Conservation Laws - Biofluidmechanics: An Introduction to Fluid Mechanics Macrocirculation and Microcirculation - page 67

the sign of each velocity component ( u , v , and w ). By knowing the signs of these two parts,
the sign of the overall product can be determined.

Example
Determine the force required to hold the brachial artery in place during peak systole
( Figure 3.13 ). Assume at the inlet the pressure is 100 mmHg and at the outlet the pressure is
85 mmHg (these are gauge pressures). The diameter of the brachial artery is 18 mm at the inflow
and 16 mm at the outflow. The blood flow velocity at the inlet is 65 cm/s. For simplicity, neglect
the weight of the blood vessel and the weight of the blood within the vessel.

Solution
Figures 3.13 and 3.14 depict what is known about the situation. The problem statement asks
to solve for F x and F y .
To solve this problem, assume that there is steady flow at the instant in time that we are inter-
ested in, that 1 atm 760 mmHg, that the blood vessel does not move and is not deformable,

and that the flow is incompressible. First, we will need to solve for the outflow velocity using
the equations for conservation of mass:

5

system 5 @

ð
V ρ

ð
area ρ

dm
dt

d - - U dV 0 ð 3 : 24 Þ

1 5

@ t

ð
area ρ

5

d - -

0

- ρ 1 v 1 A 1 5 ρ 2 v 2 A 2

U

2

ð 1050 kg = m 3 Þð 65 cm = s Þ π

18 mm
2

ð 2 j 5 v 2 5 82 : 27 cm = s Þ

2

16 mm
2

ð 1050 kg = m 3 Þ π

Note that the velocity accelerates due to the step down nature of the geometry. This is not
representative of what occurs in physiology, but this problem illustrates how to use conservation
of mass and momentum together.
Solve for the x-component of the force needed to hold the brachial artery in place:

ð ð

F sx 5 @
@

d - - U F x 5 F bx 1 u ρ dV u ρ

1

t

V area

F bx 5 0 ; u 2 5 0

F sx 5 p inflow A inflow 1 p atm A 1 2 p atm ð A inflow 1 A 1 Þ 1 F x 5 A inflow ð p inflow 2 p atm Þ 1 F x

ð

d - - A inflow ð p inflow 2 p atm Þ 1 F x 5 u ρ u 1 ð 2 ρ v inflow A inflow Þ

U 5

area

Note that the u component of the velocity is positive, but the flux is negative because the
velocity vector and the normal area vector act in opposite directions.