Fluid Mechanics 10

Control Volume Approach

Intensive and Extensive Properties:-

Extensive Properties depends on mass like mass, m, momentum, mv and energy, E.The intensive properties does not depend on mass, The intensive property for momentum is velocity v, and for energy is e, weight is g

Reynolds Transport Theorem

Take B is general extensive property and b is the intensive one for itAt t time B(t)=Bcv(t)+∆Bsc(t)At t+∆t B(t+∆t)=Bcv(t+∆t)+∆Bsc(t+∆t)Change in B for ∆t = + For ∆t ≈ 0

So =+Take B=m*b where B is extensive and b is intensive=For steady flow =

Newton's Laws

First Law:- An object remains at rest or at a constant velocity unless force acted upon it

Second Law:- F=m*a

Third Law:- When one body creates a force on a second body, the second body creates a force equal in magnitude and opposite in direction to the first body.

Force Diagram

Momentum

The Momentum is the product of the mass and velocity of an object.

Pm=F*v where pm is the momentum

Momentum units is Kg*m/s

Momentum equation

General Equation=

For steady flow

=

ExampleThe sketch below shows a 40 g rocket, of the type used for model rocketry, being fired on a test stand in order to evaluate thrust. The exhaust jet from the rocket motor has a diameter of d = 1 cm, a speed of ν = 450 m/s, and a density of ρ = 0.5 kg/m3. Assume the pressure in the exhaust jet equals ambient pressure, and neglect any momentum changes inside the rocket motor. Find the force Fb acting on the beam that supports the rocket.

Solution1- Select the control volume.

2- Draw the force diagram.

3- Draw the moment diagram.

4- Write the equation calculate the missing

=

neglect any momentum changes inside the rocketVi=0.0

=

Where m=0.4 kg, mʹ=ρ*A*v, d=0.01m, v=450m/s

=(Π**

ExampleAs shown in the sketch, concrete flows into a cart sitting on a scale. The stream of concrete has a density of ρ = 150 ibm/ft3, an area of A = 1 ft2, and a speed of V = 10 ft/s. At the instant shown, the weight of the cart plus the concrete is 800 lbf. Determine the tension in the cable and the weight recorded by the scale. Assume steady flow.

Solution

The problem has two direction x, z we will apply the equation into the 2-D according to θ

=Where the flow is steady , no out velocity=Direction X =x = T = 233 lbfDirection z = = N = 1200 lbf