1. BIEN 301 Individual Project Presentation
The Linear Momentum Equation
Hsuan-Min Huang

2. The Linear Momentum Equation

3. The Linear Momentum Equation
The fluid velocity relative to an inertial coordinate system
The sum of force is the vector sum off all forces acting on the system considered as a free body
The entire equation is a vector relation

4. Noninertial Reference Fram
Equation
Where

5. Pressure Condition at a Jet Exit
Only two effects could maintain a pressure difference between the atmosphere and a free exit.
Surface tension
Supersonic
In this problem, both effects are negligible

6. PROBLEM p3.98
As an extension of example 3.10, let the plate and its cart be unrestrained horizontally, with frictionless wheels. Derive (a) the equation of motion for cart velocity Vc(t) and (b) a formula for the time required for the cart to accelerate from rest to 90 percent of the jet velocity (assume the jet continues to strike the plate horizontally). (c) Compute numerical value for part (b) using the condition of example 3.10 and a cart mass of 2 kg.

7. APPROACH
To find the equation of motion for cart velocity Vc, we apply equation
To find the formula for the time required for the cart to accelerate from rest to 90 percent of the jet velocity, we use the mass analysis in the condition of uc =0.9 uj.
Compute numerical value from example 3.10 into the formula we got in

8. SKETCH
Jet stricking a moving plate normally

9. SKETCH
Control volume fixed relative to the plate

10. ASSUMPTION
The system is steady static, so the force acts on the cart would not be lost.
In fact, the sum of the force is zero.
The fluid is incompressible, so the jet force will be applied on the cart completely. However, if the fluid is compressible, the pressure will not be constant. As a result, some force may be absorbed.
The fluid’s density and viscosity is constant.
The wheels of the cart are frictionless, so there is no lost in the jet force.

11. GIVEN
From the example 3.10, we got Vj = 20 m/s ; Vc= 15 m/s ; jet density is 1000 kg/m3 ;
V1=V2=Vj-Vc=20-15=5 m/s
Σ Fx = Rx = m1u1+m2u2-mjuj
= -[ρjAj (Vj-Vc)](Vj-Vc)
Where m1= m2 =1/2 mj =1/2ρjAj (Vj-Vc)

12. Solution Part a Fx = Rx = m1u1+m2u2-mjuj
m1= m2 =1/2 mj =1/2ρjAj (Vj-Vc)
Vj-Vc = uj, and u1= u2 = 0
Fx = Rx = - mjuj =
- 1/2ρjAj (Vj-Vc)* (Vj-Vc)=
- 1/2ρjAj (Vj-Vc)2

13. Solution Part b uc =0.9 uj mass analysis, mj /t =ρjAj uj
uj = mj /ρjAj t
uc =0.9 mj /ρjAj t
t= 0.9 mj /ρjA uc

14. Solution Part c t = 0.9 mj /ρjAj uc
=0.9*2(kg)/1000(kg/m3)*(0.0003m2)*(20m/s)= 0.3s

15. Computational Results
Part a: Fx = Rx = - mjuj
=- 1/2ρjAj (Vj-Vc)2
Part b: t= 0.9 mj /ρjAj uc
Part c: t =0.3s

16. Applications to Biofluids
Blood cells move along with blood
We can apply the concept of the linear momentum to measure the erythrocytes, leukocyte, and thrombocytes move along with the blood. When our heart beats are increased, the blood cells speed will be increased by blood. It would be the same situation as the problem3.98. We can find the speed for the blood cells from the same concept.