Fluid Mechanics · Chapter 4
Fluid Mechanics · Chapter 4

Fluid Kinematics

Kinematics describes how fluids move and deform — before any forces enter the picture. This chapter contrasts the Lagrangian and Eulerian viewpoints, introduces the material derivative that connects them, surveys how flows are visualized and plotted, defines vorticity, and closes with the Reynolds transport theorem — the engine behind every control-volume equation to come.

Learning objectives

  • Distinguish particle-following and field descriptions and select the useful viewpoint.
  • Compute material acceleration from local and convective contributions.
  • Interpret streamlines, pathlines, streaklines, strain rate, and vorticity.
  • Test a velocity field for incompressibility and rotationality.
  • Construct a control volume and interpret every Reynolds transport theorem term.
  • Verify kinematic results through dimensions, signs, limits, and independent checks.
Section 4–1

Lagrangian & Eulerian Descriptions

Kinematics describes motion without assigning its cause. A Lagrangian observer follows a material particle of fixed identity; an Eulerian observer records fields at fixed coordinates. Measurements from probes and most control-volume calculations are Eulerian, while trajectories and residence times are naturally Lagrangian.

Field description and coordinates

In Cartesian coordinates, choose positive \(x,y,z\) before differentiating and write \(\vec V=u\vec i+v\vec j+w\vec k\). The values may vary with position and time. A field can be steady, \(\partial(\,)/\partial t=0\), while a moving particle still accelerates by entering regions having different velocity.

Velocity Field (Eulerian)
\[ \vec{V}(x,y,z,t) = u\,\vec{i} + v\,\vec{j} + w\,\vec{k} \]

The material derivative

For any field property \(\phi\), the chain rule along a particle path gives \(D\phi/Dt=\partial\phi/\partial t+\vec V\cdot\nabla\phi\). The first term is local change at a fixed point; the second is convective change caused by motion through spatial gradients. This relation is exact for a differentiable continuum field, not a low-speed approximation.

Material Derivative
\[ \frac{D}{Dt} = \frac{\partial}{\partial t} + (\vec{V}\cdot\nabla) \]
Material Acceleration
\[ \vec{a} = \frac{D\vec{V}}{Dt} = \frac{\partial \vec{V}}{\partial t} + (\vec{V}\cdot\nabla)\vec{V} \]

Physical meaning

A particle accelerometer reads the material acceleration, not merely the local derivative reported by a stationary probe. Convective acceleration therefore explains nozzle acceleration even when every fixed probe reports an unchanging velocity.

EXAMPLE 1Acceleration in a Calibrated Planar Field
A laser measurement gives \(u=2.0+0.40x\) and \(v=-0.40y\), with coordinates in meters and velocity in \(\text{m/s}\). Determine the particle acceleration at \(x=1.5\ \text{m}\), \(y=0.50\ \text{m}\).

Recognition strategy

The target follows a particle but the data are an Eulerian velocity field, so apply the material derivative rather than differentiating a trajectory.

Given information

  • The field is steady and two-dimensional; \(+x\) is right and \(+y\) is upward.
  • Pressure is not used; any pressure reference is immaterial. Fluid properties are not required, and the temperature/property basis is therefore not applicable.

Model and interpretation

Treat the supplied differentiable field as exact at the stated point. The calculation is kinematic; incompressible, inviscid, and fully developed assumptions are not needed.

Step-by-step solution
1. Evaluate velocity. At the point, \(u=2.0+0.40(1.5)=2.60\ \text{m/s}\) and \(v=-0.40(0.50)=-0.20\ \text{m/s}\).
2. Form the derivatives. Steadiness makes both local terms zero, while \(u_x=0.40\ \text{s}^{-1}\), \(u_y=0\), \(v_x=0\), and \(v_y=-0.40\ \text{s}^{-1}\).
3. Compute the \(x\) component. The exact material relation gives \(a_x=u u_x+v u_y=(2.60)(0.40)=1.04\ \text{m/s}^2\).
4. Compute the \(y\) component. Similarly, \(a_y=u v_x+v v_y=(-0.20)(-0.40)=0.080\ \text{m/s}^2\), directed upward.

Verification

Units: Velocity multiplied by a velocity gradient gives \((\text{m/s})(\text{s}^{-1})=\text{m/s}^2\), as acceleration requires.

Signs and direction: Both components are positive: rightward flow enters larger \(u\), while downward motion enters more-negative \(y\) where \(v\) becomes less negative.

Plausibility: The divergence \(u_x+v_y=0\) independently confirms a locally incompressible field; that condition does not require zero acceleration.

Final result: \( \vec a=1.04\,\vec i+0.080\,\vec j\ \text{m/s}^2 \) at the specified point.

Engineering interpretation

The acceleration is almost entirely rightward and is convective despite steady data. A force analysis could use this acceleration later, but kinematics alone does not identify the pressure or stress that produces it.

Section 4–2

Flow Visualization

A streamline is tangent to the instantaneous velocity, a pathline is one particle's trajectory, a streakline joins particles that passed through one release point, and a timeline tracks particles marked together. They coincide only in steady flow. Exposure time, seeding response, and projection can make measured images differ from the underlying three-dimensional motion.

Streamline equation

At a fixed instant, \(dx/u=dy/v=dz/w\). Integrate only after fixing that instant; treating an unsteady streamline as a particle path is a common modeling error.

Problem-solving strategy

  1. Identify whether the requested object is instantaneous, particle-following, or release-history based.
  2. Declare coordinates, time, initial location, and positive direction before integrating.
  3. Use the streamline differential relation or particle equations \(dx/dt=u\), \(dy/dt=v\), \(dz/dt=w\).
  4. Check that the resulting curve is tangent to the velocity and satisfies its initial condition.
Section 4–3

Plots of Fluid Flow Data

Profile, Vector & Contour Plots

A profile exposes variation along a line, vectors show direction and magnitude, and contours show equal scalar values. Always inspect axis scales, units, sampling density, and whether interpolation creates apparent gradients unsupported by measurements.

Section 4–4

Motion & Deformation of Fluid Elements

The velocity gradient separates into a symmetric strain-rate tensor and an antisymmetric rigid-rotation tensor. Translation moves an element without changing shape; normal strain changes length, shear strain changes angle, and rotation changes orientation.

Rate of Translation & Rate of Rotation

Translation is described by the velocity vector itself. The rate of rotation (angular velocity) at a point is the average rotation rate of two initially perpendicular lines through that point.

Linear Strain Rate

The rate of increase in length per unit length, defined along a chosen direction. In Cartesian coordinates the three normal components are \(\partial u/\partial x\), \(\partial v/\partial y\), \(\partial w/\partial z\).

Volumetric strain and incompressibility

The fractional volume-change rate is \(\nabla\cdot\vec V\). For constant-density material motion, mass conservation requires it to vanish. This local statement is exact; it does not mean every velocity component is spatially constant.

Volumetric Strain Rate
\[ \frac{1}{V}\frac{DV}{Dt} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} \]

Shear Strain Rate

Angular distortion depends on paired cross derivatives. It must not be confused with vorticity: a fluid element may deform strongly with zero net spin.

EXAMPLE 2Screening a Velocity Field for Deformation
For \(u=3y\) and \(v=3x\), with coordinates in meters and velocities in \(\text{m/s}\), classify planar volume change, shear deformation, and rigid rotation at every point.

Recognition strategy

The question asks about local element deformation, so derivatives of the velocity field—not particle trajectories or force balances—are the direct evidence.

Given information

  • The supplied steady two-dimensional field uses \(+x\) right and \(+y\) upward.
  • Pressure reference and temperature/property basis are not applicable because no pressure or property enters this exact kinematic calculation.

Model and interpretation

Use the exact Cartesian gradient for a differentiable continuum. No inviscid, fully developed, Reynolds-number, Froude-number, or Mach-number assumption is required.

Step-by-step solution
1. Differentiate the field. The nonzero derivatives are \(u_y=3.00\ \text{s}^{-1}\) and \(v_x=3.00\ \text{s}^{-1}\); both normal derivatives vanish.
2. Test volume change. The planar divergence is \(u_x+v_y=0+0=0\ \text{s}^{-1}\), so material area and volume are locally preserved.
3. Find angular distortion. The engineering shear-strain rate is \(u_y+v_x=6.00\ \text{s}^{-1}\), so a square element becomes a rhombus.
4. Find rigid spin. The out-of-plane vorticity is \(\zeta_z=v_x-u_y=0\ \text{s}^{-1}\), giving angular velocity \(\omega_z=\zeta_z/2=0\).

Verification

Units: Each velocity gradient has units \((\text{m/s})/\text{m}=\text{s}^{-1}\), appropriate for strain and rotation rates.

Signs and direction: Equal positive cross derivatives add in shear but cancel in curl, confirming distortion without clockwise or counterclockwise spin.

Plausibility: At a point on the positive \(x\)-axis the upward velocity increases with \(x\), while on the positive \(y\)-axis rightward velocity increases with \(y\); this symmetric stretching matches zero net rotation.

Final result: \( \nabla\cdot\vec V=0 \), \( \dot\gamma_{xy}=6.00\ \text{s}^{-1} \), and \( \zeta_z=0 \).

Engineering interpretation

An incompressible and irrotational field can still impose severe shear deformation. In a real Newtonian fluid, the given rate would be combined with a supplied viscosity to estimate shear stress.

Section 4–5

Vorticity & Rotationality

Vorticity Vector

Defined as the curl of the velocity field. It measures the local spin of fluid particles, and equals twice the angular velocity of rotation.

Vorticity
\[ \vec{\zeta} = \nabla\times\vec{V} = 2\vec{\omega} \]

Rotational vs. Irrotational Regions

Nonzero curl means local material rotation; zero curl means irrotational motion. Circulation and vorticity sign depend on the declared axis convention. Viscous wall layers commonly generate vorticity even when an outer region is approximately irrotational.

ROTATIONAL (ζ ≠ 0) element spins IRROTATIONAL (ζ = 0) no net spin
Figure 4–1. Vorticity measures particle spin: a rotational element rotates as it travels, an irrotational one does not. (Original schematic.)
Section 4–6

The Reynolds Transport Theorem

Reynolds Transport Theorem (RTT)

Relates the rate of change of any extensive property of a system to what happens inside, and across the boundary of, a control volume. It is the bridge that converts the conservation laws (written for systems) into control-volume form.

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

Here \(B\) is any extensive property and \(b = B/m\) the corresponding intensive property. The first term is the rate of accumulation inside the control volume; the second is the net flux of \(B\) across the control surface.

Physical meaning

The system term follows the same matter, while the control-volume terms watch a chosen region. Outward normals make outflow positive and inflow negative; moving boundaries require relative velocity \(\vec V_r\), not absolute fluid velocity.

Problem-solving strategy

  1. Name the conserved extensive property and its intensive counterpart \(b\).
  2. Draw the control volume, outward normals, moving boundaries, and every port.
  3. Separate accumulation from signed flux before invoking steady or one-dimensional approximations.
  4. Audit dimensions and confirm that setting \(b=1\) recovers mass conservation.
EXAMPLE 3Transient Inventory in a Mixing Vessel
A rigid vessel receives liquid at \(2.40\ \text{kg/s}\) and discharges it at \(1.65\ \text{kg/s}\) for exactly \(8.00\ \text{min}\). Find the mass accumulation and volume rise using supplied constant density \(960\ \text{kg/m}^3\).

Recognition strategy

Unequal inlet and outlet mass rates signal transient storage. Reynolds transport with \(B=m\) gives the balance without tracking individual particles.

Given information

  • Mass rates and density are supplied givens; density is treated constant at the unspecified operating temperature.
  • The vessel is the fixed control volume; outward flux is positive. Pressure is not used, so gauge versus absolute reference is immaterial.

Model and interpretation

The ports are represented by measured mass rates, making the mass balance exact over the stated interval if those rates remain constant. Incompressibility is used only to convert accumulated mass to volume.

Step-by-step solution
1. Select the RTT property. Set \(B=m\) and \(b=1\), so system mass conservation makes the left side zero.
2. Apply the signed balance. With outward normal convention, \(dm_{CV}/dt=\dot m_{\rm in}-\dot m_{\rm out}=0.750\ \text{kg/s}\).
3. Integrate accumulation. The duration is \(8.00(60)=480\ \text{s}\), hence \(\Delta m=(0.750)(480)=360\ \text{kg}\).
4. Convert to volume. Constant density gives \(\Delta V=\Delta m/\rho=360/960=0.375\ \text{m}^3\).

Verification

Units: A mass-rate difference multiplied by seconds gives kilograms, and kilograms divided by \(\text{kg/m}^3\) gives \(\text{m}^3\).

Signs and direction: Inflow exceeds outflow, so positive storage and a rising liquid inventory agree with the outward-normal convention.

Plausibility: Independent total accounting gives \(1152\ \text{kg}\) entering and \(792\ \text{kg}\) leaving; their \(360\ \text{kg}\) difference matches accumulation exactly.

Final result: Accumulated mass is \(360\ \text{kg}\), corresponding to \(0.375\ \text{m}^3\) of liquid.

Engineering interpretation

The result predicts inventory, not liquid height; vessel geometry and freeboard are still required. Uncertainty is dominated by flow-meter bias and density variation, so operations should reconcile measured level change independently.

Chapter recap

  • Choose Lagrangian variables for particle histories and Eulerian fields for fixed-location descriptions.
  • Use the material derivative to combine local and convective change.
  • Keep divergence, strain, and curl conceptually and algebraically distinct.
  • Use RTT to translate a system law into accumulation plus signed control-surface flux.

Before moving on: you should be able to declare coordinates, differentiate a velocity field, classify a visualization, draw a control volume, and verify units, signs, and conservation.