ME 7980 – Fall 16
MOENS KORTEWEG RELATIONSHIPS
Derivation of the equations of fluid motion
Geometry and assumptions
Navier-Stokes equations
Simplified continuity equation:
(1)
Simplified momentum equations along r and z :
(2)
First approximation: INVISCID FLOW
ME 7980 – Fall 16
First approximation: INVISCID FLOW
Simplified momentum equations along r and z :
Scaling:
Wave speed definition:
(3)
Comparison of wave velocity to fluid velocity:
Scaling applied to continuity equation:
(4)
ME 7980 – Fall 16
Therefore:
(5)
Scaling applied to NS equation along r (inertial terms):
Therefore:
Using (5):
Therefore:
Scaling applied to NS equation along z (inertial terms):
(6)
ME 7980 – Fall 16
Therefore:
(7)
Using (4) and above scaling (NS/ r ):
Therefore:
In both NS equations, the temporal acceleration is the leading term of the inertial terms.
Therefore, the non-linear convective terms have a smaller effect on the flow dynamics
than the temporal acceleration terms.
Based on our assumptions, approximations and scaling, the NS equations reduce to:
(8)
Inviscid flow approximation
We know the order of magnitude of all variables but the pressure gradient. Let’s look at the
magnitude of each term in those two equations:
Reduced NS equation / r :
Reduced NS equation / z :
Take the ratio term by term and look at the magnitudes:
ME 7980 – Fall 16
What is driving the flow in the z-direction?
Therefore:
(9)
Changes in pressure in the radial direction can be neglected as compared to those in the axial
direction.
Therefore:
(10)
And the only remaining equation is:
(11)
This simplified equation still includes 3 unknowns: vz , vr and P .
Let’s consider the continuity equation (1):
1  (rvr ) vz

0
r r
z
Integrate over a cross section:
Q can also be expressed as a function of the cross-sectional area and the mean velocity v z :
Therefore:
(12)
ME 7980 – Fall 16
Combine equations (11) and (12) to obtain the equations of fluid motion:
(13)
Inviscid flow assumption
Derivation of the equations of vessel wall
Solid body diagram of a tube element:
Assumptions
Let  ( z , t ) be the displacement of the tube in the radial direction.
Circumferential strain:
ME 7980 – Fall 16
Circumferential stress:
(14)
Newton’s law applied to tube element in radial direction:
Therefore:
(15)
Newton’s law for tube
Second approximation: NEGLECT INERTIAL FORCES
(using same reasoning as that to neglect p r )
Equation reduces to:
Using (14):
(16)
 is still an unknown in that equation. Therefore, we need a coupling with the equation of fluid
motion:
Boundary condition for wall motion:
Using (12):
(17)
Differentiate (16) with respect to time:
Differentiate with respect to time:
(18)
ME 7980 – Fall 16
Consider the equation of motion for the fluid (13) and integrate over a cross-section:
Differentiate with respect to z:
(19)
Substitute back into (18):
(20)
This equation is of the form:
2 p 1 2 p

(wave equation; c0 : velocity of wave propagation)
z 2 c02 t 2
The velocity of wave propagation is given by the Moens-Korteweg relationship:
(21)
Speed of pressure wave propagation
through
a
thin-walled,
elastic
tube
containing incompressible, inviscid fluid.
Remark: Based on those assumptions, the wave speed is constant, which is contrary to
experimental measurements demonstrating the dependence of the wave speed on the pulse
frequency. Therefore, inertial effects cannot be neglected.
ME 7980 – Fall 16
Let’s redo the derivation by accounting for the effects of inertial forces on the tube.
Let’s start from Newton’s second law for the tube (equation 15):
t hRd
d 2
 Rpd    hd
dt 2
Re-arrange and express the pressure as a function of the other parameters:
Using (14), expand the expression of the hoop stress:
(22)
This is a 2nd-order differential equation. In order to solve it, we’ll assume a sinusoidal pressure
pulse:
p  A sin(kz  t ) (23)
where k 
1

(  : wavelength)
 : pulse frequency
Assume a similar for for the tube displacement:
  B sin(kz  t )
Substituting (24) into Newton’s second law (22) yields:
Differentiate with respect to time:
(24)
ME 7980 – Fall 16
Differentiate again with respect to time:
Therefore:
(25)
Wave equation
This time, the speed of wave propagation is:
(26)
Remark:
By considering the inertial effects on the vessel wall, we have demonstrated that c0 , the wave
speed, is a function of the pulse frequency
However:
 2 t R 2
E
.
 1 (effect of tube elasticity). Therefore, c0
constant .
The discrepancy between the pressure pulse propagation speed predicted by this model and
that measured experimentally demonstrates the limitation of one of our assumptions:
Viscous effects have to be accounted for.