Continuity equation in integral and differential form

Why it matters

Conservation of mass is the first equation in every flow analysis. It sizes wind-tunnel contractions and test sections, tells you how fast air must move through an engine intake or a nozzle throat, and, in differential form, is one of the equations every CFD code solves at each cell. It also gives the first check on whether a proposed velocity field is physically possible.

Key ideas

System and control volume. A system is a fixed quantity of mass, so its mass cannot change. A control volume (CV) is a region in space through which fluid flows. The Reynolds transport theorem converts "mass of the system is constant" into a statement about a CV: the rate of increase of mass inside the CV equals the net mass inflow across its surface.

Integral form. For a fixed CV, ∂/∂t ∭ρ dV + ∯ρ(V·n) dA = 0, where n is the outward unit normal. Outflow (V·n > 0) counts positive, inflow negative. For steady flow the first term vanishes: mass in equals mass out.

One-dimensional (uniform) flow. If velocity and density are uniform across each inlet and outlet, the steady form becomes ṁ = ρAV = constant along a stream tube or duct, with several inlets and outlets adding up: Σṁ_in = Σṁ_out. For non-uniform profiles, use the mean velocity V̄ = Q/A.

Differential form. Applying the same balance to an infinitesimal cube gives ∂ρ/∂t + ∇·(ρV) = 0, or equivalently Dρ/Dt + ρ∇·V = 0. The second version shows that ∇·V is the rate of volumetric dilatation (fractional rate of volume change) of a fluid element.

Special cases.

• Steady compressible flow: ∇·(ρV) = 0.
• Incompressible flow (ρ constant following a particle): ∇·V = 0, i.e. ∂u/∂x + ∂v/∂y + ∂w/∂z = 0. This holds even if the flow is unsteady. In 1-D it reduces to Q = AV = constant.
• 2-D incompressible flow: continuity is satisfied automatically by a stream function (previous topic).
• Cylindrical coordinates: (1/r)∂(r·u_r)/∂r + (1/r)∂u_θ/∂θ + ∂u_z/∂z = 0 for incompressible flow.

Compressible flow in ducts. Taking logs of ρAV = constant gives dρ/ρ + dA/A + dV/V = 0. At low Mach number dρ/ρ is negligible, so a smaller area means a higher velocity. At supersonic speed density falls faster than velocity rises, which is why supersonic nozzles diverge (see compressible flow).

Assumptions to state. Steady or unsteady; uniform or non-uniform properties at sections; compressible or incompressible. Continuity itself needs no assumption beyond the continuum; the simplified forms do.

Formulas

∂/∂t ∭ρ dV + ∯ρ·(V·n) dA = 0

• ρ: density (kg/m³); V: velocity (m/s); n: outward unit normal; dV: volume element (m³); dA: area element (m²). Any fixed control volume.

ṁ = ρ·A·V and steady ρ₁·A₁·V₁ = ρ₂·A₂·V₂

• ṁ: mass flow rate (kg/s); A: area normal to flow (m²); V: mean velocity (m/s).

Q = A·V and incompressible A₁·V₁ = A₂·V₂, Σ Q_in = Σ Q_out

• Q: volume flow rate (m³/s).

dM_cv/dt = Σ ṁ_in − Σ ṁ_out

• M_cv: mass inside the CV (kg). Use for filling and emptying tanks.

∂ρ/∂t + ∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z = 0

• General differential form (Cartesian).

∂u/∂x + ∂v/∂y + ∂w/∂z = 0

• Incompressible flow, steady or unsteady.

dρ/ρ + dA/A + dV/V = 0

• Steady 1-D compressible flow in a duct.

Worked examples

Example 1 (standard): branching pipe. Given: a 0.3 m pipe carrying water at 2 m/s splits into a 0.2 m branch at 3 m/s and a 0.15 m branch. Find the velocity in the 0.15 m branch.

1. Q₁ = A₁·V₁ = (π/4)(0.3²)(2) = 0.14137 m³/s.
2. Q₂ = A₂·V₂ = (π/4)(0.2²)(3) = 0.09425 m³/s.
3. Q₃ = Q₁ − Q₂ = 0.04712 m³/s.
4. A₃ = (π/4)(0.15²) = 0.017671 m², so V₃ = 0.04712/0.017671 = 2.667 m/s. Answer: V₃ ≈ 2.67 m/s.

Example 2 (standard): filling tank (unsteady). Given: a tank of plan area 2 m² receives 0.05 m³/s of water and drains through a 0.1 m pipe at 3 m/s. Find the rate of rise of the level.

1. Outflow Q_out = (π/4)(0.1²)(3) = 0.02356 m³/s.
2. A_tank·dh/dt = Q_in − Q_out = 0.05 − 0.02356 = 0.02644 m³/s.
3. dh/dt = 0.02644/2 = 0.01322 m/s. Answer: level rises at ≈ 13.2 mm/s.

Example 3 (GATE level): compressible nozzle and a velocity field. (a) Air enters a nozzle with ρ₁ = 1.2 kg/m³, V₁ = 50 m/s, A₁ = 0.1 m², and leaves with ρ₂ = 0.9 kg/m³ through A₂ = 0.05 m².

1. ṁ = ρ₁·A₁·V₁ = 1.2 × 0.1 × 50 = 6.0 kg/s.
2. V₂ = ṁ/(ρ₂·A₂) = 6.0/(0.9 × 0.05) = 133.3 m/s. Answer: V₂ ≈ 133 m/s (an incompressible assumption would give 100 m/s, a 25 % error). (b) Is V = (x² + y)·i − 2xy·j a possible incompressible flow? ∂u/∂x = 2x, ∂v/∂y = −2x; sum = 0. Yes. (c) If u = 2x in 2-D incompressible flow, ∂v/∂y = −2, so v = −2y + f(x).

Common mistakes

• Using Q = AV for gases at high speed. Use ṁ = ρAV whenever density changes.
• Using the pipe diameter where the area is needed, or forgetting the π/4.
• Treating the incompressible condition as steady-only. ∇·V = 0 holds for unsteady incompressible flow too.
• Sign errors in the integral form: outflow positive, inflow negative with the outward normal.
• Forgetting the storage term dM_cv/dt in filling and emptying problems.
• Using a velocity that is not normal to the area (use the normal component V·n).

For GATE AE

Common questions: velocity in a contraction or a branching duct, mass flow through a nozzle with given densities, wind-tunnel contraction ratio and test-section speed, unsteady tank level, and checking whether a given 2-D or 3-D velocity field satisfies continuity or finding the missing component. Practise moving quickly between ṁ, Q and V, and recognising when density changes must be kept.

Quick check

1. A wind tunnel settling chamber of area 4 m² has air at 5 m/s. What is the incompressible test-section speed if its area is 0.5 m²?
2. Does u = 3x, v = −3y, w = 0 satisfy incompressible continuity?
3. Air at 1.2 kg/m³ flows at 3 m/s through 0.05 m². What is ṁ?
4. For u = x², find v (2-D, incompressible) with v = 0 on y = 0.

Answers: 1. 40 m/s 2. Yes (3 − 3 = 0) 3. 0.18 kg/s 4. v = −2xy