1. Fluid Mechanics 10

2. 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

3. Reynolds Transport Theorem
Take B is general extensive property and b is the intensive one for it
At 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
B(t+∆t)−B(t) ∆𝑡 = Bcv(t+∆t)−Bcv(t) ∆𝑡 + ∆Bsc(t+∆t)−∆Bsc(t) ∆𝑡
For ∆t ≈ 0
𝑑𝐵 𝑑𝑡 = lim∆𝑡 →0 B(t+∆t)−B(t) ∆𝑡

4. 𝑑𝐵𝑐𝑣 𝑑𝑡 = lim∆𝑡 →0 Bcv(t+∆t)−Bcv(t) ∆𝑡
𝑑𝐵𝑐𝑠 𝑑𝑡 = lim∆𝑡 →0 ∆Bcs(t+∆t)−∆Bcs(t) ∆𝑡 So
𝑑𝐵 𝑑𝑡 = 𝑑𝐵𝑐𝑣 𝑑𝑡 + 𝑑𝐵𝑠𝑐 𝑑𝑡
Take B=m*b where B is extensive and b is intensive
𝑑𝐵 𝑑𝑡 = ρ∗𝑏∗𝑑𝑉 + 𝑐𝑠 𝑚ʹ𝑜∗𝑏𝑜 − 𝑐𝑠 𝑚ʹ𝑖∗𝑏𝑖
For steady flow
𝑑𝐵 𝑑𝑡 = 𝑐𝑠 𝑚ʹ𝑜∗𝑏𝑜 − 𝑐𝑠 𝑚ʹ𝑖∗𝑏𝑖

5. 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.

6. Force Diagram

7. 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
𝑑𝑝𝑚 𝑑𝑡 =𝐹

8. Momentum equation General Equation 𝐹 = ρ∗𝑣∗𝑑𝑉 + 𝑐𝑠 𝑚ʹ𝑜∗𝑣𝑜 − 𝑐𝑠 𝑚ʹ𝑖∗𝑣𝑖
𝐹 = ρ∗𝑣∗𝑑𝑉 + 𝑐𝑠 𝑚ʹ𝑜∗𝑣𝑜 − 𝑐𝑠 𝑚ʹ𝑖∗𝑣𝑖
For steady flow
𝐹 = 𝑐𝑠 𝑚ʹ𝑜∗𝑣𝑜 − 𝑐𝑠 𝑚ʹ𝑖∗𝑣𝑖

9. Example
The 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.

11. Solution 1- Select the control volume. 2- Draw the force diagram.
3- Draw the moment diagram.
4- Write the equation calculate the missing

12. 𝐹 = ρ∗𝑣∗𝑑𝑉 + 𝑐𝑠 𝑚ʹ𝑜∗𝑣𝑜 − 𝑐𝑠 𝑚ʹ𝑖∗𝑣𝑖
neglect any momentum changes inside the rocket
Vi=0.0
𝐹 = 𝑐𝑠 𝑚ʹ𝑜∗𝑣𝑜
−𝐹𝑏−𝑚∗𝑔=−𝑚ʹ∗𝑣𝑜
Where m=0.4 kg, mʹ=ρ*A*v, d=0.01m, v=450m/s
𝐹𝑏=𝑚ʹ∗𝑣𝑜−𝑚∗𝑔
=(Π* * ∗0.5)−(0.4∗9.81)=7.56𝑁

14. Example
As 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.

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

17. 𝐹 = ρ∗𝑣∗𝑑𝑉 + 𝑐𝑠 𝑚ʹ𝑜∗𝑣𝑜 − 𝑐𝑠 𝑚ʹ𝑖∗𝑣𝑖
Where the flow is steady , no out velocity
𝐹 =− 𝑐𝑠 𝑚ʹ𝑖∗𝑣𝑖
Direction X
𝐹𝑥 =− 𝑐𝑠 𝑚ʹ𝑖∗𝑣𝑖 x
−𝑇=−ρ∗𝐴∗ 𝑣 2 ∗cos⁡(60) = −150∗ ∗ 10 2 ∗cos⁡(60)
T = 233 lbf
Direction z
𝐹𝑧 =− 𝑐𝑠 𝑚ʹ𝑖∗𝑣𝑖 z
𝑁−𝑊=−ρ∗𝐴∗ 𝑣 2 ∗−sin⁡(60) = −150∗ ∗ 10 2 ∗ sin 60 =403𝑙𝑏𝑓
N = 1200 lbf