Chapter 6

MOMENTUM PRINCIPLE

Fluid Mechanics, Spring Term 2009

Newton’s 2nd Law:

Definition of (linear) momentum:

or

Conservation of momentum:

The momentum conserved is that of the system:

Particles move about in a box. No particles leave or enter, hence the box is a system.

Particles can exchange momentum with each other, but that does not affect the total momentum of the system.

The only way to change the total momentum is to apply a force to the box.

Therefore:

Notice the similarities and differences between conservation of mass and momentum:

(conservation of mass)

(conservation of momentum)

1) Both equations are applied to the system.

2) Mass is conserved absolutely (never changes in classical physics); Momentum is conserved unless a force is applied.

3) Mass conservation is a scalar equation; Momentum conservation is a vector equation (3 equations).

Forces acting on a control volume (Fig. 6.1)

cv consists of fluid only.

Cv includes the section of the pipe.

Choice 1: control volume is entirely within the fluid.

where is a viscous shear stress acting between the fluid and the pipe wall (its direction depends on whether the fluid is moving up or down).

Choice 2: The control volume includes the entire section of the pipe.

where Wp is the weight of the pipe.

Systematic Approach to Solving Problems:

1) Problem Setup• Select appropriate control volume.• Select inertial reference frame.

2) Force analysis and diagram• Sketch body forces on force diagram (gravity).• Sketch surface forces: pressures, shear stresses,

supports and structures…

3) Momentum analysis and diagram• Evaluate momentum accumulation term. If the flow

is steady, this term is zero. Otherwise evaluate volume integral and add to momentum diagram.

• Sketch momentum flow vectors on momentum diagram. For uniform velocity, each vector is

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Example 6.1:Rocket on test stand.

Exhaust jet has:Diameter d = 1 cmSpeed v = 450 m/sDensity = 0.5 kg/m3

Assume p in jet is atmospheric p.

Neglect momentum changes inside rocket motor.

What is the force Fb acting on support beam?

Example 6.1: Solution

Only involves vertical momentum.

1) Forces:

2) Momentum change:

There is no momentum accumulation because the structure is stationary and because we neglect momentum changes in the rocket motor.

Example 6.1 (continued)

Substitute forces and momentum into the momentum equation:

or

Example 6.2: Concrete flowing onto cart on a scale

Stream of concrete:

Given:Density = Area = ASpeed = v

Cart + concrete:Weight = W

Determine tension in cable and weight recorded by scale.

Example 6.2: Solution

Forces and momenta involve x and z directions.

Forces in x: Forces in z:

Example 6.2 (continued)

Since flow is steady and cart does not move, there is no accumulation of momentum inside the cv.

Momentum changes in x and z: (with )

Example 6.2 (continued)

The momentum equations in x and z give

thus providing T and N as the answer. (Notice in this problem the cv is stationary; hence, the velocities v and V are the same).

Example 6.4: Water jet deflected by a vane

Given:

Speed of incoming jet = v1

Speed of outgoing jet = v2

Diameter of jet = D

Note that mass flow rates have to be equal:

Example 6.4: Solution

1) Forces:

2) Momenta:

Net momentum change:

Example 6.4 (continued)

Again, substitute into momentum equation:

Where is given by

Example 6.7: Water flow through a 180o reducing bend

Given:

Discharge = Q

Pressure at center ofinlet = p1

Volume of bend = V

Weight of bend = W

What force is required to hold the bend in place?

Example 6.7: Solution

Additional difficulties:

1) We don’t know the velocities.

2) We don’t know the outlet pressure.

1) Get velocities from continuity equation (incompressible!):

2) Get p2 from Bernoulli equation:

Example 6.7 (continued)

Notice that there is no momentum change in the vertical.

In the vertical we only have the force balance

Forces in x-direction:

Momentum in x-direction:

Momentum balance in x:

Using continuity and Bernoulli equations (previous slide),all variables are known except Rx.