#### 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 ( Vd ) 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 ( Vd ) 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 ( Vd ) 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.