Fluid Mechanics · Chapter 6
Fluid Mechanics · Chapter 6

Momentum Analysis of Flow Systems

Where the energy equation tracks scalars, the momentum equation tracks vectors — and it is how engineers compute the forces and torques a flow exerts on the hardware that contains it. Built from the Reynolds transport theorem, this chapter develops the linear and angular momentum equations for control volumes, the momentum-flux correction factor, and the disciplined sign-keeping that makes them reliable.

Learning objectives

  • Choose control volumes that expose required forces and moments.
  • Resolve pressure, weight, support, and momentum-flux vectors with consistent signs.
  • Apply linear momentum to jets, bends, and multiple-port devices.
  • Use momentum-flux correction factors only with compatible average velocities.
  • Apply angular momentum and interpret turbomachine torque and power signs.
  • Verify mass closure, units, directions, regimes, and practical load uncertainty.
Section 6–1

Newton’s Laws & the Conservation of Momentum

Momentum is vector-valued. A control-volume balance relates external forces on the fluid to momentum accumulation and net outward flux. The force the fluid exerts on hardware is equal and opposite to the hardware force on the fluid, so the requested object must be named before signs are interpreted.

Newton’s Second Law (momentum form)

The rate of change of the linear momentum of a body equals the net force acting on it. The product \(m\vec V\) is the linear momentum; conservation of momentum follows directly.

Physical meaning

A flow produces force when its momentum magnitude or direction changes. Energy may remain nearly constant through a smooth turn while momentum changes substantially, which is why energy alone cannot determine support loads.

Newton’s Second Law
\[ \sum \vec{F} = m\vec{a} = \frac{d(m\vec{V})}{dt} \]
Section 6–2

Choosing a Control Volume

Wise Control-Volume Selection

Half the work in a momentum problem is choosing the control volume. Slice the control surface normal to the inflow and outflow velocities, and cut through supports or connectors where the unknown reaction force acts — this exposes the force you want while avoiding unknown internal pressure distributions.

Problem-solving strategy

  1. Enclose the fluid and cut ports normal to known section-average velocities.
  2. Choose axes, outward normals, and positive moment direction before writing components.
  3. Close mass flow and list pressure, weight, viscous, and hardware forces on the fluid.
  4. Write one scalar momentum equation per direction and solve for the hardware-on-fluid reaction.
  5. Reverse that reaction only if the requested load is fluid on hardware, then audit signs and magnitude.
Section 6–3

Forces Acting on a Control Volume

Body & Surface Forces

The total force on a control volume is the sum of body forces (acting throughout the volume — chiefly gravity) and surface forces (acting on the control surface — pressure and viscous stresses, plus reaction forces from contact with solids).

Total Force
\[ \sum\vec{F} = \sum\vec{F}_{\text{body}} + \sum\vec{F}_{\text{surface}} = \vec{F}_{\text{gravity}} + \vec{F}_{\text{pressure}} + \vec{F}_{\text{viscous}} + \vec{F}_{\text{reaction}} \]
Gage-pressure trick: since atmospheric pressure acts all around a control surface and cancels, you may use gage pressures at inlets and outlets and ignore atmospheric pressure elsewhere.

Gauge pressure is convenient only where a common atmospheric pressure acts on the remaining exterior boundary. Absolute pressure is required for cavitation or compressible-state calculations. Weight and wall shear may be neglected only after their likely scales are compared with pressure and momentum forces.

Section 6–4

The Linear Momentum Equation

Linear Momentum Equation (control volume)

From the Reynolds transport theorem with \(B = m\vec V\): the net force on a control volume equals the rate of accumulation of momentum inside plus the net rate at which momentum leaves through the control surface.

\[ \sum\vec{F} = \frac{d}{dt}\!\int_{CV}\!\rho\vec{V}\,dV + \int_{CS}\!\rho\vec{V}(\vec{V}_r\cdot\vec{n})\,dA \]

Momentum-Flux Correction Factor, β

Because real velocity profiles are not uniform, the momentum flux written with the average velocity needs a correction factor \(\beta\). It is always \(\geq 1\), equalling 1 for uniform flow.

Momentum-Flux Correction Factor
\[ \beta = \frac{1}{A_c}\int_{A_c}\!\left(\frac{V}{V_{\text{avg}}}\right)^{2}\,dA_c \]
Steady Flow, One Inlet & One Outlet
\[ \sum\vec{F} = \dot{m}\big(\beta_2\vec{V}_2 - \beta_1\vec{V}_1\big) \]
ṁV₁ ṁV₂ ∑F = ṁ(β₂V₂ − β₁V₁) reaction force holds the elbow
Figure 6–1. A control volume around a reducing elbow: the change in momentum between inlet and outlet, plus the pressure forces, sets the reaction force the connectors must supply. (Original schematic.)
EXAMPLE 1Jet Turned by a Smooth Guide Vane
A water jet of supplied mass flow \(6.00\ \text{kg/s}\) enters a stationary guide vane at \(12.0\ \text{m/s}\) in \(+x\) and leaves at \(12.0\ \text{m/s}\) in \(+y\). Find the force of the water on the vane.

Recognition strategy

The speed is unchanged but direction changes, so a vector linear-momentum balance—not an energy balance—determines the vane load.

Given information

  • \(+x\) is right, \(+y\) is upward; uniform free jets have supplied \(\beta=1.00\).
  • Both jet sections are at zero gauge pressure. Temperature/property basis is unnecessary because mass flow is supplied.

Model and interpretation

A fixed control volume encloses the jet turn. Steady flow, negligible contained weight, and negligible ambient-pressure resultant are approximations; the momentum equation itself is exact.

Step-by-step solution
1. Close mass flow. The single steady inlet and outlet carry the same supplied \(\dot m=6.00\ \text{kg/s}\).
2. Write vector momentum. The vane-on-water force is \(\vec F_{v\to w}=\dot m(\vec V_2-\vec V_1)\).
3. Resolve components. \(F_{v\to w,x}=6.00(0-12.0)=-72.0\ \text{N}\) and \(F_{v\to w,y}=6.00(12.0-0)=72.0\ \text{N}\).
4. Reverse the interaction. Newton's third law gives \(\vec F_{w\to v}=72.0\,\vec i-72.0\,\vec j\ \text{N}\), with magnitude \(101.8\ \text{N}\).

Verification

Units: \((\text{kg/s})(\text{m/s})=\text{kg}\!\cdot\!\text{m/s}^2=\text{N}\), so each momentum-flux component is a force.

Signs and direction: The vane pushes water left and upward to turn it; water therefore pushes the vane right and downward.

Plausibility: Each component equals \(\dot mV=72.0\ \text{N}\). Unchanged speed is consistent with the smooth-vane idealization but does not imply zero force.

Final result: Water exerts \(72.0\ \text{N}\) rightward and \(72.0\ \text{N}\) downward; magnitude \(102\ \text{N}\).

Engineering interpretation

The support must resist both components, not merely the resultant magnitude. A real design must add vane weight, transient start-up, imperfect turning, splash, vibration, and uncertainty in flow rate.

Section 6–5

Review of Rotational Motion & Angular Momentum

Angular Momentum of a Particle

Angular momentum about an origin is \(\vec r\times m\vec V\); only the tangential velocity component contributes to torque about a shaft axis. Choose positive rotation with the right-hand rule and use the same origin for every force moment and momentum flux.

EXAMPLE 2Anchoring Load on a Reducing Elbow
Water at \(20^\circ\text{C}\) flows steadily through a horizontal \(90^\circ\) elbow at \(15.0\ \text{kg/s}\). It enters in \(+x\) at \(3.00\ \text{m/s}\), \(180\ \text{kPa}\) gauge through \(0.00500\ \text{m}^2\), and exits in \(+y\) at \(5.00\ \text{m/s}\), \(90.0\ \text{kPa}\) gauge through \(0.00300\ \text{m}^2\). Find the wall force on the fluid.

Recognition strategy

Pressure and direction changes across a fixed bend require component momentum balances after independently checking mass consistency.

Given information

  • All pressures are gauge; \(+x\) follows the inlet and \(+y\) follows the outlet. Supplied \(\beta_1=\beta_2=1.00\).
  • Water density is supplied as \(1000\ \text{kg/m}^3\) at \(20^\circ\text{C}\); the elbow is horizontal.

Model and interpretation

The fixed control volume cuts both ports. Steady incompressible section-average flow is approximate; contained weight and viscous traction at the cuts are neglected, while wall interaction remains unknown.

Step-by-step solution
1. Audit mass. \(\rho A_1V_1=1000(0.00500)(3.00)=15.0\ \text{kg/s}\) and \(\rho A_2V_2=15.0\ \text{kg/s}\), so continuity closes.
2. Resolve pressure forces. Inlet pressure acts \(+x\): \(P_1A_1=900\ \text{N}\). Outlet pressure acts opposite its outward normal, giving \(-P_2A_2=-270\ \text{N}\) in \(y\).
3. Balance \(x\) momentum. \(F_{w,x}+900=15.0(0-3.00)=-45.0\ \text{N}\), hence \(F_{w,x}=-945\ \text{N}\).
4. Balance \(y\) momentum. \(F_{w,y}-270=15.0(5.00-0)=75.0\ \text{N}\), hence \(F_{w,y}=345\ \text{N}\).

Verification

Units: Pressure-area products and mass-rate times velocity both reduce to newtons, so they can appear in one component balance.

Signs and direction: Inlet pressure pushes right, outlet pressure pushes downward, and the solved wall force is leftward and upward on the fluid.

Plausibility: Substitution gives net forces \(-45.0\ \text{N}\) in \(x\) and \(75.0\ \text{N}\) in \(y\), exactly matching the two momentum changes.

Final result: \( \vec F_{w\to f}=-945\,\vec i+345\,\vec j\ \text{N} \); fluid-on-elbow force is opposite.

Engineering interpretation

The anchor load requires the opposite interaction plus elbow and contained-fluid weight. Pressure uncertainty dominates here, and transient water-hammer loads could greatly exceed this steady estimate.

Section 6–6

The Angular Momentum Equation

Angular Momentum Equation (control volume)

The net moment on a control volume equals the rate of accumulation of angular momentum inside plus the net rate at which it leaves through the control surface — the rotational counterpart of the linear momentum equation, obtained from the RTT with \(B = \vec H\).

\[ \sum\vec{M} = \frac{d}{dt}\!\int_{CV}\!(\vec{r}\times\vec{V})\rho\,dV + \int_{CS}\!(\vec{r}\times\vec{V})\rho(\vec{V}_r\cdot\vec{n})\,dA \]
Steady Flow, Single Inlet & Outlet (scalar form)
\[ \sum M = \dot{m}\,(r_2 V_{2,t} - r_1 V_{1,t}) \]

where \(V_t\) is the tangential velocity component and \(r\) the radial arm. This is the working relation for turbomachinery — pumps, turbines, sprinklers, and centrifugal blowers — linking shaft torque to the change in the fluid's angular momentum.

Physical meaning

Euler's turbomachine relation is an angular-momentum balance: increasing the fluid's \(rV_t\) requires positive shaft torque on the fluid, while decreasing it permits the fluid to drive a turbine. Radial and axial components do not directly contribute about the shaft axis.

EXAMPLE 3Torque Transferred by a Pump Rotor
A rotor processes liquid at \(22.0\ \text{kg/s}\). The supplied section-average tangential components are \(V_{t1}=1.50\ \text{m/s}\) at \(r_1=0.0800\ \text{m}\) and \(V_{t2}=14.0\ \text{m/s}\) at \(r_2=0.220\ \text{m}\). Find shaft torque on the fluid and ideal transferred power at \(1450\ \text{rpm}\).

Recognition strategy

Radius and tangential velocity at rotor inlet and outlet identify a steady angular-momentum problem; linear momentum alone would not expose shaft torque.

Given information

  • Positive tangential direction follows rotor rotation, and positive torque adds angular momentum to the fluid.
  • Mass flow and velocity components are supplied averages. Pressure reference and temperature/property basis do not enter this torque calculation.

Model and interpretation

A stationary control volume encloses the rotor. Steady one-inlet/one-outlet flow and negligible external moment other than shaft torque are approximations; no efficiency is assumed for ideal transferred power.

Step-by-step solution
1. Compute inlet swirl moment. \(r_1V_{t1}=(0.0800)(1.50)=0.120\ \text{m}^2/\text{s}\).
2. Compute outlet swirl moment. \(r_2V_{t2}=(0.220)(14.0)=3.080\ \text{m}^2/\text{s}\).
3. Apply angular momentum. \(T_{s\to f}=\dot m(r_2V_{t2}-r_1V_{t1})=22.0(2.960)=65.1\ \text{N}\!\cdot\!\text{m}\).
4. Convert speed and power. \(\omega=1450(2\pi/60)=151.8\ \text{rad/s}\), so \(\dot W_{\rm ideal}=T\omega=9.89\ \text{kW}\).

Verification

Units: \((\text{kg/s})(\text{m}^2/\text{s})=\text{N}\!\cdot\!\text{m}\), and torque times \(\text{rad/s}\) gives watts because radians are dimensionless.

Signs and direction: Outlet \(rV_t\) exceeds inlet \(rV_t\), so shaft torque and ideal power are positive into the fluid, consistent with pump operation.

Plausibility: Dividing \(9.89\ \text{kW}\) by \(151.8\ \text{rad/s}\) recovers \(65.1\ \text{N}\!\cdot\!\text{m}\); reversing the swirl change would reverse torque.

Final result: Shaft-on-fluid torque is \(65.1\ \text{N}\!\cdot\!\text{m}\); ideal transferred power is \(9.89\ \text{kW}\).

Engineering interpretation

Actual shaft input would exceed ideal fluid power after mechanical and hydraulic losses. Blade slip, nonuniform profiles, leakage, transient torque, and manufacturer performance data remain essential for equipment and shaft design.

Chapter recap

  • Control-volume placement determines which forces are exposed and which distributions are avoided.
  • Linear momentum is a vector balance of external force, accumulation, and signed flux.
  • Pressure acts inward, and fluid-on-hardware force opposes hardware-on-fluid force.
  • Angular momentum converts changes in \(rV_t\) into shaft torque and power.

Before moving on: you should be able to close mass, draw a force diagram, resolve momentum components, reverse interaction forces correctly, calculate rotor torque, and verify units, directions, regimes, and uncertainty.