mech 5810 module 5: conservation of linear...

MECH 5810 Module 5: Conservation of LinearMomentum

D.J. Willis

Department of Mechanical EngineeringUniversity of Massachusetts, Lowell

MECH 5810 Advanced Fluid DynamicsFall 2017

Outline

1 Differential Conservation of Momentum

Summation of forces = Mass × acceleration or Change in

momentum

Forces Per Unit volume

Euler Equation in s− n Coordinates

Examples

Equation of motion

Newton’s Law applied to a point mass/constant mass differential

element of fluid, expressed in per unit volume form:

m~aVol

= ρ~a =∑(

~FVol

)

~a is acceleration in a Lagrangian description:

~a =∂~ua

∂t

Which means, in an Eulerian Description we must use the

Material or Total Derivative to determine the acceleration:

ρ~a =D~uDt

=∂~u∂t

+ (~u · ∇)~u =∑(

~FVol

)

Outline



momentum



Examples

Pressure force per unit volume acting on a fluidelement

Free body diagram of a differential element

Fxpressure

Vol=

Net x−pressure force︷︸︸︷− (P2 − P1) · A

Vol

=− (P2 − P1) · dy · dz

dx · dy · dz

=− (P2 − P1)

dx

→ −∂p∂x

Fxpressure

Vol= −∂p

∂x;

Fypressure

Vol= −∂p

∂y;

Fzpressure

Vol= −∂p

∂z

Pressure force per unit volume acting on a fluidelement

What really matters is the gradient of the pressure:

~Fpressure

Vol= −∇p

Body force per unit volume acting on a fluidelement

Let’s look at gravity forces as an example:

~Fgravity = m~g = mg~iz

But we express everything per unit volume:

~Fgravity

Vol= ρg~iz

We follow similar procedures for other body forces.

Viscous forces per unit volume acting on a fluidelement

For now, we will simply write ~FviscVol

Later in the course we will return to this

General equations of motion

Equate change in momentum per unit volume with the forces per

unit volume:

ρ~a = ρD~uDt

= −∇p + ρ~g +~Fvisc

Vol

Expanded in three dimensions:

x− direction : ρ

(∂u∂t

+ u∂u∂x

+ v∂u∂y

+ w∂u∂z

)= −∂p

∂x+ ρgx +

Fxvisc

Vol

y− direction : ρ

(∂v∂t

+ u∂v∂x

+ v∂v∂y

+ w∂v∂z

)= −∂p

∂y+ ρgy +

Fyvisc

Vol

z− direction : ρ

(∂w∂t

+ u∂w∂x

+ v∂w∂y

+ w∂w∂z

)= −∂p

∂z+ ρgz +

Fzvisc

Vol

IF ~FviscVol = 0, then the above equations are the Euler Equations for an

incompressible fluid.

Outline



momentum



Examples


ρ~a = ρD~uDt

= −∇p + ρ~g��

+~Fvisc

Vol

Assumptions:

Steady flow (∂~u∂t = 0)Inviscid flow (~Fvisc

Vol = 0)Conservative bodyforces only(~g = −∇V(r))

(~u · ∇)~u =−∇pρ−∇V


Let’s start with the term:

(~u · ∇)~u

We know ~u is, by definition, tangent to the streamline:

~u = |u|is

(~u · ∇)~u = us∂

∂s

(|u|is

)→ Chainrule

(~u · ∇)~u = us

∂u∂s· is + u · ∂ is

∂s︸︷︷︸=?

But, we need to know what ∂∂s

(is)

is.

Euler Equation in s− n Coordinates: ∂∂s

(is)

We use the following picture to determine what ∂∂s

(is)

is:

dis = −indθ

= −indSR

∂ is∂s

=−inR


Knowing ∂ is∂s = −in

R we can continue to apply the chain rule to the

material derivative:

(~u · ∇)~u = v∂

∂s

(uis)

= u

∂u∂s· is + u · ∂ is

∂s︸︷︷︸−in

R

= u

∂u∂s

is +[−u2

Rin

](~u · ∇)~u = u

∂u∂s

is −u2

Rin

Euler Equation in s− n Coordinates: s-Direction

Each direction (is, in, il) can be examined separately:

In the is-direction: u∂u∂s︸︷︷︸

∂∂s

(u22

)is =

−1ρ

∂p∂s− ∂V

∂s︸︷︷︸V=gz

∂

∂s

[u2

2+ gz

]=−1ρ

∂p∂s

If we assume ρ = const along streamlines (not as rigorous asρ = const. everywhere), then:

∂

∂s

[v2

2+

pρ+ gz

]= 0 or

v2

2+

pρ+ gz = const.

Euler Equation in s− n Coordinates: n-direction

In the in-direction−u2

R=−1ρ

∂p∂n− ∂V

∂n︸︷︷︸V=gz

If ρ = const, then, the expression for the n-direction is:

∂

∂n(p + ρgz) =

ρu2

R

If we have gz = 0 then the expression states that the pressure

gradient in the direction normal to the streamline is inversely

proportional to the radius of curvature (R) of that streamline and

proportional to the square of the velocity. This is a very useful

result (as we shall see in shortly).

Euler Equation in s− n Coordinates: l-direction

In the il-direction

0 =−1ρ

∂p∂l− ∂V

∂l︸︷︷︸∂V∂S =gil

If ρ = const, then, in the l-direction:

∂

∂l(p + ρgz) = 0

This result simply says that any pressure gradient in the l-direction

is directly related to the body forces, and that no fluid acceleration

effects are present.

Outline



momentum



Examples

Bernoulli Equation & Streamline curvatureexamples

Assumptions:

Steady flow (∂~u∂t = 0)Inviscid flow (~Fvisc

Vol = 0)Conservative body forces only (~g = −∇V(r))

In the s-direction

∂

∂s

[u2

2+

pρ+ gz

]= 0 or

u2

2+

pρ+ gz = const.

In the n-direction ∂

∂n(p + ρgz) =

ρu2

R

In the l-direction ∂

∂l(p + ρgz) = 0

mech 5810 module 5: conservation of linear...

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