Conservation Laws - Biofluidmechanics: An Introduction to Fluid Mechanics Macrocirculation and Microcirculation - page 68

F x 52 u ρ vA inflow 2 A inflow ð p inflow 2 p atm Þ 52 u ρ v inflow A inflow 2 A inflow ð p inflow 2 gauge Þ

! !

2 2

18 mm
2

18 mm
2

m 3 F x 52 ð 65 cm = s Þð 1050 kg = Þð 65 cm = s Þ π

2 π

ð 100 mmHg Þ 52 3 : 5N

This means that this force acts toward the left because the flow/pressure is pushing toward
the right. Now solve for the y-component of force:

ð ð

F sy 5 @
@

d -

- U

F y 5 F by 1 v ρ dV v ρ

1

t

area

V

F by 5 0 ; v 1 5 0

F sy 5 p outflow A outflow 1 p atm A 2 2 p atm ð A outflow 1 A 2 Þ 1 F y 5 A outflow ð p outflow 2 p atm Þ 1 F y

ð

d - - U A outflow ð p outflow 2 p atm Þ 1 F y 5 v ρ v 2 ðρ V outflow A outflow Þ

5

area

Note that the v 2 -velocity component is negative (this will be accounted for later) and the flux
term is positive because the area and the velocity vectors act in the same direction.
F y 5 v 2 ðρ v outflow A outflow Þ 2 A outflow ð p outflow 2 p atm Þ 5 v 2 ðρ v outflow A outflow Þ 2 A outflow ð p outflow 2 gauge Þ

! !

2 2

16mm
2

16mm
2

m 3

F y 5 ð 2 82 : 27 cm = s Þð 1050 kg = Þð 82 : 27 cm = s Þ π

2 π

ð 85mmHg Þ 52 2 : 42 N

This means that this force acts downward. The overall force is 4.25 N acting at an angle of
214

from the positive x -axis.

FIGURE 3.13

Brachial artery schematic for example problem.

Inflow

Outflow

FIGURE 3.14

P atm Free body

diagram for the preceding
example problem.

Volume of interest

v inflo w

P inflow

A inflow

P atm

F y

P atm

A 1

F x

Y

v outflow

A 2

A outflow

P atm

P outflow

X