Fluid Mechanics · Chapter 9
Fluid Mechanics · Chapter 9

Differential Analysis of Fluid Flow

Control volumes give the big picture; differential analysis gives the detail. By applying conservation of mass and momentum to an infinitesimal element, this chapter derives the continuity and Navier–Stokes equations — the point-by-point governing laws of fluid motion — introduces the stream function, and shows how a few classic flows can be solved exactly.

Learning objectives

  • Understand how the differential equations of mass and momentum are derived and what they mean.
  • Calculate the stream function and pressure field for a known velocity field.
  • Obtain and verify analytical solutions for justified simplified flows.
  • Declare coordinates, pressure reference, properties, boundary conditions, and signs.
  • Distinguish exact governing equations from approximate reduced models.
  • Audit local mass and momentum conservation and CFD convergence evidence.
Section 9–1

Introduction

The control-volume analysis of earlier chapters gives overall forces and flow rates but no detail inside the flow. The differential approach instead applies the conservation laws to an infinitesimal element, producing equations valid at every point — and therefore the full field of velocity and pressure everywhere.

Differential vs. Control-Volume Analysis

Control-volume analysis treats a finite region as a black box and tracks what crosses its boundary. Differential analysis shrinks the region to a point, yielding partial differential equations whose solution is the complete flow field. The trade-off is that those equations are usually hard to solve.

Physical meaning

A differential balance is the limit of a conservation statement on a shrinking control volume. Its solution must still satisfy finite control-volume mass and momentum balances, so local and global audits are complementary rather than competing methods.

Problem-solving strategy

  1. Sketch the domain and declare coordinates, positive velocity, and pressure reference.
  2. State property basis, constitutive law, and steady or unsteady character.
  3. Apply continuity before momentum and count field unknowns and equations.
  4. Remove terms only when geometry, symmetry, scaling, or boundary data justify it.
  5. Integrate and apply all boundary conditions without inventing missing data.
  6. Check units, residuals, conservation, limiting behavior, and regime assumptions.
Section 9–2

Conservation of Mass: The Continuity Equation

Continuity Equation (differential form)

Mass conservation applied to a fluid element: the rate of density change plus the net rate of mass efflux per unit volume is zero.

\[ \frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\vec{V}) = 0 \]
Incompressible Continuity
\[ \nabla\cdot\vec{V} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0 \]

For incompressible flow the divergence of velocity vanishes — a direct statement that the volume of a fluid element cannot change. It is the single most-used constraint when constructing or checking a velocity field.

EXAMPLE 1Audit a planar velocity field
For \(\vec V=(2.0x)\hat i+(-2.0y)\hat j\) in SI units, determine whether the steady two-dimensional field conserves mass for an incompressible fluid and find local acceleration at \((0.30,0.20)\ \text{m}\).

Recognition strategy

A prescribed field calls for differential continuity followed by the material derivative; no integral control-volume unknown is needed.

Given information

  • The coefficient \(2.0\ \text{s}^{-1}\) is supplied; \(x\) is rightward, \(y\) upward, and component signs follow these axes.
  • Fluid density is constant at the stated reference temperature; pressure is not required and its gauge or absolute reference is therefore immaterial.

Model and interpretation

The field is exactly steady and two-dimensional as given. Incompressibility is tested, not assumed in the continuity calculation; acceleration is exact for the stated differentiable field.

Step-by-step solution
1. Test divergence. \(\partial u/\partial x+\partial v/\partial y=2.0-2.0=0\ \text{s}^{-1}\).
2. Remove local acceleration. Because the field is steady, \(\partial\vec V/\partial t=\vec0\).
3. Evaluate convection. \(a_x=u\,u_x+v\,u_y=(2x)(2)=4x\) and \(a_y=u\,v_x+v\,v_y=(-2y)(-2)=4y\).
4. Substitute location. \(\vec a=(1.20\hat i+0.800\hat j)\ \text{m/s}^2\).

Verification

Units: A velocity gradient has \(\text{s}^{-1}\), while velocity times gradient gives \(\text{m/s}^2\).

Signs and direction: Both acceleration components are positive at the specified first-quadrant point even though \(v\) itself is downward.

Plausibility: Zero divergence confirms local and finite-area volume conservation; nonzero acceleration remains possible because particles move through a spatially varying steady field.

Final result: \(\nabla\cdot\vec V=0\) and \(\vec a=(1.20\hat i+0.800\hat j)\ \text{m/s}^2\).

Engineering interpretation

Steady does not mean zero particle acceleration. The extensional field preserves area while stretching in \(x\) and compressing in \(y\).

Section 9–3

The Stream Function

Stream Function

For 2-D incompressible flow, a single scalar function \(\psi\) can be defined whose derivatives give the two velocity components. It is built so that continuity is satisfied automatically, replacing two unknowns with one.

Stream Function (2-D, incompressible)
\[ u = \frac{\partial \psi}{\partial y}, \qquad v = -\frac{\partial \psi}{\partial x} \]

Physical Meaning

Curves of constant \(\psi\) are streamlines. Moreover, the difference in \(\psi\) between two streamlines equals the volume flow rate (per unit depth) passing between them — so the stream function both maps and measures the flow.

EXAMPLE 2Flow between measured streamlines
A two-dimensional incompressible field has stream function \(\psi=3.0xy\) in \(\text{m}^2/\text{s}\). Find velocity at \((0.40,0.20)\ \text{m}\) and flow per unit depth between \(\psi=0.12\) and \(0.48\ \text{m}^2/\text{s}\).

Recognition strategy

The supplied scalar automatically enforces continuity; differentiate using the declared sign convention and use stream-function difference for flux.

Given information

  • The supplied coefficient is \(3.0\ \text{s}^{-1}\); \(x\) is rightward, \(y\) upward, \(u=\psi_y\), and \(v=-\psi_x\).
  • Properties are constant at the stated reference temperature; pressure reference is unnecessary because no momentum equation is evaluated.

Model and interpretation

The domain is planar and incompressible exactly as stated, with unit out-of-plane depth. The reported volume flux is signed from the lower to higher stream-function value.

Step-by-step solution
1. Differentiate. \(u=\partial\psi/\partial y=3.0x\) and \(v=-\partial\psi/\partial x=-3.0y\).
2. Evaluate velocity. At the point, \(\vec V=(1.20\hat i-0.600\hat j)\ \text{m/s}\).
3. Confirm continuity. \(\partial u/\partial x+\partial v/\partial y=3.0-3.0=0\).
4. Find flux. \(q'=\psi_2-\psi_1=0.48-0.12=0.36\ \text{m}^2/\text{s}\).

Verification

Units: Differentiating \(\text{m}^2/\text{s}\) by metres gives \(\text{m/s}\); a stream-function difference gives flow per depth.

Signs and direction: The negative \(v\) points downward, while positive stream-function difference follows the declared cross-stream orientation.

Plausibility: Exact zero divergence independently verifies mass conservation, and multiplying \(0.36\ \text{m}^2/\text{s}\) by physical depth would produce \(\text{m}^3/\text{s}\).

Final result: \(\vec V=(1.20\hat i-0.600\hat j)\ \text{m/s}\) and \(q'=0.36\ \text{m}^2/\text{s}\).

Engineering interpretation

The stream function simultaneously encodes a mass-compatible velocity field and measurable planar flux, but momentum and boundary conditions are still needed to determine whether the field is physically realizable.

Section 9–4

Conservation of Linear Momentum: Cauchy’s Equation

Cauchy’s Equation of Motion

Newton’s second law for a fluid element: density times acceleration equals the body force plus the divergence of the stress tensor. It holds for any continuum, before any assumption about the fluid’s stress–strain behavior.

\[ \rho\,\frac{D\vec{V}}{Dt} = \rho\vec{g} + \nabla\cdot\sigma_{ij} \]

The acceleration on the left is the material derivative from Chapter 4. To turn Cauchy’s equation into something solvable, the stress tensor must be related to the motion — which is where the Newtonian-fluid assumption enters.

Section 9–5

The Navier–Stokes Equation

Navier–Stokes Equation (incompressible, constant viscosity)

Substituting the Newtonian stress law into Cauchy’s equation gives the governing equation of viscous fluid motion: acceleration equals pressure gradient, gravity, and viscous diffusion of momentum.

\[ \rho\,\frac{D\vec{V}}{Dt} = -\nabla P + \rho\vec{g} + \mu\,\nabla^2\vec{V} \]

Nature of the Equation

It is a nonlinear, second-order partial differential equation — the nonlinearity coming from the advective term in the material derivative. Together with continuity it forms a closed set for \(\vec V\) and \(P\), but exact analytical solutions exist only for a handful of simplified geometries.

Solution strategy: set up the problem, simplify with justified assumptions (steady, fully developed, one velocity component), apply boundary conditions — chiefly the no-slip condition at walls — and integrate. The classic exact solutions are Couette flow (between moving plates) and Poiseuille flow (pressure-driven in a channel or pipe).

EXAMPLE 3Pressure-driven microchannel profile
A Newtonian liquid flows between fixed plates separated by \(2h=1.00\ \text{mm}\). For supplied \(dp/dx=-1200\ \text{Pa/m}\) and \(\mu=0.0500\ \text{Pa}\!\cdot\!\text{s}\), find centerline and mean speeds.

Recognition strategy

Parallel walls, constant pressure gradient, and fully developed motion reduce Navier–Stokes to a one-dimensional exact boundary-value problem.

Given information

  • At \(25^\circ\text{C}\), supplied \(\mu=0.0500\ \text{Pa}\!\cdot\!\text{s}\), \(\rho=900\ \text{kg/m}^3\), \(h=0.000500\ \text{m}\).
  • \(x\) is downstream, \(y=0\) at centerline, gauge pressure gradient is negative downstream, and both walls satisfy no slip.

Model and interpretation

A steady incompressible differential control element has \(u=u(y)\), zero body force in \(x\), constant viscosity, and negligible edge effects. The solution is exact under these assumptions and approximate for a finite-width channel.

Step-by-step solution
1. Reduce momentum. \(0=-dp/dx+\mu d^2u/dy^2\), while continuity is satisfied by fully developed \(u(y)\).
2. Integrate and apply symmetry. \(du/dy=(dp/dx)y/\mu\) because \(du/dy=0\) at \(y=0\).
3. Apply wall no slip. \(u(y)=[-dp/dx](h^2-y^2)/(2\mu)\), giving \(u_{\max}=0.00300\ \text{m/s}\).
4. Average exactly. \(V=(2h)^{-1}\int_{-h}^{h}u\,dy=(2/3)u_{\max}=0.00200\ \text{m/s}\).

Verification

Units: \((\text{Pa/m})\text{m}^2/(\text{Pa}\!\cdot\!\text{s})\) reduces to \(\text{m/s}\).

Signs and direction: A negative downstream pressure gradient yields positive \(u\); the symmetric profile reaches zero at both walls.

Plausibility: Using hydraulic diameter \(D_h=4h=0.00200\ \text{m}\), \(\mathrm{Re}=\rho VD_h/\mu=0.072\), supporting laminar flow; mass flux from the profile equals \(2hV\) per unit width.

Final result: \(u_{\max}=3.00\times10^{-3}\ \text{m/s}\) and \(V=2.00\times10^{-3}\ \text{m/s}\) downstream.

Engineering interpretation

The low Reynolds number supports the exact laminar reduction. Manufacturing gap uncertainty is important because both speeds scale with \(h^2\).

From equations to CFD

When analytical reduction is unavailable, CFD discretizes the same mass and momentum balances. A credible result reports mesh and time-step refinement, boundary-condition sensitivity, iterative residuals, global mass imbalance, and a quantity-of-interest convergence study. Small algebraic residuals alone do not establish discretization independence or model validity.

Instructor Note

Don’t be intimidated by the vector form. In every solvable problem you’ll cross out most terms first — steady kills \(\partial/\partial t\), fully developed kills streamwise derivatives, and a single-direction flow kills the others. What survives is usually a simple ODE you can integrate twice. The skill being tested is knowing which terms vanish and why.

Chapter recap

  • Continuity and momentum are local forms of mass and force balances.
  • A stream function enforces two-dimensional incompressible continuity automatically.
  • Every vanished Navier–Stokes term needs a physical or scaling justification.
  • Before moving on, be able to reduce equations, apply boundary conditions, audit conservation, and distinguish iterative convergence from discretization verification.