الجامعة التكنولوجية

قسم هندسة المكائن و المعدات

Fluid Mechanics II )2(ميكانيك الموائع

)ملخص(

3rd year Mechanical Engineering سيارات/تكييف/ عام-المرحلة الثالثة

2012/2011 المادةمدرسو

محمد إدريس محسن.إخالص محمد فياض د.د أستاذ مساعدأستاذ مساعد

Syllabus فرع الميكانيك العام(مفردات المادة(

653/ همك: رقم الموضوع II ميكانيك الموائع : الموضوع 5:الوحدات

2:نظري: الساعات اإلسبوعية 1: مناقشة

1:ي عمل

Subject No.: ME\653 Subject: Fluid Mechanics II Units:5 Weekly Hours: Theoretical:2 Tutorial:1 Practical:1

Contents Week المحتويات االسبوع 1

ستوك–معادالت نافير مقدمة- االشتقاق-

Navier-Stokes equations - Introduction - Derivation

1

2

الجريان الطباقي بين صفيحتين متوازيتين- جريان آوتا-

- Laminar flow between parallel plates - Couette flow

2

Hydrodynamic lubrication 3 - التزييت الهيدروديناميكي- 3 مل االنزالقي المح- 4

المحمل ذو أطواق- - Sliding bearing - Collar bearing

4

الجريان الطباقي بين اسطوانتين متمرآزتين - 5 دوارتين

- Laminar flow between coaxial rotating cylinders

5

6

نظرية الطبقة المتاخمة مقدمة- الطاقة سمك اإلزاحة ، سمك الزخم ، سمك-

Boundary layer theory - Introduction - Displacement, Momentum, and Energy thicknesses.

6

7

Momentum equation for the - معادلة الزخم للطبقة المتاخمة- boundary layer.

7

Laminar boundary layer 8 - الطبقة المتاخمة الطباقية- 89

الطبقة المتاخمة االضطرابية- التحول من الجريان الطباقي إلى االضطرابي-

- Turbulent boundary layer - Transition from laminar to turbulent flow

9

تأثير انحدار الضغط- 10 االنفصال و الكبح الناشئ عن الضغط-

- Effect of pressure gradient - Separation and pressure drag

10

11

المكائن الهيدروليكية مقدمة عامة- التصنيف-

Hydraulic machines - General introduction - Classification

11

12

المضخة الطاردة المرآزية معادلة عزم الزخم - ة مثلثات السرع- الكفاءات-

Centrifugal pump - The moment of momentum equation - Velocity diagrams - Efficiencies

12

أنواع الدفاعات- 13 منحنيات األداء-

- Types of impellers - Performance curves

13

نقطة االشتغال- 14 )توازي/ توالي ( ربط المضخات -

- Operating point - Pump connection (series / parallel)

14

15

المضخات متعددة المراحل نظرة عامة- أنواع المضخات متعددة المراحل-

Multistage pumps - General view - Types of multistage pumps

15

16

اختيار المضخات خرائط اختيار المضخات- خرائط األداء-

Pump selection - Pump selection charts - Performance charts

16

17

التوربينات المائية )توربين بلتون( التوربين الدفعي -

Water turbines - Impulse turbine (Pelton turbine)

17

Reaction turbine (Francis turbine) 18 - )توربين فرانسس( توربين رد الفعل - 18 Axial flow reaction turbine (Kaplan - )توربين آابالن( توربين رد الفعل المحوري - 19

turbine) 19

20

المكائن الهيدروليكية السرعة النوعية- التحليل البعدي و قوانين التشابه للمكائن -

الهيدروليكية

Hydraulic machines - Specific speed - Dimensional analysis and similarity laws for hydraulic machines

20

21

التكهف في المكائن الهيدروليكية التكهف في المضخات- التكهف في التوربينات المائية-

Cavitation in hydraulic machines - Cavitation in pumps - Cavitation in water turbines

21

22

الجريان ثنائي الطور مقدمة- أنماط الجريان-

Two-phase flow - Introduction - Flow patterns

22

23

)سائل+ غاز (الجريان ثنائي الطور نموذج الجريان المتجانس-

Two-phase flow(gas + liquid) - Homogeneous flow model

23

Separated flow model 24 - فصل نموذج الجريان المن- 24 Applications 25 - تطبيقات- 25 26

)المائع المثالي( نظرية الجريان الجهدي مقدمة- معادلة االستمرارية- معادلة الدوامات-

Potential flow theory (Ideal fluid) - Introduction

- Continuity equation - Vorticity equation

26

27

مفاهيم أساسية في الجريان الجهدي دالة االنسياب- دالة الجهد- الدوران-

Basic concepts in potential flow - Stream function - Potential function - Circulation

27

28

أنماط الجريان األساسية الجريان المنتظم- بع ، الغور المن- القطب المزدوج- الدوامة الحرة-

Basic flow patterns - Uniform flow - Source , Sink - Doublet - Free vortex

28

29

جمع أنماط الجريان جريان حول نصف الجسم- رانكني جريان حول بيضاو-

Combination of basic flows - Flow past a half body - Flow past a Rankine oval

29

30

جريان حول اسطوانة- جريان حول اسطوانة مع دوران-

- Flow past a cylinder - Flow past a cylinder with circulation

30

Syllabus )فرعي التكييف و السيارات(مفردات المادة 663/ همك: رقم الموضوع

II ميكانيك الموائع : الموضوع 5:الوحدات

2:نظري: الساعات اإلسبوعية 1: مناقشة

1: عملي

Subject No.: ME\663 Subject: Fluid Mechanics II Units:5 Weekly Hours: Theoretical:2 Tutorial:1 Practical:1

Contents Week المحتويات االسبوع 1

ستوك–معادالت نافير مقدمة- االشتقاق-

Navier-Stokes equations - Introduction - Derivation

1

2

يحتين متوازيتين الجريان الطباقي بين صف- جريان آوتا-

- Laminar flow between parallel plates - Couette flow

2

Hydrodynamic lubrication 3 - التزييت الهيدروديناميكي- 3 المحمل االنزالقي- 4

المحمل ذو أطواق- - Sliding bearing - Collar bearing

4

الطباقي بين اسطوانتين متمرآزتين الجريان- 5 دوارتين

- Laminar flow between coaxial rotating cylinders

5

6

نظرية الطبقة المتاخمة مقدمة- سمك اإلزاحة ، سمك الزخم ، سمك الطاقة-

Boundary layer theory - Introduction - Displacement, Momentum, and Energy thicknesses.

6

7

Momentum equation for the - معادلة الزخم للطبقة المتاخمة- boundary layer.

7

Laminar boundary layer 8 - الطبقة المتاخمة الطباقية- 89

الطبقة المتاخمة االضطرابية- التحول من الجريان الطباقي إلى االضطرابي-

- Turbulent boundary layer - Transition from laminar to turbulent flow

9

تأثير انحدار الضغط- 10

االنفصال و الكبح الناشئ عن الضغط- - Effect of pressure gradient - Separation and pressure drag

10

Separation of flow inside pipes - ء انفصال الجريان في األنابيب و مجاري الهوا- 11 and ducts

11

12

المضخات مقدمة عامة- التصنيف-

Pumps - General introduction - Classification

12

13

المضخة الطاردة المرآزية معادلة عزم الزخم - ة مثلثات السرع- الكفاءات-

Centrifugal pump - The moment of momentum equation - Velocity diagrams - Efficiencies

13

أنواع الدفاعات- 14 منحنيات األداء-

- Types of impellers - Performance curves

14

نقطة االشتغال- 15 )توازي/ توالي ( ربط المضخات -

- Operating point - Pump connection (series / parallel)

15

16

المضخات متعددة المراحل نظرة عامة- أنواع المضخات متعددة المراحل-

Multistage pumps - General view - Types of multistage pumps

16

17

اختيار المضخات خرائط اختيار المضخات- خرائط األداء-

Pump selection - Pump selection charts - Performance charts

17

18

المضخات السرعة النوعية- التحليل البعدي و قوانين التشابه للمضخات-

Pumps - Specific speed - Dimensional analysis and similarity laws for pumps

18

19

التكهف التكهف في المضخات-

Cavitation - Cavitation in pumps

19

20

المراوح مقدمة عامة- التصنيف-

Fans - General introduction - Classification

20

21

أنواع المراوح المروحة الطاردة المرآزية- المروحة المحورية-

Types of fans - Centrifugal fan - Axial flow fan

21

22

خصائص المروحة )المنظومة+المروحة( خصائص - )توازي/ توالي ( ربط المراوح -

Fan characteristics - Fan and system characteristics - Fan arrangements (series / parallel)

22

23

الجريان ثنائي الطور مقدمة- أنماط الجريان-

Two-phase flow - Introduction - Flow patterns

23

24

)سائل+ غاز (الجريان ثنائي الطور نموذج الجريان المتجانس-

Two-phase flow(gas + liquid) - Homogeneous flow model

24

Separated flow model 25 - نموذج الجريان المنفصل- 25 Applications 26 - تطبيقات- 26 27

)المائع المثالي( نظرية الجريان الجهدي مقدمة- معادلة االستمرارية- معادلة الدوامات-

Potential flow theory (Ideal fluid) - Introduction

- Continuity equation - Vorticity equation

27

28

مفاهيم أساسية في الجريان الجهدي النسياب دالة ا- دالة الجهد- الدوران-

Basic concepts in potential flow - Stream function - Potential function - Circulation

28

29

أنماط الجريان األساسية الجريان المنتظم- المنبع ، الغور- القطب المزدوج-

Basic flow patterns - Uniform flow - Source , Sink - Doublet

29

الدوامة الحرة- 30 الجريان في زاوية- الجريان في مجرى منحني-

- Free vortex - Flow in the vicinity of a corner - Flow in a bend

30

References المصادر 1- Fluid Mechanics By Streeter & Wylie 2- Fluid Mechanics with Engineering Applications By Daugherty & Franzini 3- Introduction to Fluid Mechanics By Fox & McDonald 4- Engineering Fluid Mechanics By Roberson & Crowe

:ملحوظةللمرحلة الثالثة II مادة ميكانيك الموائع لهذه المحاضرات تتضمن ملخص

دول ي ج ررة ف ة المق ساعات النظري ضور ال ن ح ب ع ي الطال ي ال تغن وه .الدروس األسبوعي

Chapter One Navier – Stokes Equations

Contents 1- Navier-Stokes equations. 2- Steady laminar flow between parallel flat plates. 3- Hydrodynamic lubrication. 4- Laminar flow between concentric rotating cylinders. 5- Examples. 6- Problems; sheet No. 1 1- Navier-Stokes equations: The general equations of motion for viscous incompressible, Newtonian fluids may be written in the following form: x- direction:

-(1)----- 2

2

2

2

2

2

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+∂∂

−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+∂∂

zu

yu

xu

xpg

zuw

yuv

xuu

tu

x μρρ

y- direction:

-(2)----- 2

2

2

2

2

2

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+∂∂

−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+∂∂

zv

yv

xv

ypg

zvw

yvv

xvu

tv

y μρρ

Equations (1) and (2) are called: Navier Stokes equations. 2- Steady laminar flow between parallel flat plates:

The fluid moves in the x- direction without acceleration. v= 0 , w= 0 , 0=

∂∂t

the Navier-Stokes equation in the x- direction (eq. 1) reduces to:

(3)--------- 2

2

dyud

dxdpgx μρ =+−

dxdhgggx −== θsin.

eq. 3 will be

( ) -(4)--------- 12

2

dxhpd

dyud γ

μ+

=

Integration of eq. 4:

( ) Aydx

hpddydu

++

=γ

μ1

( ) (5)----------- 21 2 BAyy

dxhpdu ++

+=

γμ

B.C (Two fixed parallel plates)

B( )

dxhpdaAuay

uyγ

μ+−

=⇒==

=⇒==

2 0

0 0 0

eq. 5 will be

( ) ( ) -(6)----------- 21 2 ayy

dxhpdu −

+=

γμ

B.C (One plate is fixed and the other plate moves with a constant velocity U) (Couette flow)

( )dx

hpdaaUAUuay

Buyγ

μ+

−=⇒==

=⇒==

2

0 0 0

eq. 5 will be

( ) ( ) (7)----------- a

Uy21 2 +−

+= ayy

dxhpdu γ

μ

For the case of horizontal parallel plates:

0=dxdh

eq. 7 will be

( ) (8)----------- a

Uy21 2 +−= ayy

dxdpu

μ

The location of maximum velocity umax may be found by evaluating dydu and setting it to

zero. The volume flow rate is

∫=a

dyuwQ0

-(9)----------- .

3- Hydrodynamic lubrication: Sliding bearing Large forces are developed in small clearance when the surfaces are slightly inclined and one is in motion so that fluid is wedged into the decreasing space. Usually the oils employed for lubrication are highly viscous and the flow is of laminar nature.

Assumptions: The acceleration is zero. The body force is small and can be neglected.

Also 2

2

2

2

2

2

2

2

and zu

yu

xu

yu

∂∂

⟩⟩∂∂

∂∂

⟩⟩∂∂

The Navier-Stokes equation in the x-direction (eq. 1) reduces to:

dxdp

dyud

μ1

2

2

=

Integration:

BAyydxdpu ++⎟

⎠⎞

⎜⎝⎛= 2

21μ

B.C

x

xx h

UdxdphAuhy

UBUuy

−−

=⇒==

=⇒==

μ2 0

0

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−+−=

xx h

yUyhydxdpu 1

21 2

μ

The volume flow rate in every section will be constant.

1 wassume .0

== ∫xh

dyuwQ

dxdphUhQ xx

μ122

3

−=∴ --------(*)

** For a constant taper bearing:

lhh 21 −=δ

( )xhhx δ−=∴ 1

Sub in eq.(*) and solving for dxdp produces:

( ) ( )312

1

126xh

Qxh

Udxdp

δμ

δμ

−−

−=

Integration gives:

( ) ( )C

xhQ

xhUxp +

−−

−= 2

11

66)(δδ

μδδ

μ ---------(**)

B.C

0 0 0

======

o

o

pplxppx

( )2121

21 6 and hhUC

hhhUhQ

+−

=+

=⇒δ

μ

With these values inserted in eq.(**) we obtain the pressure distribution inside the bearing.

( )( )21

226)(

hhhhhUxxp

x

x

+−

=μ

The load that the bearing will support per unit width is:

∫=l

dxxpF0

).(

( ) ⎥⎦⎤

⎢⎣⎡

+−

−−

=1

)1(2ln62

21

2

kkk

hhUlF μ

where

2

1

hhk =

4- Laminar flow between concentric rotating cylinders: Consider the purely circulatory flow of a fluid contained between two long concentric rotating cylinders of radius R1 and R2 at angular velocities ω1 and ω2.

In this case the Navier-Stokes equations in cylindrical coordinates are used. r- direction:

rrrrrrrr

rr g

zuu

ru

rru

rur

rrrp

zuw

ruu

ru

ruu

tu

+⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−∂∂

+−⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+∂∂−

=∂∂

+−∂∂

+∂∂

+∂∂

2

2

22

2

22

2 2111θθρ

μρθ

θθθ

θ- direction:

θθθθθθθθθθθ

θθρμ

θρθg

zuu

ru

rru

rur

rrp

rzuw

ruuu

ru

ruu

tu rr

r +⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+−⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+∂∂−

=∂∂

++∂∂

+∂∂

+∂∂

2

2

22

2

222111

In the above equations: ur = 0 w = 0

0 , 0 , 0 =∂∂

=∂∂

=∂∂

θθθ pu

t

body force = 0 The equation in θ- direction reduces to:

02

2

=⎟⎠⎞

⎜⎝⎛+

ru

drd

drud θθ

Integration: ( ) Aru

drd

r=θ

1

rBAru +=θ ---------(i)

B.C

( )

( )1221

22

22

21

1221

22

22

1

222

111

ωω

ωωω

ωω

θ

θ

−−

−=

−−

+=⇒

====

RRRRB

RRRA

RuRrRuRr

Sub. in eq.(i) yields:

( ) ( ⎥⎦

⎤⎢⎣

⎡−−−

−= 12

22

212

112222

122

1 ωωωωθ rRRrRR

RRu ) -------(ii)

The shear stress may be evaluated by the equation:

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

ru

drdr θμτ

By using eq.(ii):

( )122

22

21

21

22

2 ωωμτ −−

=rRR

RR

5- Examples: 1- Using the Navier-Stokes equation in the flow direction, calculate the power required to pull (1m × 1m) flat plate at speed (1 m/s) over an inclined surface. The oil between the surfaces has (ρ = 900 kg/m3 , μ = 0.06 Pa.s).The pressure difference between points 1 and 2 is (100 kN/m2) . Solution:

The Navier-Stokes equation in x- direction

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+∂∂

−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

2

zu

yu

xu

xpg

zuw

yuv

xuu

tu

x μρρ

We have: Acceleration =0 , v=0 , w=0 , 2

2

2

2

, 0zu

xu

∂∂

=∂∂

The equation reduces to:

xgdxdp

dyud

μρ

μ−=

12

2

Integration

BAyygydxdpu

Aygydxdp

dydu

x

x

++−=

+−=

22

221

1

μρ

μ

μρ

μ

B.C (b=10 mm) B

xgbdxdpb

bUAUuby

uy

μρ

μ 22

0 0 0

+−−

=⇒−==

=⇒==

xx gbdxdpb

bUygy

dxdp

dydu

μρ

μμρ

μ 221 +−−−=∴

The shearing force on the moving plate:

areadyduF

areaF

by

o

×⋅=

×=

=

μ

τ

area=1 m2

xgbdxdpb

bUF ρμ

22−+−=

We have l

pdxdpgg x

Δ−=⋅= , sinθ

(Ans) 5281528

528

30sin81.9900201.0

110100

201.0

01.0106.0 3

WPowerUFPower

NF

F

=×=⋅=

−=

×××−⎟⎟⎠

⎞⎜⎜⎝

⎛ ×−

×−=

2- The external and internal diameters of a collar bearing are (20 cm and 15 cm) respectively. An oil film (0.3 mm) thick is maintained between the collar surface and the bearing. Find the power lost in overcoming the viscous resistance of the oil when the shaft is running at (240 rpm). Take μ=0.1 Pa.s . Solution: The figure shows two views of a collar bearing.

Consider an elemental angular circular strip of the bearing surface of radius r and width dr. Circumferential resistance offered by the oil film over the strip = area×τ

=h

drrrdrhr 222 πμωπωμ =×

Torque required to overcome this resistance:

hdrrr

hdrrT

32 22 πμωπμω=×=

Total torque required ( )∫ −==1

2

42

41

3

22R

R

RRhh

drrT πμωπμω

Power absorbed by the bearing= ( )42

41

2

2RR

hT −=

πμωω

rad/s 860

240260

2 πππω =×

==N

( ) W6.22075.01.00003.02

)8(1.0Power 442

=−××

=ππ (Ans)

University of Technology Sheet No. 1 Mechanical Engineering Dep. Navier-Stokes Equations Fluid Mechanics II (3 rd year) 2011/2012 1- Using the Navier-Stokes equations, determine the pressure gradient along flow, the average velocity, and the discharge for an oil of viscosity 0.02 N.s/m2 flowing between two stationary parallel plates 1 m wide maintained 10 mm apart. The velocity midway between the plates is 2 m/s. [-3200 N/m2 per m ; 1.33 m/s ; 0.0133 m3/s] 2- An incompressible, viscous fluid is placed between horizontal, infinite, parallel plates as shown in figure. The two plates move in opposite directions with constant velocities U1 and U2. The pressure gradient in the x-direction is zero. Use the Navier-Stokes equations to derive expression for the velocity distribution between the plates. Assume laminar flow. [ ( ) 221 UUU

byu −+= ]

3- Two parallel plates are spaced 2 mm apart, and oil (μ = 0.1 N.s/m2 , S = 0.8) flows at a rate of 24×10-4 m3/s per m of width between the plates. What is the pressure gradient in the direction of flow if the plates are inclined at 60o with the horizontal and if the flow is downward between the plates? [-353.2 kPa/m] 4- Using the Navier-Stokes equations, find the velocity profile for fully developed flow of water (μ = 1.14×10-3 Pa.s) between parallel plates with the upper plate moving as shown in figure. Assume the volume flow rate per unit depth for zero pressure gradient between the plates is 3.75×10-3 m3/s. Determine: a- the velocity of the moving plate. b- the shear stress on the lower plate. c- the pressure gradient that will give zero shear stress at y = 0.25b. (b = 2.5 mm) d- the adverse pressure gradient that will give zero volume flow rate between the plates. [3 m/s ; 1.37 N/m2 ; 2.19 kN/m2 per m ; -3.28 kN/m2 per m] 5- A vertical shaft passes through a bearing and is lubricated with an oil (μ = 0.2 Pa.s) as shown in figure. Estimate the torque required to overcome viscous resistance when the shaft is turning at 80 rpm. (Hint: The flow between the shaft and bearing can be treated as laminar flow between two flat plates with zero pressure gradient). [0.355 N.m] 6- Determine the force on the piston of the figure due to shear, and the leakage from the pressure chamber for U = 0. [295.1 N ; 1.636×10-8 m3/s]

7- A layer of viscous liquid of thickness b flows steadily down an inclined plane. Show that, by using the Navier-Stokes equations that velocity distribution is:

( ) θμγ sin2

22ybyu −= and that the discharge per unit width is: θ

μγ sin

33bQ =

8- A wide moving belt passes through a container of a viscous liquid. The belt moving vertically upward with a constant velocity Vo, as illustrated in figure. Because of viscous forces the belt picks up a film of fluid of thickness h. Gravity tends to make the fluid drain down the belt. Use the Navier-Stokes equations to determine an expression for the average velocity vav of the fluid film as it is dragged up the belt. Assume the flow is laminar,

steady, and uniform. [μ

γ3

2hVoav −=v ]

9- Determine the formulas for shear stress on each plate and for the velocity distribution for flow in the figure when an adverse pressure gradient exists such that Q = 0.

[byU

byUu

bU

bU

by 23 ; 4 ; 22

2

0 −==−

= ==μτμ

yτ ]

10- A plate 2 mm thick and 1 m wide is pulled between the walls shown in figure at speed of 0.4 m/s. The space over and below the plate is filled with glycerin (μ = 0.62 N.s/m2). The plate is positioned midway between the walls. Using the Navier-Stokes equations, determine the force required to pull the plate at the speed given for zero pressure gradient; and the pressure gradient that will give zero volume flow rate. [496 N ; 372 kN/m2.m] 11- A slider plate 0.5 m wide constitutes a bearing as shown in figure. Estimate: a- the load carrying capacity. b- the drag. c- the power lost in the bearing. d- the maximum pressure in the oil and its location. [739.6 kN ; 348.6 N ; 348.6 W ; 12500 kN/m2 ; 150 mm] 12- Consider a shaft that turns inside a stationary cylinder, with a lubricating fluid in the annular region. Using the Navier-Stokes equation in θ-direction, show that the torque per

unit length acting on the shaft is given by: 1

42

2

1

21

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

RR

RT πμω

Where: ω = angular velocity of the shaft. R1 = radius of the shaft. R2 = radius of the cylinder.

Problem No. 2 Problem No. 4

Problem No. 5 Problem No. 6

Problem No. 8 Problem No. 9

Problem No. 10 Problem No. 11