University Physics — Year 2 · Bachelor Year 2

3Perfect Fluids: Euler and Bernoulli

Hold a sheet of paper by its edge and blow over its top: it lifts. Open the tap in the shower and the curtain leans in. A wing holds up three hundred tonnes because the air flows faster over its upper surface than under it. All three are the same statement: where a fluid moves faster, its pressure is lower. This chapter writes Newton’s second law for a fluid particle in the simplest model — the perfect fluid, which feels pressure but no friction — and draws from it Bernoulli’s theorem, the energy balance of a streamline, whose applications run from the flow meter in a pipe to the speed indicator of an aircraft.

3.1 The Euler equation

Proposition 3.1 (Pressure force on a fluid particle)

The resultant of the pressure forces exerted by the surrounding fluid on a particle of volume $\dd\tau$ is

\dd\vect F_P = -\operatorname{\vect{grad}}P\,\dd\tau :

pressure acts as a volume force of density $-\operatorname{\vect{grad}}P$ , pushing from high toward low pressure.

Proof. On the box of Theorem 2.8, the faces at $x$ and $x + \dd x$ receive $+P(x)\dd y\dd z$ and $-P(x + \dd x)\dd y\dd z$ along $x$ : net $-\partial_xP\,\dd x\dd y\dd z$ ; likewise for $y$ and $z$ . ∎

Definition 3.2 (Perfect fluid)

A perfect fluid is one in which the only contact force between neighbouring particles is the pressure — normal to every surface, no tangential (viscous) stress. It is an idealization valid, as the next chapter explains, far from walls and at high Reynolds number: water in a pipe away from the wall, air around a wing outside the thin boundary layer, a wave on the sea.

Theorem 3.3 (Euler’s equation)

In a Galilean frame, a perfect fluid of density $\rho$ subject to gravity obeys, at every point,

\rho\Bigl(\frac{\partial\vect v}{\partial t} + (\vect v\cdot\operatorname{\vect{grad}}) \vect v\Bigr) = -\operatorname{\vect{grad}}P + \rho\vect g ,

to which any other volume force density $\vect f_v$ (electric, inertial in a non-Galilean frame) is added on the right. At rest it reduces to the hydrostatic law $\operatorname{\vect{grad}}P = \rho\vect g$ of the Year 1 volume.

Proof. Newton’s second law for the particle of mass $\rho\,\dd\tau$ : its acceleration is the material derivative (Theorem 2.4), the forces are the pressure resultant and the weight $\rho\vect g\,\dd\tau$ ; divide by $\dd\tau$ . ∎

Example 3.4 (A tank that accelerates)

Water in a tank carried by a truck accelerating at $\vect a$ is at rest in the truck’s frame; there, Euler with the inertial force $-\rho\vect a$ gives $\operatorname{\vect{grad}}P = \rho(\vect g - \vect a)$ : the pressure grows along the "effective gravity" $\vect g - \vect a$ , the free surface (an isobar) is perpendicular to it and tilts backward by the angle $\arctan(a/g)$ — $11{}^{\circ}$ for $a = 2\,\mathrm{m}/\mathrm{s}^{2}$ . In a bucket spinning at $\Omega$ , the centrifugal force $\rho\Omega^2r\,\vect e_r$ gives $P = P_0 + \tfrac12\rho\Omega^2r^2 - \rho gz$ and the free surface is the paraboloid $z = \Omega^2r^2/2g$ .

Left: the pressure forces on the faces of a fluid box do not cancel when P varies — their resultant is - gradP per unit volume. Middle and right: two fluids at rest in an accelerated frame; the free surface is everywhere perpendicular to the effective gravity. — Left: the pressure forces on the faces of a fluid box do not cancel when $P$ varies — their resultant is $-\operatorname{\vect{grad}}P$ per unit volume. Middle and right: two fluids at rest in an accelerated frame; the free surface is everywhere perpendicular to the effective gravity.

3.2 Bernoulli’s theorem

Theorem 3.5 (Bernoulli)

For a perfect, incompressible fluid of uniform density in steady flow, in a Galilean frame with uniform gravity, the quantity

P + \tfrac12\rho v^2 + \rho gz

is constant along each streamline. If the flow is moreover irrotational, it is the same constant throughout the fluid. The term $\tfrac12\rho v^2$ is the dynamic pressure; $P + \tfrac12\rho v^2$ is the stagnation pressure, the pressure the fluid would reach if brought to rest along its streamline.

Proof. Steady Euler with $(\vect v\cdot\operatorname{\vect{grad}})\vect v = \operatorname{\vect{grad}}(v^2/2) + \vect\omega\wedge\vect v$ and $\rho\vect g = -\operatorname{\vect{grad}}(\rho gz)$ , $\rho$ uniform:

\operatorname{\vect{grad}}\bigl(P + \tfrac12\rho v^2 + \rho gz\bigr) = -\rho\, \vect\omega\wedge\vect v .

The right side is perpendicular to $\vect v$ , so the gradient of the Bernoulli quantity has no component along the streamline: the quantity is constant along it. If $\vect\omega = \vect 0$ the gradient vanishes everywhere. ∎

Remark 3.6 (What Bernoulli says and does not say)

Divided by $\rho$ , the theorem reads: the mechanical energy per unit mass, $v^2/2 + gz$ , plus $P/\rho$ is conserved along a particle’s path. The pressure term is the work done on the particle by the pressure of the fluid behind it, minus the work it does on the fluid ahead: the energy theorem for a particle pushed through a pressure field without friction. Four conditions: perfect fluid (no viscous loss), steady, incompressible, and along a streamline. Comparing two points on different streamlines is allowed only in irrotational flow. It does not say "fast air sucks": it says that a particle that has been accelerated (by a pressure drop) is now at lower pressure — the cause is the pressure field, the speed is the effect.

Remark 3.7 (Compressible fluids)

For a gas the density varies, but if the flow is steady, perfect and isentropic the same proof gives the conservation of $h + v^2/2 + gz$ along a streamline, where $h$ is the enthalpy per unit mass ( $h = c_PT$ for a perfect gas): the form used for nozzles and turbines in Chapter 27. At speeds well below the speed of sound ( $v \lesssim 100\,\mathrm{m}/\mathrm{s}$ in air) the density changes by less than $5\%$ and the incompressible theorem is accurate.

3.3 Applications

Proposition 3.8 (The Venturi meter)

In a horizontal pipe narrowing from section $S_1$ to $S_2 < S_1$ , the speed rises ( $v_2 = v_1S_1/S_2$ , mass conservation) and the pressure drops:

P_1 - P_2 = \tfrac12\rho(v_2^2 - v_1^2) = \tfrac12\rho v_1^2\Bigl(\frac{S_1^2}{S_2^2} - 1\Bigr) , \qquad D_V = S_1v_1 = S_1S_2\sqrt{\frac{2(P_1 - P_2)}{\rho(S_1^2 - S_2^2)}} .

Measuring the pressure difference with a manometer gives the flow rate: a Venturi meter. The same effect drives a carburettor, a paint spray, a Bunsen burner’s air intake and the "suction" of a passing train.

Proof. Bernoulli along a central streamline at constant height, with $v_2 = v_1S_1/S_2$ ; solve for $v_1$ . ∎

Proposition 3.9 (The Pitot tube)

A tube facing the flow, closed at the far end, brings the fluid to rest at its mouth (a stagnation point); a second opening on the side, parallel to the flow, reads the static pressure $P$ . The difference is the dynamic pressure:

P_{\text{stag}} - P = \tfrac12\rho v^2 , \qquad v = \sqrt{2(P_{\text{stag}} - P)/\rho} .

Every aircraft measures its airspeed this way.

Proof. Bernoulli along the streamline that ends at the stagnation point, where $v = 0$ ; the side holes do not disturb the flow, which passes at the undisturbed $v$ and $P$ . ∎

Proposition 3.10 (Torricelli’s formula)

A large open tank drains through a small hole at depth $h$ below the free surface; the jet leaves at

v = \sqrt{2gh} ,

the speed of free fall from the surface. The flow rate is $D_V \approx \alpha s\sqrt{2gh}$ with $s$ the hole’s area and $\alpha \approx 0.6$ the contraction coefficient of a sharp-edged orifice (the streamlines keep converging past the hole).

Proof. Bernoulli from the free surface (at rest, since the tank is large: $v_{\text{surf}} = v\,s/S \ll v$ ; pressure $P_0$ ) to the jet (pressure $P_0$ just outside the hole): $P_0 + \rho gh = P_0 + \tfrac12\rho v^2$ . The flow is quasi-steady: $h$ varies slowly. ∎

Three applications of Bernoulli’s theorem. Left: the Venturi meter — the manometer reads the pressure drop at the throat, hence the flow rate. Middle: the Pitot tube — the stagnation pressure at the mouth exceeds the static pressure at the side holes by the dynamic pressure. Right: Torricelli’s efflux at the free-fall speed.

Example 3.11 (Numbers)

A Venturi of $5\,\mathrm{cm}$ narrowing to $2.5\,\mathrm{cm}$ on a water pipe, with $\Delta h = 20\,\mathrm{cm}$ of water ( $2.0\,\mathrm{kPa}$ ): $v_1 = \sqrt{2 \times 2000/ (1000 \times 15)} = 0.52\,\mathrm{m}/\mathrm{s}$ , $D_V = 1.0\,\mathrm{L}/\mathrm{s}$ . An airliner at $11\,\mathrm{km}$ ( $\rho = 0.36\,\mathrm{kg}/\mathrm{m}^{3}$ ) at $250\,\mathrm{m}/\mathrm{s}$ : dynamic pressure $11\,\mathrm{kPa}$ , $11\%$ of the ambient pressure — the Pitot tube still works, with a compressibility correction. A water tower $40\,\mathrm{m}$ above the taps: $4\,\mathrm{bar}$ when closed, a jet at $28\,\mathrm{m}/\mathrm{s}$ if opened wide (in practice far less: the pipes are not perfect).

Proposition 3.12 (Lift on a wing)

In the flow around a wing the air passes faster over the upper surface than under the lower one; the pressure is therefore lower above than below, and the resultant is the lift $F_L = \tfrac12\rho v^2SC_L$ ( $S$ the wing area, $C_L \approx 0.3$ – $1.5$ the lift coefficient, depending on the profile and the angle of attack). The speed difference is equivalent to a circulation $\Gamma$ of the velocity around the profile, and the lift per unit span is $F_L' = \rho v\Gamma$ (Kutta–Joukowski, admitted). A spinning ball drags air round with it and is lifted sideways the same way: the Magnus effect of a sliced tennis ball or a curving free kick.

Proof. Bernoulli on the streamlines just above and below, which start from the same upstream conditions: $P_{\text{low}} - P_{\text{up}} = \tfrac12\rho (v_{\text{up}}^2 - v_{\text{low}}^2)$ ; integrate over the chord. Why the upper flow is faster — the sharp trailing edge forces the rear stagnation point there and fixes the circulation (the Kutta condition) — is beyond the perfect-fluid model alone; it needs the boundary layer of the next chapter. ∎

Left: streamlines around a wing — crowded and fast above, slow below; the pressure difference is the lift. Right: a liquid column of total length L oscillating in a U-tube, the simplest unsteady application of Euler’s equation. — Left: streamlines around a wing — crowded and fast above, slow below; the pressure difference is the lift. Right: a liquid column of total length $L$ oscillating in a U-tube, the simplest unsteady application of Euler’s equation.

Proposition 3.13 (An unsteady flow: the U-tube)

A liquid column of total length $L$ fills a U-tube of uniform section; displaced by $z$ from equilibrium, it oscillates with

\ddot z + \frac{2g}{L}\,z = 0 , \qquad T = 2\pi\sqrt{\frac{L}{2g}} ,

like a pendulum of length $L/2$ . More generally, for an unsteady but irrotational flow with potential $\varphi$ , Euler integrates to $\rho\,\partial_t\varphi + P + \tfrac12\rho v^2 + \rho gz = C(t)$ , the same constant throughout the fluid (unsteady Bernoulli).

Proof. The speed $v = \dot z$ is uniform along the tube (incompressible, uniform section). Project Euler on the tube’s axis and integrate along the column from one free surface to the other: the pressure terms cancel ( $P_0$ at both ends), $\int\rho\,\partial_tv\,\dd s = \rho L\ddot z$ , the convective term $\rho\,\Delta(v^2/2)$ vanishes (same speed at both ends), and the gravity term gives $\rho g\cdot2z$ . Hence $\rho L\ddot z = -2\rho gz$ . The general statement follows from $\partial_t\vect v = \operatorname{\vect{grad}} \partial_t\varphi$ for $\vect v = \operatorname{\vect{grad}}\varphi$ . ∎

Example 3.14 (Starting a pipe)

A horizontal pipe of length $\ell$ leads from a reservoir of constant head $h$ to an open end; the valve is opened at $t = 0$ . With $v$ uniform along the pipe, Euler integrated from the reservoir surface to the outlet gives $\ell\,\dd v/\dd t = gh - \tfrac12v^2$ , whose solution is $v = V\tanh(Vt/2\ell)$ with $V = \sqrt{2gh}$ : the flow reaches Torricelli’s speed with the time constant $2\ell/V$ — $13\,\mathrm{s}$ for a $400\,\mathrm{m}$ penstock under $200\,\mathrm{m}$ of head.

Method 3.15 (Using Bernoulli)

(1) Check the four conditions; decide whether the flow is irrotational (uniform upstream flow around an obstacle: yes; flow in a pipe with a velocity profile: no — stay on one streamline). (2) Choose two points on a streamline where as many quantities as possible are known: a free surface ( $P = P_0$ , $v \approx 0$ if the section is large), a jet in the open air ( $P = P_0$ ), a stagnation point ( $v = 0$ ). (3) Add mass conservation $vS =$ const to relate the speeds. (4) For unsteady flows, integrate Euler along the line instead.

3.4 Exercises

Exercise 3.1 ★

Water flows in a horizontal pipe of diameter $8\,\mathrm{cm}$ at $1.5\,\mathrm{m}/\mathrm{s}$ and pressure $3.0\,\mathrm{bar}$ . It passes into a $4\,\mathrm{cm}$ section: speed and pressure there; into a $2\,\mathrm{cm}$ section: pressure — what happens if it would come out negative?

Solution

Solution of Exercise 3.1.

$v_2 = 1.5 \times 4 = 6.0\,\mathrm{m}/\mathrm{s}$ , $P_2 = 3.0 \times 10^5 - 500(36 - 2.25) = 2.83\,\mathrm{bar}$ . At $2\,\mathrm{cm}$ : $v = 24\,\mathrm{m}/\mathrm{s}$ , $P = 3.0 \times 10^5 - 500 (576 - 2.25) = 0.13\,\mathrm{bar}$ — close to the vapour pressure. A negative result means the assumed flow rate is impossible: the water cavitates and the flow is throttled.

Exercise 3.2 ★

The Pitot tube of an aircraft at $5\,\mathrm{km}$ ( $\rho = 0.74\,\mathrm{kg}/\mathrm{m}^{3}$ ) reads a dynamic pressure of $15\,\mathrm{kPa}$ . True airspeed. The airspeed indicator is calibrated for sea-level density ( $1.22\,\mathrm{kg}/\mathrm{m}^{3}$ ): what "indicated airspeed" does it show, and why do pilots find the difference useful rather than annoying?

Solution

Solution of Exercise 3.2.

$v = \sqrt{2 \times 15000/0.74} = 201\,\mathrm{m}/\mathrm{s}$ ; indicated $\sqrt{2 \times 15000/ 1.22} = 157\,\mathrm{m}/\mathrm{s}$ . The indicated airspeed measures the dynamic pressure, which is what the wing feels: the stall speed is a fixed indicated speed at any altitude.

Exercise 3.3 ★

A tank holds water $3.0\,\mathrm{m}$ deep; a $1.0\,\mathrm{cm}^{2}$ hole is pierced $1.0\,\mathrm{m}$ above the floor. Exit speed; flow rate (contraction $0.6$ ); where does the jet hit the floor? At what depth should a second hole be pierced to reach the same point?

Solution

Solution of Exercise 3.3.

Depth $2.0\,\mathrm{m}$ : $v = \sqrt{2g \times 2} = 6.3\,\mathrm{m}/\mathrm{s}$ ; $D_V = 0.6 \times 10^{-4} \times 6.3 = 0.38\,\mathrm{L}/\mathrm{s}$ ; fall time $\sqrt{2/g} = 0.45\,\mathrm{s}$ , range $2.8\,\mathrm{m}$ . Range $= 2\sqrt{h(H - h)}$ is symmetric in $h \leftrightarrow H - h$ : a hole at depth $1.0\,\mathrm{m}$ ( $2\,\mathrm{m}$ above the floor) hits the same point.

Exercise 3.4 ★

A wind of $30\,\mathrm{m}/\mathrm{s}$ blows over a flat roof of $100\,\mathrm{m}^{2}$ ; the air under the roof is still at atmospheric pressure ( $\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}$ ). Pressure difference and net force on the roof; compare with the weight of the roof ( $50\,\mathrm{kg}/\mathrm{m}^{2}$ ). Which way is it pushed?

Solution

Solution of Exercise 3.4.

$\Delta P = \tfrac12 \times 1.2 \times 900 = 540\,\mathrm{Pa}$ , outside pressure lower: net force $54\,\mathrm{kN}$ upward; the roof weighs $49\,\mathrm{kN}$ : it lifts off. (Real roofs fail at their edges first, where the flow separates.)

Exercise 3.5 ★★

A closed rectangular tank $2.0\,\mathrm{m}$ long, half full of water, sits on a truck. (a) Accelerating at $3.0\,\mathrm{m}/\mathrm{s}^{2}$ : angle of the free surface, and the difference of water height between the rear and the front walls. (b) Pressure difference between the two bottom corners. (c) Braking at $6\,\mathrm{m}/\mathrm{s}^{2}$ — does the water reach the lid $0.8\,\mathrm{m}$ above the bottom? (d) Same truck going round a bend of radius $40\,\mathrm{m}$ at $15\,\mathrm{m}/\mathrm{s}$ : tilt of the surface.

Solution

Solution of Exercise 3.5.

(a) $\tan\theta = 3/9.81$ , $\theta = 17{}^{\circ}$ ; $2 \times 0.306 = 0.61\,\mathrm{m}$ (higher at the rear). (b) $\Delta P = \rho aL = 6.0\,\mathrm{kPa}$ . (c) $\tan\theta = 0.61$ , difference $1.22\,\mathrm{m}$ : the front rises $0.61\,\mathrm{m}$ above the mean $0.40\,\mathrm{m}$ : $1.01\,\mathrm{m}$ $>$ $0.8\,\mathrm{m}$ , it hits the lid. (d) $a = v^2/R = 5.6\,\mathrm{m}/\mathrm{s}^{2}$ : $\theta = 30{}^{\circ}$ , outward side higher.

Exercise 3.6 ★★

A bucket of radius $15\,\mathrm{cm}$ containing water $20\,\mathrm{cm}$ deep is spun about its axis at $2.0\,\mathrm{turns}/\mathrm{s}$ . (a) Shape and equation of the free surface. (b) Height difference between rim and centre. (c) Volume being conserved, height of the water at the centre and at the rim. (d) At what rotation rate does the bottom centre dry out? (e) Why does the same paraboloid make a perfect telescope mirror (a spinning pool of mercury)?

Solution

Solution of Exercise 3.6.

(a) Paraboloid $z = z_0 + \Omega^2r^2/2g$ , $\Omega = 4\pi = 12.6\,\mathrm{rad}/\mathrm{s}$ . (b) $\Omega^2R^2/2g = 158 \times 0.0225/19.6 = 0.18\,\mathrm{m}$ . (c) The mean height of a paraboloid over the disk is halfway: centre $0.20 - 0.09 = 0.11\,\mathrm{m}$ , rim $0.29\,\mathrm{m}$ . (d) Centre dry when $\Omega^2R^2/4g = h_0$ : $\Omega = \sqrt{4gh_0}/R = 18.7\,\mathrm{rad}/\mathrm{s} = 3.0$ turns per second. (e) A paraboloid focuses parallel rays to one point, at $f = g/2\Omega^2$ ; liquid-mirror telescopes spin mercury.

Exercise 3.7 ★★

Siphon. A tube carries water from a tank over a crest $2.0\,\mathrm{m}$ above the free surface to an outlet $3.0\,\mathrm{m}$ below it. (a) Exit speed and, for a $2\,\mathrm{cm}$ tube, flow rate. (b) Pressure at the crest. (c) Maximum height of the crest for the siphon to work (the water’s vapour pressure, $2.3\,\mathrm{kPa}$ , must not be reached). (d) Why does the siphon not work with the tube full of air?

Solution

Solution of Exercise 3.7.

(a) $v = \sqrt{2g \times 3} = 7.7\,\mathrm{m}/\mathrm{s}$ ; $D_V = \pi \times 10^{-4} \times 7.7 = 2.4\,\mathrm{L}/\mathrm{s}$ . (b) $P = P_0 - \rho g(2) - \tfrac12\rho v^2 = 100 - 19.6 - 29.4 = 51\,\mathrm{kPa}$ . (c) $P_0 - \rho gh_c - \tfrac12\rho v^2 \ge 2.3\,\mathrm{kPa}$ : $h_c \le 7.0\,\mathrm{m}$ . (d) Air is compressible and light: no continuous liquid column to transmit the pressure, the water on both sides just stays.

Exercise 3.8 ★★

A light aircraft of $1000\,\mathrm{kg}$ cruises at $60\,\mathrm{m}/\mathrm{s}$ at sea level on a wing of $15\,\mathrm{m}^{2}$ . (a) Mean pressure difference between the two faces of the wing; lift coefficient. (b) If the speed under the wing is $58\,\mathrm{m}/\mathrm{s}$ , what is it over the wing (take the same upstream conditions)? (c) Circulation around the profile from the Kutta–Joukowski formula, for a span of $10\,\mathrm{m}$ . (d) Stalling speed if $C_L$ cannot exceed $1.4$ .

Solution

Solution of Exercise 3.8.

(a) $\Delta P = mg/S = 654\,\mathrm{Pa}$ ; $\tfrac12\rho v^2 = 2.2\,\mathrm{kPa}$ , $C_L = 0.30$ . (b) $v_{\text{up}}^2 = 58^2 + 2 \times 654/1.22$ : $66.6\,\mathrm{m}/\mathrm{s}$ . (c) $F_L' = 981\,\mathrm{N}/\mathrm{m} = \rho v\Gamma$ : $\Gamma = 13.4\,\mathrm{m}^{2}/\mathrm{s}$ . (d) $v = \sqrt{2mg/ \rho SC_L} = \sqrt{19620/25.6} = 28\,\mathrm{m}/\mathrm{s}$ .

Exercise 3.9 ★★

Stenosis. Blood ( $\rho = 1060\,\mathrm{kg}/\mathrm{m}^{3}$ ) flows at $0.50\,\mathrm{m}/\mathrm{s}$ in an artery of section $0.50\,\mathrm{cm}^{2}$ at $13\,\mathrm{kPa}$ above atmospheric pressure. A plaque narrows it to $0.10\,\mathrm{cm}^{2}$ . (a) Speed and pressure in the stenosis. (b) The tissue outside the artery is at about $1\,\mathrm{kPa}$ : can the artery collapse? (c) If it narrows, the speed rises further: explain the runaway and why such a narrowing can flutter.

Solution

Solution of Exercise 3.9.

(a) $v_2 = 2.5\,\mathrm{m}/\mathrm{s}$ ; $\Delta P = \tfrac12 \times 1060 \times (6.25 - 0.25) = 3.2\,\mathrm{kPa}$ , $P_2 = 9.8\,\mathrm{kPa}$ . (b) Still above $1\,\mathrm{kPa}$ : no. (c) At $0.05\,\mathrm{cm}^{2}$ , $v = 5\,\mathrm{m}/\mathrm{s}$ and $\Delta P = 13\,\mathrm{kPa}$ : $P_2 \approx 0$ , below the tissue pressure — the wall collapses, the flow stops, the pressure recovers, it reopens: a flutter (the murmur a physician hears).

Exercise 3.10 ★★★

Stagnation temperature. For a perfect gas in steady isentropic flow, $c_PT + v^2/2$ is conserved along a streamline. (a) Show that the air brought to rest at the nose of an aircraft reaches $T_0 = T(1 + \tfrac{\gamma - 1}{2}M^2)$ , with $M = v/c$ the Mach number and $c = \sqrt{\gamma RT/M_{\text{mol}}}$ the speed of sound. (b) Nose temperature of an airliner at $M = 0.85$ in air at $220\,\mathrm{K}$ ; of a supersonic aircraft at $M = 2$ ; of a re-entry capsule at $M = 25$ (is the formula still meaningful?). (c) Show that the incompressible Pitot formula overestimates the speed at $M = 0.85$ , and by roughly how much (expand $P_0/P = (T_0/T)^{\gamma/(\gamma - 1)}$ to second order in $M^2$ ).

Solution

Solution of Exercise 3.10.

(a) $c_PT_0 = c_PT + v^2/2$ with $c_P = \gamma R/(\gamma - 1)M_{\text{mol}}$ and $c^2 = \gamma RT/M_{\text{mol}}$ : $v^2/2c_PT = \tfrac12(\gamma - 1)M^2$ . (b) $M = 0.85$ : $T_0 = 220 \times 1.145 = 252\,\mathrm{K}$ ; $M = 2$ : $220 \times 1.8 = 396\,\mathrm{K}$ ; $M = 25$ : $220 \times 126 = 28\,000\,\mathrm{K}$ — meaningless: the air dissociates and ionizes, $c_P$ is not constant (real stagnation temperatures are several thousand kelvin). (c) $P_0/P = (1 + \tfrac{\gamma - 1}2M^2)^{\gamma/(\gamma - 1)} \approx 1 + \tfrac{\gamma}2M^2 + \tfrac{\gamma}8M^4$ , and $\gamma PM^2 = \rho v^2$ : $P_0 - P = \tfrac12\rho v^2(1 + M^2/4)$ . Reading $v$ from $\sqrt{2(P_0 - P)/\rho}$ overestimates it by $\sqrt{1 + M^2/4} - 1 \approx 9\%$ at $M = 0.85$ .

Exercise 3.11 ★★★

Starting transient. A pipe of length $\ell = 50\,\mathrm{m}$ and uniform section leads from a large reservoir (head $h = 5.0\,\mathrm{m}$ above the outlet) to a valve at the open end. The valve is opened at $t = 0$ . (a) Integrate Euler along the pipe to get $\ell\,\dd v/\dd t = gh - v^2/2$ . (b) Solve it: $v = V\tanh(Vt/2\ell)$ with $V = \sqrt{2gh}$ . (c) Time to reach $90\%$ of the final speed. (d) The valve is now shut abruptly in $0.5\,\mathrm{s}$ from full flow: estimate the mean deceleration and the over-pressure at the valve from $\rho\ell\,\dd v/\dd t$ — and why a real closure produces far more (next chapters).

Solution

Solution of Exercise 3.11.

(a) From the surface ( $v \approx 0$ , $P_0$ , height $h$ ) to the outlet ( $v$ , $P_0$ , height $0$ ), with $\int\partial_tv\,\dd s = \ell\,\dd v/\dd t$ . (b) $\dd v/(V^2 - v^2) = \dd t/2\ell$ , $V^2 = 2gh$ : $\operatorname{artanh}(v/V) = Vt/2\ell$ . (c) $V = 9.9\,\mathrm{m}/\mathrm{s}$ ; $\tanh x = 0.9$ at $x = 1.47$ : $t = 2\ell x/V = 15\,\mathrm{s}$ . (d) $\dd v/\dd t \approx -20\,\mathrm{m}/\mathrm{s}^{2}$ : $\Delta P = \rho\ell \times 20 = 9.9\,\mathrm{bar}$ . Water is slightly compressible: a fast closure launches a pressure wave, $\Delta P = \rho c\Delta v \approx 140\,\mathrm{bar}$ .

Exercise 3.12 ★★★

The free surface of a vortex. Water rotates as the point vortex of Chapter 2, $v_\theta = \Gamma/2\pi r$ , irrotational, with a free surface at $z = 0$ far away. (a) Why may Bernoulli be applied between any two points? (b) Shape of the free surface $z(r)$ . (c) Inside a Rankine core ( $r < a$ , $v_\theta = \Omega r$ , rotational) use instead Euler in the rotating frame or directly: show $z = z(a) + \Omega^2(r^2 - a^2)/2g$ . (d) Total depth of the funnel for a bathtub vortex with $\Gamma = 0.03\,\mathrm{m}^{2}/\mathrm{s}$ and $a = 5\,\mathrm{mm}$ ; for a tornado ( $\Gamma = 3 \times 10^{4}\,\mathrm{m}^{2}/\mathrm{s}$ , $a = 60\,\mathrm{m}$ , in air: express the result as a pressure drop at the centre instead).

Solution

Solution of Exercise 3.12.

(a) Steady, perfect, incompressible and irrotational: one constant for the whole flow. (b) On the surface $P = P_0$ : $\tfrac12\rho v^2 + \rho gz = 0$ , $z = -\Gamma^2/8\pi^2gr^2$ . (c) Radial Euler: $\partial_rP = \rho\Omega^2r$ , vertical hydrostatics: the surface satisfies $z(r) - z(a) = \Omega^2(r^2 - a^2)/2g$ . (d) Bath: $\Omega = \Gamma/2\pi a^2 = 191\,\mathrm{rad}/\mathrm{s}$ ; $z(a) = -\Omega^2a^2/ 2g = -4.7\,\mathrm{cm}$ , and the core adds another $4.7\,\mathrm{cm}$ : $9.3\,\mathrm{cm}$ deep. Tornado: $\Delta P = \tfrac12\rho v_{\max}^2$ outside plus the same inside: $\rho v_{\max}^2 = 1.2 \times 6400 = 7.7\,\mathrm{kPa}$ below ambient at the centre.

Two penstocks carry the water of a mountain reservoir down to a turbine house: a head of two hundred metres, which Bernoulli’s theorem turns into a jet at sixty metres per second.

3.5 Problem: The dam, the penstock and the turbine

Problem 3.1

Weekend problem — a hydroelectric plant from the reservoir to the jet: pressures, speeds, powers, losses and the water hammer

A reservoir holds water $150\,\mathrm{m}$ deep behind a dam; its free surface is $H = 200\,\mathrm{m}$ above the nozzles of a turbine. A penstock (a steel pipe) of diameter $D = 3.0\,\mathrm{m}$ and length $\ell = 400\,\mathrm{m}$ carries the design flow $D_V = 60\,\mathrm{m}^{3}/\mathrm{s}$ down to the turbine house, where a nozzle turns it into a free jet at atmospheric pressure $P_0 = 1.0\,\mathrm{bar}$ which strikes the buckets of a Pelton wheel. $\rho = 1000\,\mathrm{kg}/\mathrm{m}^{3}$ , $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$ ; the fluid is perfect unless stated otherwise.

Part I — At rest and at design flow.

Pressure at the base of the dam, and the force on a $1\,\mathrm{m}$ -wide vertical strip of the dam from top to bottom.
With the turbine valves closed, pressure in the penstock at the turbine house.
Speed in the penstock at design flow; dynamic pressure there; pressure at the turbine house end of the penstock (before the nozzle), by Bernoulli from the reservoir surface.
Speed of the free jet; section and diameter of the nozzle.
Kinetic power carried by the jet; show it equals $\rho gD_VH$ and compute it.
Mass and kinetic energy of the water contained in the penstock at design flow; transit time of a particle down the pipe.
Show that the head $H$ is shared, at design flow, between the pressure at the bottom of the penstock and the kinetic energy of the water, and that the nozzle converts the first into the second: what is the pressure just after the nozzle?
A perfect fluid has no head loss, a real one does: the loss in a pipe is $\Delta P = \lambda(\ell/D)\,\tfrac12\rho v^2$ with $\lambda \approx 0.015$ for smooth steel. Head loss in metres of water, and the fraction of the power lost.
An engineer proposes a $2\,\mathrm{m}$ penstock to save steel. Redo question 7; comment.

Part II — Measuring the flow. A Venturi is built into the penstock, with a throat of $2.0\,\mathrm{m}$ .

Speed at the throat and pressure drop between the pipe and the throat at design flow.
The drop is read on a mercury manometer ( $\rho_{\text{Hg}} = 13\,600\,\mathrm{kg}/\mathrm{m}^{3}$ ): height difference.
Show that the flow rate is proportional to the square root of the manometer reading, and give the reading at half the design flow.
Why must the pressure taps be flush with the wall and not protrude into the flow?
A section of the pipe runs, for topographic reasons, over a crest $6\,\mathrm{m}$ above the reservoir surface. Pressure there at design flow; what happens, and what must the designer do?

Part III — Unsteady regimes. The reservoir has a surface area $A = 2.0\,\mathrm{km}^{2}$ .

Rate at which the reservoir level falls at design flow, in centimetres per hour.
The turbine is stopped and the nozzle (section from question 4) is left open to the air: the reservoir drains by gravity. Write the mass balance and find the level $h(t)$ (quasi-steady Torricelli flow).
Time for the level to fall by $10\,\mathrm{m}$ ; compare with the time at constant design flow.
Starting the plant: the penstock is full, the nozzle is opened at $t = 0$ . Taking the speed uniform along the penstock and neglecting the nozzle’s geometry, integrate Euler along the pipe to obtain $\ell\,\dd v/\dd t = gH - v^2/2$ , solve it, and give the time constant.
During this start-up, at the instant when $v$ is half its final value, what is the pressure at the bottom of the penstock? Explain the sign of the difference with the steady value.

Part IV — Closing the valve.

The valve at the turbine is shut in $\tau = 10\,\mathrm{s}$ ; assume the water in the penstock decelerates uniformly. Using Euler along the pipe, estimate the over-pressure at the valve, in bars.
For a fast closure the water is not incompressible: a pressure wave travels up the pipe at $c = 1400\,\mathrm{m}/\mathrm{s}$ (the speed of sound in water in a steel pipe — Chapter 7), and the over-pressure is $\Delta P = \rho c\,\Delta v$ (Joukowsky). Value for a sudden stop from design flow; compare with the static pressure. How long does the wave take to reach the reservoir and back, and what does that say about the meaning of "fast"?
To protect the penstock a surge tank — a vertical open shaft of section $A_s = 100\,\mathrm{m}^{2}$ — is connected where a horizontal tunnel of length $L = 2000\,\mathrm{m}$ and section $A = 7.0\,\mathrm{m}^{2}$ from the reservoir joins the penstock. After a sudden closure, the water in the tunnel keeps moving and the level $z$ in the shaft rises. Write mass conservation between the tunnel and the shaft.
Integrate Euler along the tunnel (perfect fluid) to show that $z$ obeys $\ddot z + \dfrac{gA}{LA_s}z = 0$ ; period of the oscillation.
Maximum rise of the level in the shaft after a closure from design flow (energy argument: the kinetic energy of the tunnel water becomes potential energy in the shaft).
Sum up: the five pressures met in this plant (static at the dam, static at the turbine, dynamic in the pipe, water hammer, surge) and the law that gives each.

Solution

Solution of Problem 3.1.

1. $\rho g \times 150 = 14.7\,\mathrm{bar}$ above atmospheric; $\tfrac12\rho gH_d^2 \times 1\,\mathrm{m} = 1.1 \times 10^{8}\,\mathrm{N}$ .

2. $\rho gH = 19.6\,\mathrm{bar}$ gauge, $20.6\,\mathrm{bar}$ absolute.

3. $v = 60/7.07 = 8.5\,\mathrm{m}/\mathrm{s}$ ; $\tfrac12\rho v^2 = 0.36\,\mathrm{bar}$ ; $P = P_0 + \rho gH - \tfrac12\rho v^2 = 20.3\,\mathrm{bar}$ absolute.

4. $v_j = \sqrt{2gH} = 62.6\,\mathrm{m}/\mathrm{s}$ ; $s = 60/62.6 = 0.96\,\mathrm{m}^{2}$ , $d = 1.1\,\mathrm{m}$ .

5. $\tfrac12\rho D_Vv_j^2 = \tfrac12\rho D_V(2gH) = \rho gD_VH = 118\,\mathrm{MW}$ .

6. $m = \rho S\ell = 2.8 \times 10^{6}\,\mathrm{kg}$ , $E_k = \tfrac12mv^2 = 1.0 \times 10^{8}\,\mathrm{J}$ ; transit $\ell/v = 47\,\mathrm{s}$ .

7. $\rho gH = (P - P_0) + \tfrac12\rho v^2$ : $19.3\,\mathrm{bar}$ of pressure and $0.36\,\mathrm{bar}$ of kinetic energy; in the nozzle the pressure falls to $P_0$ while $\tfrac12\rho v^2$ rises to $\rho gH$ : just after it, $P = P_0$ .

8. $\Delta h = \lambda(\ell/D)v^2/2g = 0.015 \times 133 \times 3.67 = 7.3\,\mathrm{m}$ , $3.7\%$ of the power.

9. $v = 19.1\,\mathrm{m}/\mathrm{s}$ , $v^2/2g = 18.6\,\mathrm{m}$ , $\Delta h = 0.015 \times 200 \times 18.6 = 56\,\mathrm{m}$ : $28\%$ lost — the saving in steel is paid every second in energy.

10. $v_t = 60/\pi = 19.1\,\mathrm{m}/\mathrm{s}$ ; $\Delta P = 500(19.1^2 - 8.49^2) = 146\,\mathrm{kPa}$ .

11. $\Delta P = (\rho_{\text{Hg}} - \rho)g\,\Delta h$ : $\Delta h = 146000/(12600 \times 9.81) = 1.18\,\mathrm{m}$ .

12. $\Delta P \propto v^2 \propto D_V^2$ : $D_V \propto \sqrt{\Delta h}$ ; half the flow gives a quarter, $0.30\,\mathrm{m}$ .

13. A protruding tap faces or disturbs the flow and reads part of the dynamic pressure (a Pitot), not the static pressure.

14. $P = P_0 - \rho g \times 6 - \tfrac12\rho v^2 = 100 - 59 - 36 = 5\,\mathrm{kPa}$ : near the vapour pressure, dissolved air comes out and the water may boil (cavitation), breaking the column. Lower the crest, or fit an air valve and a vacuum pump, or reduce the flow.

15. $D_V/A = 60/2 \times 10^6 = 3 \times 10^{-5}\,\mathrm{m}/\mathrm{s} = 11\,\mathrm{cm}/\mathrm{h}$ .

16. $A\,\dd h/\dd t = -s\sqrt{2gh}$ : $\sqrt h = \sqrt{h_0} - (s/A)\sqrt{g/2}\,t$ .

17. $\Delta t = (A/s)\sqrt{2/g}(\sqrt{200} - \sqrt{190}) = 2.09 \times 10^6 \times 0.45 \times 0.36 = 3.4 \times 10^{5}\,\mathrm{s} = 3.9\,\mathrm{days}$ ; at constant design flow $10A/D_V = 3.3 \times 10^{5}\,\mathrm{s}$ : nearly the same, the nozzle having been sized for Torricelli’s flow at $h \approx H$ .

18. $\ell\,\dd v/\dd t = gH - v^2/2$ : $v = V\tanh(Vt/2\ell)$ , $V = 62.6\,\mathrm{m}/\mathrm{s}$ , time constant $2\ell/V = 13\,\mathrm{s}$ .

19. At $v = V/2$ : $\dd v/\dd t = (gH - V^2/8)/\ell = 3.7\,\mathrm{m}/\mathrm{s}^{2}$ ; $P = P_0 + \rho gH - \tfrac12\rho v^2 - \rho\ell\,\dd v/\dd t = 1.0 + 19.6 - 4.9 - 14.7 = 1.0\,\mathrm{bar}$ : most of the head is busy accelerating the column, so the pressure at the bottom is far below its steady value.

20. $|\dd v/\dd t| = 8.49/10 = 0.85\,\mathrm{m}/\mathrm{s}^{2}$ : $\Delta P = \rho\ell \times 0.85 = 3.4\,\mathrm{bar}$ .

21. $\Delta P = 1000 \times 1400 \times 8.49 = 119\,\mathrm{bar}$ , six times the static pressure. The wave’s round trip is $2\ell/c = 0.57\,\mathrm{s}$ : a closure faster than that is "sudden" and gets the full Joukowsky over-pressure; slower closures get less.

22. $Av = A_s\,\dd z/\dd t$ .

23. Euler along the tunnel, from the reservoir (pressure $P_0 + \rho g\times$ depth) to the foot of the shaft (pressure $P_0 + \rho g (\text{depth} + z)$ ): $\rho L\,\dd v/\dd t = -\rho gz$ ; with $v = (A_s/A)\dot z$ : $\ddot z + (gA/LA_s)z = 0$ ; $T = 2\pi\sqrt{LA_s/gA} = 2\pi\sqrt{2000 \times 100/ 68.7} = 340\,\mathrm{s} \approx 5.7\,\mathrm{min}$ .

24. $\tfrac12\rho ALv^2 = \tfrac12\rho gA_sz_{\max}^2$ with $v = 60/7 = 8.6\,\mathrm{m}/\mathrm{s}$ : $z_{\max} = v\sqrt{AL/gA_s} = 32\,\mathrm{m}$ .

25. Dam: hydrostatics, $\rho gh$ . Turbine, valves closed: the same, $\rho gH$ . Penstock in operation: Bernoulli, $\tfrac12\rho v^2$ taken from the static head. Water hammer: compressibility, $\rho c\Delta v$ . Surge: unsteady Euler along the tunnel, an oscillation of period $2\pi\sqrt{LA_s/gA}$ .

Terms defined in this chapter

Perfect fluid

See all 393 terms in the glossary