Euler’s equation of motion for inviscid flow

At a Glance

Why This Chapter Matters

Euler’s equation is the foundation of inviscid fluid mechanics: it appears in Section B (optional), carries up to 20 marks, and connects directly to Bernoulli’s theorem — the primary computational tool in both questions. The 2024 question (20 marks) asks for pressure in a 2D doublet-flow field after verifying incompressibility and irrotationality, while the 2017 question (15 marks) requires the compressible-flow version of Bernoulli to prove a velocity ratio formula. Both question types are entirely mechanical once you know the two Bernoulli forms and one algebraic identity: (x2−y2)2+4x2y2=(x2+y2)2(x^2-y^2)^2 + 4x^2y^2 = (x^2+y^2)^2.

Minimum Theory

Euler’s Equation. For an inviscid fluid (viscosity μ=0\mu = 0) the Navier-Stokes equation reduces to Euler’s equation:

ρ ⁣[∂v∂t+(v⋅∇)v]=−∇p+ρF.\rho\!\left[\frac{\partial\mathbf{v}}{\partial t} + (\mathbf{v}\cdot\nabla)\mathbf{v}\right] = -\nabla p + \rho\mathbf{F}.

There is no viscous term. For steady flow (∂t=0\partial_t = 0) this becomes ρ(v⋅∇)v=−∇p+ρF\rho(\mathbf{v}\cdot\nabla)\mathbf{v} = -\nabla p + \rho\mathbf{F}.

Bernoulli’s Theorem (incompressible, irrotational). For steady, inviscid, incompressible flow that is also irrotational (∇×v=0\nabla \times \mathbf{v} = 0), Bernoulli’s equation holds throughout the fluid (not just along a streamline):

p+12ρ∣v∣2+ρgz=p∞=const.p + \frac{1}{2}\rho|\mathbf{v}|^2 + \rho gz = p_\infty = \text{const.}

Equivalently p=p∞−12ρ∣v∣2p = p_\infty - \tfrac{1}{2}\rho|\mathbf{v}|^2 (neglecting gravity or setting z=constz = \text{const}).

Bernoulli’s Theorem (compressible, isothermal). For steady compressible flow with p=Kρp = K\rho (isothermal, KK constant), the compressible Bernoulli integral along a streamline is ∣v∣22+∫dp/ρ=const\tfrac{|\mathbf{v}|^2}{2} + \int dp/\rho = \text{const}. With p=Kρp = K\rho, ∫dp/ρ=Kln⁡ρ\int dp/\rho = K\ln\rho, giving:

∣v∣22+Kln⁡ρ=const.\frac{|\mathbf{v}|^2}{2} + K\ln\rho = \text{const.}

Continuity for steady compressible flow: ρAq=const\rho A q = \text{const} (mass flux), where AA is the cross-sectional area.

Euler's equation anatomy: fluid element with pressure forces, Bernoulli forms, and key conditions.

Question Archetypes

ArchetypeRecognition
bernoulli-pressureGiven a 2D velocity field; verify incompressible and irrotational; find pressure
compressible-bernoulliSteady flow in a pipe; K=p/ρK = p/\rho constant; prove a velocity or density ratio

bernoulli-pressure (1 question(s); 2024)

Recognition Cues — The question hands you u(x,y)u(x,y) and v(x,y)v(x,y) with a rational form involving (x2+y2)n(x^2+y^2)^n in the denominator. You are asked to “verify” the flow satisfies the equations of motion and to “determine the pressure”. The phrase “inviscid incompressible” tells you Euler applies; you must first check ∇⋅v=0\nabla\cdot\mathbf{v} = 0 (incompressibility) and ωz=0\omega_z = 0 (irrotationality) before invoking Bernoulli.

Solution Template

  1. Compute ∂u/∂x+∂v/∂y\partial u/\partial x + \partial v/\partial y using the quotient rule. Show the sum is zero (incompressibility).
  2. Compute ωz=∂v/∂x−∂u/∂y\omega_z = \partial v/\partial x - \partial u/\partial y. Show it is zero (irrotationality).
  3. Compute ∣v∣2=u2+v2|\mathbf{v}|^2 = u^2 + v^2. Simplify using the identity (x2−y2)2+4x2y2=(x2+y2)2(x^2-y^2)^2 + 4x^2y^2 = (x^2+y^2)^2 if applicable.
  4. Apply Bernoulli: p=p∞−12ρ∣v∣2p = p_\infty - \tfrac{1}{2}\rho|\mathbf{v}|^2.

Worked Example(s)

2024 Paper 2, 2024-P2-Q7c (20 marks)

Let u=B(x2−y2)/(x2+y2)2u = B(x^2-y^2)/(x^2+y^2)^2, v=2Bxy/(x2+y2)2v = 2Bxy/(x^2+y^2)^2, w=0w = 0 satisfy the equations of motion for inviscid incompressible flow. Determine the pressure.

Step 1 — Incompressibility. Compute ∂u/∂x\partial u/\partial x by quotient rule:

∂u∂x=B⋅2x(x2+y2)2−B(x2−y2)⋅4x(x2+y2)(x2+y2)4=2Bx(3y2−x2)(x2+y2)3.\frac{\partial u}{\partial x} = \frac{B\cdot 2x(x^2+y^2)^2 - B(x^2-y^2)\cdot 4x(x^2+y^2)}{(x^2+y^2)^4} = \frac{2Bx(3y^2 - x^2)}{(x^2+y^2)^3}.

Compute ∂v/∂y\partial v/\partial y:

∂v∂y=2Bx(x2+y2)2−2Bxy⋅4y(x2+y2)(x2+y2)4=2Bx(x2−3y2)(x2+y2)3.\frac{\partial v}{\partial y} = \frac{2Bx(x^2+y^2)^2 - 2Bxy\cdot 4y(x^2+y^2)}{(x^2+y^2)^4} = \frac{2Bx(x^2 - 3y^2)}{(x^2+y^2)^3}.

Sum: 2Bx[(3y2−x2)+(x2−3y2)](x2+y2)3=0.\dfrac{2Bx[(3y^2-x^2)+(x^2-3y^2)]}{(x^2+y^2)^3} = 0. Incompressible. ✓

Step 2 — Irrotationality. Compute ∂v/∂x\partial v/\partial x:

∂v∂x=2By(y2−3x2)(x2+y2)3.\frac{\partial v}{\partial x} = \frac{2By(y^2 - 3x^2)}{(x^2+y^2)^3}.

Compute ∂u/∂y\partial u/\partial y:

∂u∂y=−2By(x2+y2)2−B(x2−y2)⋅4y(x2+y2)(x2+y2)4=−2By[x2+y2+2(x2−y2)](x2+y2)3=2By(y2−3x2)(x2+y2)3.\frac{\partial u}{\partial y} = \frac{-2By(x^2+y^2)^2 - B(x^2-y^2)\cdot 4y(x^2+y^2)}{(x^2+y^2)^4} = \frac{-2By[x^2+y^2+2(x^2-y^2)]}{(x^2+y^2)^3} = \frac{2By(y^2-3x^2)}{(x^2+y^2)^3}.

So ∂v/∂x=∂u/∂y\partial v/\partial x = \partial u/\partial y, hence ωz=0\omega_z = 0. Irrotational. ✓

Step 3 — Compute ∣v∣2|\mathbf{v}|^2.

u2+v2=B2(x2−y2)2+4B2x2y2(x2+y2)4=B2[(x2−y2)2+4x2y2](x2+y2)4.u^2 + v^2 = \frac{B^2(x^2-y^2)^2 + 4B^2x^2y^2}{(x^2+y^2)^4} = \frac{B^2[(x^2-y^2)^2 + 4x^2y^2]}{(x^2+y^2)^4}.

Apply the identity (x2−y2)2+4x2y2=x4+2x2y2+y4=(x2+y2)2(x^2-y^2)^2 + 4x^2y^2 = x^4 + 2x^2y^2 + y^4 = (x^2+y^2)^2:

∣v∣2=B2(x2+y2)2.|\mathbf{v}|^2 = \frac{B^2}{(x^2+y^2)^2}.

Step 4 — Pressure via Bernoulli.

p=p∞−ρB22(x2+y2)2.\boxed{p = p_\infty - \frac{\rho B^2}{2(x^2+y^2)^2}.}

As ∣z∣→∞|z| \to \infty, ∣v∣→0|\mathbf{v}| \to 0 and p→p∞p \to p_\infty (far-field pressure). ✓

Common Traps

compressible-bernoulli (1 question(s); 2017)

Recognition Cues — The problem involves “steady” flow through a pipe or nozzle of varying cross-section. A density–pressure relation is given, typically K=p/ρK = p/\rho (isothermal) stated as a constant. You are asked to prove a ratio formula relating velocities at the two ends. The key is that density is not constant (compressible), so mass continuity keeps ρAq\rho A q rather than just AqAq.

Solution Template

  1. Write the mass continuity equation ρ1A1q1=ρ2A2q2\rho_1 A_1 q_1 = \rho_2 A_2 q_2. Note A∝D2A \propto D^2 (diameter squared). Express v/V=(ρ1/ρ2)(D2/d2)v/V = (\rho_1/\rho_2)(D^2/d^2).
  2. Write the compressible Bernoulli integral: q22+∫dp/ρ=const\tfrac{q^2}{2} + \int dp/\rho = \text{const}.
  3. Use the given thermodynamic relation to evaluate ∫dp/ρ\int dp/\rho. For p=Kρp = K\rho (isothermal): ∫dp/ρ=Kln⁡ρ\int dp/\rho = K\ln\rho.
  4. Apply Bernoulli at both ends; solve for ρ1/ρ2\rho_1/\rho_2 in terms of the velocities.
  5. Substitute into the continuity result to obtain the required formula.

Worked Example(s)

2017 Paper 2, 2017-P2-Q7c (15 marks)

A stream rushes through a conical pipe with end diameters DD and dd. If VV and vv are the corresponding velocities and K=p/ρK = p/\rho is constant, prove v/V=(D2/d2) e(v2−V2)/2Kv/V = (D^2/d^2)\,e^{(v^2-V^2)/2K}.

Step 1 — Continuity. For steady flow, mass flux is constant: ρ1A1V=ρ2A2v\rho_1 A_1 V = \rho_2 A_2 v. With A∝D2A \propto D^2:

ρ1D2V=ρ2d2v⇒vV=ρ1ρ2⋅D2d2.(1)\rho_1 D^2 V = \rho_2 d^2 v \quad\Rightarrow\quad \frac{v}{V} = \frac{\rho_1}{\rho_2}\cdot\frac{D^2}{d^2}. \qquad(1)

Step 2 — Compressible Bernoulli. Along a streamline with p=Kρp = K\rho:

q22+∫dpρ=const,∫dpρ=∫K dρρ=Kln⁡ρ.\frac{q^2}{2} + \int\frac{dp}{\rho} = \text{const}, \quad \int\frac{dp}{\rho} = \int\frac{K\,d\rho}{\rho} = K\ln\rho.

Applying at both ends:

V22+Kln⁡ρ1=v22+Kln⁡ρ2⇒Kln⁡ρ1ρ2=v2−V22.\frac{V^2}{2} + K\ln\rho_1 = \frac{v^2}{2} + K\ln\rho_2 \quad\Rightarrow\quad K\ln\frac{\rho_1}{\rho_2} = \frac{v^2 - V^2}{2}.

Hence:

ρ1ρ2=e(v2−V2)/2K.(2)\frac{\rho_1}{\rho_2} = e^{(v^2-V^2)/2K}. \qquad(2)

Step 3 — Combine (1) and (2):

vV=D2d2 e(v2−V2)/2K.\boxed{\frac{v}{V} = \frac{D^2}{d^2}\,e^{(v^2-V^2)/2K}.}

Common Traps

Marks-Aware Writing

15-mark answer (compressible-bernoulli). Three clear steps: (1) continuity with the D2/d2D^2/d^2 ratio explicitly derived; (2) compressible Bernoulli integral with the isothermal substitution shown step by step; (3) algebraic combination. Each step should display one key equation. The result must match the boxed formula exactly — leave nothing implicit.

20-mark answer (bernoulli-pressure). Four steps, each needing workings: (1) full quotient-rule computation of ∂u/∂x\partial u/\partial x and ∂v/∂y\partial v/\partial y; (2) full computation of ∂v/∂x\partial v/\partial x and ∂u/∂y\partial u/\partial y; (3) algebra showing ∣v∣2=B2/(x2+y2)2|\mathbf{v}|^2 = B^2/(x^2+y^2)^2 via the (x2±y2)(x^2\pm y^2) identity; (4) Bernoulli giving the pressure formula. The derivations in steps 1 and 2 are the main marks-earners — a final answer without the algebra will lose most of the credit.

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