Integral Analysis for a Control Volume: Reynolds Transport Theorem, Mass and Linear Momentum

This chapter answers Section XE1.4 of XE1 Fluid Mechanics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam: “Reynolds Transport Theorem (RTT) for conservation of mass and linear momentum.” It teaches the difference between a system and a control volume, the Reynolds Transport Theorem and how it produces the mass and linear-momentum equations, continuity for steady and unsteady flow (mass flow, volume flow, average velocity), and the momentum equation for fixed, moving and accelerating control volumes. It also says where the energy equation takes over (the Bernoulli equation itself belongs to Section XE1.5). The numericals are tank filling and draining, average velocity from a profile, forces on pipe bends, nozzles and sudden expansions, jets on plates and vanes, and rocket thrust and speed.

1. System, control volume and the Reynolds Transport Theorem

A system is a fixed identity of matter: the same fluid particles, however they move or deform, and no mass crosses its boundary. A control volume (CV) is a region of space chosen by the analyst, bounded by a control surface (CS) through which fluid may cross. The basic laws (mass is conserved, F = d(mV)/dt, the first law) are stated for a system, but in a pipe, pump or nozzle we cannot follow particles; we can only watch what crosses fixed boundaries. The Reynolds Transport Theorem converts the system statement into a control-volume statement.

Let B be any extensive property of the fluid (mass, momentum, energy) and b = B per unit mass its intensive form. For a CV that is fixed in space:
dBsys/dt = (d/dt) ∫_CV ρ b dV + ∫_CS ρ b (V · n) dA
The first term is storage: the rate at which B accumulates inside the CV (zero in steady flow). The second is the net efflux: B carried out minus B carried in across the CS, where n is the outward unit normal, so V · n > 0 on outflow faces and < 0 on inflow faces. For a CV moving at constant velocity, V in the flux term is replaced by the velocity relative to the control surface, Vr = V − Vcv.

Same theorem, three choices of b
LawBbResult for the CV
Massm10 = d/dt ∫ρ dV + ∫ρ V · n dA
Linear momentummVVΣF = d/dt ∫ρ V dV + ∫ρ V (V · n) dA
EnergyEeEnergy equation (first law for a CV)
ℹ️ Where the energy equation is used
Mass and momentum balances cannot give the pressure change along a flow without knowing the force of the walls. The energy equation (b = e) supplies that missing relation. For steady, incompressible, frictionless flow along a streamline it reduces to the Bernoulli equation, which is Section XE1.5; with pump work and head loss it is the energy equation of internal flows. In this chapter, when a numerical needs a pressure that would come from it, the stem gives that pressure.

2. Conservation of mass: continuity, mass flow, volume flow and average velocity

With B = m and b = 1, a system has dm/dt = 0, so 0 = (d/dt) ∫_CV ρ dV + ∫_CS ρ V · n dA. For ports where velocity and density are uniform, the flux across a port is the mass flow rate ṁ = ρ V A (kg/s), and the volume flow rate is Q = V A (m³/s). If the velocity varies over the section, V stands for the average velocity Vavg = (1/A) ∫ u dA = Q/A. Steady flow gives Σṁin = Σṁout. For an incompressible fluid and a CV of fixed volume this becomes ΣQin = ΣQout, even if the flow is unsteady.

Average velocity from a profile. For laminar pipe flow u = umax (1 − r²/R²), Q = ∫₀ᴿ umax (1 − r²/R²) 2πr dr = umax · 2π (R²/2 − R²/4) = umax πR²/2, so Vavg = umax/2. For the turbulent power law u = umax (1 − r/R)^(1/n) the same integral gives Vavg/umax = 2n²/[(n + 1)(2n + 1)], which is 0.817 for n = 7. A velocity profile is never a mass-flow trap if you integrate it; the trap is multiplying the centreline velocity by the area.

Unsteady flow. For a tank, (d/dt)(ρ Vol) = ṁin − ṁout. With constant density and plan area At, At dh/dt = Qin − Qout. Worked line: At = 1.5 m², Qin = 0.012 m³/s, Qout = 0.005 m³/s give dh/dt = 0.007/1.5 = 4.67 × 10⁻³ m/s. For a rigid gas tank of volume Vol, Vol dρ/dt = ṁin − ṁout, so a tank of 2 m³ charged at 0.5 kg/s with no outflow gains density at 0.25 kg/m³ per second.

⚠️ Compressible flow conserves ṁ, not Q or V
In a steady gas flow through a duct of changing area, ρ₁A₁V₁ = ρ₂A₂V₂. Halving the area does not double the velocity unless the density is constant. Use volume flow only for liquids or for gases at low Mach number (below about 0.3).

3. Linear momentum equation for a fixed control volume

With B = mV and b = V, Newton’s second law for the system becomes ΣF = (d/dt) ∫_CV ρ V dV + ∫_CS ρ V (V · n) dA, a vector equation written in an inertial frame. ΣF is the total force on the fluid in the CV: surface forces (pressure, shear, the reaction of solid walls) plus body forces (weight). For steady flow with uniform ports it becomes ΣF = Σṁout Vout − Σṁin Vin: force equals the net rate of outflow of momentum, component by component, with signs fixed by the chosen axes.

  • Draw the CV so that its inlet and outlet planes cut the flow where pressure and velocity are known, and note that the unknown wall force R is the force of the solid on the fluid.
  • Use gauge pressures everywhere. A uniform atmospheric pressure over a closed surface has zero resultant, so only gauge pressure produces a net force. A pressure force on a port is pA, directed into the CV.
  • Write each component, then reverse the sign of R: the force of the fluid on the support is −R (Newton’s third law).

Momentum-flux correction factor. The exact flux through a section is ∫ρ u² dA. Writing it as β ρ A V² defines β = (1/(A V²)) ∫ u² dA. β = 1 for a uniform profile; for laminar pipe flow u = 2V (1 − r²/R²) the integral gives ∫u² dA = 4V² · 2πR²/6 = (4/3) V² πR², so β = 4/3. Turbulent profiles are flat and β is only slightly above 1, so use β = 1 unless the stem states a profile or a factor.

Worked example: reducing bend. Water (ρ = 1000 kg/m³) flows at Q = 0.2 m³/s through a horizontal 90° reducing bend, inlet 0.30 m (flow along +x), outlet 0.15 m (flow along +y). Gauge pressures are p₁ = 150 kPa and p₂ = 90 kPa. A₁ = 0.07069 m², A₂ = 0.01767 m², V₁ = 2.829 m/s, V₂ = 11.318 m/s, ṁ = 200 kg/s.
x: p₁A₁ + Rx = ṁ(0 − V₁), so Rx = −10,603 − 566 = −11,169 N.
y: Ry − p₂A₂ = ṁ(V₂ − 0), so Ry = 1,590 + 2,264 = +3,854 N.
The water pushes on the bend with (+11,169, −3,854) N, magnitude 11.8 kN. Note that the larger share is the pressure force, not the momentum change.

4. Jets on plates and vanes; moving control volumes

A free jet is at atmospheric pressure all round, so on a smooth (frictionless) surface its speed relative to the surface is unchanged (a consequence of the Bernoulli equation of Section XE1.5) and only its direction changes. The force on the surface is then ṁ times the change of velocity component. Take a jet of area A and speed V, density ρ, and plate or vane speed u along the jet.

Jet forces, frictionless and neglecting gravity
CaseForce along the jet direction
Fixed flat plate normal to the jetρAV²
Fixed smooth plate inclined at θ to the jetNormal force ρAV² sin θ; component along the jet ρAV² sin²θ
Fixed vane turning the jet through δρAV² (1 − cos δ); 2ρAV² for a 180° cup
Single plate receding at uρA (V − u)²
Series of plates or vanes (all the jet intercepted)ρAV (V − u) for flat plates; ρAV (V − u)(1 − cos δ) for vanes

Moving CV at constant velocity. Fix the CV to the plate and use relative velocities: the jet arrives at V − u, so the mass reaching a single receding plate per second is ρA(V − u), not ρAV, and F = ρA(V − u)². For a wheel with many vanes, new vanes keep arriving and the whole jet flow ρAV is eventually intercepted, so the relative speed V − u multiplies ρAV. Power is F · u. For a single plate, P = ρA(V − u)² u is maximum at u = V/3. For a series of 180° vanes, F = 2ρAV(V − u) and P is maximum at u = V/2, where P = ½ρAV³ equals the whole kinetic energy flux of the jet.

Worked example. A water jet of diameter 50 mm and speed 20 m/s (A = 1.963 × 10⁻³ m²) strikes a fixed normal plate: F = 1000 × 1.963 × 10⁻³ × 20² = 785 N. If the plate recedes at 5 m/s: F = 1000 × 1.963 × 10⁻³ × 15² = 442 N. If instead it is one of a train of plates all intercepting the jet: F = 1000 × 1.963 × 10⁻³ × 20 × 15 = 589 N. Same jet, three answers: the difference is only the mass flow that reaches the surface.

5. Accelerating control volumes, thrust and rocket problems

If the CV accelerates with acv (translation only), Newton’s law holds only in an inertial frame, so a fictitious body force is added. With Vr measured relative to the CV: ΣF − ∫_CV acv ρ dV = (d/dt) ∫_CV ρ Vr dV + ∫_CS ρ Vr (Vr · n) dA. Measuring velocities relative to the CV does not remove the inertial term; the extra term is needed whenever acv ≠ 0.

Thrust. For a rocket whose exhaust leaves at speed Ve relative to the rocket with mass flow ṁ, exit area Ae and exit pressure pe in ambient pressure pa, the CV around the rocket gives T = ṁVe + (pe − pa) Ae. The first term is the momentum thrust, the second the pressure thrust. For a jet engine, ignoring the fuel mass, T = ṁ(Vexit − Vflight) + (pe − pa)Ae. A rocket needs no air to push against and works better in vacuum, because pa falls to zero.

Vertical flight. With pe = pa, no drag and constant g, the rocket mass is M = M₀ − ṁt and M dU/dt = ṁVe − Mg. Dividing by M and integrating gives U = Ve ln(M₀/M) − g t. In gravity-free space the final speed depends only on the mass ratio M₀/M and Ve, not on how fast the propellant burns. Worked line: M₀ = 500 kg, ṁ = 5 kg/s, Ve = 2000 m/s give T = 10 kN, initial acceleration 10,000/500 − 9.81 = 10.19 m/s², and after 40 s (M = 300 kg) U = 2000 ln(500/300) − 9.81 × 40 = 1021.7 − 392.4 = 629.3 m/s.

6. Choosing the equation: mass, momentum or energy

Matching the question to the balance
You wantUseWhy
Speed or flow rate at another sectionContinuityNeeds no force or pressure
Force on a bend, nozzle, plate, vane or rocketMomentumNeeds only port velocities and pressures, not the losses inside
Pressure change along the flowEnergy (Bernoulli if frictionless)Relates p, V and elevation; the head-loss term covers friction
Head loss in a sudden changeMomentum, then energyMomentum gives the pressure; energy then gives the loss

Worked example: sudden expansion. Water leaves a pipe at V₁ = 4 m/s into a pipe of four times the area, so continuity gives V₂ = 1 m/s. Take p₁ as the pressure on the whole step plane and ignore wall shear: (p₁ − p₂)A₂ = ρQ(V₂ − V₁), so p₂ − p₁ = ρV₂(V₁ − V₂) = 1000 × 1 × 3 = 3000 Pa. The frictionless Bernoulli rise would be ½ρ(V₁² − V₂²) = 7500 Pa. The shortfall of 4500 Pa equals ½ρ(V₁ − V₂)², the head loss of the eddies at the step: momentum gave the pressure, and energy then gave the loss.

⚠️ Four slips that cost marks
(1) Using absolute pressure in one port and gauge in another.
(2) Forgetting that the pressure force at an outlet acts opposite to the outward normal, so its sign flips relative to the momentum flux.
(3) Using the force on the fluid as the force on the support without reversing the sign.
(4) For a moving plate, using ρAV instead of ρA(V − u) as the mass flow.

Key takeaways

  • A system follows fixed matter; a control volume is a region with a control surface. The RTT is dBsys/dt = (d/dt)∫ρ b dV + ∫ρ b (V · n) dA: storage plus net efflux, and b = 1, V, e gives mass, momentum, energy.
  • Continuity: ṁ = ρVA and Q = VA, with V the average velocity ∫u dA/A. Steady flow conserves ṁ (and Q if ρ is constant); a tank obeys At dh/dt = Qin − Qout.
  • Steady momentum: ΣF = Σṁout Vout − Σṁin Vin, with gauge pressures, pA directed inward, and the force on the support equal to −R. For laminar pipe flow β = 4/3.
  • Jet forces: ρAV² on a fixed normal plate, 2ρAV² on a fixed 180° cup, ρA(V − u)² on a single receding plate, ρAV(V − u) per series of plates. Max power is at u = V/3 for one plate and u = V/2 for a 180° vane series.
  • An accelerating CV needs the extra term −∫acv ρ dV. Thrust T = ṁVe + (pe − pa)Ae; with pe = pa the rocket speed is U = Ve ln(M₀/M) − g t.
  • Use continuity for speeds, momentum for forces, and energy for pressure change and losses. Momentum and energy together give the sudden-expansion loss ½ρ(V₁ − V₂)².

Practice questions (18)

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. In the Reynolds Transport Theorem for a fixed control volume, the surface integral ∫_CS ρ b (V · n) dA represents:

    1. the rate at which the property accumulates, that is, is stored, inside the control volume itself
    2. the net rate at which the property is carried out across the control surface (outflow minus inflow)
    3. the rate at which the property is created by internal sources and sinks within the control volume
    4. the rate at which the control surface is displaced relative to the fluid particles it encloses
    Show answer

    Answer: B — the net rate at which the property is carried out across the control surface (outflow minus inflow)

    The surface integral adds up, over every face of the control surface, the property carried by the flow. V · n is positive on outflow faces and negative on inflow faces, so the integral is the net efflux (out minus in). The accumulation inside the volume is a different term, the time derivative (d/dt)∫ρ b dV; it is the tempting alternative but it vanishes in steady flow while the efflux does not. The theorem has no source term, since mass is neither created nor destroyed and momentum changes only through the force term of the law.

  2. Which of the following statements about a system, a control volume and the Reynolds Transport Theorem are correct?

    1. A system always contains the same fluid particles, whereas the fluid inside a control volume is replaced as flow crosses its surface
    2. The theorem converts laws written for a system, such as conservation of mass and Newton’s second law, into statements about a control volume
    3. In steady flow the theorem shows that every extensive property of the system stays constant in time
    4. The theorem can be used for linear momentum but not for energy, because energy has no per-unit-mass form
    Show answer

    Answer: A — A system always contains the same fluid particles, whereas the fluid inside a control volume is replaced as flow crosses its surface; B — The theorem converts laws written for a system, such as conservation of mass and Newton’s second law, into statements about a control volume

    A system is a fixed set of particles and a control volume is a region of space; the theorem is the bridge from one to the other, which is why the statements about particles versus region and about converting system laws hold. In steady flow only the storage term vanishes; the efflux term remains, so the momentum of the system still changes when a net force acts (for example on a bend). Energy has the per-unit-mass form e, so b = e gives the energy equation.

  3. A control volume moves at a constant velocity of 4 m/s along +x. Water (density 1000 kg/m³) moves at 10 m/s along +x, measured in the ground frame, and crosses a 0.01 m² inlet face of the control volume that is normal to x. The mass flow rate into the control volume is:

    1. 100 kg/s, using the ground-frame velocity of the water
    2. 140 kg/s, adding the water velocity and the control volume velocity
    3. 60 kg/s, using the velocity of the water relative to the control surface
    4. 40 kg/s, using only the velocity of the control volume itself
    Show answer

    Answer: C — 60 kg/s, using the velocity of the water relative to the control surface

    For a control volume moving at constant velocity the flux term uses the relative velocity Vr = V − Vcv = 10 − 4 = 6 m/s. Then ṁ = ρ Vr A = 1000 × 6 × 0.01 = 60 kg/s. Using the ground-frame 10 m/s gives 100 kg/s, the tempting slip, but the inlet face is itself running away from the water at 4 m/s; adding the speeds is wrong because the face moves in the same direction as the water.

  4. Air flows steadily through a duct whose cross-section area and the air density both change along its length. Which quantity has the same value at every cross-section?

    1. the mass flow rate ρ V A
    2. the volume flow rate V A at the section
    3. the average velocity V over the section
    4. the mass flux ρ V per unit area at the section
    Show answer

    Answer: A — the mass flow rate ρ V A

    In steady flow nothing accumulates between sections, so the mass entering per second equals the mass leaving: ρ₁A₁V₁ = ρ₂A₂V₂, i.e. ρ V A is constant. The volume flow rate V A is conserved only when density is constant, which is the tempting assumption carried over from liquids. Velocity changes with area, and the mass flux ρ V changes with area as well.

  5. An open cylindrical tank of diameter 2 m receives water at 0.02 m³/s. Water leaves through a 50 mm diameter pipe flowing full with an average velocity of 1.2 m/s. Water is incompressible. The rate of rise of the water level, in mm/s, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 5.62

    Outflow Qout = V A = 1.2 × π(0.05)²/4 = 1.2 × 1.9635 × 10⁻³ = 2.356 × 10⁻³ m³/s. Net inflow = 0.02 − 0.002356 = 0.017644 m³/s. Tank area At = π(2)²/4 = 3.1416 m². Then dh/dt = 0.017644/3.1416 = 5.616 × 10⁻³ m/s = 5.62 mm/s. The usual slips are using the tank radius as the diameter (4 times too large) or forgetting to subtract the outflow.

  6. In a circular pipe of diameter 0.2 m the turbulent velocity profile is u = umax (1 − r/R)^(1/7), where R is the pipe radius, r the radial distance from the axis and umax = 3 m/s. The volume flow rate, in litres per second, rounded to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 77

    Q = ∫₀ᴿ u 2πr dr. Put s = 1 − r/R: Q = 2πR² umax ∫₀¹ s^(1/7) (1 − s) ds = 2πR² umax [7/8 − 7/15] = 2πR² umax × 49/120 = 0.8167 umax πR². So Vavg = 0.8167 × 3 = 2.450 m/s. With A = π(0.1)² = 0.031416 m², Q = 2.450 × 0.031416 = 0.07697 m³/s = 77 L/s (numerical integration gives 76.97). Using umax × A = 94 L/s is the trap.

  7. A rigid tank of volume 2 m³ is being charged with air at 0.6 kg/s while 0.2 kg/s is drawn off. The air in the tank is uniform in density. The rate of increase of its density is:

    1. 0.10 kg/m³ per second
    2. 0.30 kg/m³ per second
    3. 0.40 kg/m³ per second
    4. 0.20 kg/m³ per second
    Show answer

    Answer: D — 0.20 kg/m³ per second

    Mass balance on the rigid tank: Vol dρ/dt = ṁin − ṁout = 0.6 − 0.2 = 0.4 kg/s, so dρ/dt = 0.4/2 = 0.20 kg/m³ per second. The value 0.40 comes from forgetting to divide the net mass rate by the volume, 0.30 from ignoring the outflow, and 0.10 from dividing the outflow rather than the net rate by the volume.

  8. Fully developed laminar flow in a circular pipe has the profile u = 2V (1 − r²/R²), where V is the average velocity. The momentum flux through a section is written as β ρ A V². The momentum-flux correction factor β is:

    1. 1, because the average velocity already accounts for the profile
    2. 4/3, because ∫u² dA = (4/3) V² A for this parabola
    3. 3/2, because the centreline velocity is 3/2 of the average
    4. 2, because the centreline velocity is twice the average
    Show answer

    Answer: B — 4/3, because ∫u² dA = (4/3) V² A for this parabola

    β = (1/(A V²)) ∫u² dA. With s = r/R, ∫u² dA = 4V² · 2πR² ∫₀¹ (1 − s²)² s ds = 4V² · 2πR² · (1/6) = (4/3) V² πR², so β = 4/3. The value 1 holds only for a uniform profile: the average velocity gives the right mass flux, but momentum flux weights the fast core by u², so it is larger. The value 2 is the kinetic-energy factor of the same profile, a different quantity.

  9. Water (density 1000 kg/m³) flows at 0.1 m³/s through a horizontal 90° bend of uniform internal diameter 0.2 m. The gauge pressure is 100 kPa at both the inlet and the outlet. Neglect friction on the bend and the weight of the water. The magnitude of the resultant force exerted by the water on the bend, in N, rounded to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 4893

    A = π(0.2)²/4 = 0.031416 m², V = Q/A = 3.183 m/s, ṁ = 100 kg/s. Take the inlet flow along +x and the outlet flow along +y. Force on the water: x: p A + Rx = ṁ(0 − V) so Rx = −(pA + ṁV); y: Ry − pA = ṁ(V − 0) so Ry = pA + ṁV. Here pA = 100,000 × 0.031416 = 3141.6 N and ṁV = 318.3 N, so each component has magnitude 3459.9 N. The force on the bend has magnitude √2 × 3459.9 = 4893 N. The common slip is to use only the momentum change (450 N) or to omit the pressure force.

  10. Water flows steadily at V₁ = 4 m/s from a pipe into a coaxial pipe of four times the area. Assume the pressure acting on the annular step face equals p₁, the pressure upstream of the step, ignore wall shear, and take the water density as 1000 kg/m³. The pressure rise p₂ − p₁ between the two pipes, far from the step on each side, is:

    1. 3.0 kPa
    2. 7.5 kPa
    3. 12 kPa
    4. 4.5 kPa
    Show answer

    Answer: A — 3.0 kPa

    Continuity gives V₂ = V₁/4 = 1 m/s. Momentum on the fluid between the two planes: (p₁ − p₂)A₂ = ρ V₂A₂ (V₂ − V₁), so p₂ − p₁ = ρ V₂ (V₁ − V₂) = 1000 × 1 × 3 = 3000 Pa. The value 7.5 kPa is the frictionless Bernoulli rise ½ρ(V₁² − V₂²), which ignores the loss from the eddies at the step, and 12 kPa comes from writing ρV₁ in place of ρV₂ in the outflow term. The loss itself is 7.5 − 3.0 = 4.5 kPa = ½ρ(V₁ − V₂)².

  11. A steady water flow passes through a horizontal pipe bend, and the momentum equation is applied to a control volume enclosing the water in the bend. Which statements are correct?

    1. Gauge pressures can be used on every port, because a uniform atmospheric pressure acting over a closed surface has zero resultant force
    2. The force of the water on the bend is equal in magnitude and opposite in direction to the force of the bend on the water
    3. The force can be found from the port velocities and pressures alone, without knowing the energy loss inside the bend
    4. The force depends only on the change of velocity, so the same bend at a much higher line pressure carries the same load
    Show answer

    Answer: A — Gauge pressures can be used on every port, because a uniform atmospheric pressure acting over a closed surface has zero resultant force; B — The force of the water on the bend is equal in magnitude and opposite in direction to the force of the bend on the water; C — The force can be found from the port velocities and pressures alone, without knowing the energy loss inside the bend

    The three correct statements are the standard features of the method: gauge pressure suffices, Newton’s third law links the two forces, and the control-volume form avoids the details inside the bend, including friction and eddy losses. The claim that force depends only on the velocity change is false because the pressure force pA on each port is added to the momentum term, and at line pressures of hundreds of kilopascals it is usually the larger part (3,142 N against 318 N in a typical case).

  12. Two identical water jets (same density, area and speed) strike two fixed surfaces. One surface is a flat plate normal to the jet, which spreads the water sideways. The other is a smooth cup that turns the jet through 180° and returns it parallel to the incident jet. In both cases the water speed relative to the surface is unchanged and gravity is ignored. The ratio of the force on the cup to the force on the plate is:

    1. 0.5, because the cup only turns the flow gently and so gives less force than the flat plate
    2. 1, because the jet mass flow rate and the jet speed are the same for both surfaces
    3. 2, because the jet loses all its momentum on the plate but reverses its momentum in the cup
    4. 4, because the reversal of the jet squares the effect found for the plate
    Show answer

    Answer: C — 2, because the jet loses all its momentum on the plate but reverses its momentum in the cup

    Plate: the water leaves with no component along the jet, so F = ṁ(V − 0) = ρAV². Cup: the water leaves with velocity −V along the jet axis, so F = ṁ(V − (−V)) = 2ρAV². The ratio is 2. The same mass flow and speed do not give equal forces, because force is mass flow times the change in velocity, and the change is V for the plate but 2V for the cup.

  13. A water jet of diameter 30 mm and speed 30 m/s strikes a single flat plate that is normal to the jet and moves away from the nozzle along the jet axis at 10 m/s. Take the water density as 1000 kg/m³ and neglect gravity. The force exerted by the jet on the plate, in N, rounded to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 283

    Work in a frame attached to the plate. The jet arrives at the relative speed V − u = 30 − 10 = 20 m/s, and the mass reaching the plate per second is ρA(V − u), not ρAV. A = π(0.03)²/4 = 7.0686 × 10⁻⁴ m². F = ρA(V − u)² = 1000 × 7.0686 × 10⁻⁴ × 20² = 282.7 N, i.e. 283 N. Using the fixed-plate value ρAV² = 636 N, or the series-of-plates value ρAV(V − u) = 424 N, is wrong for a single plate.

  14. A water jet of speed V drives a wheel carrying many identical vanes, each turning the jet through 180° relative to the vane, so that the entire jet flow is always intercepted. The vanes are smooth and move at speed u along the jet; the speed of the water relative to a vane is unchanged and the power delivered is F u. Power is greatest when u equals:

    1. V/3
    2. V/2
    3. 2V/3
    4. V
    Show answer

    Answer: B — V/2

    With all the flow intercepted the force is F = ρAV (V − u)(1 − cos 180°) = 2ρAV (V − u), so P = 2ρAV (V − u) u. Setting dP/du = 0 gives V − 2u = 0, so u = V/2, and then P = ½ρAV³, the whole kinetic energy flux of the jet. The value V/3 belongs to a single receding plate, where the force is ρA(V − u)², and at u = V the power is zero because the jet no longer catches the vanes.

  15. A water jet (density ρ, area A, speed V) strikes a fixed smooth flat plate inclined at 30° to the jet axis. The water leaves along the plate with its speed unchanged, and gravity is ignored. The component of the force on the plate in the direction of the original jet is:

    1. ρAV²/2
    2. ρAV²
    3. (√3/4) ρAV²
    4. ρAV²/4
    Show answer

    Answer: D — ρAV²/4

    A smooth plate can exert force only along its normal. The jet’s momentum component normal to the plate is ρAV · V sin 30°, and the water leaves with none, so the normal force is Fn = ρAV² sin 30° = ρAV²/2. This normal makes 60° with the jet axis, so the component along the jet is Fn cos 60° = Fn sin 30° = ρAV²/4. The value ρAV²/2 is the normal force itself, and (√3/4)ρAV² is the component perpendicular to the jet.

  16. A rocket motor ejects 30 kg/s of exhaust at 2000 m/s relative to the rocket, through an exit of area 0.1 m², at an absolute exit pressure of 150 kPa. The ambient absolute pressure is 100 kPa. The thrust is:

    1. 60 kN, the momentum thrust alone
    2. 75 kN, adding the full absolute exit pressure force
    3. 65 kN, the momentum thrust plus the pressure thrust
    4. 55 kN, subtracting the pressure thrust from the momentum thrust
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    Answer: C — 65 kN, the momentum thrust plus the pressure thrust

    T = ṁVe + (pe − pa)Ae = 30 × 2000 + (150,000 − 100,000) × 0.1 = 60,000 + 5,000 = 65,000 N. The exhaust leaves at a pressure above ambient, so the unbalanced exit pressure pushes the rocket forward. Adding pe Ae = 15 kN ignores the ambient pressure that acts on the rest of the outer surface; subtracting gives the wrong sign.

  17. Which of the following statements about thrust and rocket motion are correct?

    1. When the exit pressure differs from the ambient pressure, the thrust contains an additional term (pe − pa)Ae on top of ṁVe
    2. In the frame of an accelerating control volume the inertial term may be dropped provided all velocities are measured relative to the control volume
    3. A rocket produces thrust only by pushing its exhaust against the surrounding air, so it cannot operate in vacuum
    4. In gravity-free vacuum with constant exhaust speed relative to the rocket, the gain in speed depends on that exhaust speed and on the ratio of initial to final mass, not on how fast the propellant is burnt
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    Answer: A — When the exit pressure differs from the ambient pressure, the thrust contains an additional term (p_{e} − p_{a})A_{e} on top of ṁV_{e}; D — In gravity-free vacuum with constant exhaust speed relative to the rocket, the gain in speed depends on that exhaust speed and on the ratio of initial to final mass, not on how fast the propellant is burnt

    The pressure thrust term and the speed result U = Ve ln(M₀/M) are both correct: the speed result follows from M dU/dt = ṁVe with M = M₀ − ṁt, and the burn rate cancels on integration. The inertial term cannot be dropped, because measuring velocities relative to the control volume does not remove the need for the fictitious force −acv ρ per unit volume. Rockets need no air; with pa = 0 the pressure thrust only adds to the thrust.

  18. A rocket of initial mass 500 kg is fired vertically upward from rest. It burns propellant at a constant 5 kg/s and the exhaust leaves at 2000 m/s relative to the rocket with exit pressure equal to ambient pressure. Take g = 9.81 m/s² as constant and neglect drag. The upward speed after 40 s of burning, in m/s, rounded to the nearest integer, is ____.

    Numerical answer — type the value.

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    Answer: 629

    M dU/dt = ṁVe − Mg with M = M₀ − ṁt integrates to U = Ve ln(M₀/M) − g t. At t = 40 s, M = 500 − 5 × 40 = 300 kg. ln(500/300) = 0.5108, so Ve ln(M₀/M) = 2000 × 0.5108 = 1021.7 m/s. Gravity loss = 9.81 × 40 = 392.4 m/s. U = 1021.7 − 392.4 = 629.3, i.e. 629 m/s. Forgetting the gravity term gives 1022; using ln with the burnt mass 200 in place of the final mass is another slip.