Fluid Mechanics: Statics, Energy, Pipe Flow and Similitude

Water Resources opens with the mechanics that the other three sub-items spend, and the Civil emphasis is different from the Mechanical one: the questions are about gates, rising mains, networks and meters, not about compressors and boundary layers on aerofoils. Start with statics, where the whole of it is two results. The force on a submerged plane surface is F = ρg h̄ A, using the depth of the centroid — and the force acts not at the centroid but lower, at the centre of pressure h_cp = h̄ + I_G/(A h̄), because pressure grows with depth so the lower half of a gate carries more than the upper. Getting that offset wrong is the difference between a gate that opens and a hinge that fails. Then the energy equation, which for an incompressible steady flow says p/ρg + V²/2g + z is constant along a streamline in the absence of losses — three heads that trade against one another, so a contraction that raises the velocity head must lower the pressure head by the same amount, and that is the entire explanation of a venturi, an aspirator and cavitation at a pump inlet. For real pipes the losses go back in through Darcy-Weisbach, h_f = fLV²/2gD, whose V² is worth respecting: doubling the flow quadruples the friction loss and takes eight times the pumping power. A pipe network then follows from two rules only — continuity at each junction and a single-valued head at each point — which is all Hardy Cross ever does. And dimensional analysis closes the chapter by making a model of a spillway meaningful: choose the right dimensionless group, and a 1:25 model tells you the prototype discharge to within the accuracy of the measurement.

Fluid Statics: Where the Force Acts, Not Just How Big It Is

Pressure in a static fluid grows linearly with depth, p = ρgh, and acts equally in all directions at a point. Integrate it over a submerged plane surface and the total force is F = ρg h̄ A, in which h̄ is the depth of the centroid of the area — a compact result that hides the fact that the pressure is not uniform. That non-uniformity reappears when you ask where the force acts, and for anything with a hinge, a bearing or an overturning check, where is as important as how much.

🎯 A sluice gate, and the third of the way down that is not the middle
Take a vertical rectangular sluice gate 2 m wide, its top edge 1 m below the water surface and its bottom 3 m down. The area is 4 m² and the centroid sits at h̄ = 2 m, so the total force is F = 9810 × 2 × 4 = 78.48 kN. Now the position. The centre of pressure is h_cp = h̄ + I_G/(A h̄), where I_G = bd³/12 = 2(2)³/12 = 1.3333 m⁴, giving h_cp = 2 + 1.3333/8 = 2.1667 m below the surface. Measure that from the gate itself: the gate runs from 1 m to 3 m, so the resultant acts 1.1667 m down a 2 m gate — 0.583 of the way down, not halfway. The offset is small in absolute terms and decisive in its effect, because it is the lever arm about the hinge: put the resultant at mid-height and the moment on the hinge is understated by around 17 per cent. Two properties of the offset are worth carrying. It shrinks with depth: the term I_G/(A h̄) has h̄ in the denominator, so the same gate lowered to h̄ = 10 m has its centre of pressure only 0.033 m below the centroid — deep gates behave almost as though the pressure were uniform, because the variation over the gate is small compared with the mean. And it is always below the centroid, never above and never at it, because pressure increases downward without exception. The limiting case makes the pattern clear: a gate whose top edge is exactly at the water surface has h̄ = d/2 and h_cp = 2d/3, the familiar third-point rule, and that is the one case where the answer is a fixed fraction of the gate height rather than a function of the depth.

Continuity and Energy: Three Heads That Trade

Continuity for an incompressible fluid is A₁V₁ = A₂V₂, so velocity is inversely proportional to area — halve the diameter and the area falls to a quarter, so the velocity rises four times. Energy, in the absence of losses, is Bernoulli’s p/ρg + V²/2g + z = constant, three terms all measured in metres and therefore directly comparable. Put the two together across a contraction and the conclusion is forced: if the velocity head rises, the pressure head must fall by the same amount. Take a 300 mm main reducing to 150 mm at Q = 0.05 m³/s. The velocities are 0.707 and 2.829 m/s, the velocity heads 0.0255 m and 0.408 m, so the pressure head falls by 0.383 m across the contraction. Nothing has been lost; energy has changed form. Run the same reasoning far enough and you get cavitation — keep raising the velocity and the pressure head falls until the absolute pressure reaches the vapour pressure, at which point the water boils cold, and the collapsing bubbles pit the metal downstream. That is the single most common cause of damage at a pump inlet, and it is Bernoulli read to its limit.

Pipe Flow, Losses, and Networks

Real flow in pipes loses head to friction, and Darcy-Weisbach is the expression to work from: h_f = fLV²/(2gD). Take a 300 mm main, 1000 m long, carrying 1 m/s with f = 0.02: h_f = 3.398 m, the discharge is 0.0707 m³/s, and the power lost is ρgQh_f = 2.36 kW. The V² is the term to respect. Double the velocity and the head loss quadruples to 13.59 m — and since power is ρgQh_f and Q has doubled too, the pumping power goes up eight-fold. That cubic relationship between flow and pumping power is why a slightly larger pipe is almost always cheaper over a design life than a slightly bigger pump, and why velocities in water mains are held to around 1 to 2 m/s rather than pushed higher.

🧠 Two pipes in parallel are one pipe fifteen per cent wider
Pipes in series carry the same discharge and their head losses add. Pipes in parallel share the same head loss between the same two junctions and their discharges add — which is the more interesting case, because it produces a result most people guess wrongly. Rearranging Darcy-Weisbach in terms of discharge gives h_f ∝ Q²/D⁵, so for a fixed head loss Q ∝ D^2.5, and replacing several parallel pipes by one equivalent pipe of the same length requires D_e⁵ = Σ D_i⁵. Two 300 mm pipes therefore behave as a single pipe of D_e = 2^(1/5) × 300 = 344.6 mm. Twice the capacity for a diameter only 14.9 per cent larger. Three pipes give 373.7 mm and four 395.9 mm, so the returns diminish but the point stands: capacity is extraordinarily sensitive to diameter, going as D^2.5, and comparatively insensitive to everything else. The practical consequences run through water supply design. Doubling a main by laying a second pipe alongside is usually poor value against simply laying one larger pipe, once trenching is counted. A small reduction in bore from scaling or tuberculation costs far more capacity than it appears to — losing 10 per cent of the diameter loses 23 per cent of the flow. And a network needs nothing more than these ideas plus two rules: continuity at every junction, and a single-valued head at every point, so the algebraic sum of head losses round any closed loop is zero. Hardy Cross is just an iterative way of satisfying the second when the first has already been guessed.

One more question always arises with f: is the flow laminar or turbulent? The Reynolds number Re = VD/ν answers it, with the transition in a pipe at about Re = 2000. Run the numbers for a water main and the answer is not close. At V = 1 m/s in a 300 mm pipe with ν = 10⁻⁶ m²/s, Re = 300 000. To bring that below 2000 the velocity would have to fall to 0.0067 m/s — 6.7 millimetres per second. So flow in a water supply pipe is turbulent under every condition a civil engineer will meet, which is why the friction factor is read from a Moody chart or a Colebrook-White fit rather than from f = 64/Re, and why pipe roughness matters at all. Laminar pipe flow is a laboratory condition and a viscous-oil condition; it is not a water condition.

Flow Measurement, and Making a Model Mean Something

A venturi meter is Bernoulli plus continuity turned into an instrument: contract the pipe, measure the pressure difference, and infer the discharge from Q = C_d A₂ √[2gΔh/(1 − (A₂/A₁)²)]. For a 300 → 150 mm venturi with a head difference of 0.5 m of water and C_d = 0.98, Q = 0.056 m³/s, or 56 litres per second. The coefficient is close to 1 because a well-made venturi has a gradual divergent cone that recovers most of the pressure it borrowed, so the permanent head loss is small — perhaps 10 per cent of the differential. An orifice plate measuring the same flow has C_d around 0.62 and loses far more permanently, because the jet separates and the energy goes into turbulence. That is the trade the two instruments represent: the orifice is cheap, short and lossy; the venturi is long, expensive and efficient, which is why the venturi is chosen for a permanent installation on a main and the orifice for a temporary measurement.

Before either instrument, the properties of fluids the whole section rests on are three. Density sets the inertia and therefore every dynamic pressure. Viscosity sets the resistance to shearing, τ = μ (du/dy), and it is what makes a boundary layer exist at all — the fluid at a solid surface does not slip, so a layer near the wall is retarded and grows in thickness along the surface. Surface tension matters only at small scales and is why a model of a weir can mislead if the head on it is a few millimetres. The boundary layer is where forces on immersed bodies originate. Total drag splits into skin friction, the integral of the wall shear, and form drag, the pressure difference between front and back when the boundary layer separates. A streamlined body is one shaped to postpone separation, so its drag is mostly friction; a bluff body separates early and its drag is mostly form. That distinction decides real civil quantities — the force on a bridge pier in a flood, on a chimney in a gale, on a submerged trash rack — and it is also why a cylinder’s drag coefficient falls abruptly at a high enough Reynolds number: the boundary layer turns turbulent, sticks to the surface further round, and the wake narrows. Lift is the same integral resolved perpendicular to the flow, and for a civil structure it is usually unwanted rather than useful.

💡 Why a spillway model runs on Froude and a pipe model on Reynolds
Dimensional analysis groups the variables of a problem into dimensionless numbers, and each number is a ratio of two forces. Re = VD/ν is inertia over viscosity. Fr = V/√(gL) is inertia over gravity. We is inertia over surface tension, Eu inertia over pressure. Complete similarity would require every relevant group to match between model and prototype at once, which is generally impossible — matching Re at a 1:25 scale needs the model velocity 25 times higher while matching Fr needs it 5 times lower, and no fluid will do both. So the engineer chooses the group that governs and accepts a scale effect in the others. For anything with a free surface — a spillway, a weir, an open channel, a hydraulic jump — the dominant balance is inertia against gravity, so the model is run on Froude similarity. Setting Fr_m = Fr_p at a length ratio L_r = 25 gives velocity and time ratios √25 = 5, a discharge ratio L_r^2.5 = 3125, and a force ratio L_r³ = 15 625. So 1 m/s measured on the model means 5 m/s in the prototype, and 1 litre/s on the model means 3.125 m³/s in the real spillway. For fully enclosed flow with no free surface — flow in a pipe, around a submerged body, through a valve — gravity does no work on the flow pattern and the governing group is Reynolds instead. Knowing which to use is the examinable skill, and the test is simply whether there is a free surface for gravity to act on. The residual error from the group you could not match is the scale effect, and it is why a Froude-scaled model reproduces the shape of a hydraulic jump faithfully while getting the friction losses somewhat wrong.

Key takeaways

  • F = ρg h̄ A uses the centroid, but the force acts lower, at h_cp = h̄ + I_G/(A h̄). On the worked gate that is 0.583 of the way down, not halfway — and the offset shrinks as the gate goes deeper.
  • Velocity goes as 1/A, so halving the diameter quadruples the velocity. Bernoulli then forces the pressure head down by exactly the rise in velocity head — 0.383 m across the worked contraction, and cavitation when it goes far enough.
  • h_f = fLV²/2gD — the V² means doubling the flow quadruples the loss and takes eight times the pumping power. That cubic is why a bigger pipe usually beats a bigger pump.
  • Capacity goes as D^2.5, so two parallel 300 mm pipes equal one of 344.6 mm — twice the flow for 15% more diameter. The same sensitivity means losing 10% of a bore loses 23% of the flow.
  • Laminar flow in a 300 mm water main would need 6.7 mm/s. Pipe flow in water supply is always turbulent, which is why roughness matters and f = 64/Re does not apply.
  • Choose the dimensionless group by asking whether there is a free surface: Froude for spillways and channels, Reynolds for enclosed flow. At 1:25, Froude gives velocity ×5, discharge ×3125, force ×15 625.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A vertical rectangular sluice gate 2 m wide has its top edge 1 m and its bottom edge 3 m below a water surface. The total hydrostatic force on the gate, in kN, is ______ (take ρg = 9.81 kN/m³)

    Numerical answer — type the value.

    Show answer

    Answer: 78.48

    78.48 kN. F = ρg h̄ A, where h̄ is the depth of the centroid of the gate, not of the water surface or the bottom edge. The gate is 2 m wide and 2 m tall, so A = 4 m², and its centroid is midway between 1 m and 3 m, so h̄ = 2 m. Then F = 9.81 × 2 × 4 = 78.48 kN. The related question, which is where most marks are lost, is where that force acts. The centre of pressure is h_cp = h̄ + I_G/(A h̄) with I_G = bd³/12 = 2(2)³/12 = 1.3333 m⁴, giving h_cp = 2 + 1.3333/8 = 2.167 m below the surface — which is 1.167 m down a 2 m gate, or 0.583 of the way down rather than halfway. Taking the resultant at mid-height understates the moment about a bottom hinge by about 17 per cent. Two features of the offset are worth knowing. It is always below the centroid, because pressure rises with depth without exception, so I_G/(A h̄) is strictly positive. And it shrinks with depth: the same gate at h̄ = 10 m has its centre of pressure only 33 mm below the centroid, because what matters is the variation of pressure across the gate compared with the mean, and deep gates see almost uniform pressure. The special case worth recognising is a gate whose top edge is at the surface: then h̄ = d/2 and h_cp = 2d/3, the familiar third-point result, which is the only case where the answer is a fixed fraction of the gate height.

  2. Water flows in a pipe that reduces from 300 mm to 150 mm diameter. The velocity in the smaller pipe is

    1. twice that in the larger pipe
    2. four times that in the larger pipe
    3. half that in the larger pipe
    4. the same, since the discharge is the same
    Show answer

    Answer: B — four times that in the larger pipe

    Four times. Continuity for an incompressible fluid is A₁V₁ = A₂V₂, and area goes as the square of the diameter — so halving the diameter quarters the area and the velocity must rise four-fold to pass the same discharge. Option A is the answer that comes from tracking the diameter instead of the area, and it is the commonest slip here. Option D is right about the discharge and wrong about what follows from it: the volume per second is indeed unchanged, which is exactly why the speed must increase when the pipe narrows. Run the numbers at Q = 0.05 m³/s: V₁ = 0.707 m/s and V₂ = 2.829 m/s, a ratio of 4.000. The consequence worth carrying is what happens to the pressure. The velocity heads are V²/2g = 0.0255 m and 0.408 m, so Bernoulli requires the pressure head to fall by 0.383 m across the contraction — energy has not been lost, it has changed form from pressure to velocity. Push that far enough and the absolute pressure reaches the vapour pressure of water, which boils cold, and the collapsing vapour bubbles pit the metal downstream: that is cavitation, the most common cause of damage at a pump inlet and in a partly closed valve, and it is nothing more than this same calculation taken to its limit. The reverse happens in the diffuser of a venturi or a pump volute, where the flow is slowed deliberately to convert velocity head back into pressure.

  3. A 300 mm diameter pipe 1000 m long carries water at 1 m/s with a friction factor f = 0.02. The head loss due to friction, in m, is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 3.40

    3.40 m. Darcy-Weisbach: h_f = fLV²/(2gD) = 0.02(1000)(1²)/(2 × 9.81 × 0.3) = 20/5.886 = 3.398 m. The discharge is Q = VA = 1 × 0.0707 = 0.0707 m³/s, so the power dissipated is ρgQh_f = 9810 × 0.0707 × 3.398 = 2.36 kW — the pumping power that friction alone consumes over a kilometre of main. The term to respect is V². Double the velocity and the head loss quadruples, to 13.59 m; and since power is ρgQh_f and Q has doubled as well, the pumping power rises eight-fold. That cube relationship between flow and power governs a great deal of water supply economics: a slightly larger pipe is almost always cheaper over a design life than a slightly larger pump, and design velocities in mains are held near 1 to 2 m/s rather than pushed higher. One check the question invites: is f = 0.02 even plausible? Re = VD/ν = 1 × 0.3/10⁻⁶ = 300 000, comfortably turbulent, where a friction factor of 0.02 is typical for a moderately smooth pipe — so yes. The related trap is f = 64/Re, which is the laminar result and would give 0.00021 here. Laminar pipe flow needs Re < 2000, which at this diameter means a velocity below 6.7 mm/s: water supply pipes never run laminar, so the friction factor always comes from a Moody chart or a Colebrook-White fit, and pipe roughness always matters.

  4. Two identical pipes of 300 mm diameter run in parallel between the same two points. The diameter of a single pipe of the same length that would carry the same total discharge under the same head loss, in mm, is ______ (round to one decimal)

    Numerical answer — type the value.

    Show answer

    Answer: 344.6

    344.6 mm. Write Darcy-Weisbach in terms of discharge: substituting V = Q/A = 4Q/πD² gives h_f = 8fLQ²/(π²gD⁵), so for a fixed head loss and length, Q ∝ D^2.5 and the equivalent single pipe satisfies D_e⁵ = Σ D_i⁵. With two pipes of the same size, D_e = 2^(1/5) × 300 = 1.1487 × 300 = 344.6 mm. So twice the capacity is bought with a diameter only 14.9 per cent larger, which is a result almost nobody guesses correctly — the intuition that two pipes need "one pipe twice as big" is out by a factor of nearly six in area. Three pipes give 373.7 mm and four 395.9 mm, so the returns diminish, but the direction is consistent: capacity is extraordinarily sensitive to bore. Three practical consequences follow. Duplicating a main by laying a second pipe beside it is usually poor value against laying one larger pipe, once the second trench, the second set of fittings and the second maintenance liability are counted. A modest loss of bore is expensive: scaling or tuberculation that takes 10 per cent off the diameter takes 23 per cent off the flow, which is why old cast-iron mains lose capacity long before they lose integrity. And the same D^2.5 explains why pipe sizes in a distribution network step up so slowly as the demand accumulates. Note the assumption the question relies on: the pipes are the same length and have the same f. If lengths differ, the equivalence uses D⁵/L rather than D⁵.

  5. A 1:25 scale model of a spillway is to be tested. The appropriate similitude and the resulting discharge ratio are

    1. Reynolds similitude, with discharge ratio 25
    2. Froude similitude, with discharge ratio 3125
    3. Froude similitude, with discharge ratio 625
    4. Both simultaneously, for complete similarity
    Show answer

    Answer: B — Froude similitude, with discharge ratio 3125

    Froude, with a discharge ratio of 3125. A spillway has a free surface, so the dominant balance is inertia against gravity and the group to match is the Froude number Fr = V/√(gL). Setting Fr_m = Fr_p with g the same at both scales gives V_r = √L_r = √25 = 5, and since Q = AV ∝ L²V, the discharge ratio is L_r^2.5 = 25^2.5 = 3125. So 1 litre/s on the model corresponds to 3.125 m³/s on the prototype, and a model velocity of 1 m/s means 5 m/s in the real structure. Option C uses L_r², forgetting that the velocity scales too. Option A picks the wrong group: Reynolds governs flow with no free surface — inside a pipe, around a submerged body, through a valve — where gravity does no work on the flow pattern. Option D is the one worth understanding rather than merely rejecting, because it describes something genuinely impossible with the same fluid. Matching Re at 1:25 requires the model velocity to be 25 times higher than the prototype; matching Fr requires it 5 times lower. No single fluid can satisfy both, so the engineer picks the group that governs the phenomenon and accepts a scale effect in the others. For a Froude-scaled spillway model that residual error sits mainly in the friction and viscous losses, which is why such a model reproduces the shape and position of a hydraulic jump faithfully while getting the energy dissipation somewhat wrong — a limitation to state rather than to hide.

  6. A venturi meter with inlet 300 mm and throat 150 mm has a coefficient of discharge of 0.98 and registers a head difference of 0.5 m of water. The discharge in litres per second is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 56.02

    56.02 litres/s. A₁ = π(0.3)²/4 = 0.070686 m² and A₂ = π(0.15)²/4 = 0.017671 m², so A₂/A₁ = 0.25 and 1 − (A₂/A₁)² = 0.9375. Then Q = C_d A₂ √[2gΔh/(1 − (A₂/A₁)²)] = 0.98 × 0.017671 × √(9.81/0.9375) = 0.98 × 0.017671 × 3.2348 = 0.056021 m³/s. The derivation is worth reconstructing rather than memorising: it is Bernoulli between the inlet and the throat, giving the velocity difference from the pressure difference, closed with continuity to eliminate one of the two velocities — the whole meter is those two equations turned into an instrument. Two observations. C_d = 0.98 is close to 1 because a well-made venturi has a gradual divergent cone downstream of the throat, which decelerates the flow without separation and recovers most of the pressure it borrowed; the permanent head loss is only around 10 per cent of the measured differential. An orifice plate measuring the same flow has C_d ≈ 0.62 and loses far more permanently, because the jet separates from the sharp edge and the energy goes irrecoverably into turbulence. That is the trade between the two: the orifice is cheap, compact and lossy, the venturi long, expensive and efficient — so a venturi is chosen for a permanent installation on a main where the pumping cost of the loss is paid every day, and an orifice for a temporary or occasional measurement. Note also that the head difference must be expressed in metres of the flowing fluid; a manometer reading in mercury needs converting by the ratio of specific gravities first, which is a routine source of lost marks.

  7. The friction factor relation f = 64/Re applies to

    1. all pipe flow, laminar and turbulent
    2. laminar pipe flow only, which water supply pipes essentially never exhibit
    3. turbulent flow in smooth pipes only
    4. open channel flow
    Show answer

    Answer: B — laminar pipe flow only, which water supply pipes essentially never exhibit

    Laminar pipe flow only — and water mains essentially never run laminar. The relation follows from integrating the parabolic velocity profile of fully developed laminar flow, so it holds strictly below the transition, conventionally Re ≈ 2000. Test whether that regime is reachable in practice. In a 300 mm main carrying water at 1 m/s with ν = 10⁻⁶ m²/s, Re = VD/ν = 300 000 — two orders of magnitude past the transition. To bring the same pipe into the laminar range the velocity would have to fall below 2000 × 10⁻⁶/0.3 = 0.0067 m/s, that is 6.7 millimetres per second, which is not a flow anyone designs for. So option A is wrong twice over: the relation is not universal, and the regime it describes is not the one civil pipes are in. Option C names the right regime for the wrong formula — turbulent flow in smooth pipes has its own correlations, such as Blasius f = 0.316 Re^(−0.25), and beyond that the Colebrook-White equation or the Moody chart, which bring in the relative roughness k/D. That dependence on roughness is the practical point: in laminar flow the wall roughness does not appear at all, because the flow near the wall is orderly and the asperities sit inside the viscous layer, whereas in turbulent flow they project into it and dominate the loss at high Re. Option D belongs to a different family altogether — open channels use Manning or Chezy, in which the resistance coefficient carries different dimensions and the hydraulic radius replaces the diameter. Laminar flow is worth knowing for viscous oils, groundwater seepage and the laboratory, not for a rising main.

  8. Water is pumped through a main. Which of the following are true? (One or more)

    1. Doubling the flow rate quadruples the friction head loss
    2. Doubling the flow rate multiplies the pumping power by eight
    3. Reducing the bore by 10% reduces the capacity by about 23% at the same head loss
    4. The centre of pressure on a submerged gate lies above its centroid
    Show answer

    Answer: A — Doubling the flow rate quadruples the friction head loss; B — Doubling the flow rate multiplies the pumping power by eight; C — Reducing the bore by 10% reduces the capacity by about 23% at the same head loss

    A, B and C. All three come out of Darcy-Weisbach once it is written in terms of discharge, h_f = 8fLQ²/(π²gD⁵). A: h_f ∝ Q², so doubling the flow quadruples the loss — on the worked main, from 3.40 m to 13.59 m. B: pumping power is ρgQh_f, and if Q doubles while h_f quadruples, the power goes up by eight. That cube law is the single most consequential relation in water supply economics — it is why design velocities are held near 1 to 2 m/s, and why a slightly larger pipe usually beats a slightly larger pump over a design life. C: at fixed head loss Q ∝ D^2.5, so a bore reduced to 0.9D carries 0.9^2.5 = 0.768 of the flow, a loss of 23.2 per cent for a 10 per cent loss of diameter. That extreme sensitivity is why tuberculation and scaling in old mains destroy capacity long before they threaten integrity, and it is the same exponent that makes two parallel 300 mm pipes equivalent to a single 344.6 mm one. D is false and belongs to the first section of this chapter: the centre of pressure is always below the centroid, at h_cp = h̄ + I_G/(A h̄), because pressure increases with depth without exception, so the added term is strictly positive. It approaches the centroid as the gate goes deeper — 167 mm below at h̄ = 2 m and only 33 mm below at h̄ = 10 m — but it never reaches it and never rises above it.