6.3 Conservation of Momentum

Conservation of Momentum in Fluid Mechanics

Conservation of momentum extends Newton's second law to fluid systems flowing through a control volume. It connects the forces acting on a fluid to the rate at which momentum changes inside the control volume and the net momentum carried across its boundaries. This is the tool you'll use to find forces on pipe bends, nozzles, turbine blades, and many other engineering components.

Conservation of Momentum Equation

The starting point is Newton's second law applied to a system of fixed identity (a fixed collection of fluid particles):

$\sum F = \frac{d}{dt}(mV)_{system}$

where $F$ is the net force, $m$ is the system mass, and $V$ is velocity. To make this useful for flow problems, you convert from a system (which moves with the fluid) to a control volume (a fixed region in space) using the Reynolds Transport Theorem. That conversion gives two terms:

• Rate of change of momentum stored inside the CV: $\frac{\partial}{\partial t} \int_{CV} \rho \, V \, d\mathcal{V}$ This term captures unsteady effects. If the flow is steady, it drops to zero.

• Net momentum flux across the control surface (CS): $\int_{CS} \rho \, V \, (V \cdot \hat{n}) \, dA$ This accounts for momentum carried in and out by the flow. The outward unit normal $\hat{n}$ means outflow is positive and inflow is negative.

Combining these, the integral form of the momentum equation is:

$\sum F = \frac{\partial}{\partial t} \int_{CV} \rho \, V \, d\mathcal{V} \; + \; \int_{CS} \rho \, V \, (V \cdot \hat{n}) \, dA$

Key variables:

• $\rho$ = fluid density (mass per unit volume)
• $\hat{n}$ = outward-pointing unit normal on the control surface
• $d\mathcal{V}$ = differential volume element inside the CV
• $dA$ = differential area element on the CS

Note that this is a vector equation. In practice, you'll write separate scalar equations for the $x$-, $y$-, and $z$-directions and solve each one independently.

Forces on Control Volumes

The left-hand side, $\sum F$, includes every external force acting on the fluid inside the control volume. These fall into three categories:

Pressure forces act perpendicular to the control surface at every point:

$F_p = -\int_{CS} p \, \hat{n} \, dA$

The negative sign appears because pressure pushes inward on the fluid while $\hat{n}$ points outward. At inlets and outlets with uniform pressure, this simplifies to $p \cdot A$ applied in the appropriate direction.

Shear (viscous) forces act tangentially along the control surface, caused by velocity gradients in the fluid:

$F_s = \int_{CS} \boldsymbol{\tau} \, dA$

where $\boldsymbol{\tau}$ is the shear stress vector. In many problems you choose the control surface so that it cuts through regions of negligible shear (like at inlets and outlets with uniform flow), which lets you ignore this term.

Body forces act on the entire volume of fluid inside the CV. Gravity is the most common:

$F_b = \int_{CV} \rho \, \mathbf{g} \, d\mathcal{V} = m_{CV} \, \mathbf{g}$

where $\mathbf{g}$ is gravitational acceleration. For horizontal-flow problems with small fluid weight relative to other forces, this term is often negligible.

Momentum Equation Applications

Steady-state problems are the most common type you'll encounter. When the flow doesn't change with time, the storage term vanishes and the equation reduces to:

$\sum F = \int_{CS} \rho \, V \, (V \cdot \hat{n}) \, dA$

For discrete inlets and outlets with uniform velocity at each port, this simplifies further to:

$\sum F = \sum_{out} \dot{m} \, V \; - \; \sum_{in} \dot{m} \, V$

where $\dot{m} = \rho \, V_{avg} \, A$ is the mass flow rate at each port. This is the form you'll use most often.

Steps for solving a steady momentum problem:

1. Sketch the physical setup and choose a control volume. Draw it so that inlets and outlets are at locations where you know (or can find) pressure and velocity.
2. Draw a free-body diagram showing all external forces on the fluid inside the CV: pressure forces at every opening, the reaction force from any solid surface (pipe wall, vane, etc.), and gravity if relevant.
3. Establish a coordinate system and write the momentum equation in each relevant direction.
4. Substitute known values for $\dot{m}$, velocities, pressures, and areas. Use the continuity equation if you need to relate velocities at different ports.
5. Solve for the unknown force or velocity.

Unsteady problems retain the full time-derivative term. Examples include flow during valve opening/closing, water hammer, and filling or draining of tanks. These require tracking how conditions inside the CV change with time and are generally more involved.

Common simplifying assumptions (apply only when justified):

• Uniform velocity profiles at inlets/outlets (plug flow)
• Incompressible flow ($\rho$ = constant)
• Negligible viscous forces on the control surface
• Steady flow
• Atmospheric pressure at free surfaces or open jets

Momentum-Flux Correction Factor

Real velocity profiles are not perfectly uniform across a cross-section. The momentum-flux correction factor $\beta$ corrects for this non-uniformity so you can still use the average velocity $V_{avg}$ in your calculations.

It's defined as:

$\beta = \frac{1}{A} \int_A \left(\frac{V}{V_{avg}}\right)^2 dA$

This is the ratio of the true momentum flux to what you'd calculate assuming a flat (uniform) profile at $V_{avg}$.

Typical values:

Because turbulent flow profiles are relatively flat, $\beta$ is close to 1, and many turbulent-flow problems simply set $\beta = 1$. For laminar flow, the 33% correction is significant and shouldn't be ignored.

With $\beta$ included, the momentum equation for discrete inlets and outlets becomes:

$\sum F = \sum_{out} \beta \, \dot{m} \, V_{avg} \; - \; \sum_{in} \beta \, \dot{m} \, V_{avg}$

Or in full integral form:

$\sum F = \frac{\partial}{\partial t} \int_{CV} \rho \, V \, d\mathcal{V} \; + \; \int_{CS} \beta \, \rho \, V_{avg} \, (V_{avg} \cdot \hat{n}) \, dA$