University Physics — Year 2 · Bachelor Year 2

5Momentum and Energy Balances in Flows

A firefighter braces against the hose: the water leaving at thirty metres per second pushes back with hundreds of newtons, though nothing touches the nozzle but water. A jet engine hangs from a wing and pulls an aircraft through the sky by throwing air backward. A garden sprinkler spins with no motor. In each case a fluid enters a region, leaves it with a different momentum, and the difference is a force. This chapter writes the laws of mechanics — momentum, angular momentum, energy — not for a fixed mass of fluid but for the fluid crossing a fixed region, an open system: the form in which engineers size turbines, rockets, pumps and pipes.

5.1 Balances for an open system

Definition 5.1 (Control volume; open system)

A control volume (or control surface) is a fixed region of space $\Sigma$ through which fluid flows; the fluid inside it at any instant is an open system, which exchanges mass with the outside through the inlets and outlets of $\Sigma$ . The closed system to which Newton’s laws apply is the fluid that is inside $\Sigma$ at time $t$ , followed to $t + \dd t$ : it then occupies $\Sigma$ minus what has left plus what has entered.

Theorem 5.2 (Momentum balance in steady flow)

For a steady flow through a control volume with one inlet (section $S_1$ , uniform velocity $\vect v_1$ ) and one outlet ( $S_2$ , $\vect v_2$ ), the mass flow rate $D_m$ being the same at both,

\sum\vect F_{\text{ext}} = D_m\,(\vect v_2 - \vect v_1) ,

where $\sum\vect F_{\text{ext}}$ is the sum of all external forces on the fluid inside $\Sigma$ : weight, forces from walls and bodies in contact with it, and the pressure forces on the inlet and outlet sections ( $P_1S_1\vect n_1$ pushing inward, $-P_2S_2\vect n_2$ at the outlet). With several inlets and outlets, the right side is the sum of the outgoing momentum fluxes $D_{m,i}\vect v_i$ minus the incoming ones.

Proof. Consider the closed system $\mathcal S$ made of the fluid in $\Sigma$ at time $t$ plus the slab $\delta m = D_m\dd t$ about to enter through $S_1$ . At $t + \dd t$ it consists of the fluid in $\Sigma$ plus the slab $\delta m$ that has left through $S_2$ . Its momentum went from $\vect p_\Sigma(t) + \delta m\,\vect v_1$ to $\vect p_\Sigma(t + \dd t) + \delta m\,\vect v_2$ ; in steady flow $\vect p_\Sigma$ is constant, so $\dd\vect p_{\mathcal S} = D_m\dd t\,(\vect v_2 - \vect v_1)$ . Newton’s second law for $\mathcal S$ , whose external forces are (to first order) those on the fluid in $\Sigma$ , gives the result. ∎

Left: a control volume in a steady flow — the momentum leaving through S_2 minus that entering through S_1 equals the sum of the external forces on the fluid inside, pressure forces on the two sections included. Right: a jet stopped by a wall pushes it with D_mv; a jet turned back by a Pelton bucket moving at u pushes with up to 2D_m(v - u). — Left: a control volume in a steady flow — the momentum leaving through $S_2$ minus that entering through $S_1$ equals the sum of the external forces on the fluid inside, pressure forces on the two sections included. Right: a jet stopped by a wall pushes it with $D_mv$ ; a jet turned back by a Pelton bucket moving at $u$ pushes with up to $2D_m(v - u)$ .

Example 5.3 (Jet on a wall; the fire hose)

A jet of section $s$ and speed $v$ strikes a wall perpendicularly and spreads along it: the outgoing momentum along the jet is zero, so the wall receives $F = D_mv = \rho sv^2$ — $1.1\,\mathrm{kN}$ for a $5\,\mathrm{cm}$ jet at $24\,\mathrm{m}/\mathrm{s}$ . By the same balance applied to the fluid in the hose and nozzle, the nozzle pushes back on the firefighter with $D_mv$ : $20\,\mathrm{L}/\mathrm{s}$ at $30\,\mathrm{m}/\mathrm{s}$ give $600\,\mathrm{N}$ , the weight of a person. If the wall recedes at $u$ , only $D_m' = \rho s(v - u)$ reaches it per unit time and it arrives at the relative speed $v - u$ : $F = \rho s(v - u)^2$ .

Proposition 5.4 (Pelton bucket; rocket thrust)

(i) A jet of speed $v$ and mass flow rate $D_m$ is turned back through $180^\circ$ by a bucket moving at $u$ in the jet’s direction: the force on the bucket is $F = 2D_m(v - u)$ (with $D_m$ the flow rate actually intercepted by the moving buckets of a wheel), the power $Fu$ is maximal for $u = v/2$ and then equals $\tfrac12D_mv^2$ : the jet’s whole kinetic power. (ii) A rocket of mass $m(t)$ ejecting gas backward at the relative speed $v_e$ and mass flow rate $D_m = -\dd m/\dd t$ feels the thrust $F = D_mv_e$ ; in free space its speed increases by $\Delta v = v_e\ln(m_0/m_1)$ (Tsiolkovsky’s rocket equation); under gravity, $m\,\dd v/\dd t = D_mv_e - mg$ .

Proof. (i) In the bucket’s frame the jet arrives at $v - u$ and leaves at $-(v - u)$ ; the momentum balance gives $2D_m(v - u)$ (the bucket is perfect: no loss of relative speed). Power $2D_mu(v - u)$ , maximal at $u = v/2$ . (ii) Momentum of rocket plus gas ejected in $\dd t$ : $(m + \dd m) (v + \dd v) + (-\dd m)(v - v_e) = mv$ in free space, so $m\,\dd v = -\dd m\,v_e$ and $v_1 - v_0 = v_e\ln(m_0/m_1)$ ; with gravity the total momentum changes by $-mg\,\dd t$ . ∎

Example 5.5 (A launcher)

First stage: $D_m = 2500\,\mathrm{kg}/\mathrm{s}$ at $v_e = 3000\,\mathrm{m}/\mathrm{s}$ : thrust $7.5\,\mathrm{MN}$ , which lifts a $600\,\mathrm{t}$ rocket at $7.5 \times 10^6/6 \times 10^5 - g = 2.7\,\mathrm{m}/\mathrm{s}^{2}$ off the pad. Burning $500\,\mathrm{t}$ in $200\,\mathrm{s}$ gives $\Delta v = 3000\ln6 - 9.81 \times 200 = 5400 - 2000 = 3.4\,\mathrm{km}/\mathrm{s}$ — the gravity loss is why launchers climb fast, and the logarithm why they are staged.

Example 5.6 (Force on a pipe bend)

Water at $P$ and $v$ in a pipe of section $S$ turns through $90^\circ$ . The balance on the fluid in the elbow: $\vect F_{\text{wall}} + PS\vect e_x - PS \vect e_y = D_m(v\vect e_y - v\vect e_x)$ , so the wall pushes the fluid with $\vect F_{\text{wall}} = -(PS + \rho Sv^2)(\vect e_x - \vect e_y)$ and the fluid pushes the elbow outward with $(PS + \rho Sv^2)\sqrt2$ . For $D = 20\,\mathrm{cm}$ , $P = 3\,\mathrm{bar}$ , $v = 3\,\mathrm{m}/\mathrm{s}$ : $PS = 9.4\,\mathrm{kN}$ , $\rho Sv^2 = 0.28\,\mathrm{kN}$ , force $13.7\,\mathrm{kN}$ along the bisector — the pressure term dominates, and it is what the anchor blocks of a pipeline hold.

5.2 Angular momentum: turbines and sprinklers

Theorem 5.7 (Angular momentum balance; Euler’s turbine equation)

In steady flow, the total moment about a point $O$ of the external forces on the fluid in $\Sigma$ equals the flux of angular momentum out minus in: $\sum\vect{\mathcal M}_O = D_m(\vect r_2\wedge\vect v_2 - \vect r_1 \wedge\vect v_1)$ . For a rotating machine (turbine or pump) of axis $Oz$ , with the fluid entering at radius $r_1$ with azimuthal velocity $v_{\theta1}$ and leaving at $r_2$ with $v_{\theta2}$ , the torque exerted by the fluid on the rotor and the power it delivers are

\Gamma = D_m(r_1v_{\theta1} - r_2v_{\theta2}) , \qquad \mathcal P = \Gamma\omega = D_m\omega(r_1v_{\theta1} - r_2v_{\theta2}) .

A turbine takes the swirl out of the fluid; a pump puts it in.

Proof. Same closed-system argument with $\vect r\wedge\vect v$ in place of $\vect v$ . The torque on the rotor is the opposite of the torque of the rotor on the fluid; the pressure forces on the inlet and outlet surfaces (which are surfaces of revolution) have no moment about the axis. ∎

Example 5.8 (The lawn sprinkler)

Two arms of length $R$ eject water tangentially at the relative speed $w$ ; the sprinkler turns at $\omega$ . Water enters on the axis with no angular momentum and leaves with $r v_\theta = R(w - R\omega)$ (absolute azimuthal speed $w - R\omega$ , opposite to the rotation): the torque on the arms is $D_mR(w - R\omega)$ . At rest it is $D_mRw$ ; with no friction the sprinkler accelerates until $R\omega = w$ — the water leaves with zero absolute speed and the torque vanishes; with a friction torque $\Gamma_f$ it settles at $\omega = (w - \Gamma_f/D_mR)/R$ .

Left: the lawn sprinkler — water leaves the arms tangentially at the relative speed w; the outgoing angular momentum turns the arms the other way. Right: a radial-inflow turbine (a Francis runner seen along its axis): the fluid enters at the rim with a large swirl and leaves near the axis with little; the difference of angular momentum flux is the torque on the runner. — Left: the lawn sprinkler — water leaves the arms tangentially at the relative speed $w$ ; the outgoing angular momentum turns the arms the other way. Right: a radial-inflow turbine (a Francis runner seen along its axis): the fluid enters at the rim with a large swirl and leaves near the axis with little; the difference of angular momentum flux is the torque on the runner.

5.3 Energy balance: pumps, turbines and losses

Theorem 5.9 (Mechanical energy balance of a steady flow)

For an incompressible fluid in steady flow through a machine (pump or turbine) between an inlet 1 and an outlet 2, with $\mathcal P_u$ the mechanical power delivered to the fluid by moving parts (positive for a pump, negative for a turbine) and $\mathcal P_{\text{diss}} \ge 0$ the power dissipated by viscosity,

D_V\Bigl[\bigl(P_2 + \tfrac12\rho v_2^2 + \rho gz_2\bigr) - \bigl(P_1 + \tfrac12\rho v_1^2 + \rho gz_1\bigr)\Bigr] = \mathcal P_u - \mathcal P_{\text{diss}} .

Dividing by $\rho gD_V$ gives the balance in heads (metres of fluid): $H_2 - H_1 = H_{\text{pump}} - H_{\text{loss}}$ , with $H = P/\rho g + v^2/2g + z$ . For a perfect fluid with no machine this is Bernoulli; the hydraulic power of a pump raising the head by $H_{\text{pump}}$ is $\rho gD_VH_{\text{pump}}$ .

Proof. Kinetic energy theorem for the closed system of the proof of Theorem 5.2: in $\dd t$ its kinetic energy changes by $D_m\dd t\,(v_2^2 - v_1^2)/2$ ; the work done on it is that of gravity, $-D_m\dd t\,g(z_2 - z_1)$ , of the pressure forces on the moving end faces, $(P_1S_1v_1 - P_2S_2v_2)\dd t = (P_1 - P_2)D_V\dd t$ , of the moving parts, $\mathcal P_u\dd t$ , and of the internal viscous forces, $-\mathcal P_{\text{diss}}\dd t$ (the walls, at rest, do no work). Divide by $\dd t$ . ∎

Example 5.10 (Sizing a pump)

Water is to be lifted $30\,\mathrm{m}$ at $20\,\mathrm{L}/\mathrm{s}$ through $100\,\mathrm{m}$ of $10\,\mathrm{cm}$ pipe ( $\lambda = 0.02$ ): $v = 2.5\,\mathrm{m}/\mathrm{s}$ , head loss $\lambda(L/D) v^2/2g = 0.02 \times 1000 \times 0.32 = 6.4\,\mathrm{m}$ , exit kinetic head $0.32\,\mathrm{m}$ : $H_{\text{pump}} = 36.7\,\mathrm{m}$ , hydraulic power $\rho gD_VH = 7.2\,\mathrm{kW}$ , and $10\,\mathrm{kW}$ of electricity at $70\%$ efficiency. The pressure at the pump outlet is $\rho gH_{\text{pump}} \approx 3.6\,\mathrm{bar}$ above the inlet.

Proposition 5.11 (Sudden expansion: the Borda–Carnot loss)

When a pipe of section $S_1$ opens suddenly into a pipe of section $S_2 > S_1$ , the jet spreads in a turbulent zone and the pressure rises downstream, but less than Bernoulli would say: the mechanical energy lost per unit volume is

\Delta P_{\text{loss}} = \tfrac12\rho(v_1 - v_2)^2 ,

the kinetic energy of the "lost" relative velocity. A flow leaving a pipe into a large tank ( $v_2 \to 0$ ) loses all its kinetic energy.

Proof. Momentum balance on the fluid between the expansion and a section downstream where the flow is again uniform; experiment shows the pressure on the annular wall at the expansion to be $P_1$ (the jet has not yet spread). Hence $(P_1 - P_2)S_2 = D_m(v_2 - v_1) = \rho S_2v_2(v_2 - v_1)$ , i.e. $P_2 - P_1 = \rho v_2(v_1 - v_2)$ . Bernoulli would give $\tfrac12\rho(v_1^2 - v_2^2)$ ; the difference is $\tfrac12\rho(v_1 - v_2)^2$ . ∎

Left: a sudden expansion — the jet spreads in a turbulent dead zone and part of its kinetic energy is dissipated. Right: the fractions of the upstream kinetic energy lost and recovered as pressure, against the section ratio; the loss is largest for a jet into a tank.

Method 5.12 (Setting up a balance)

(1) Draw the control volume: its boundary should cut the flow only where the velocity and pressure are known or wanted (uniform sections, free jets at $P_0$ , free surfaces). (2) List the external forces on the fluid inside: weight, wall forces (unknown, the usual target), pressure on the cut sections. (3) Write the momentum flux out minus in; for moving parts, work in the frame of the part if the flow is steady there. (4) For torques, the angular-momentum balance; for powers, the energy balance in heads. (5) The force on the wall is minus the force of the wall on the fluid; add the atmospheric pressure acting on the outside of the device if the net force on the device is wanted.

5.4 Exercises

Exercise 5.1 ★

A fire hose delivers $25\,\mathrm{L}/\mathrm{s}$ through a nozzle of $3.0\,\mathrm{cm}$ diameter. Exit speed; reaction force on the nozzle; force of the jet on a wall it hits perpendicularly; on a wall receding at $10\,\mathrm{m}/\mathrm{s}$ .

Solution

Solution of Exercise 5.1.

$s = 7.1\,\mathrm{cm}^{2}$ , $v = 0.025/7.1 \times 10^{-4} = 35\,\mathrm{m}/\mathrm{s}$ ; $F = D_mv = 25 \times 35 = 880\,\mathrm{N}$ on the nozzle and on the wall; receding wall: $\rho s(v - u)^2 = 1000 \times 7.1 \times 10^{-4} \times 25.4^2 = 460\,\mathrm{N}$ .

Exercise 5.2 ★

A rocket engine ejects $250\,\mathrm{kg}/\mathrm{s}$ at $3.0\,\mathrm{km}/\mathrm{s}$ . Thrust; initial acceleration of a $60\,\mathrm{t}$ rocket; acceleration when $40\,\mathrm{t}$ of propellant have burned; speed gained after that burn (free space, then with gravity, duration $160\,\mathrm{s}$ ).

Solution

Solution of Exercise 5.2.

$F = 250 \times 3000 = 750\,\mathrm{kN}$ ; $a_0 = 750/60 - 9.81 = 2.7\,\mathrm{m}/\mathrm{s}^{2}$ ; at $20\,\mathrm{t}$ : $37.5 - 9.8 = 28\,\mathrm{m}/\mathrm{s}^{2}$ . $\Delta v = 3000\ln3 = 3.3\,\mathrm{km}/\mathrm{s}$ in free space; minus $g\tau = 1.6\,\mathrm{km}/\mathrm{s}$ : $1.7\,\mathrm{km}/\mathrm{s}$ .

Exercise 5.3 ★

Water flows at $2.0\,\mathrm{m}/\mathrm{s}$ and $4.0\,\mathrm{bar}$ in a $15\,\mathrm{cm}$ pipe that turns through $90{}^{\circ}$ . Force exerted by the water on the elbow (magnitude and direction); which part comes from the pressure, which from the momentum change? Same elbow with the pipe open to the air just after the bend.

Solution

Solution of Exercise 5.3.

$S = 177\,\mathrm{cm}^{2}$ : $PS = 7.1\,\mathrm{kN}$ , $\rho Sv^2 = 71\,\mathrm{N}$ . Each component of the force on the elbow is $PS + \rho Sv^2 = 7.1\,\mathrm{kN}$ : $10\,\mathrm{kN}$ along the outward bisector, $99\%$ of it pressure. Open outlet (gauge pressure zero there): $7.1$ kN along the inlet direction and $71\,\mathrm{N}$ along the outlet: $7.1\,\mathrm{kN}$ , almost along the inlet axis.

Exercise 5.4 ★

A jet engine on a test stand swallows $80\,\mathrm{kg}/\mathrm{s}$ of still air and exhausts it at $550\,\mathrm{m}/\mathrm{s}$ (fuel mass negligible). Thrust. In flight at $240\,\mathrm{m}/\mathrm{s}$ , same exhaust speed relative to the engine: thrust, propulsive power, and the kinetic power given to the air in the engine frame; propulsive efficiency.

Solution

Solution of Exercise 5.4.

$F = 80 \times 550 = 44\,\mathrm{kN}$ . Flight: $F = 80(550 - 240) = 25\,\mathrm{kN}$ , $Fv = 6.0\,\mathrm{MW}$ ; $\mathcal P_{\text{kin}} = \tfrac12 \times 80 \times (550^2 - 240^2) = 9.8\,\mathrm{MW}$ ; $\eta_p = 0.61 = 2 \times 240/790$ .

Exercise 5.5 ★★

Pelton wheel. A jet of $0.50\,\mathrm{m}^{3}/\mathrm{s}$ at $60\,\mathrm{m}/\mathrm{s}$ hits the buckets of a wheel of radius $0.80\,\mathrm{m}$ . (a) Force and power on the buckets at bucket speed $u$ (perfect $180^\circ$ deflection); optimal $u$ and the corresponding rotation speed in rpm. (b) Power and efficiency at the optimum. (c) Real buckets deflect the jet by $165{}^{\circ}$ : force and power at $u = v/2$ ; efficiency. (d) Torque at the optimum; what happens to the torque at standstill and at the runaway speed?

Solution

Solution of Exercise 5.5.

$D_m = 500\,\mathrm{kg}/\mathrm{s}$ . (a) $F = 1000(60 - u)$ N, $\mathcal P = 1000u(60 - u)$ ; $u = 30\,\mathrm{m}/\mathrm{s}$ , $\omega = 37.5$ rad/s $= 360\,\mathrm{rpm}$ . (b) $900\,\mathrm{kW}$ $= \tfrac12D_mv^2$ : $100\%$ . (c) $F = D_m(v - u)(1 + \cos15^\circ) = 29.5\,\mathrm{kN}$ , $\mathcal P = 885\,\mathrm{kW}$ , $98\%$ . (d) $\Gamma = FR = 24\,\mathrm{kN}\,\mathrm{m}$ ; at standstill the force doubles ( $\Gamma = 48\,\mathrm{kN}\,\mathrm{m}$ ); at $u = v$ force and torque vanish.

Exercise 5.6 ★★

Sprinkler. Two arms of $15\,\mathrm{cm}$ end in $3.0\,\mathrm{mm}$ nozzles; the total flow is $0.20\,\mathrm{L}/\mathrm{s}$ . (a) Relative exit speed. (b) Torque at rest. (c) Rotation rate with a bearing friction torque of $5.0 \times 10^{-3}\,\mathrm{N}\,\mathrm{m}$ ; the limiting rate with no friction. (d) Absolute speed of the water leaving, in each case.

Solution

Solution of Exercise 5.6.

(a) $s = 7.1\,\mathrm{mm}^{2}$ per nozzle, $0.10\,\mathrm{L}/\mathrm{s}$ each: $w = 14\,\mathrm{m}/\mathrm{s}$ . (b) $\Gamma_0 = D_mRw = 0.2 \times 0.15 \times 14 = 0.42\,\mathrm{N}\,\mathrm{m}$ . (c) $D_mR(w - R\omega) = 5 \times 10^{-3}$ : $w - R\omega = 0.17\,\mathrm{m}/\mathrm{s}$ , $\omega = 93\,\mathrm{rad}/\mathrm{s} \approx 890\,\mathrm{rpm}$ ; without friction $R\omega = w$ : $900\,\mathrm{rpm}$ . (d) $0.17\,\mathrm{m}/\mathrm{s}$ backward, then zero: the water drops straight down.

Exercise 5.7 ★★

Hovering. A helicopter of $2.0\,\mathrm{t}$ hovers with a rotor of $10\,\mathrm{m}$ diameter. Model the rotor as a disk that takes in still air and pushes it down through a uniform speed $v_i$ at the disk, $2v_i$ far below (admit the factor 2). (a) Thrust in terms of $\rho$ , $A$ , $v_i$ ; value of $v_i$ . (b) Power given to the air (its kinetic power far below). (c) Why does a larger rotor need less power? (d) Same calculation for a $1\,\mathrm{kg}$ drone with four $25\,\mathrm{cm}$ propellers; power-to-weight ratio compared.

Solution

Solution of Exercise 5.7.

(a) $D_m = \rho Av_i$ , momentum flux $D_m\cdot2v_i$ : $T = 2\rho Av_i^2$ ; $A = 78.5\,\mathrm{m}^{2}$ , $T = 19.6\,\mathrm{kN}$ : $v_i = 10\,\mathrm{m}/\mathrm{s}$ . (b) $\mathcal P = \tfrac12D_m(2v_i)^2 = Tv_i = 200\,\mathrm{kW}$ . (c) $\mathcal P = T^{3/2}/\sqrt{2\rho A}$ : doubling the area divides the induced power by $\sqrt2$ — a bigger rotor moves more air more slowly. (d) $A = 0.20\,\mathrm{m}^{2}$ , $v_i = \sqrt{9.81/0.47} = 4.6\,\mathrm{m}/\mathrm{s}$ , $\mathcal P = 45\,\mathrm{W}$ : $45\,\mathrm{W}/\mathrm{kg}$ against $100\,\mathrm{W}/\mathrm{kg}$ — the disk loading $T/A$ is lower.

Exercise 5.8 ★★

Sudden expansion. Water at $4.0\,\mathrm{m}/\mathrm{s}$ in a $5\,\mathrm{cm}$ pipe enters a $10\,\mathrm{cm}$ pipe. (a) Downstream speed; pressure rise according to Bernoulli and according to Borda–Carnot; loss in pascals and in metres of head. (b) The same expansion made gradual (a diffuser) recovers $80\%$ of the Bernoulli rise: loss. (c) Loss when the $5\,\mathrm{cm}$ pipe discharges into a large tank. (d) Power lost in case (a) and (c) at the given flow.

Solution

Solution of Exercise 5.8.

(a) $v_2 = 1.0\,\mathrm{m}/\mathrm{s}$ ; Bernoulli $\tfrac12\rho(16 - 1) = 7.5\,\mathrm{kPa}$ ; actual $\rho v_2(v_1 - v_2) = 3.0\,\mathrm{kPa}$ ; loss $4.5\,\mathrm{kPa}$ $= 0.46\,\mathrm{m}$ . (b) $6$ kPa recovered, $1.5\,\mathrm{kPa}$ lost. (c) $\tfrac12\rho v_1^2 = 8\,\mathrm{kPa}$ , $0.82\,\mathrm{m}$ . (d) $D_V = 7.9\,\mathrm{L}/\mathrm{s}$ : $35\,\mathrm{W}$ and $63\,\mathrm{W}$ .

Exercise 5.9 ★★

A pump draws water from a well $5.0\,\mathrm{m}$ below it and delivers $10\,\mathrm{L}/\mathrm{s}$ to a tank $25\,\mathrm{m}$ above it through $60\,\mathrm{m}$ of $6\,\mathrm{cm}$ pipe ( $\lambda = 0.025$ , plus a singular loss of $2\,v^2/2g$ for the fittings). (a) Speed and losses. (b) Pump head and hydraulic power; shaft power at $65\%$ efficiency. (c) Pressure at the pump inlet (suction side) and outlet. (d) The water’s vapour pressure is $2.3\,\mathrm{kPa}$ : maximum height of the pump above the well.

Solution

Solution of Exercise 5.9.

(a) $v = 0.01/2.83 \times 10^{-3} = 3.5\,\mathrm{m}/\mathrm{s}$ , $v^2/2g = 0.64\,\mathrm{m}$ ; friction $0.025 \times 1000 \times 0.64 = 16\,\mathrm{m}$ , fittings $1.3\,\mathrm{m}$ : $17.3\,\mathrm{m}$ . (b) $H = 30 + 17.3 + 0.64 = 48\,\mathrm{m}$ ; $\rho gD_VH = 4.7\,\mathrm{kW}$ ; $7.2\,\mathrm{kW}$ at the shaft. (c) Inlet (no suction losses): $P_0 - \rho g(5 + 0.64) = 45\,\mathrm{kPa}$ ; outlet $45 + 470 = 515\,\mathrm{kPa}$ . (d) $P_{\text{in}} \ge 2.3\,\mathrm{kPa}$ : $h \le (100 - 2.3)/9.81 - 0.64 = 9.3\,\mathrm{m}$ (less with suction losses; $7$ – $8\,\mathrm{m}$ in practice).

Exercise 5.10 ★★★

Water rocket. A $2.0\,\mathrm{L}$ bottle holds $1.0\,\mathrm{L}$ of water and air at $5.0\,\mathrm{bar}$ (absolute); the nozzle has a diameter of $2.2\,\mathrm{cm}$ ; empty bottle $0.10\,\mathrm{kg}$ . (a) Initial exit speed (Bernoulli, water surface at rest) and thrust; initial acceleration. (b) As water leaves the air expands isothermally: pressure and exit speed when half the water is gone. (c) Estimate the burn duration and the speed reached (neglect gravity and drag, treat the thrust as its mean value). (d) What does the empty bottle do once the water is gone, and why does a little water (not a lot) give the best height?

Solution

Solution of Exercise 5.10.

(a) $v = \sqrt{2 \times 4 \times 10^5/1000} = 28\,\mathrm{m}/\mathrm{s}$ ; $s = 3.8\,\mathrm{cm}^{2}$ , $D_m = 10.8\,\mathrm{kg}/\mathrm{s}$ , $F = D_mv = 2\Delta P\,s = 300\,\mathrm{N}$ ; $a = 300/1.1 - 9.8 = 260\,\mathrm{m}/\mathrm{s}^{2}$ . (b) Air at $1.5\,\mathrm{L}$ : $P = 3.3\,\mathrm{bar}$ , $v = \sqrt{2 \times 2.3 \times 10^5/1000} = 22\,\mathrm{m}/\mathrm{s}$ . (c) Mean exit speed $\approx 22\,\mathrm{m}/\mathrm{s}$ : burn $\approx 10^{-3} /(3.8 \times 10^{-4} \times 22) = 0.12\,\mathrm{s}$ ; rocket equation with $v_e \approx 22\,\mathrm{m}/\mathrm{s}$ : $\Delta v \approx 22\ln(1.1/0.1) = 50\,\mathrm{m}/\mathrm{s}$ . (d) The remaining $2.5\,\mathrm{bar}$ of air blows out and adds a little; too much water leaves little air and the pressure collapses early, too little leaves little mass to throw — about a third of the volume is best.

Exercise 5.11 ★★★

Francis turbine. Water enters a runner at $r_1 = 1.0\,\mathrm{m}$ with $v_{\theta1} = 20\,\mathrm{m}/\mathrm{s}$ and leaves at $r_2 = 0.50\,\mathrm{m}$ with no swirl; flow $30\,\mathrm{m}^{3}/\mathrm{s}$ ; rotation $150\,\mathrm{rpm}$ . (a) Torque and power. (b) Head extracted, $H = \mathcal P/\rho gD_V$ . (c) If the runner ran at $200\,\mathrm{rpm}$ with the same inlet velocity and the exit swirl became $v_{\theta2} = 3\,\mathrm{m}/\mathrm{s}$ in the direction of rotation: power and the fraction of the head turned into useless exit swirl. (d) Why is the Pelton wheel used for high heads and small flows, the Francis for the opposite?

Solution

Solution of Exercise 5.11.

(a) $\Gamma = D_mr_1v_{\theta1} = 3 \times 10^4 \times 20 = 600\,\mathrm{kN}\,\mathrm{m}$ , $\omega = 15.7$ rad/s, $\mathcal P = 9.4\,\mathrm{MW}$ . (b) $H = \mathcal P/\rho gD_V = 32\,\mathrm{m}$ . (c) $\Gamma = 3 \times 10^4(20 - 1.5) = 555\,\mathrm{kN}\,\mathrm{m}$ , $\mathcal P = 11.6\,\mathrm{MW}$ (head $39\,\mathrm{m}$ ); the exit swirl carries $v_{\theta2}^2/2g = 0.46\,\mathrm{m}$ , about $1\%$ . (d) A Pelton jet is at atmospheric pressure: it needs a high head to be fast, and its flow is limited by the nozzle; a Francis runner is full of water under pressure and passes large flows at moderate heads.

Exercise 5.12 ★★★

Hydraulic jump. In a horizontal channel of unit width, water of depth $h_1$ flowing at $v_1$ jumps abruptly to depth $h_2$ and speed $v_2$ (the foamy step below a weir or in a kitchen sink). (a) Mass conservation. (b) Momentum balance on the fluid between the two sides of the jump, the pressure being hydrostatic on each side: show $\tfrac12gh_1^2 + v_1^2h_1 = \tfrac12gh_2^2 + v_2^2h_2$ . (c) Deduce $h_2/h_1 = \tfrac12\bigl(-1 + \sqrt{1 + 8\,\mathrm{Fr}_1^2}\bigr)$ with $\mathrm{Fr}_1 = v_1/\sqrt{gh_1}$ (the Froude number); a jump needs $\mathrm{Fr}_1 > 1$ . (d) Head lost in the jump, $\Delta H = (h_2 - h_1)^3/4h_1h_2$ ; numbers for $h_1 = 0.20\,\mathrm{m}$ , $v_1 = 4.0\,\mathrm{m}/\mathrm{s}$ , and the power dissipated per metre of width.

Solution

Solution of Exercise 5.12.

(a) $v_1h_1 = v_2h_2 = q$ . (b) Hydrostatic force per unit width on a section $\int_0^h\rho g(h - z)\dd z = \tfrac12\rho gh^2$ ; momentum: $\tfrac12\rho g(h_1^2 - h_2^2) = \rho q(v_2 - v_1)$ ; divide by $\rho$ and use $q = v_1h_1 = v_2h_2$ . (c) With $v_2 = v_1h_1/h_2$ and $h_1 \ne h_2$ : $\tfrac12g(h_1 + h_2)h_2 = v_1^2h_1$ , i.e. $(h_2/h_1)^2 + h_2/h_1 - 2\mathrm{Fr}_1^2 = 0$ ; $h_2 > h_1$ needs $\mathrm{Fr}_1 > 1$ . (d) $\Delta H = (h_1 + v_1^2/2g) - (h_2 + v_2^2/2g) = (h_2 - h_1)^3/4h_1h_2$ . $\mathrm{Fr}_1 = 2.86$ , $h_2/h_1 = 3.57$ , $h_2 = 0.71\,\mathrm{m}$ , $v_2 = 1.1\,\mathrm{m}/\mathrm{s}$ , $\Delta H = 0.514^3/0.57 = 0.24\,\mathrm{m}$ ; $\rho gq\,\Delta H = 1.9\,\mathrm{kW}$ per metre of width.

The runner of a Pelton turbine: each bucket turns the jet back through nearly 180, and the change of the water’s momentum is the force on the wheel. — The runner of a Pelton turbine: each bucket turns the jet back through nearly $180^\circ$ , and the change of the water’s momentum is the force on the wheel.

The penstocks of a hydroelectric plant feed the turbines below: the momentum balance of the jet on the buckets of a Pelton wheel turns the head of water into shaft work.

5.5 Problem: Jets, rockets and runners

Problem 5.1

Weekend problem — what pushes a jet aircraft, what lifts a rocket, what turns a turbine, and what it costs to pump water uphill

Part I — The turbojet. An engine on a test stand takes in $D_a = 50\,\mathrm{kg}/\mathrm{s}$ of still air, burns $D_f = 1.0\,\mathrm{kg}/\mathrm{s}$ of fuel, and exhausts the mixture at $v_e = 600\,\mathrm{m}/\mathrm{s}$ at atmospheric pressure.

Thrust on the test stand.
In flight at $v = 250\,\mathrm{m}/\mathrm{s}$ (same $v_e$ relative to the engine): thrust, and the propulsive power $Fv$ .
Kinetic power given to the gas, computed in the engine’s frame. Propulsive efficiency $\eta_p = Fv/\mathcal P_{\text{kin}}$ ; show that, neglecting the fuel mass, $\eta_p = 2v/(v + v_e)$ .
The fuel releases $43\,\mathrm{MJ}/\mathrm{kg}$ : thermal efficiency of the engine (kinetic power over heat power) and overall efficiency.
A turbofan uses the same kinetic power to accelerate $500\,\mathrm{kg}/\mathrm{s}$ of air: exhaust speed, thrust and $\eta_p$ at $250\,\mathrm{m}/\mathrm{s}$ . Why do airliners have huge fans?
Why can a turbojet not work at rest on a rocket’s job, above the atmosphere?
Reverse thrust on landing deflects the fan flow forward at $45{}^{\circ}$ from the axis: braking force for the turbofan of question 5 at $70\,\mathrm{m}/\mathrm{s}$ , the fan exhaust being then $150\,\mathrm{m}/\mathrm{s}$ relative to the engine; compare with the forward thrust the same flow would give.

Part II — The rocket. A single-stage rocket: initial mass $m_0 = 300\,\mathrm{t}$ , propellant $200\,\mathrm{t}$ , exhaust speed $v_e = 3.0\,\mathrm{km}/\mathrm{s}$ , burn time $\tau = 150\,\mathrm{s}$ at constant mass flow rate.

Apply the momentum balance to rocket plus the gas ejected in $\dd t$ to derive $m\,\dd v/\dd t = D_mv_e - mg$ for a vertical flight.
Mass flow rate, thrust, and the acceleration at lift-off and at burn-out.
Integrate to get the burn-out speed; what is the "gravity loss"?
Burn-out speed with the same rocket split into two stages of $100\,\mathrm{t}$ of propellant each, the first stage’s empty mass ( $20\,\mathrm{t}$ ) being dropped before the second ignites (neglect gravity here): compare with the single stage.
What exhaust speed would a single stage need to reach orbital speed ( $7.8\,\mathrm{km}/\mathrm{s}$ , neglecting gravity) with the same mass ratio? Comment on the choice of propellants.
At lift-off the exhaust jet ( $1333\,\mathrm{kg}/\mathrm{s}$ at $3\,\mathrm{km}/\mathrm{s}$ ) hits the flame deflector of the pad and is turned horizontally: force on the deflector.
The rocket’s exhaust is also a fluid flow: why does the thrust increase with altitude for a real nozzle (think of the pressure on the nozzle exit section)?

Part III — The Pelton wheel. The jet of the previous chapter’s plant: $D_V = 60\,\mathrm{m}^{3}/\mathrm{s}$ at $v = 62.6\,\mathrm{m}/\mathrm{s}$ , split between two wheels; each wheel has a radius $R = 1.5\,\mathrm{m}$ and buckets deflecting the jet by $165{}^{\circ}$ .

Force on the buckets of one wheel at bucket speed $u$ , and the power; optimal $u$ .
Rotation speed at the optimum, in rpm; torque on the shaft.
The generator must turn at a multiple of $50\,\mathrm{Hz}$ divided by its number of pole pairs: choose the number of pole pairs.
Power of one wheel and efficiency with respect to the jet’s kinetic power; total electrical power at $95\%$ generator efficiency.
Using the angular-momentum balance (Euler’s turbine equation) on a bucket at the optimum: angular momentum of the water entering and leaving, per kilogram, and check the torque.
The load drops suddenly and the wheel runs away toward $u = v$ : what happens to the force and to the torque, and why is a deflector that diverts the jet part of every Pelton plant?

Part IV — Pumping uphill. The plant is also used as storage: at night $40\,\mathrm{m}^{3}/\mathrm{s}$ are pumped back up through the penstock ( $400\,\mathrm{m}$ , $3\,\mathrm{m}$ , $\lambda = 0.015$ ) to the reservoir $200\,\mathrm{m}$ above.

Head losses in the penstock at $40\,\mathrm{m}^{3}/\mathrm{s}$ ; pump head needed.
Hydraulic power, and electrical power drawn at $88\%$ pump efficiency.
Pressure at the bottom of the penstock while pumping; compare with its value in turbine mode (previous chapter).
Energy stored per night of $8\,\mathrm{h}$ (potential energy of the water raised); round-trip efficiency of the storage if the turbine mode recovers $90\%$ of the available head and the generator $95\%$ .
Sum up the four balances used in this problem (momentum, angular momentum, energy, mass) and the device each one sized.

Solution

Solution of Problem 5.1.

1. $F = (D_a + D_f)v_e = 51 \times 600 = 30.6\,\mathrm{kN}$ .

2. $F = 51 \times 600 - 50 \times 250 = 18.1\,\mathrm{kN}$ ; $Fv = 4.5\,\mathrm{MW}$ .

3. $\mathcal P_{\text{kin}} = \tfrac12 \times 51 \times 600^2 - \tfrac12 \times 50 \times 250^2 = 7.6\,\mathrm{MW}$ ; $\eta_p = 0.59$ . With $D_f = 0$ : $\eta_p = 2v(v_e - v)/(v_e^2 - v^2) = 2v/(v_e + v) = 500/850$ .

4. Heat $43\,\mathrm{MW}$ : thermal $7.6/43 = 18\%$ ; overall $4.5/43 = 10\%$ .

5. $\tfrac12 \times 500(v_e'^2 - 250^2) = 7.6 \times 10^6$ : $v_e' = 305\,\mathrm{m}/\mathrm{s}$ ; $F = 500 \times 55 = 27.5\,\mathrm{kN}$ ; $\eta_p = 500/555 = 0.90$ . Moving more air more slowly gives more thrust for the same power: hence the fans.

6. It needs the outside air both as working fluid and as oxidizer; a rocket carries both.

7. Air enters at $70\,\mathrm{m}/\mathrm{s}$ (backward in the engine’s frame) and leaves with a forward component $150\cos45^\circ = 106\,\mathrm{m}/\mathrm{s}$ : the momentum given to the air is forward, $500(106 + 70) = 88\,\mathrm{kN}$ of braking; the same flow exhausted backward would give $500(150 - 70) = 40\,\mathrm{kN}$ of thrust.

8. Over $\dd t$ the system (rocket $m$ + gas $D_m\dd t$ ejected at $v - v_e$ ): $(m - D_m\dd t)(v + \dd v) + D_m\dd t(v - v_e) - mv = -mg\,\dd t$ , hence $m\,\dd v/\dd t = D_mv_e - mg$ .

9. $D_m = 1333\,\mathrm{kg}/\mathrm{s}$ ; $F = 4.0\,\mathrm{MN}$ ; $a = 13.3 - 9.8 = 3.5\,\mathrm{m}/\mathrm{s}^{2}$ at lift-off, $40 - 9.8 = 30\,\mathrm{m}/\mathrm{s}^{2}$ at burn-out.

10. $v = v_e\ln(m_0/m) - gt$ : $3000\ln3 - 9.81 \times 150 = 3296 - 1472 = 1.8\,\mathrm{km}/\mathrm{s}$ ; the gravity loss is $1.5\,\mathrm{km}/\mathrm{s}$ .

11. Stage 1: $3000\ln(300/200) = 1.2\,\mathrm{km}/\mathrm{s}$ ; drop $20\,\mathrm{t}$ ; stage 2: $3000\ln(180/80) = 2.4\,\mathrm{km}/\mathrm{s}$ : $3.6\,\mathrm{km}/\mathrm{s}$ against $3.3\,\mathrm{km}/\mathrm{s}$ for the single stage (with gravity the gap widens).

12. $v_e = 7800/\ln3 = 7.1\,\mathrm{km}/\mathrm{s}$ : no chemical propellant reaches it (hydrogen–oxygen gives $4.5\,\mathrm{km}/\mathrm{s}$ ); staging is compulsory.

13. The jet’s vertical momentum flux $D_mv = 4.0\,\mathrm{MN}$ is removed and an equal horizontal one created: a $4\,\mathrm{MN}$ downward force plus $4\,\mathrm{MN}$ sideways, $5.7\,\mathrm{MN}$ in all.

14. The balance on the engine includes $(P_e - P_0)S_e$ on the exit section; as $P_0$ falls with altitude this term grows: the thrust rises by some $10$ – $15\%$ in vacuum.

15. $D_m = 3 \times 10^{4}\,\mathrm{kg}/\mathrm{s}$ per wheel: $F = D_m(v - u)(1 + \cos15^\circ) = 1.97 \times 3 \times 10^4(62.6 - u)$ ; $\mathcal P = Fu$ , maximal at $u = v/2 = 31.3\,\mathrm{m}/\mathrm{s}$ .

16. $\omega = u/R = 20.9$ rad/s $= 199\,\mathrm{rpm}$ ; $F = 1.85\,\mathrm{MN}$ , $\Gamma = FR = 2.8\,\mathrm{MN}\,\mathrm{m}$ .

17. $f = pn/60$ : $p = 3000/199 \approx 15$ pole pairs, $n = 200\,\mathrm{rpm}$ .

18. $\mathcal P = Fu = 58\,\mathrm{MW}$ ; jet power $\tfrac12D_mv^2 = 59\,\mathrm{MW}$ : $98\%$ ; $2 \times 58 \times 0.95 = 110\,\mathrm{MW}$ .

19. In: $Rv = 94\,\mathrm{m}^{2}/\mathrm{s}$ per kg. Out: absolute azimuthal speed $u - (v - u)\cos15^\circ = 1.1\,\mathrm{m}/\mathrm{s}$ , i.e. $1.6\,\mathrm{m}^{2}/\mathrm{s}$ . $\Gamma = D_m(94 - 1.6) = 2.8\,\mathrm{MN}\,\mathrm{m}$ — as in 16.

20. As $u \to v$ the relative speed, the force and the torque vanish: nothing brakes the wheel, which overspeeds; the deflector throws the jet off the buckets within a second while the needle valve closes slowly enough to avoid water hammer.

21. $v = 5.7\,\mathrm{m}/\mathrm{s}$ , $v^2/2g = 1.6\,\mathrm{m}$ ; loss $0.015 \times 133 \times 1.6 = 3.3\,\mathrm{m}$ ; pump head $200 + 3.3 + 1.6 = 205\,\mathrm{m}$ .

22. $\rho gD_VH = 9810 \times 40 \times 205 = 80\,\mathrm{MW}$ ; $91\,\mathrm{MW}$ electrical.

23. $P = P_0 + \rho g(200 + 3.3) + \tfrac12\rho v^2 \approx 21\,\mathrm{bar}$ absolute, against $20.3\,\mathrm{bar}$ in turbine mode: the losses now add to the static head instead of subtracting.

24. $V = 40 \times 28800 = 1.15 \times 10^{6}\,\mathrm{m}^{3}$ , $E = \rho gV \times 200 = 2.3 \times 10^{12}\,\mathrm{J} = 630\,\mathrm{MWh}$ ; consumed $91 \times 8 = 730\,\mathrm{MWh}$ ; recovered $630 \times 0.9 \times 0.95 = 540\,\mathrm{MWh}$ : round trip $74\%$ .

25. Momentum: the jet engine’s and the rocket’s thrust, the Pelton force. Angular momentum: the turbine torque (Euler). Energy: the pump head and power, the storage efficiency. Mass: every flow rate, the rocket’s mass loss.

Terms defined in this chapter

Control volume; open system

See all 393 terms in the glossary