12.4 Impulse-Momentum Equation and Forces on Control Volumes

12.4 Impulse-Momentum Equation and Forces on Control Volumes

The NCEES sub-topic is "Energy, impulse, and momentum equations (e.g., Bernoulli equation)." The energy half is covered in the continuity and Bernoulli section; this section covers the momentum half, which answers a different question: not how fast the fluid moves, but what force it exerts on the hardware. Anchor blocks, thrust blocks, nozzle reactions, and turbine blade forces are all momentum problems.

The Steady-Flow Momentum Equation

$\sum \vec{F} = \dot{m}\left(\vec{V}_{\text{out}} - \vec{V}_{\text{in}}\right), \qquad \dot{m} = \rho Q = \rho A V$

Applied component by component:

$\sum F_x = \dot{m}(V_{2x} - V_{1x}), \qquad \sum F_y = \dot{m}(V_{2y} - V_{1y})$

The forces on the left side include every external force on the control volume:

Use gauge pressures. Atmospheric pressure acts uniformly on the outside of the fitting and cancels, so working in gauge pressure automatically accounts for it. Mixing absolute and gauge is a reliable way to get a wrong anchor force.

Why Momentum Succeeds Where Bernoulli Fails

This is the key insight of the topic. Across a sudden expansion, a hydraulic jump, or a partly closed valve, the internal flow is violently turbulent and energy is destroyed in ways you cannot compute from geometry. Bernoulli's equation is invalid there. The momentum equation does not care: it needs only what crosses the control surface, so it works across the most dissipative device imaginable. When a question asks for a force across a lossy transition, momentum is the only usable tool.

Forces on Pipe Bends

For a bend turning flow through angle $\theta$ in the horizontal plane, with the inlet along $+x$:

$R_x = \dot{m}(V_2\cos\theta - V_1) - p_1A_1 + p_2A_2\cos\theta$ $R_y = \dot{m}(V_2\sin\theta - 0) + p_2A_2\sin\theta$

A bend generates force even at constant speed. In a constant-diameter bend $V_1 = V_2$ in magnitude, so nothing changes energetically — yet the momentum vector rotates, and that requires a force. This is why a 90° elbow in a large water main needs a thrust block even though the pressure and velocity are the same on both sides.

Worked Example: Anchor Force on a 90° Reducing Bend

A horizontal 90° bend reduces from $D_1 = 300$ mm to $D_2 = 200$ mm. Water ($\rho = 1{,}000$ kg/m³) flows at $Q = 0.15$ m³/s. Inlet gauge pressure is 250 kPa; outlet gauge pressure is 220 kPa. Inlet flow is along $+x$, outlet along $+y$. Find the anchoring force.

Areas and velocities:

$A_1 = \frac{\pi(0.300)^2}{4} = 0.07069\ \text{m}^2, \qquad V_1 = \frac{0.15}{0.07069} = 2.122\ \text{m/s}$ $A_2 = \frac{\pi(0.200)^2}{4} = 0.03142\ \text{m}^2, \qquad V_2 = \frac{0.15}{0.03142} = 4.775\ \text{m/s}$ $\dot{m} = \rho Q = 1{,}000(0.15) = 150\ \text{kg/s}$

$x$ direction (outlet has no $x$ velocity, and $\theta = 90°$ so $\cos\theta = 0$):

$R_x + p_1A_1 = \dot{m}(0 - V_1)$ $R_x = -\dot{m}V_1 - p_1A_1 = -150(2.122) - 250{,}000(0.07069)$ $R_x = -318.3 - 17{,}672 = -17{,}990\ \text{N}$

$y$ direction (inlet has no $y$ velocity):

$R_y - p_2A_2 = \dot{m}(V_2 - 0)$ $R_y = \dot{m}V_2 + p_2A_2 = 150(4.775) + 220{,}000(0.03142)$ $R_y = 716.3 + 6{,}912 = 7{,}628\ \text{N}$

Resultant:

$R = \sqrt{(17{,}990)^2 + (7{,}628)^2} = \sqrt{3.236\times10^8 + 5.819\times10^7} = \sqrt{3.818\times10^8} = \boxed{19{,}540\ \text{N} \approx 19.5\ \text{kN}}$

$\phi = \arctan\!\left(\frac{7{,}628}{17{,}990}\right) = 23.0° \text{ from the } -x \text{ axis}$

Note the dominance of the pressure terms. The momentum flux contributions are only 318 N and 716 N, while the pressure-area terms are 17{,}672 N and 6{,}912 N — over 20 times larger. For pressurized liquid systems, the anchor force is essentially a pressure-times-area problem, and neglecting the momentum flux entirely would introduce under 5% error. The reverse is true for high-velocity, low-pressure jets, where momentum flux dominates. Recognizing which regime you are in is a fast sanity check on the magnitude of your answer.

Jets Striking Surfaces

For a jet of area $A$ and velocity $V$, with $\dot{m} = \rho A V = \rho Q$:

The 180° result is worth remembering: reversing a jet gives twice the force of stopping it, because the momentum change is from $+V$ to $-V$, a total of $2V$. This is the principle of the Pelton turbine bucket, and it is why the buckets are split to turn the jet nearly back on itself.

Worked Example: Jet on a Moving Vane and Maximum Power

A 40 mm diameter water jet at 30 m/s strikes a single vane that deflects it 180° and moves away at 12 m/s. Find the force and power, then the velocity that maximizes power.

$A = \frac{\pi(0.040)^2}{4} = 1.257\times10^{-3}\ \text{m}^2$

Relative velocity governs the interaction: $V_{\text{rel}} = 30 - 12 = 18$ m/s.

$F = 2\rho A(V-u)^2 = 2(1{,}000)(1.257\times10^{-3})(18)^2 = 2(1{,}000)(1.257\times10^{-3})(324) = 814.6\ \text{N}$

$P = Fu = 814.6(12) = 9{,}775\ \text{W} = \boxed{9.78\ \text{kW}}$

Maximizing power. With $P = 2\rho A(V-u)^2u$, differentiate with respect to $u$:

$\frac{dP}{du} = 2\rho A\left[(V-u)^2 - 2u(V-u)\right] = 2\rho A(V-u)\left[(V-u) - 2u\right] = 0$

$V - 3u = 0 \;\Rightarrow\; u = \frac{V}{3} = 10\ \text{m/s}$

$P_{\max} = 2(1{,}000)(1.257\times10^{-3})(30-10)^2(10) = 2(1{,}000)(1.257\times10^{-3})(400)(10) = 10{,}056\ \text{W} = 10.06\ \text{kW}$

So $u = V/3$ for a single vane, giving 10.06 kW versus 9.78 kW at 12 m/s — only 3% better, showing the optimum is quite flat.

A caution the exam exploits: for a series of vanes on a turbine wheel, where the jet is continuously intercepted, the force is $\rho AV(V-u)$ rather than $\rho A(V-u)^2$, and the optimum shifts to $u = V/2$ with 50% theoretical efficiency for a flat vane. The $V/2$ result is the one usually quoted for Pelton wheels. Read whether the problem describes one vane or a wheel.