraulic Institute, formed in 1917, exists partly for that reason — its acceptance-test standards, today the ANSI/HI 14.6 family, fix where the gauges go and what the total head derived from them means, so that "this pump makes 33 m at 60 L/s" is a claim a witnessed test can falsify. TDH is physics. It is also a term in a contract.
Textbooks drop the velocity-head term from TDH with a wave of the hand, and
for a long force main that is fair: at 1.2 m/s, V²/2g is 0.073 m
against maybe 30 m of TDH. But the term scales with the square of velocity while
friction over a short main scales with length, so on a short, fast
discharge the ratio inverts. At 3 m/s the velocity head is 0.46 m; on a station
lifting 3 m into an adjacent gravity manhole, that single dropped term is 15% of
the total.
Compute V²/2g once and compare it to your loss total before
deciding whether to drop it. It costs one line. Dropping it silently on a low-lift,
high-velocity station is one of the more common ways a pump ends up running off the
right-hand end of its curve.
Keep the alignment and boundary elevations from the station §1.5 read off a drawing, but make the training well 0.60 m deeper so its floor is at elevation 0.00 m: arriving invert at 3.20 m, high water alarm at 2.75 m, discharge water surface at 27.30 m, and 1 200 m (3 937 ft) of 250 mm (9.84 in) cement-lined ductile iron force main between. Two levels this lesson adds, because TDH cannot be assembled without them: pumps stopping at 1.35 m and starting at 2.60 m. Where those two come from is not hydraulics — submergence over the suction bell sets the stop level and permitted motor starts set the volume between them, §8.1 and §8.3. The duty: 60 L/s (951 gpm, 1.369 MGD).
Velocity first, because three of the four terms hang off it. A 250 mm bore has an area of 0.0491 m², so 60 L/s is 1.222 m/s (4.010 ft/s) and the velocity head is v²/2g = 0.076 m (0.25 ft). Note what that needed: no coefficient, no roughness, no chart. Flow and bore determine it completely — the only loss-side term you can never be wrong about.
Static lift next, and here you must choose a condition. With the well drawn down to the stop level the lift is 27.30 − 1.35 = 25.95 m (85.14 ft). That is the worst case, and worst case is what a pump is sized against.
Friction is a black box in this module, so take it the way an engineer in an existing station takes it — from measurement. The drawing's loss schedule states 6.50 m (21.33 ft) for this main at 60 L/s, and a pump test recovers the same figure — two gauges, one at each end of the main, and a flow meter give it directly (§2.2). A single gauge on the discharge flange does not: it gives you TDH at best, out of which friction comes only after the static lift and the fittings are subtracted back off. One measured point carries the term to other flows, because the shape is known: hf = R·Q², with R = 6.50/0.060² = 1 806 s²/m⁵. That shortcut sits within about 4% of a full Darcy-Weisbach calculation across the 40–80 L/s band, and is worth nothing at 10 L/s. Module 3 turns the measurement into a prediction you can make before the pipe exists.
Minor losses are a count of velocity heads. Every fitting costs K velocity heads, so the whole term is ΣK · v²/2g and the work is arithmetic on a schedule:
| Fitting | K | No. | ΣK |
|---|---|---|---|
| Suction bell on the pump | 0.05 | 1 | 0.05 |
| 90° elbows — pump discharge and valve vault | 0.6 | 2 | 1.20 |
| Swing check valve | 2.0 | 1 | 2.00 |
| Plug valve, open | 1.0 | 1 | 1.00 |
| Tee into the common header, through the branch | 1.3 | 1 | 1.30 |
| Magnetic flow meter | 0.2 | 1 | 0.20 |
| 45° bends along the force main | 0.4 | 4 | 1.60 |
| ΣK, exit excluded | 7.35 | ||
So the fittings cost 7.35 × 0.076 m = 0.560 m (1.84 ft). Those K values are this course's table, in src/core/hydraulics.js; a real design uses the manufacturer's figure for the specific valve, and Module 4 is where they are chosen rather than quoted. Notice that 0.560 m is only 8.6% of the friction term, and that the ratio is an accident of having 1 200 m of pipe: shorten the main to 406 m — 2.2 m of friction, because friction is linear in length — and the fittings are 25% of it, while inside the station's own pipework they are all of it. "Minor" describes a long force main, not a pump station.
Add the column: 25.95 + 6.50 + 0.560 + 0.076 = 33.09 m (108.55 ft). Static lift is 78.4% of it, friction 19.6%, fittings 1.7%, velocity head 0.23%. That last figure is the honest reason the fourth term gets so little respect; §2.5.4 is why it has to be there anyway.
Interactive 3D instrument
The ledger — four terms, one number, and an energy line that has to land
A 3D instrument you drive yourself, one variable at a time. It needs JavaScript and WebGL, so it is not shown in this static copy of the page.
The fourth term is the one that produces double counting, and the cure is to say where the system ends before you add anything up.
If the main ends under water in the receiving structure, the jet dissipates into still water and the whole velocity head is destroyed there: book it as an exit loss with K = 1.0, take the receiving water surface as the endpoint, and there is no residual, because the water has stopped. If the main ends in free discharge, spilling into a manhole above the water, it leaves at 1.222 m/s and takes that kinetic energy with it. Nothing recovers it, so the pump must supply it: book a residual velocity head, with the outlet as the endpoint.
Those are two descriptions of the same duty, and they give the same answer — with the outlet at the receiving water surface, both come to 33.09 m. One velocity head is spent either way; the only question is which column it is written in. Write it in both, and TDH is over-stated by exactly v²/2g.
The double count, and how to see it
Here the error is 0.076 m in 33 m: invisible, and never caught by looking at the total. Catch it structurally. Draw the energy line — it leaves the pump at TDH above the wet well surface, falls by each loss in turn, and must land exactly on the receiving water surface. Count the velocity head twice and the line finishes above the water it is supposed to be filling: not a small error but an impossible one. The instrument above draws that line; set the far end to "both (wrong)" and it stops closing. The gap it opens is one velocity head — 0.076 m against a 26 m lift, so the instrument draws that last gap at twenty times its true size and labels the exaggeration. You look for the miss; you do not look at the total. The same check catches the other classic double count, a fitting entered once as a K value and again as an equivalent length, which §4.2 takes apart.Then name the case where the term is not decoration. Velocity head goes as v², so it is negligible in a high-head force main at 1.2 m/s and it is not negligible where velocity is high or lift is low. A drainage station lifting 3.0 m into an adjacent manhole at 3.0 m/s (9.84 ft/s) spends 0.459 m (1.51 ft) — 15% of its static lift. Round that away and you have mis-specified the pump by 0.459 m: four fifths of the entire fittings budget of §2.5.3, on a station lifting a ninth as far. The rule is not "the velocity head is small"; it is "compute it, because it is the cheapest term you will ever compute".
Now push the same station harder. At 90 L/s (1 427 gpm) the friction term is not 1.5 × 6.50 = 9.75 m but 6.50 × (90/60)² = 14.63 m (47.98 ft), the fittings reach 1.26 m and the velocity head 0.171 m — while the static lift sits exactly where it was. TDH goes from 29.96 m at 45 L/s to 42.01 m at 90 L/s: double the flow for a 40% rise in head. Since the loss terms scale as Q² and the static term does not scale at all, TDH(2Q) = 4·TDH(Q) − 3·static.
So TDH is not a number. It is a curve — head demanded against flow — and a station has as many TDHs as it has operating conditions. Two of those belong on every specification. The first is flow, because that is what the curve is drawn against. The second is the wet well level: between start at 2.60 m and stop at 1.35 m the static term travels 1.25 m (4.10 ft), so at 60 L/s this station demands anywhere from 31.84 m to 33.09 m depending on nothing but what minute it is. The band is exactly the working depth, because nothing else follows the level. That is why a pump is selected against a range, and §7.2 is where the range meets the pump's own curve and settles into a flow.
Which term dominates is the first question to ask about any station, and this one answers it differently depending on where you stand. At 60 L/s it is 78% static, so a variable-speed drive can trim very little and a bigger main buys almost nothing. At 100 L/s it is 57% static and 39% friction, and the bigger main starts to pay. Shorten the main to 406 m and it is 90% static; let it tuberculate until it measures 11.9 m at 60 L/s and it is 67% static. That ratio is worth more to you than any number in this lesson.
Four things this lesson has deliberately borrowed and not paid for:
Take the form from Module 2, and the habit of writing it as a column. Static lift, friction, minor losses, one velocity head; the condition beside it; the flow above it. Every module after this one argues about the size of one of those entries. None of them adds a fifth.
Check your understanding
Check your understanding
3 auto-graded questions with an explanation for every wrong answer. Requires JavaScript. (m2-l5-q1)
Lab 2.5
Assemble a TDH, and make it a function of flow
Build the ledger as code. Two functions; between them they are the calculation you will run on every station for the rest of your career, and everything Modules 3 to 7 add is a better value for one of the inputs. velocityHead(qM3s, dM) — return the velocity head in metres for a flow in m³/s through a circular pipe of inside diameter dM metres. Use G (provided, 9.80665 m/s²). Return 0 at zero flow. tdh(duty) — assemble the four terms and return { staticLiftM, frictionM, minorM, velocityHeadM, tdhM } , all in metres, from duty = { dischargeWaterLevelM, wellLevelM, qM3s, dM, hfRatedM, qRatedM3s, sumK, endpoint } : staticLiftM is the discharge water level minus the wet well water level. frictionM scales one measured point as the square of flow: hfRatedM · (qM3s/qRatedM3s)² . minorM is sumK velocity heads — plus one more (the exit loss, K = 1.0) when endpoint === 'submerged' . velocityHeadM is the residual : one velocity head when endpoint === 'free' , and zero when the outlet is submerged, because the exit loss has already taken it. Default endpoint to 'free' . tdhM is the sum of the four. Get the last two right and the two bookkeepings agree; get them wrong and you have the commonest double count in the subject. Graded in the browser against 8 assertions; the editor and harness require JavaScript.
Every number in this lesson belongs to one station — the one §1.5 read off a drawing — and each is checked in src/scenes/m2/assemble-tdh.js against a closed form, hand arithmetic, the same term worked from US-customary inputs, a second leg-by-leg assembly, or Colebrook-White standing in for the friction term this module has not yet earned.