Introduction to Mechanical Engineering...Introduction to Mechanical Engineering •Chapter 1 The...

Introduction to Mechanical Engineering

• Chapter 1 The Mechanical Engineering Profession

• Chapter 2 Problem-Solving and Communication Skills

• Chapter 3 Forces in Structures and Machines

• Chapter 4 Materials and Stresses

• Chapter 5 Fluids Engineering

• Chapter 6 Thermal and Energy Systems

• Chapter 7 Motion and Power Transmission

• Chapter 8 Mechanical Design

1

5-2

• Hydrodynamics • Aerodynamics • Steam – Water recycling • Biomedical Engineering • Buoyancy, drag and lift by fluid

3

Properties of Fluids

1. Not equilibrium on shear stress

2. Liquid(incompressible)

3. Gas(compressible)

4

5-5

Properties of Fluids

Continuous motion (flow) by shear stress

No-Slip Condition at microscopic level; only several molecules thick , adheres to solid surface

Properties of Fluids

time)-thmass/(leng

; viscosityof units

fluid) (Newtonian

velocityflow ; v

friction stickness, of measure also

and flow tofluid a of resistance

);Viscosity(

kg/m.s

h

v

A

F

6

P 0.01cP 1

.1.0)(P 1

cP P, ;ity for viscosunit special

sm

kgPoise

7

Example 5.1 Machine Tool Guideways

• oil viscosity ; 240 cP • width of ways ; 8 cm • length of ways ; 40 cm • force ; 90 N • velocity ; 15 cm/s

8

Example 5.1 Machine Tool Guideways

• oil viscosity ; 240 cP • width of ways ; 8 cm • length of ways ; 40 cm • force ; 90 N • velocity ; 15 cm/s • What is thickness of oil film?

mmx

smkg

smsmkgm

F

vA h

h

v

A

F

6.25 1056.2

)/.(90

)/ 15.0)(./ 24.0)()(4.0)(08.0)(2(

5

2

2

9

Pressure and Buoyancy Force

fluid of volume; V

fluid ofdensity ;

gVw

ghPP o 1

10

Buoyancy and pressure

ghpp

gAhApAp

balanceforcemequilibriu

Ahmmassliquid

01

01 0)(

,

11

Units

• 1 pascal = 1 Pa(N/m2)

• psi = lb/in2

• psf = lb/ft2

• 1 atm = 1.013 x 105 Pa

12

objectfluidB gVF

• Pressure and Buoyancy Force

13

14

Example 5.2 Aircraft’s Fuel Capacity

• Fuel capacity 90,000 L • Density ; 840 kg/m3

• What is the weight of fuel?

kNs

mkgx

s

m

L

mL

m

kg

Vgmgw

6.741).

(10416.7

)81.9)(001.0)(000,90)(840(

2

5

2

3

3

15

Example 5.3 Deep Submergence Rescue Vehicle

• Max. dive depth ; 5000 ft • What is water pressure in psi?

psiinx

ft

ft

lbx

fts

ftslugx

sft

slugx

fts

ft

ft

slug

ghppp

ghpp

2225) 1212

)((10204.3

)1

)(.

(10204.3).

(10204.3

)5000)(2.32)(99.1(

2

2

2

5

22

5

2

5

23

01

01

16

Example 5.4 Great White Shark Attack

• 55 gal barrel weighing 35 lb • What force Shark to overcome ?

lblbs

ftslug

lbgal

ftgal

s

ft

ft

slugs

lbgVwFT

wTF

objectfluidB

B

2.436)(35).

(2.471

)(35)1337.0)(55)(2.32)(99.1(

)(35

0

2

3

23

17

5.4 Laminar and Turbulent Fluid Flows

) ,viscosity,,(Re

number Reynolds

lengthsticcharacterispeeddensityvl

18

Example 5.5 Reynolds Number

5000

)./(100.1

)01.0)(/5.0)(/1000(

pipe in the flowingfor water Re )2

10679.3

./5^108.1

)3^106.7)(/720)(3^/(2.1

6.7

/720:.

Re

number Reynolds

3

3

5

smkgx

msmmkgR

x

smkgx

mxsmmkgR

mmd

smvspeedWinchester

vl

e

e

19

Example 5.5 Reynolds Number

7

3

3

3

10861021

1081026Re

sm/ 8 of speed and m 10 ofdiameter hull with Submarine 4)

63.17)./(26.0

)01.0)(/5.0)(/917(Re

... 30 3)

Re

number Reynolds

x.kg/(m.s) x .

m) m/s)()( kg/m

smkg

msmmkg

waterofinsteadoilSAE

vl

20

5.5 Fluid Flows in Pipes

21

In pipe, Not with any disturbances and low enough Re number(<2000), then flow is laminar

)2000Re of case ( 16

)2000Re of case ( ))(1(

2

max

2

max

inL

pdv

inR

rvv

22

)2000Re (128

82

)2

(

2

)2000Re (2

1

;

/

4

max

2

max

2

max

max

casespecialL

pd

vd

v

d

vA

q

casespecialvv

Avq

rateflowVolumetric

txv

xAV

avg

avg

avg

Poiseuille’s law

23

)2000Re of case ( 16

2

max

inL

pdv

24

constant2

2

2211

ghvp

vAvA

25

Bernoulli’s equation

constant2

2

ghvp

1st; Work of pressure force 2nd; Kinetic energy of flow fluid 3rd; Gravitational potential energy of it

26

27

Example 5.6 Automotive Fuel Line

1. A car speed ; 40 mph 2. With fuel economy rate; 28 miles/gallon 1. Fuel line inside dia.; 3/8 in.

8.467)./(101.6

.)/)(12/1.)()(8/3)(/10917.6)(/32.1(Re )

10917.6

).12

1(.)375.0(

4

)(10306.5

v

is rate flow line)

)(10306.5)1337.0)(10968.3(

10968.3)(3600

1

/28

/40

/28

/40

rate fuel

is rate flow c volumetri)

6

23

2

22

35

avg

35

34

4

sftslugx

inftinsftxftslugvlc

s

ftx

in

ftin

s

ftx

A

q

b

s

ftx

gal

ft

s

galx

s

galx

s

hx

galmi

hmi

galmi

hmispeedcarq

a

a) Volumetric flow of oil in ft3/s? b) Fuel avg. velocity in in./s? c) What Re number?

28

Drag forces and Viscosity

dragoftcoefficienC

speedrelativev

objectofareafrontalA

densitysfluid

CAvF

D

DD

:

:

:

':

2

1 2

29

30

Nearly constant

31

Drag forces and Viscosity

pois) , ty(kg/m.s viscosisfluid' ;μ

diameter ssphere' ; d

)1Re ; spherefor case Special( 3 dvFD

32

Example 5.7 Golf Ball in Flight

1. 1.68 in. dia. golf ball speed 70 mph 2. Drag force? A) smooth ball B) CD=0.27

lbxs

ftslugxF

insteadCWithB

lbxs

ftslugx

s

ftftx

ft

slugxF

Csohighveryxvl

CdecidingfornumberknowToA

CAvF

D

D

D

D

D

DD

2

2

2

2

2

2

222

3

3

4

2

10104.5.

1010.5

, 27.0 )

10452.9.

10452.9

)5.0()7.102)(10538.1)(1033.2(2

1

)apply , (10813.8Re

Re )

2

1

33

34

)1(Re )(18

)6/( ,6/

,3

0

2

33

fluidobject

objectobject

objectfluidBD

DB

ρρgd

v

ρdmdV

mgWgVρπμdv , FF

-wFFbalanceforce

Terminal Velocity

Drag and Lift forces in fluids

35

Wind tunnel

36

37

38

39

equation sBernoulli'

constant2

2

ghvp

40

constant2

2

ghvp

Lift force by pressure difference on the airfoil

bottombottom ppthenvv

vp

toptop

2

, If

constant2

lift oft coefficien ;

2

1 2

L

LL

C

CAvF

42

43

44

45

1. Buoyancy in fluids

2. Drag forces in fluids

3. Lift forces in fluids

objectfluidB gVF

)1(Re 3 dvFD

2

1 2

DD CAvF

LLCAvF 2

2

1

vlRe

46

Summary

Problem 4.36: (a) A luxury sports car has a frontal area of 22.4 ft2 and a 0.29 coefficient of drag at 60 mi/hr. What is the drag force on the vehicle at this speed? (b) A sport utility vehicle has CD = 0.45 at 60 mi/hr, and the slightly larger frontal area of 29.1 ft2. What is the drag force in this case?

DD CAvF 2

2

1

47

s

ft

s

hr

mi

ft

hr

miv 8810778.2528060 4

lbs

ftftftslugF

D6.5829.0884.22/1033.2

2

12

233

lbs

ftftftslugF

D11845.0881.29/1033.2

2

12

233

Approach: Apply Equation using the density of air

listed in Table 4.3. Convert velocity to consistent units when calculating

the drag force. Solution:

a) Luxury sports car Velocity:

b) Sport utility vehicle

33 /1033.2 ftslug

48

Problem 4.28: A steel storage tank is filled with gasoline. The tank has partially corroded on its inside, and small particles of rust have contaminated the fuel. The rust particles are spherical, have diameter 25 µm, and density 5.3 g/cm3. (a) What is the terminal velocity of the particles as they fall through the gasoline? (b) How long would it take the particles to fall 5 m and settle out of the tank?

• Approach: Apply terminal velocity Equation to find the terminal velocity using the density(680 kg/m3) and viscosity(2.9x10-4 kg/m.s) of gasoline listed in Table 4.3. Calculate the Reynolds number to confirm that the velocity is low enough that the equation was correctly applied.

3

3

35300001.01003.5

m

kg

g

kg

m

cm

cm

gsphere

)/(0054.0

6805300/109.218

1025/81.9)(

18 334

2622

sm

m

kg

m

kg

smkg

msmgdv gp

Solution: a) Terminal velocity

49

which is dimensionless. Since the numerical value of above Re is less than one, the velocity was properly calculated.

31.0/109.2

1025/0054.0/680Re

4

63

smkg

msmmkgvd

Check Reynolds number,

b) Fall time

min4.15926/0054.0

5 s

sm

m

50

Problem 4.39: Submarines dive by opening vents which allow air to escape from ballast tanks, and water to flow in and fill them. In addition, diving planes located at the bow are angled downward to help push the boat below the surface. Calculate the diving force produced by a 20 ft2 hydroplane that is inclined by 3o as the boat cruises at 15 knots (1 knot = 1.152 mi/hr).

Approach: Apply Equation (4.19) using consistent units for velocity. From Figure 4.33

at an angle of attack of 3o, CL = 0.32. Use the density of sea water(1.99 slut/ft3)

s

ft

s

hr

mi

ft

kt

hrmiktv

Velocity

3.2510778.25280/

1516.115 4

lbs

ftft

ft

slugCAvL

Downward

L 408832.03.25201099.12

1

2

1

force diving

2

2

3

32

51

Reference

• Wickert J., Lewis K., “An Introduction to Mechanical Engineering”, 3rd Edition, 2013.

• www.slideshare.com

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