The linear momentum equation

A reducing elbow held down by four bolts is not, in the sense Newton meant it, a body. Water arrives through one flange and leaves through another, and the fluid you would need to track to apply \(\sum F = ma\) directly is somewhere in the municipal supply a moment later. What the bolts actually feel is not the weight of anything sitting still. It is the rate at which the elbow redirects momentum that keeps passing through it, and that quantity is exactly what this topic gives you a way to compute.

From the transport theorem to momentum

Newton's second law is honest about what it is a law for: a system, meaning a fixed collection of matter, tracked wherever it goes.

The Reynolds transport theorem is the bridge from that statement to one written for a control volume — a region of space you draw once and never move, chosen so its boundary cuts cleanly across every pipe, jet or surface where matter crosses it. Applied to momentum, with the extensive property \(\mathbf{V}\) playing the role the theorem asks for, it reads

The first term on the right is the momentum currently stored inside the box; the second is the net rate at which momentum is carried across its boundary. For the steady flows this course spends most of its time on, the box's own contents stop changing and the first term drops out entirely, leaving a statement about the boundary alone. Reduce the surface integral to a handful of inlet and outlet ports, each carrying a uniform velocity and a known mass flow rate, and the whole thing collapses to arithmetic:

That "uniform velocity" is doing some quiet work. A real profile is not flat — it is closer to what you will meet in laminar and turbulent flow, and the true momentum flux through a port is somewhat larger than \(\dot m \bar V\) because faster fluid carries a disproportionate share of it. The full accounting multiplies each term by a momentum-flux correction factor \(\beta\) — around 1.03 for a turbulent profile, 4/3 for a laminar one — but for the worked cases in this article \(\beta \approx 1\) is close enough, and stating it plainly rather than quietly assuming it is the more honest habit.

One bookkeeping point pays for itself immediately. Atmospheric pressure acts on every exposed part of a control surface's exterior, and where that exterior is open to the same atmosphere throughout, its net force is zero — it pushes in from every side equally and cancels itself. That is why the pressure terms in every worked example below are gauge pressures, not absolute ones. It is not that absolute pressure would be wrong; it is that the atmospheric part of it never survives the sum, so there is no reason to carry it.

Force on a pipe bend

Take a reducing elbow lying flat, turning the flow through 90° while narrowing it: inlet diameter 15 cm, outlet 10 cm, carrying 0.1 m³/s of water. Continuity fixes both velocities before anything else is asked:

Draw the control volume so its boundary cuts across the inlet flange, the outlet flange, and nowhere else — the pipe wall and the fluid inside are both part of what is enclosed. With the losses small enough to neglect, Bernoulli between the cuts gives the outlet gauge pressure from the inlet one: a 300 kPa inlet becomes 235 kPa at the narrower outlet, the pressure spent accelerating the flow rather than lost to friction.

Set +x along the inlet and +y along the outlet, since the bend is square. Each axis then keeps exactly one velocity:

The forces on the left are the pressure pushing in at the inlet cut, the pressure pushing back at the outlet cut, and the reaction \(\mathbf{R}\) supplied by everything solid touching the fluid — the pipe wall, transmitted out through the flanges to the anchor bolts. Solve for the one unknown on each axis:

That is the force the bend must exert on the water to turn it — about 6.6 kN, mostly pushing back against the incoming flow and partly deflecting it upward. By Newton's third law the water pushes back on the bend with the same magnitude, reversed, and it is that reaction the flanges and anchor bolts are sized to carry. Nothing about the number is exotic; it is simply what turning 100 kg of water a second through a right angle costs.

Thrust from a jet

A turbojet clamped to a test stand is the same equation with the geometry stripped down to a straight line. Air is drawn in nearly at rest — the test is static, so whatever momentum the intake air carries is small enough to treat as zero — and leaves the nozzle hundreds of metres a second faster, having picked up a little mass from the fuel along the way.

With \(V_1 \approx 0\) and the fuel's own contribution to the mass flow small enough to fold in or ignore depending on how precise the answer needs to be, this reduces to \(F \approx \dot m_a V_e\). Take \(\dot m_a = 45\) kg/s and \(V_e = 500\) m/s and the engine must push 22 500 N of momentum into the exhaust every second. That force has to come from somewhere, and the only thing touching the air inside the engine is the engine itself — so by Newton's third law the air pushes back on the engine with the same 22.5 kN, and that is what the test stand's load cell reads as thrust.

A fully expanded nozzle — exit pressure equal to ambient — is what makes this the whole story. If the exhaust leaves at a pressure above or below atmospheric, an extra pressure-thrust term \((P_e - P_{\text{atm}})A_e\) belongs on the right, and a rocket operating in vacuum is the extreme case of that term refusing to vanish, since there is no ambient pressure to subtract at all.

Force on a moving vane

Every example so far used a control volume fixed in space. Nothing in the derivation actually required that — the transport theorem only needs an inertial frame, one moving at constant velocity counts exactly as well as one standing still — and letting the box move is what a turbine bucket needs.

A single vane retreating from a jet at speed \(V_c\) does not feel the jet's own speed \(V\). It feels the closing speed, the relative velocity \(V_r = V - V_c\), and — this is the detail that catches people out — only the mass flow rate corresponding to that relative speed actually reaches it:

not the nozzle's own \(\rho AV\). Some of what the nozzle emits is spent simply catching up to a vane that is running away from it. A series of vanes on a wheel is the opposite case — as one bucket moves on, the next is already there to take its place, so the full nozzle flow rate is intercepted after all, just still at the relative speed \(V_r\). Mixing the two up, using the full flow rate for a single vane or the reduced one for a wheel, is a specific, common way to get this topic wrong.

The vane turns the relative jet through an angle \(\theta\) without changing its speed — there is no pressure difference across a free jet to slow it down — so the momentum equation in the vane's own frame gives the force on the vane directly:

With a 30 m/s jet of area 0.003 m² striking a vane retreating at 10 m/s and turning the flow through 150°, \(V_r = 20\) m/s, \(\dot m_r = 60\) kg/s, and \(F \approx 2.24\) kN. Only the component along the vane's direction of travel does any work; multiply it by \(V_c\) and 2.24 kN becomes 22.4 kW delivered to the vane, which is the entire operating principle of an impulse turbine.

The sign conventions

Almost every mistake with this equation is a sign mistake, and almost every one of them has the same root: the formula already does some of the sign-tracking for you, and re-doing it by intuition on top double-counts it.

Fix positive x and y once, at the start of a problem, in whatever directions are convenient — usually along the inlet and outlet — and never revisit that choice. Every velocity, in or out, is then a signed number along those fixed axes. \(\sum F_x = \dot m(V_{2x} - V_{1x})\) already subtracts the inlet term; it does that regardless of which way the inlet flow happens to point, because the vector \(V_{1x}\) carries its own sign. Flipping the sign again because the flow "enters" rather than "leaves" is applying the same correction twice.

The force \(\sum \mathbf{F}\) on the left is the net force on the fluid: gauge pressure at every cut, weight, and whatever reaction the walls, a vane or a mount supply. It is never the force on the hardware directly. Solve the equation for the reaction on the fluid first, and only at the very last step invoke Newton's third law and reverse it to get the force the fluid exerts back — folding that reversal into the middle of the algebra is the single most common way this topic goes wrong on an exam.

The gauge-pressure trick from the first section has a matching limit worth stating plainly: it works because atmospheric pressure cancels on the parts of the control surface exposed to it evenly. A control volume that is not uniformly exposed — part of it inside a pressurised chamber, say, or cutting through the working section of a wind tunnel where the surrounding pressure is not atmospheric — does not get that cancellation for free, and absolute pressures have to be carried through instead.