CHAPTER 6 MOMENTUM PRINCIPLE Dr. Ercan Kahya Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved 

MOMENTUM PRINCIPLE Newton’s second law; for a single particle in terms of momentum; for a single particle For a system of particles – Lagrangian form b: intensive property is momentum per unit mass = m v / m = v Momentum Principle Reynolds Transport Theorem for momentum Combining the last two equations:

This equation states that the sum of the external forces acting on the material in the control volume equals the rate of momentum change inside the control volume (aka momentum accumulation) plus net rate at which momentum flows out of the control volume. If there is no momentum accumulation; Zero momentum accumulation is common for many problems in fluid mechanics MOMENTUM PRINCIPLE

Reynolds Transport Theorem: Simplified form If the flow crossing the control surface occurs through a number of inlet and outlet ports, and the velocity v is uniformly distributed (constant) across each port; then In Cartesian coordinate (x,y,z) system, the component form of the momentum equation in x-direction :

FORCE DIAGRAMS Forces associated with flow in a pipe: (a) pipe schematic, (b) control volume situated inside the pipe, (c) control volume surrounding the pipe. Force Terms In (b): The fluid within the control volume has been isolated from its surroundings and the effects of the surroundings are shown as forces. Forces in z direction: pressure, shear, and weight In (c): The control volume cuts through the pipe wall. Forces in z direction: pressure, tension (F 1, F 2 ), and weight (fluid +wall)

FORCE TYPE Body Force: A force that acts on mass elements within the body and acts at a distance without any physical contact. – Gravitational – Electrostatic – Magnetic Surface Force: A force that requires physical contact, acting at the control surfaces – All forces except weight in figure (b) and (c) – For example: hydrostatic pressure – pressure integrated over the area of control surface

Momentum Equation: Useful Form for Steady Flow This is the key equation to Chapter 6. Sign convention is important ! Note the flow direction, and include that in the velocity term y x For each flow stream: is it going “in” or “out” of the CS? In Out

Systematic Approach Problem setup Select a control volume. Select coordinate axes. Select an inertial reference frame. Identify governing equations (scalar vs. vector), and the other equations (Bernoulli, continuity) may also be applied. Force analysis and diagram Sketch body forces on the force diagram Sketch surface forces on the force diagram Momentum analysis and diagram Evaluate the momentum accumulation term. If the flow is steady and other materials in the CV are stationary, it is zero. Sketch momentum flow vectors on the momentum diagram. For uniform velocity, each vector is d mV / dt

Typical Momentum Applications Fluid Jets Nozzles Vanes Pipe Bends

EXAMPLE : Momentum Application (Q 6.8) A 15 m/s jet of water (diameter 30 mm) is filling a tank. The tank has a mass of 5 kg, and contains 20 liters of water as shown. The water temp is 15 deg C. Find: ‐ Force acting on the stop block. ‐ Force acting on the bottom of the tank

SOLUTION

SOLUTION………….cont.

Moving Control Volumes Important reminder: V : fluid velocity relative to the control surface where the flow crosses the surface. It is always measured w.r.t. the control surface because it relates to the mass flux across the surface. v : velocity used to define momentum “mv” of any fluid particle of mass “m” in the system & is relative to an inertial reference frame. This frame does not rotate and can be either stationary or moving at a constant velocity. It is evaluated at the control surface w.r.t. the inertial reference frame selected.

Moving Control Volumes Up to this point, we have been going through the applications of the momentum equation involving a stationary control volume (CV) In some problems, to attach CV to a moving body is more useful... For example, jet impinging on moving block (see Example 6.10) Inertial reference frame on the block V x = 0 zero accumulation term V ix = V j – V b at station 1 V ox = 0 at station 2 Flow is steady w.r.t. the block, so Mass flow rate = ρ A (V j – V b )

EXAMPLE: Momentum Application-Vane (Q6.29) A horizontal jet strikes a vane that is moving at a speed v v = 7 m/s. Diameter of the jet is 6 cm. Speed of the fluid jet is 20 m/s, relative to a fixed frame. What components of force are exerted on the vane by the water in the x and y directions? Assume negligible friction between the water and the vane.

SOLUTIONS

SOLUTIONS cont……..

Class Exercises: (Problem 6.26)

Class Exercises: (Problem 6.55)