Transcript Huang_-_Individual_Presentation_P3.98

BIEN 301
Individual Project Presentation
The Linear Momentum Equation
Hsuan-Min Huang
The Linear Momentum Equation
Equation
d
F  (  Vd )   V (V  n)dA
dt CV
CS
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
Noninertial Reference Fram
• Equation
d
F   adm  (  Vd )   V (Vr  n)dA
dt CV
CV
CS
• Where
d R d
a 2 
 r  2  V    (  r )
dt
dt
2
Pressure Condition at a Jet Exit
•
1.
2.
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
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.
APPROACH
• To find the equation of motion for cart velocity Vc, we
apply equation
F 
d
 adm  dt (  Vd )   V (Vr  n)dA
CV
CV
CS
• 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.
mj
  Aju j
t
j
• Compute numerical value from example 3.10 into the
formula we got in
SKETCH
• Jet stricking a moving plate normally
SKETCH
• Control volume fixed relative to the plate
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.
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)
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
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
Solution
Part c
• t = 0.9 mj /ρjAj uc
=0.9*2(kg)/1000(kg/m3)*(0.0003m2)*(20m/s)
= 0.3s
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
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.