Linear Momentum Equation: Fluid Mechanics

Newton's second law for a control volume — from the Reynolds Transport Theorem with b = V, to the steady F = ṁ(V₂ − V₁).

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⏱️ About 18 min

Continuity (Module 4) told you how fast a fluid goes; the energy equation (Module 5) will tell you why. But neither of them says anything about <em>force</em>. When a flowing fluid rounds a bend, slams through a nozzle, or blasts out the back of a rocket, something has to hold it on course — and the size of that something is exactly what the <strong>linear momentum equation</strong> predicts. It is Newton's <em>F = ma</em> rewritten for a control volume, and like continuity and energy it drops straight out of the Reynolds Transport Theorem — this time by setting the intensive property <em>b = V</em>. The catch, and the single biggest source of mistakes in the whole subject, is sign: the force the equation hands you is the force <em>on the fluid</em>, and the force on the pipe or vane is its reaction. Get that flip right and the rest is arithmetic.

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The big idea: Setting <em>b = V</em> (so <em>B = mV</em>, linear momentum) in the Reynolds Transport Theorem yields the <strong>momentum equation for a control volume</strong>: the sum of all external forces acting <em>on the fluid inside the CV</em> equals the rate at which momentum accumulates inside plus the net momentum flux out — <em>ΣF = ∂/∂t ∫_CV ρV dV + ∫_CS V ρ(V·n̂) dA</em>. For <strong>steady</strong> flow the accumulation term vanishes, and for one inlet and one outlet this collapses to the compact vector form <em>ΣF = ṁ(V₂ − V₁)</em>. The forces ΣF include both the <strong>pressure</strong> acting on each port of the control surface (use <em>gauge</em> pressure) and the force the walls or structure exert on the fluid. The force on the bend, nozzle, or vane is the <em>reaction</em> to that wall force — a sign flip.
🎯 By the end, you'll be able to
  • Derive the linear momentum equation for a control volume as the b = V (momentum) case of the Reynolds Transport Theorem
  • Reduce it to the steady one-inlet/one-outlet vector form ΣF = ṁ(V₂ − V₁) and its x- and y-component forms
  • Include pressure forces on the control-surface ports using gauge pressure, and explain why absolute pressure is not used
  • Distinguish the force on the fluid (what the equation gives) from the force on the structure (the reaction), and compute an anchor force on a reducing bend
📎 Helpful to know first

Momentum is the Reynolds Transport Theorem with b = V

Two modules ago we built a single generic tool, the Reynolds Transport Theorem, and turned it into a factory for conservation laws. The recipe was: pick the extensive property B, divide by mass to get the intensive property b = dB/dm, plug into RTT. With b = 1 we got continuity (mass); with b = e we get energy (Module 5). Now set b = V, so that B = mV — the system's linear momentum — and out drops Newton's second law written for a region of space rather than for a blob of matter.

Here is the move in one line. The left side of RTT is dB/dt of a system; Newton's second law says the rate of change of a system's momentum equals the net external force on it, d(mV)/dt = ΣF. Setting b = V in RTT and applying this identity replaces the abstract dB/dt with a concrete sum of forces. What survives is the momentum equation for a control volume: the net external force on the fluid inside the CV equals the rate at which momentum piles up inside it plus the net momentum streaming out across its boundary.

Read that sentence again, slowly. It says force balances a flow of momentum, not a stock of mass. That is the whole conceptual shift from particle mechanics to fluid mechanics: the same F = ma, but 'ma' has become a momentum-flux balance over a region. Everything in this module — anchor forces, jet thrust, sprinkler torque — is one or another specialization of that single sentence.

\[ \sum\mathbf{F}=\frac{\partial}{\partial t}\int_{CV}\rho\mathbf{V}\,d\mathcal{V}+\int_{CS}\mathbf{V}\,\rho(\mathbf{V}\cdot\hat{\mathbf{n}})\,dA \]
The momentum equation for a control volume — the b = V case of the Reynolds Transport Theorem, using Newton's second law d(mV)/dt = ΣF. The left side is the sum of ALL external forces acting on the fluid inside the CV. The first term on the right is the rate at which momentum accumulates inside the CV (zero for steady flow); the second is the net momentum flux out across the control surface. It is a vector equation, so it holds component-by-component.
🔑 Three siblings, one theorem

This is the same Reynolds Transport Theorem you already trust, specialized a third way. Continuity (M4) was b = 1; the energy equation (M5) is b = e; momentum is b = V. The factory produces a new conservation law for each choice of b, and the structure — accumulation inside the CV plus flux across its boundary — is identical every time. If you understood continuity's derivation, you already understand this one; only the bookkeeping quantity (mass, energy, momentum) changes.

The steady, one-port form: ΣF = ṁ(V₂ − V₁)

The integral form is fully general but, like continuity, it is rarely used as-is. Two simplifications cover almost every engineering problem.

The first is steady flow: nothing changes with time at any fixed point, so the accumulation term ∂/∂t ∫_CV ρV dV vanishes. The second is the single-inlet, single-outlet geometry of a length of pipe, a nozzle, or a vane. When the flow through each port is steady and roughly uniform, the surface integral collapses to a difference between the outlet and inlet momentum fluxes, giving the compact vector form

That little equation does an enormous amount of work. ṁ is the mass flow rate (the same ṁ = ρAV from continuity, identical at inlet and outlet for steady flow); V₁ and V₂ are the average velocity vectors at the inlet and outlet. The equation is a vector equation, so it splits cleanly into components: ΣF_x = ṁ(V₂ₓ − V₁ₓ) and ΣF_y = ṁ(V₂ᵧ − V₁ᵧ). Pick your axes to align with the flow directions and most of the bookkeeping disappears.

Notice what is and is not in the equation. The velocities and the mass flow are kinematics — continuity's job. What momentum adds is the force: it tells you the net push or pull the surroundings must supply to make the fluid change direction or speed by that amount. A jet that merely speeds up demands one force; one that also turns ninety degrees demands a very different force. Both come from the same equation, applied component by component.

\[ \sum\mathbf{F}=\dot m\,(\mathbf{V}_{2}-\mathbf{V}_{1})\qquad\Longleftrightarrow\qquad \sum F_x=\dot m(V_{2x}-V_{1x}),\quad \sum F_y=\dot m(V_{2y}-V_{1y}) \]
Steady, one-inlet/one-outlet form. The net external force on the fluid equals the mass flow rate times the change in velocity vector from inlet to outlet. It is a vector equation: apply it once per coordinate direction. ṁ is the same at both ports (continuity); only the direction and magnitude of V differ.
⚠️ ΣF is the force ON THE FLUID — flip the sign for the structure

This is the trap that catches almost everyone, and it is worth a warning box of its own. The ΣF on the left side of ΣF = ṁ(V₂ − V₁) is the sum of all external forces acting on the fluid inside the control volume. When the question asks for the force on the pipe bend, noz­zle, or vane — which it usually does — you must take the reaction: the force the fluid exerts on the structure is equal and opposite to the force the structure exerts on the fluid. Solve for the wall force R in the momentum balance (a force on the fluid), then report −R as the force on the structure, and R again (with its sign) as the force an anchor must supply to hold the structure still. The arithmetic is trivial; the sign discipline is everything.

A control volume (dashed outline) around a horizontal 90-degree reducing elbow. Water enters along +x at port 1 (area A1, gauge pressure p1, velocity V1) and leaves along +y at port 2 (area A2 smaller than A1, gauge pressure p2, velocity V2). Three force families act ON THE FLUID: the inlet pressure p1 A1 pushing in the +x direction, the outlet pressure p2 A2 pushing back in the minus-y direction, and the wall reaction R with components Rx and Ry that the elbow exerts on the fluid. The force the anchor must apply to the elbow is minus R. V₁ inlet (1): A₁, p₁ V₂ outlet (2) A₂, p₂ CV = fluid in elbow p₁A₁ (→ +x) p₂A₂ (↓ −y) R (wall → fluid) components Rx, Ry ΣF on fluid = ṁ(V₂ − V₁); force on elbow = −R
A reducing elbow as a control volume. The forces on the fluid are the inlet pressure p₁A₁ (pushing in, +x), the outlet pressure p₂A₂ (pushing back in, −y), and the wall reaction R = (Rx, Ry). The momentum equation ΣF = ṁ(V₂ − V₁) solves for R; the anchor must then supply −R to hold the elbow. Note the outlet area is smaller than the inlet — V₂ is larger than V₁ by continuity.

Pressure on the boundary: why we use gauge pressure

The 'external forces' in ΣF are not only the wall reaction. The fluid at every port of the control surface is acted on by the pressure there, and those pressure forces must enter the balance alongside the momentum-flux terms. Omit them and you will get the wall force badly wrong — usually by an order of magnitude, because pressure forces on a pipe cross-section are often larger than the momentum-flux terms themselves.

The clean way to include them is with gauge pressure, not absolute. Here is why. Atmospheric pressure acts on the outside of the entire structure as well as on the fluid, and a uniform pressure all around a closed surface produces zero net force. So the atmospheric part cancels in the real force balance on the structure; what remains is only the part of the pressure that differs from atmosphere — which is exactly what gauge pressure measures. Using gauge pressure at every port does the cancellation for you automatically. (When a port discharges to the open atmosphere, its gauge pressure is simply zero.)

So the full force balance on the fluid reads: the pressure force at each port (gauge pressure × port area, directed into the CV — that is, opposite to the outward normal at that port), plus the wall reaction, plus weight if the CV is not in a horizontal plane, equals ṁ(V₂ − V₁). The anchor force follows by flipping the sign of the wall reaction.

\[ \underbrace{p_{1}A_{1}(\text{into CV})+p_{2}A_{2}(\text{into CV})+\mathbf{R}+m\mathbf{g}}_{\text{all forces ON THE FLUID}}=\dot m(\mathbf{V}_{2}-\mathbf{V}_{1}) \]
The full steady force balance on the fluid in a CV. Each port contributes a pressure force (gauge pressure × area) directed into the control volume — opposite to the outward normal, so an inlet port's pressure pushes with the flow and an outlet port's pressure pushes against it. R is the force the walls exert on the fluid; the force on the structure (and the anchor force) is −R. Weight mg enters only if the CV has vertical extent.
📝 Worked example: Water (ρ = 1000 kg/m³) flows steadily through a horizontal 90° reducing elbow. Inlet: A₁ = 0.008 m², V₁ = 5 m/s (along +x), p₁ = 100 kPa gauge. Outlet: A₂ = 0.002 m² (along +y), p₂ = 50 kPa gauge. Find the force the anchor must exert on the elbow to hold it stationary (components and magnitude).
  1. Mass flow rate (continuity): ṁ = ρA₁V₁ = 1000(0.008)(5) = 40 kg/s. (Check: ρA₂V₂ with V₂ = (A₁/A₂)V₁ = 4(5) = 20 m/s gives 1000(0.002)(20) = 40 kg/s. ✓)
  2. Choose axes: inlet along +x, outlet along +y. So V₁ = (5, 0) and V₂ = (0, 20) m/s.
  3. Momentum balance on the FLUID, x: ΣF_x = ṁ(V₂ₓ − V₁ₓ) = 40(0 − 5) = −200 N.
  4. Forces on the fluid in x: inlet pressure p₁A₁ = (100000)(0.008) = 800 N in +x; outlet pressure has no x-component; wall reaction Rx. So 800 + Rx = −200 → Rx = −1000 N.
  5. Momentum balance, y: ΣF_y = ṁ(V₂ᵧ − V₁ᵧ) = 40(20 − 0) = 800 N.
  6. Forces on the fluid in y: outlet pressure p₂A₂ = (50000)(0.002) = 100 N directed INTO the CV, i.e. −y (opposite the +y outflow); wall reaction Ry. So −100 + Ry = 800 → Ry = 900 N.
  7. Wall force on the fluid: R = (Rx, Ry) = (−1000, 900) N. The elbow pushes the fluid in −x and +y — exactly what is needed to remove its x-momentum and give it +y momentum.
  8. Force on the elbow by the fluid = −R = (1000, −900) N. To hold the elbow still, the anchor must supply the opposite: F_anchor = (−1000, 900) N.
  9. Magnitude: |F_anchor| = √(1000² + 900²) = √1 810 000 ≈ 1345 N.
✓ Anchor force = (−1000 N, +900 N); magnitude ≈ 1345 N. The fluid tends to push the elbow in +x and −y; the anchor must pull back in −x and +y.
✏️ Practice: A straight horizontal reducing nozzle has inlet A₁ = 0.010 m², outlet A₂ = 0.002 m², inlet velocity V₁ = 4 m/s (both ports along +x). The inlet gauge pressure is p₁ = 50 kPa and the outlet discharges to atmosphere (p₂ = 0). Find the magnitude of the horizontal force the fluid exerts on the nozzle, in N. (The nozzle accelerates the flow; expect a backward reaction.)
N
Solution
  1. Mass flow: ṁ = ρA₁V₁ = 1000(0.010)(4) = 40 kg/s. By continuity V₂ = (A₁/A₂)V₁ = 5(4) = 20 m/s.
  2. Both ports along +x: ΣF_x on fluid = ṁ(V₂ − V₁) = 40(20 − 4) = 640 N.
  3. Forces on fluid in x: inlet pressure p₁A₁ = 50000(0.010) = 500 N (+x); outlet p₂ = 0; wall reaction Rx. So 500 + Rx = 640 → Rx = +140 N (the nozzle pushes the fluid forward).
  4. Force on the nozzle by the fluid = −Rx = −140 N: a 140 N reaction pointing in −x. The accelerating fluid drags the nozzle backward — the reaction-thrust idea we develop fully in Lesson 3.
✏️ Practice: Water flows steadily through a 60° reducing bend in a horizontal plane. Inlet A₁ = 0.005 m², V₁ = 4 m/s along +x, p₁ = 60 kPa gauge. Outlet A₂ = 0.002 m², p₂ = 0 (to atmosphere), leaving at 60° above the +x axis. Find the x-component of the force the fluid exerts on the bend, in N (take cos 60° = 0.5).
N
Solution
  1. Continuity: ṁ = ρA₁V₁ = 1000(0.005)(4) = 20 kg/s; V₂ = (A₁/A₂)V₁ = 2.5(4) = 10 m/s.
  2. Outlet velocity components: V₂ₓ = V₂ cos 60° = 10(0.5) = 5 m/s; V₂ᵧ = V₂ sin 60° = 8.66 m/s.
  3. Momentum on fluid, x: ΣF_x = ṁ(V₂ₓ − V₁ₓ) = 20(5 − 4) = 20 N.
  4. Forces on fluid in x: inlet pressure p₁A₁ = 60000(0.005) = 300 N (+x); outlet p₂ = 0; wall Rx. So 300 + Rx = 20 → Rx = −280 N.
  5. Force on bend by fluid, x = −Rx = +280 N. The incoming jet's recoil dominates: the fluid pushes the bend in +x.

Check your understanding

1. The linear momentum equation for a control volume is obtained from the Reynolds Transport Theorem by setting the intensive property b to:
Linear momentum per unit mass is the velocity V, so b = V and B = mV. With Newton's second law d(mV)/dt = ΣF, RTT gives the momentum equation ΣF = accumulation + net momentum flux. (b = 1 → continuity; b = e → energy; b = r × V → angular momentum, Lesson 4.)
2. In the steady one-inlet/one-outlet form ΣF = ṁ(V₂ − V₁), the ΣF on the left is:
ΣF is the net external force on the fluid: pressure (gauge) on each port plus the wall/anchor reaction plus weight. Solve for the wall force R; the force on the structure and the anchor force are −R. Confusing 'force on the fluid' with 'force on the structure' is the single most common error here.
3. When applying the momentum equation to a control volume, pressure forces on the ports are evaluated using:
A uniform pressure on a closed surface produces zero net force, so the atmospheric contribution cancels between the fluid and the outside of the structure. Gauge pressure subtracts it automatically. A port open to atmosphere therefore has gauge pressure zero and contributes no pressure force.
🔎 Related lessons
The Reynolds Transport Theorem · soon The Continuity Equation Forces on Bends, Nozzles & Vanes Jet Propulsion & Reaction Forces · soon
✅ Key takeaways
  • Momentum is the Reynolds Transport Theorem with b = V (so B = mV): ΣF = ∂/∂t ∫_CV ρV dV + ∫_CS V ρ(V·n̂) dA, using Newton's second law.
  • For steady, one-inlet/one-outlet flow this reduces to the vector form ΣF = ṁ(V₂ − V₁), applied component by component.
  • ΣF is the force ON THE FLUID: gauge pressure on each port plus the wall reaction (plus weight). The force on the structure and the anchor is the reaction, −R.
  • Use gauge pressure so the uniform atmospheric contribution cancels; a port open to atmosphere has gauge pressure zero.
➡️ You now hold the momentum equation in its general and steady forms, and you have survived the sign-discipline trap. The next lesson applies that equation to its most visual case — a free jet striking a stationary vane — and watches the force on the vane swing as the deflection angle changes, all the way to the classic 'twice the momentum flux' limit when the jet is turned fully back on itself.
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