Continuity Equation: Fluid Mechanics Explained
Conservation of mass for a control volume — derived from the Reynolds Transport Theorem by setting one symbol to 1.
In the final lesson of Module 3 we built a single generic tool — the Reynolds Transport Theorem — and made a promise: set its intensive property <em>b</em> to 1 and out drops the continuity equation, the most-used equation in all of fluid mechanics. This lesson collects on that promise. We will not re-derive RTT; we will specialize it to mass, read off the result, and then simplify it down to the compact <em>A₁V₁ = A₂V₂</em> that engineers actually carry in their heads. Along the way you will see exactly why a nozzle speeds a fluid up (and why that fact has <em>nothing</em> to do with pressure).
Continuity is the Reynolds Transport Theorem with b = 1
In the final lesson of Module 3 we derived a single, generic equation — the Reynolds Transport Theorem (RTT) — and then made a promise: set its intensive property b to 1, and out drops the continuity equation. This lesson collects on that promise. We will not re-derive RTT; we will simply specialize it to mass and read off the most-used equation in fluid mechanics.
Recall the move. The extensive property B = mass m has intensive property b = dm/dm = 1. Plugging b = 1 into RTT, the left side becomes dm/dt of a system — and a system, by definition, always contains the same matter, so dm/dt = 0. Setting the right-hand side to zero gives the integral continuity equation: the mass inside any control volume can change only because mass flows across its boundary. Nothing else. Mass is neither created nor destroyed; it is only transported.
That one line — the accumulation of mass inside the control volume balances the net mass flowing across its surface — is the whole of mass conservation written for a region of space. Every special case we use, from a contracting nozzle to a branching pipe network, is this same equation with some terms dropped. The rest of this lesson is about which terms to drop, and when.
This is the moment the Reynolds Transport Theorem was built for. Last module we carried the generic theorem all the way to the statement 'with b = 1, the accumulation of mass plus the net mass outflow is zero' — and stopped there, one line short. This lesson simply names that line the continuity equation and starts using it. The same template, with b = V or b = e, will hand us the momentum equation (M6) and the energy equation (M5) later — continuity is just the first of three siblings.
The port form: ṁ = ρAV and Q = AV
Real hardware — pumps, valves, lengths of pipe — connects to the rest of a system through a small number of openings, or ports. When the flow through each port is steady and roughly uniform across the opening, the surface integral collapses into a sum over ports, and each port contributes a single number: the mass flow rate
where ρ is the density, A the port's cross-sectional area, and V the average speed normal to that area. This port form was introduced in the RTT lesson; here it becomes our working currency. Pair it with the volume flow rate Q = A V (cubic metres per second), and note that ṁ = ρ Q — mass flow is just density times volume flow.
A useful bookkeeping habit concerns sign. At an outlet the velocity points outward (V·n̂ > 0) so the port's ρAV is positive; at an inlet the velocity points inward (V·n̂ < 0) so the same product enters with a minus sign. Keeping those signs straight is what turns a jumble of connected pipes into a solvable mass balance.
Two special cases that do most of the work
The integral form is completely general but is rarely used as-is, because two simplifications apply to almost every engineering flow.
The first is steady flow. 'Steady' means nothing changes with time at any fixed point, so the accumulation term ∂/∂t ∫_CV ρ dV vanishes. What remains says: the net mass flowing out across the control surface is zero — equivalently, the total mass flow rate in equals the total mass flow rate out. In port form, Σṁ_out = Σṁ_in. This steady balance is the form used for the great majority of design calculations.
The second is steady plus incompressible. A liquid such as water is, for practical purposes, incompressible — its density ρ is constant and cancels out of the balance. The mass balance then becomes a volume balance: the total volume flow in equals the total volume flow out, ΣQ_out = ΣQ_in. For the commonest geometry of all — a single inlet and a single outlet — this collapses to the beautifully compact
which says that the product of area and velocity is the same at every cross-section. That single equation is the continuity equation as most engineers actually use it. Squeeze the area (a nozzle) and the velocity rises to keep A·V constant; widen it (a diffuser) and the velocity falls. The duct simulator below lets you feel that trade-off directly.
Here is the trap that catches almost everyone first. You squeeze a nozzle, the fluid speeds up, and the tempting conclusion is that pressure is pushing it harder. It is not — and continuity cannot even discuss pressure, because pressure never appears in the equation. Continuity is purely kinematic: given the areas and one velocity, it fixes the other velocity, full stop. The fluid speeds up in a nozzle because the area falls and A·V must stay constant — no more, no less. Why the fluid obligingly accelerates, and what happens to the pressure when it does, is the energy equation's job (Module 5). Until then, do not let continuity tempt you into a pressure claim.
- Steady, incompressible, one inlet and one outlet: A₁V₁ = A₂V₂.
- Solve for V₂: V₂ = A₁V₁/A₂ = (0.008)(3)/(0.002) = 0.024/0.002 = 12 m/s. The area fell 4:1, so the velocity rose 4:1.
- Volume flow rate (same at every section): Q = A₁V₁ = (0.008)(3) = 0.024 m³/s (= 24 L/s).
- Check with the outlet: Q = A₂V₂ = (0.002)(12) = 0.024 m³/s. ✓ Same value.
- Mass flow rate: ṁ = ρQ = (1000)(0.024) = 24 kg/s. Equivalently ṁ = ρA₁V₁ = ρA₂V₂ = 24 kg/s at both ports.
- One inlet, one outlet, incompressible: A₁V₁ = A₂V₂.
- V₂ = A₁V₁/A₂ = (0.005)(2)/(0.001) = 0.010/0.001 = 10 m/s.
- The 5:1 area contraction produces a 5:1 velocity increase.
- Mass flow rate through a port: ṁ = ρ A V.
- ṁ = (1.2)(0.02)(5) = 0.12 kg/s.
- This is the quantity that must balance (in = out) at any steady junction — the subject of the next lesson.
Check your understanding
- Continuity is the Reynolds Transport Theorem with b = 1 (so B = mass): ∂/∂t ∫_CV ρ dV + ∫_CS ρ(V·n̂) dA = 0.
- The port forms are ṁ = ρAV (mass flow rate) and Q = AV (volume flow rate), related by ṁ = ρQ.
- Steady flow drops the accumulation term (Σṁ_out = Σṁ_in); steady incompressible flow cancels ρ and becomes a volume balance (ΣQ_out = ΣQ_in).
- For one inlet and one outlet, A₁V₁ = A₂V₂. A nozzle accelerates the fluid because its area falls — continuity is kinematic and silent on pressure.
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