Topic 08 · 12 min · 5 figures

The energy equation for fluid flow

Bernoulli's equation was built on three assumptions this course is now ready to give up, one at a time. Put friction back, and a pump and a turbine into the pipe run, and the tidy conserved total becomes a running account — spent here, topped up there — which is a far more honest description of every pumped system anyone actually builds.

Nobody designs a real pipe system to the equation the previous topic derived. That equation was deliberately built without friction, and without anywhere for a pump or a turbine to sit, because the point of deriving it that way was to see the underlying trade of pressure, elevation and velocity with nothing else in the way. Almost every system worth building has a pump in it, and almost every pipe run has friction in it, so the working equation of this topic restores both — and turns a statement about what stays constant into a statement about what is spent, added, and taken back.

From Bernoulli to the energy equation

Begin from the head form of Bernoulli's equation between two stations on a streamline, and add back, one at a time, exactly the effects that were assumed away to get it:

\[\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2\]

A real pipe has friction, and friction only ever removes mechanical energy from the flow, never adds it. That belongs on the downstream side of the balance, as a positive quantity:

\[\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_L\]

A pump, wherever it sits between the two stations, does the opposite — it adds mechanical energy to the fluid, at the cost of whatever drives its shaft, so its head goes on the upstream side. A turbine takes mechanical energy back out, converting it to shaft work, so its head joins the loss on the downstream side. Put all three together and the full mechanical energy equation for a control volume between two stations is

\[\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 + h_{\text{pump}} = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_{\text{turbine}} + h_L\]

Every term still carries units of length, still reads as a head, and the equation is still nothing more than an energy balance written per unit weight of fluid passing through. What has changed is the accounting: instead of a total that has to stay fixed, there is now a total that is free to rise wherever work is added and fall wherever work is removed or friction spends it — which is a far closer description of what a pump station, a pipeline and a turbine hall actually do to the fluid passing through them.

It is worth being honest about one simplification carried over silently: \(V\) at each station is, as in conservation of mass, the average velocity, not the true profile, which strictly needs a kinetic-energy correction factor close to but not exactly one. The correction matters for precise work and is routinely dropped for turbulent flow, where the profile is flat enough that the error is smaller than most of the other uncertainties in the calculation.

Three honest terms — step 1 of 4

Start where Bernoulli left off: the same total, station to station.

\(\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2\)
\(\cdots = \cdots + h_L\)
\(\cdots + h_{\text{pump}} = \cdots\)
\(\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h_{\text{pump}} = \frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_{\text{turbine}}+h_L\)
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Head loss: major and minor

\(h_L\) is not one effect but the sum of two physically distinct ones, and telling them apart matters for how each is calculated, even though this article stops short of the calculation itself. Major losses are distributed — friction against the pipe wall, acting continuously along every metre the fluid travels, growing in rough proportion to the length of the run:

\[h_{L,\text{major}} = f\,\frac{L}{D}\,\frac{V^2}{2g}\]

with the friction factor \(f\) carrying the dependence on roughness and on the flow regime — laminar or turbulent — that governs how strongly the fluid drags against the wall. Minor losses, despite the name, are not always small; they are simply local, spent almost entirely at one fitting — a valve, an elbow, a sudden contraction, an entrance or exit — where the flow is forced to separate, swirl or accelerate sharply over a short distance:

\[h_{L,\text{minor}} = \sum K_L \, \frac{V^2}{2g}\]

with a loss coefficient \(K_L\) specific to each fitting, found experimentally and tabulated rather than derived. A pipe run with several valves in quick succession can easily lose more head to its fittings than to the straight pipe between them, which is exactly why the distinction earns a name rather than being folded silently into one number.

The two add directly, because both are honest losses of the same mechanical energy, spent by the same irreversibility, just distributed differently along the route:

\[h_L = h_{L,\text{major}} + h_{L,\text{minor}} = f\frac{L}{D}\frac{V^2}{2g} + \sum K_L \frac{V^2}{2g}\]

Working out \(f\) and each \(K_L\) in practice — the Moody chart, the effect of relative roughness, the coefficients for standard fittings — is deliberately kept for flow in pipes and head losses, where the detail belongs. What this section is for is recognising the two kinds of loss on sight, and knowing which formula each one calls for once the numbers are needed.

Continuous or all at once — step 1 of 4

A real pipe run has both kinds of loss, everywhere along it.

\(h_L = h_{L,\text{major}} + h_{L,\text{minor}}\)
\(h_{L,\text{major}} = f\,\frac{L}{D}\,\frac{V^2}{2g}\)
\(h_{L,\text{minor}} = \sum K_L\,\frac{V^2}{2g}\)
\(h_L = f\frac{L}{D}\frac{V^2}{2g} + \sum K_L \frac{V^2}{2g}\)
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Pump head and turbine head

\(h_{\text{pump}}\) and \(h_{\text{turbine}}\) are heads in exactly the same sense as the pressure, elevation and velocity heads already in the equation — energy per unit weight of fluid — which is what lets them sit in the same balance without any conversion. Converting a head into a power just needs the weight flow rate, \(\rho g Q\), multiplied through:

\[\dot W_{\text{pump},u} = \rho g Q\, h_{\text{pump}}\]

This is the useful pump power — the power actually delivered to the fluid, which is smaller than the power drawn at the shaft by whatever the pump's own mechanical and hydraulic losses take, a gap accounted for by its efficiency, \(\dot W_{\text{shaft}} = \dot W_{\text{pump},u}/\eta_{\text{pump}}\). A turbine runs the same relation the other way, extracting energy from the flow and delivering less of it as usable shaft or electrical output than it took from the fluid:

\[\dot W_{\text{turbine},e} = \rho g Q\, h_{\text{turbine}}\, \eta_{\text{turbine}}\]

Plotted against distance along the pipe, the effect of each machine is a discrete jump in the total head line, right at the station where it sits — up at a pump, down at a turbine — sitting on top of the gentle, continuous decline that friction alone would produce. Nothing about the surrounding pipe changes what the jump looks like; the machine is a local source or sink of head, exactly the way a heat exchanger is a local source or sink of enthalpy in a thermodynamic cycle, and the two ideas are drawn the same way for exactly that reason.

It is the same physical device met already in vapour power cycles, only tracked here in head rather than in specific enthalpy — a reminder that a pump does not care which course is describing it, only that energy is being added to a fluid crossing its boundary.

Machinery on the grade line — step 1 of 4

Left to friction alone, the total head line only ever falls.

\(h_L \ge 0\)
\(\dot W_{\text{pump},u} = \rho g Q\, h_{\text{pump}}\)
\(\dot W_{\text{turbine},e} = \rho g Q\, h_{\text{turbine}}\, \eta_{\text{turbine}}\)
\(h_{\text{pump}}, -\,h_{\text{turbine}}\)
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The energy and hydraulic grade lines

Two lines are conventionally drawn over a pipe's elevation profile, and confusing them is a reliable way to misread a system diagram. The energy grade line, EGL, is the total head at every station — exactly the running total this article has been building:

\[\text{EGL} = z + \frac{P}{\rho g} + \frac{V^2}{2g}\]

Left alone, it only ever falls, because friction is the only thing acting on it in an ordinary run of pipe; it climbs abruptly only where a pump adds head and drops abruptly only where a turbine removes it. The hydraulic grade line, HGL, is the same quantity with the velocity head stripped out:

\[\text{HGL} = \text{EGL} - \frac{V^2}{2g} = z + \frac{P}{\rho g}\]

which is precisely what a piezometer tapped into the pipe at that point would show, since it has no way to register the flow's kinetic energy, only its static pressure and elevation. The gap between the two lines, at any station, is nothing but the local velocity head, which is why the gap widens through a constriction and narrows again once the pipe returns to its original bore — the EGL barely notices a short, well-rounded constriction, since little energy is actually lost there, but the HGL dips sharply through it and recovers just as sharply after, purely because velocity, and therefore the size of the term being subtracted, rose and fell.

The practical use of the HGL is a warning one, and it is the reason the two lines are drawn together rather than separately: wherever the HGL falls below the physical centreline of the pipe, the absolute pressure there has fallen below atmospheric, and if it falls far enough it approaches the fluid's vapour pressure and cavitation becomes a real risk. Reading that directly off a diagram — rather than discovering it from a pump that has started to chatter — is the entire reason the two lines are worth plotting in the first place.

Two lines, one gap — step 1 of 4

Draw the same pipe as area against distance, area changes and all.

\(A = A(x)\)
\(\text{EGL} = z + \frac{P}{\rho g} + \frac{V^2}{2g}\)
\(\text{HGL} = \text{EGL} - \frac{V^2}{2g}\)
\(\text{EGL} - \text{HGL} = \frac{V^2}{2g}\)
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A worked pipe system

Water is pumped from a lower reservoir to an upper one, 20 m higher, through a pipe whose friction and fittings together cost \(h_L = 3\) m at the operating flow rate. Both reservoirs are open to the atmosphere and large enough that their surfaces barely move. Write the energy equation between the two free surfaces, station 1 on the lower one and station 2 on the upper:

\[\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 + h_{\text{pump}} = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_L\]

Both pressures are zero gauge, since both surfaces sit open to the same atmosphere, and both velocities are taken as zero, since a reservoir large enough not to move noticeably does not carry any appreciable kinetic energy at its free surface. Every term except the elevations, the pump head and the loss drops out, leaving

\[z_1 + h_{\text{pump}} = z_2 + h_L \;\;\Longrightarrow\;\; h_{\text{pump}} = (z_2 - z_1) + h_L\]

Take the lower surface as the datum, \(z_1 = 0\), so \(z_2 = 20\) m, and

\[h_{\text{pump}} = 20 + 3 = 23\ \text{m}\]

which is a clean, physical statement before a single watt has been mentioned: the pump has to lift every kilogram of water 20 m and additionally supply the equivalent of 3 m of lift to cover what friction spends along the way. Give the system a flow rate, \(Q = 0.05\) m³/s, and the head converts directly to the power the pump must deliver to the fluid:

\[\dot W_{\text{pump},u} = \rho g Q\, h_{\text{pump}} = (1000)(9.81)(0.05)(23) \approx 11{,}280\ \text{W} \approx 11.3\ \text{kW}\]

That figure is the useful power delivered to the water alone. The motor driving the pump has to supply more than that, by whatever the pump's own efficiency costs — the last piece of the story, and exactly where a pump's performance curve, not this energy balance, takes over.

One pump, two reservoirs — step 1 of 4

Two open reservoirs, both large enough that their surfaces barely move.

\(P_1 = P_2 = 0 \text{ (gauge)}, \quad V_1 \approx V_2 \approx 0\)
\(h_{\text{pump}} = (z_2 - z_1) + h_L = 20 + 3 = 23\ \text{m}\)
\(Q = 0.05\ \text{m}^3/\text{s}\)
\(\dot W_{\text{pump},u} = \rho g Q h_{\text{pump}} \approx 11.3\ \text{kW}\)
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