Fluid Forces on Objects: Linear Momentum Approach

Hello! Welcome to your next lesson in fluid mechanics.

In our previous lessons, we've focused on the energy within a fluid system, using the extended Bernoulli equation to account for energy additions by pumps and energy losses due to friction (major losses) and components (minor losses). We were analyzing the fluid's state inside the pipes and ducts.

Today, we shift our perspective. Instead of looking only at the energy within the flow, we will analyze the forces that the moving fluid exerts on its surroundings. This is a critical topic with direct applications in your area of interest, aerospace engineering, as it forms the basis for calculating aerodynamic forces and engine thrust.

Your learning outcome for this lesson is to apply the linear momentum equation to determine forces exerted by flowing fluids on objects. By the end of our session, you'll be able to calculate the forces on pipe bends, vanes, and even estimate the thrust produced by a jet engine.

1. From Newton's Law to the Linear Momentum Equation

You'll recall from physics that Newton's Second Law states that the net force on an object equals its mass times acceleration ($\vec{F}=m\vec{a}$). A more general form of this law, which is better suited for fluid mechanics, is that force equals the time rate of change of linear momentum.

$\vec{F}=\frac{d(m\vec{v})}{dt}$

For a fluid flowing through a system (like a pipe bend or an engine), it's more practical to analyze a fixed region in space, which we call a control volume (CV), rather than trying to track individual fluid particles. By applying this principle to a control volume, we arrive at the linear momentum equation.

The video below provides a concise derivation of the equation we'll be using. You don't need to memorize the derivation, as you prefer a formula-based approach. Instead, focus on understanding the final "workhorse" equation and the components of force it includes.

2. The "Workhorse" Equation and Its Components

As summarized in the video, for a steady flow with defined inlets and outlets, the linear momentum equation is a vector equation that can be written as:

$\sum \vec{F}={\sum }_{\text{out}}\beta \dot{m}\vec{v}-{\sum }_{\text{in}}\beta \dot{m}\vec{v}$

Let's break this down:

• $\sum \vec{F}$: This is the vector sum of all external forces acting on the fluid inside the control volume. It's the left-hand side of our equation and includes:

  • Pressure forces: Caused by pressure acting on the inlet and outlet areas ($P\times A$).
  • Viscous forces: Friction forces exerted by pipe or vane walls on the fluid.
  • Body forces: Typically gravity acting on the mass of the fluid in the control volume (i.e., its weight).
  • Reaction forces (${\vec{F}}_R$): These are the forces exerted by solid surfaces (like bolts, supports, or vanes) on the fluid to change its direction or speed. This is often the force we want to find. By Newton's Third Law, the force of the fluid on the object is equal and opposite to this reaction force.

• $\dot{m}\vec{v}$: This term represents the momentum flux—the rate at which momentum is carried by the fluid across the control surface.

  • $\dot{m}$ is the mass flow rate ($\rho Av$), which you know from the continuity equation.
  • $\vec{v}$ is the average velocity vector of the flow.
  • $\beta$ is the momentum flux correction factor, which we'll usually assume is close to 1 for turbulent flow unless stated otherwise.

• The Right-Hand Side: ${\sum }_{\text{out}}\beta \dot{m}\vec{v}-{\sum }_{\text{in}}\beta \dot{m}\vec{v}$ represents the net rate of momentum flow out of the control volume. It's the change in momentum between the outlets and inlets.

A Crucial Note on Signs and Direction

The momentum equation is a vector equation. This means you must solve it for each coordinate direction (e.g., x, y, z) separately.

$\sum F_x={\sum }_{\text{out}}\beta \dot{m}{v}_x-{\sum }_{\text{in}}\beta \dot{m}{v}_x$
$\sum F_y={\sum }_{\text{out}}\beta \text{dotm}{v}_y-{\sum }_{\text{in}}\beta \text{dotm}{v}_y$

A common point of confusion is the sign convention for the momentum flux terms. The video below offers a very practical and memorable way to get the signs right every time.

3. Application 1: Force on a Stationary Vane

Let's start with a classic example: a jet of water hitting a stationary plate. This scenario isolates the momentum change, as pressure is atmospheric everywhere and we can often neglect gravity.

The general strategy is:

1. Draw a Control Volume (CV): The CV should enclose the object and cut through the fluid streams at the inlet and outlet(s).
2. Set up a Coordinate System: Define your positive x and y axes.
3. Apply the Momentum Equation: Write the equation for each direction, identifying all forces and momentum terms.

Watch how this is done for a jet hitting a vertical plate.

As the video shows, the force on the fluid in the x-direction is:
$\sum F_x=(\dot{m}{v}_x{)}_{\text{out}}-(\dot{m}{v}_x{)}_{\text{in}}$
Assuming the reaction force from the plate on the fluid is $-F_{plate}$, the outlet velocity in the x-direction is 0, and the inlet velocity is $v_j$:
$-F_{plate}=(\dot{m}\times 0)-(\dot{m}\times v_j)$
$-F_{plate}=-\dot{m}{v}_j$
The force exerted by the fluid on the plate is equal and opposite, so $F_{plate}=\dot{m}{v}_j$.

Test your understanding!

A 2 kg/s jet of water moving at 20 m/s strikes a stationary flat plate. The plate is angled so that the jet's initial direction is 30° to the plate's surface normal, as shown in the diagram. What is the magnitude of the force exerted by the jet on the plate? (Assume the water flows off the plate parallel to its surface, and neglect friction).

Show answer

The most effective way to solve this is to align one of our coordinate axes normal to the plate (let's call it the n-axis). The force on the plate is due to the change in the fluid's momentum in this normal direction.

1. Velocity Components: The initial velocity of the jet normal to the plate is $v_{n,\text{in}}=v_j\cos (30^{\circ })$. Since the water flows off parallel to the plate, the final velocity normal to the plate is $v_{n,\text{out}}=0$.

2. Mass Flow Rate: We are given $\dot{m}=2\text{ kg/s}$.

3. Momentum Equation (n-direction): The force from the plate on the fluid, $F_n$, is what causes the change in momentum.
   $\sum F_n=\dot{m}(v_{n,\text{out}}-v_{n,\text{in}})$
   $F_n=(2\text{ kg/s})\times (0-20\text{ m/s}\times \cos (30^{\circ }))$
   $F_n=-2\times (20\times 0.866)=-34.64\text{ N}$
   This is the force the plate exerts on the fluid (in the negative n-direction).

4. Reaction Force: The force the fluid exerts on the plate is equal and opposite.
   $F_{on\_plate}=-F_n=34.64\text{ N}$
   The magnitude of the force on the plate is 34.64 N.

4. Application 2: Force on a Pipe Bend

Now let's consider a more complex case where both pressure and momentum contribute to the force. When a fluid flows through a pipe bend, forces are required to both change the fluid's direction (a momentum change) and contain its pressure.

This example clearly shows the power of breaking the problem into components and systematically accounting for all forces and momentum changes.

5. Application 3: Thrust of a Jet Engine

The principles we've discussed are directly applicable to your primary interest: aerospace engineering. The thrust generated by a jet engine or rocket is a reaction force that can be calculated using the linear momentum equation.

To see how this works, we'll analyze a turbojet on a test stand.

The general thrust equation for an air-breathing jet engine is:
$F_{thrust}=({\dot{m}}_{air}+{\dot{m}}_{fuel})v_{exhaust}-{\dot{m}}_{air}{v}_{inlet}+(P_{exhaust}-P_{ambient})A_{exhaust}$

This equation is a direct application of our workhorse formula, $\sum F=(\dot{m}v{)}_{out}-(\dot{m}v{)}_{in}$, where:

• $\sum F$ includes the thrust (the reaction force) and pressure forces.
• The mass flow rate out is the sum of air and fuel (${\dot{m}}_{out}={\dot{m}}_{air}+{\dot{m}}_{fuel}$).

The following resource from Embry-Riddle Aeronautical University provides a clear explanation and a worked example.

This principle is identical for a rocket engine, as shown in the NASA diagram below. The main difference is that a rocket carries its own oxidizer, so there is no inlet mass flow rate of air.

Conclusion

Today, we've unlocked a powerful tool for fluid analysis. The linear momentum equation allows us to calculate the forces that result from fluid motion, bridging the gap between fluid dynamics and solid mechanics.

Key Takeaways:

• The linear momentum equation is an application of Newton's Second Law to a fluid control volume.
• The total external force on the fluid in a control volume equals the net rate of momentum flowing out of it.
• Forces include pressure, viscous, body (gravity), and reaction forces. The reaction force is often the unknown we want to find.
• The equation must be solved as a vector equation, typically by breaking it down into x and y components.
• This single principle is used to calculate forces on diverse objects, from simple pipe bends to the thrust of a jet engine.

Preview of the Next Lesson:
We've learned how to calculate the total force exerted by a fluid. In our next lesson, we will focus specifically on objects immersed in an external flow, like an aircraft wing. We'll learn how to decompose the total force into two critical components, lift and drag, and how to use dimensionless coefficients to analyze and predict aerodynamic performance.