MENG 575_Lecture # 3_Momentum

8/2/2019 MENG 575_Lecture # 3_Momentum

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Department of Mechanical Engineering

MENG5753

Lecture # 3

Governing Equations of Motion

Part II: Momentum Equation


– Identify the various kinds of forces and moments

acting on a control volume.

– Use control volume/integral analysis to determine

the forces associated with fluid flow.

– Derive the differential form of the conservation of

momentum using Newton‘s 2nd law.

Objectives



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I nt egral Analysis

When dealing with engineering problems, it is desirable to obtain fast

and accurate solutions at minimal cost. Most engineering problems,

including those associated with fluid flow, can be analyzed using one

of three basic approaches:

- Differential,

- Experimental,

- Control volume/Integeral.

The finite control volume approach is remarkably fast and simple

and usually gives answers that are sufficiently accurate for most

engineering purposes. Therefore, despite the approximations

involved, the basic finite control volume analysis performed with a

paper and pencil has always been an indispensable tool for

engineers.


• Newton’s laws are relations between motions ofbodies and the forces acting on them.

– First law : a body at rest remains at rest, and a body inmotion remains in motion at the same velocity in a straightpath when the net force acting on it is zero.

– Second law : the acceleration of a body is proportional tothe net force acting on it and is inversely proportional to itsmass.

– Third law : when a body exerts a force on a second body,the second body exerts an equal and opposite force on thefirst.

New ton’s Laws



3


Forces Acting on a CV

• Forces acting on CV consist of body forces that actthroughout the entire body of the CV (such as gravity,electric, and magnetic forces) and surface forcesthat act on the control surface (such as pressure andviscous forces, and reaction forces at points ofcontact).

• Body forces act on eachvolumetric portion dV of the CV.

• Surface forces act on each

portion dA of the CS.


Different ial Analysis

When dealing with engineering problems, it is desirable to obtain fast

and accurate solutions at minimal cost. Most engineering problems,

including those associated with fluid flow, can be analyzed using one

of three basic approaches:

- Differential,

- Experimental,- Control volume.

In differential approaches, the problem is formulated accurately using

differential quantities, but the solution of the resulting differential

equations is difficult, usually requiring the use of numerical methods

with extensive computer codes.



4


Forces Acting on a CV

• Surface integrals are cumbersome.

• Careful selection of CV allowsexpression of total force in terms ofmore readily available quantitieslike weight, pressure, and reactionforces.

• Goal is to choose CV to exposeonly the forces to be determined

and a minimum number of otherforces.

Total force acting on

Control Volume (CV): surfacebodyF F F ∑∑∑ +=rrr


Body Forces

• The most common body forceis gravity, which exerts adownward force on everydifferential element of the CV

• The different body force

• Typical convention is thatacts in the negative z -direction,

• Total body force acting on CV



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Sur face Forces

• Surface forces are not as simple to

analyze since they include both

normal and tangential components

• Diagonal components σ xx , σ yy , σ zz are

called normal stresses and are due

to pressure and viscous stresses

• Off-diagonal components σ xy , σ xz ,

etc., are called shear stresses and are

due solely to viscous stresses

• Total surface force acting on CS


Different ial Form of Moment um

∑ === Dt

V Ddzdydx

Dt

V DmamF

rr

r

r

ρ

Where: ∑ ∑∑ += surfacebody F F F r

∑ ∑∑

+=

==

ForceViscousForcePressure

3,2,1,

surface

ibody

F

idzdydxgF ρ



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The Reynolds St ress Tensor

τ xx τ xy τ xz

τ yx τ yy τ yz

τ zx τ zy τ zz

⎝

⎜⎜⎜

⎠

⎟⎟⎟

τ ij =


Conservat ion of Linear Momentum





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• The Navier-Stokes equations

• This results in a closed system of equations!– 4 equations (continuity and momentum equations)

– 4 unknowns (U, V, W, p)

Incompressible NSEwritten in vector form

Navier-Stokes Equat ion

Incompressible Continuity

written in vector form


Cartesian Coordinates

Continuity

X-momentum

Y-momentum

Z-momentum

Continui ty and Navier-Stokes Equation



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Tensor and Vector Notation

Continuity

Conservation of MomentumTensor notation Vector notation

Vector notationTensor notation

Tensor and Vector notation offer a more compact form of the equations.

Repeated indices are summed over j(x 1 = x, x 2 = y, x 3 = z, U 1 = U, U 2 = V, U 3 = W )

Navier-Stokes Equat ion


Conservation of Momentum for Inviscid Flow

Tensor notation

Vector notation

Euler’s Equat ions

0

0

Good approximation to the componentsof the stress tensor for many flows, especiallyfor flow away from boundary (flow around an

airfoil) are displayed by the array ⇒

Assuming a constant-density andsteady flow, the above equation canbe Integrated along a streamline to

yield Bernoulli‘s equation. ⇒

.2

2

const zg

p

g

V =++

ρ



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1.Set up the problem and geometry, identifying all relevant

dimensions and parameters

2.List all appropriate assumptions, approximations,

simplifications, and boundary conditions

3.Simplify the differential equations as much as possible

4.Integrate the equations

5.Apply BC to solve for constants of integration

6.Verify results

Procedure for solving continuity and NSE

Exact Solut ions of t he NSE


Fully Developed Couette Flow

• For the given geometry and BC’s, calculate the velocity

and pressure fields, and estimate the shear force per unit

area acting on the bottom plate

• Step 1: Geometry, dimensions, and properties

Exam ple exact solut ion (Ex. 9-15)

U



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• Step 2: Assumptions and BC’s

– Assumptions1. Plates are infinite in x and z

2. Flow is steady, ∂ / ∂t = 0

3. Parallel flow, v=0

4. Incompressible, Newtonian, laminar, constant properties

5. No pressure gradient

6. 2D, w=0, ∂ / ∂z = 0

7. Gravity acts in the -z direction,

– Boundary conditions1. Bottom plate (y=0) : u=0, v=0, w=0

2. Top plate (y=h) : u=U, v=0, w=0



• Step 3: Simplify

3 6

Note: these numbers referto the assumptions on theprevious slide

This means the flow is “fully developed”

or not changing in the direction of flow

Continuity

X-momentum

2 Cont. 3 6 5 7 Cont. 6


y



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• St ep 3: Simplif y, cont .Y-momentum

2,3 3 3 3,6 7 3 33

Z-momentum

2,6 6 6 6 7 6 66



• Step 4: Integrate

Z-momentum

X-momentum

integrate integrate

integrate




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• Step 5: Apply BC’s

– y=0, u=0=C1(0) + C2 ⇒ C2 = 0

– y=h, u=U=C1h ⇒ C1 = V/h

– This gives

– For pressure, no explicit BC, therefore C3 can remain

an arbitrary constant (recall only ∇P appears in NSE).• Let p = p0 at z = 0 (C3 renamed p0) 1. Hydrostatic pressure

2. Pressure acts independently of flow


U

U


• Step 6: Verify solution by back-substituting into

differential equations

– Given the solution (u,v,w)=(Uy/h, 0, 0)

– Continuity is satisfied0 + 0 + 0 = 0

– X-momentum is satisfied




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Shear force per unit area acting on the wall

Note that τw is equal and opposite to theshear stress acting on the fluid τyx

(Newton’s third law).



Literature

- Yunus A. Cengel and John M. Cimbala, Fluid Mechanics:

Fundamentals and Applications, NY McGraw-Hill, 2nd edit. 2010.

- Douglas, J. F., Gasiorek, J., “Fluid Mechanics,” 4th edit. 2001,Ashford Colour Press Ltd. Gosport.

- Frank White, “Fluid Mechanics,” McGraw-Hill, 2nd edit. 2010.

- Franz Durst, “Fluid Mechanics: An Introduction to the Theory ofFluid Flows,“ Springer, Berlin, 2008.

- Eric G. Paterson, “Fluid Mechanics Lectures,” Department of Mechanical and

Nuclear Engineering The Pennsylvania State University

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MENG 575_Lecture # 3_Momentum

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