4800866 introduction fluid mechanics solution chapter 07

7/31/2019 4800866 Introduction Fluid Mechanics Solution Chapter 07

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Problem 7.4



Problem 7.5

Nondimensionalizing the velocity, pressure, spatial measures, and time:

L

V t t

L

r r

L

x x

p

p p

V

uu ===

∆== *****

Hence

***** t V

Lt r Dr x L x p p puV u ===∆==

Substituting into the governing equation

∂∂

+∂

∂+

∂∂

∆−=∂∂

=∂∂

*

*

*

1

*

*1

*

*11

*

*2

2

2 r

u

r r

u

DV

x

p

L p

t

u

L

V V

t

uν

ρ

The final dimensionless equation is

∂

∂

+∂

∂

+∂

∂∆

−=∂

∂

*

*

*

1

*

*

*

*

*

*

2

2

2 r

u

r r

u

D

L

V D x

p

V

p

t

u ν

ρ

The dimensionless groups are

D

L

V DV

p ν

ρ 2

∆



Problem 7.6

Recall that the total acceleration is

V V t

V

Dt

V D rr

rr

∇⋅+∂∂

=

Nondimensionalizing the velocity vector, pressure, angular velocity, spatial measure, and

time, (using a typical velocity magnitude V and angular velocity magnitude Ω):

L

V t t

L

x x

p

p p

V

V V ==

ΩΩ

=Ω∆

== *****

r

r

r

r

Hence

***** t V

Lt x L x p p pV V V ==ΩΩ=Ω∆==

rrrr

Substituting into the governing equation

*1

**2****

* p

L

pV V V V

L

V V

t

V

L

V V ∇

∆−=×ΩΩ+⋅∇+

∂∂

ρ

rrrr

r



The final dimensionless equation is

**2****

*2

pV

pV

V

LV V

t

V ∇

∆−=×Ω

Ω+⋅∇+∂∂

ρ

rrrr

r

The dimensionless groups are

V

L

V

p Ω∆2ρ

The second term on the left of the governing equation is the Coriolis force due to a

rotating coordinate system. This is a very significant term in atmospheric studies, leading

to such phenomena as geostrophic flow.



Problem 7.7



Problem 7.8

Given: That drag depends on speed, air density and frontal area

Find: How drag force depend on speed

Apply the Buckingham Π procedure

c F V ρ A n = 4 parameters

d Select primary dimensions M , L, t

e 2

32 L

L

M

t

L

t

ML

AV F ρ

r = 3 primary dimensions

f V ρ A m = r = 3 repeat parameters

g Then n – m = 1 dimensionless groups will result. Setting up a dimensional equation,

( ) 000

2

2

3

1

t L M t

ML L

L

M

t

L

F AV

cba

cba

=

=

=Π ρ



Summing exponents,

202:

10123:

101:

−==−−

−==++−

−==+

aat

ccba L

bb M

Hence

AV

F 21

ρ=Π

h Check using F , L, t as primary dimensions

[ ]12

2

2

4

21 ==Π

Lt

L

L

Ft

F

The relation between drag force F and speed V must then be

22 V AV F ∝∝ ρ

The drag is proportional to the square of the speed.



Problem 7.10



Problem 7.11



Problem 7.12



Problem 7.13



Problem 7.14



Problem 7.15



Problem 7.16



Problem 7.17 (In Excel)

Given: That drain time depends on fluid viscosity and density, orifice diameter, and gravity

Find: Functional dependence of t on other variables

Solution

We will use the workbook of Example Problem 7.1, modified for the current problem

The number of parameters is: n = 5

The number of primary dimensions is: r = 3

The number of repeat parameters is: m = r = 3

The number of Π groups is: n - m = 2

Enter the dimensions (M, L, t) of

the repeating parameters, and of up to

four other parameters (for up to four Π groups).

The spreadsheet will compute the exponents a , b , and c for each.

REPEATING PARAMETERS: Choose , g , d

M L t

ρ 1 -3

g 1 -2d 1

GROUPS:

M L t M L t

t 0 0 1 µ 1 -1 -1

Π1: a = 0 Π2: a = -1

b = 0.5 b = -0.5

c = -0.5 c = -1.5



The following Π groups from Example Problem 7.1 are not used:

M L t M L t

0 0 0 0 0 0

Π3: a = 0 Π4: a = 0

b = 0 b = 0c = 0 c = 0

Hence and with

The final result is

d

g t =Π1 32

2

2

3

2

12 gd

d g ρ

µ

ρ

µ→=Π ( )21 Π=Π f

=

32

2

gd f

g

d t

ρ

µ



Problem 7.18



Problem 7.19



Problem 7.20



Problem 7.21



Problem 7.22



Problem 7.23




Given: That dot size depends on ink viscosity, density, and surface tension, and geometry

Find: Π groups

Solution










REPEATING PARAMETERS: Choose , V , D

M L t

ρ 1 -3

V 1 -1

D 1

GROUPS:

M L t M L t

d 0 1 0 µ 1 -1 -1

Π1: a = 0 Π2: a = -1

b = 0 b = -1

c = -1 c = -1



M L t M L t

σ 1 0 -2 L 0 1 0

Π3: a = -1 Π4: a = 0

b= -2

b= 0

c = -1 c = -1

Note that groups Π1 and Π4 can be obtained by inspection

Hence D

d =Π

1

µ

ρ

ρ

µ VD

VD→=Π 2

DV 23

ρ

σ =Π

D

L=Π 4



Problem 7.25




Given: Bubble size depends on viscosity, density, surface tension, geometry and pressure

Find: Π groups

Solution










REPEATING PARAMETERS: Choose , p , D

M L t

ρ 1 -3

∆ p 1 -1 -2 D 1

GROUPS:

M L t M L t

d 0 1 0 µ 1 -1 -1

Π1: a = 0 Π2: a = -0.5

b = 0 b = -0.5

c = -1 c = -1



M L t M L t

σ 1 0 -2 0 0 0

Π3: a = 0 Π4: a = 0

b = -1 b = 0c = -1 c = 0

Note that theΠ1 group can be obtained by inspection

Hence D

d =Π

1 2

2

2

1

2

12 pD

D p∆

→

∆

=Πρ

µ

ρ

µ

p D∆=Π

σ3




Given: Speed depends on mass, area, gravity, slope, and air viscosity and thickness

Find: Π groups

Solution










REPEATING PARAMETERS: Choose g , , m

M L t

g 1 -2δ 1

m 1

GROUPS:

M L t M L t

V 0 1 -1 µ 1 -1 -1

Π1: a = -0.5 Π2: a = -0.5

b = -0.5 b = 1.5

c = 0 c = -1



M L t M L t

θ 0 0 0 A 0 2 0

Π3: a = 0 Π4: a = 0

b = 0 b = -2c = 0 c = 0

Note that theΠ1 , Π3 and Π4 groups can be obtained by inspection

Henceδ

δ g

V

g

V 2

2

1

2

11 →=Π g m

m g

2

32

2

1

2

3

2

δµµδ→=Π θ=Π

3 24

δ

A=Π




Given: Time to speed up depends on inertia, speed, torque, oil viscosity and geometry

Find: Π groups

Solution










REPEATING PARAMETERS: Choose , D , T

M L t

ω -1

D1

T 1 2 -2

GROUPS:

Two Π groups can be obtained by inspection: / D and L / D . The others are obtained below

M L t M L t

t 0 0 1 µ 1 -1 -1

Π1: a = 1 Π2: a = 1

b = 0 b = 3

c = 0 c = -1



M L t M L t

I 1 2 0 0 0 0

Π3: a = 2 Π4: a = 0

b = 0 b = 0

c= -1

c= 0

Note that theΠ1 group can also be easily obtained by inspection

Hence the Π groups are

ωt D

δ

T

D3

µω

T

I 2ω

D

L



Problem 7.29



Problem 7.30



Problem 7.31



Problem 7.32



Problem 7.33



Problem 7.34



Problem 7.35

Given: That the cooling rate depends on rice properties and air properties

Find: The Π groups

Apply the Buckingham Π procedure

dT/dt c k L c p ρ µ V n = 8 parameters

Select primary dimensions M , L, t and T (temperature)

t

L

Lt

M

L

M

T t

L L

T t

ML

T t

L

t

T

V c Lk cdt dT p

32

2

22

2

µ ρ

r = 4 primary dimensions

V ρ L c p m = r = 4 repeat parameters

Then n – m = 4 dimensionless groups will result.

By inspection, one Π group is c/c p



Setting up a dimensional equation,

( ) 0000

2

2

3

1

t L M T t

T

T t

L L

L

M

t

L

dt

dT c LV

d

cba

d p

cba

=

=

=Π ρ

Summing exponents,

3012:

12023:

00:

101:

−==−−−

=→−=+=++−

==

==+−

ad at

ccad cba L

bb M

d d T

Hence

31V

Lc

dt

dT p=Π

By a similar process, find

pc L

k 22

ρ

=Π

and

LV ρ

µ =Π3

Hence

=

LV c L

k

c

c f

V

Lc

dt

dT

p p

p

ρ

µ

ρ

,,23



Problem 7.36



Problem 7.37



Problem 7.38



Problem 7.39



Vw 6.9m

s=

Vw Vair µw

µair

⋅ ρair

ρw

⋅ Lair

Lw

⋅= Vair µw

µair

⋅ ρair

ρw

⋅ Lratio⋅= 5 m

s⋅ 10 3−

1.8 105−×

× 1.24

999

× 20×=

Then

ρw Vw⋅ Lw⋅

µw

ρair Vair ⋅ Lair ⋅

µair

=For dynamic similarity we assume

Fw 2 kN⋅=Lratio 20=Vair 5m

s

⋅=The given data is

µw 103− N s⋅

m2

⋅=ρw 999kg

m3

⋅=

µair 1.8 105−×

N s⋅

m2

⋅=ρair 1.24kg

m3

⋅=From Appendix A (inc. Fig. A.2)

Solution

Find: Required water model water speed; drag on protype based on model drag

Given: Model scale for on balloon

Problem 7.40



For the same Reynolds numbers, the drag coefficients will be the same so

Fair 1

2ρair ⋅ Aair ⋅ Vair

2⋅

Fw1

2ρw⋅ Aw⋅ Vw

2⋅=

whereAair

Aw

Lair

Lw

2

= Lratio2=

Hence the prototype drag is

Fair Fw

ρair

ρw

⋅ Lratio2⋅

Vair

Vw

2

⋅= 2000 N⋅1.24

999

× 202×

5

6.9

2

×=

Fair 522N=



Problem 7.41



Problem 7.42



Problem 7.43



Problem 7.44



Problem 7.45



Problem 7.46



Problem 7.47



Problem 7.48



Problem 7.49



Problem 7.50



Problem 7.51



Vm 0.125m

s=Vm Vinsect

Linsect

Lm

⋅= Vinsect Lratio⋅= 1.25m

s⋅

1

10×=

Hence

ωinsect Linsect⋅

Vinsect

ωm Lm⋅

Vm

=

Vinsect Linsect⋅

νair

Vm Lm⋅

νair

=

For dynamic similarity the following dimensionless groups must be the same in the insect and

Lratio1

10

=Vinsect 1.25m

s

⋅=ωinsect 50Hz=The given data is

νair 1.5 105−

×m

2

s⋅=ρair 1.24

kg

m3

⋅=From Appendix A (inc. Fig. A.3)

Solution

Find: Required model speed and oscillation frequency

Given: 10-times scale model of flying insect

Problem 7.52



Also ωm ωinsect

Vm

Vinsect

⋅

Linsect

Lm

⋅= ωinsect

Vm

Vinsect

⋅ Lratio⋅= 50 Hz⋅0.125

1.25×

1

10×=

ωm 0.5 Hz⋅=

It is unlikely measurable wing lift can be measured at such a low wing frequency (unless the

measured lift was averaged, using an integrator circuit). Maybe try hot air for the model

For hot air try ν

hot2 10

5−×

m2

s⋅= instead of ν

air 1.5 10

5−×

m2

s⋅=

HenceVinsect Linsect⋅

νair

Vm Lm⋅

νhot

=

Vm Vinsect

Linsect

Lm

⋅

νhot

νair

⋅= 1.25m

s

⋅1

10

×2

1.5

×= Vm 0.167m

s

=

Also ωm ωinsect

Vm

Vinsect

⋅

Linsect

Lm

⋅= ωinsect

Vm

Vinsect

⋅ Lratio⋅= 50 Hz⋅0.167

1.25×

1

10×=

ω

m0.67 Hz⋅=

Hot air does not improve things much



Finally, try modeling in water νw 9 107−

×m

2

s⋅=

HenceVinsect Linsect⋅

νair

Vm Lm⋅

νw

=

Vm Vinsect

Linsect

Lm

⋅

νw

νair

⋅= 1.25m

s⋅

1

10×

9 107−

×

1.5 105−

×

×= Vm 0.0075m

s=

Also ωm ωinsect

Vm

Vinsect

⋅

Linsect

Lm

⋅= ωinsect

Vm

Vinsect

⋅ Lratio⋅= 50 Hz⋅0.0075

1.25×

1

10×=

ωm 0.03 Hz⋅=

This is even worse! It seems the best bet is hot (very hot) air for the wind tunnel.



Problem 7.53



Problem 7.54



Problem 7.55



Problem 7.56



Problem 7.57



Problem 7.58



Problem 7.59



Problem 7.60



k m1

2ρ⋅ Am⋅ CD⋅=andk p

1

2ρ⋅ A p⋅ CD⋅=where

Fm k m Vm2⋅=andF p k p V p

2⋅=or

Fm1

2ρ⋅ Am⋅ CD⋅ Vm

2⋅=andF p1

2ρ⋅ A p⋅ CD⋅ V p

2⋅=We have

The problem we have is that we do not know the area that can be used for the entire

model or prototype (we only know their chords).

CD

F p

1

2ρ⋅ A p⋅ V p

2⋅=

Fm

1

2ρ⋅ Am⋅ Vm

2⋅=

For high Reynolds number, the drag coefficient of model and prototype agree

Solution

Find: Plot of lift vs speed of model; also of prototype

Given: Data on model of aircraft

Problem 7.62



Note that the area ratio A p/ Am is given by ( L p/ Lm )2 where L p and Lm are length scales,

e.g., chord lengths. Hence

k p

A p

Amk m⋅=

L p

Lm

2

k m⋅=5

0.15

2

k m⋅= 1110 k m⋅=

We can use Excel 's Trendline analysis to fit the data of the model to find k m, and then

find k p from the above equation to use in plotting the prototype lift vs velocity curve. This

is done in the corresponding Excel workbook

An alternative and equivalent approach would be to find the area-drag coefficient AmC D for the

model and use this to find the area-drag coefficient A pC D for the prototype.




Given: Data on model of aircraft

Find: Plot of lift vs speed of model; also of prototype

Solution

V m (m/s) 10 15 20 25 30 35 40 45 50

F m (N) 2.2 4.8 8.7 13.3 19.6 26.5 34.5 43.8 54.0

This data can be fit to

From the trendline, we see that

k m = N/(m/s)2

(And note that the power is 1.9954 or 2.00 to three signifcant

figures, confirming the relation is quadratic)

Also,k

p = 1110k

m

Hence,

k p = 24.3 N/(m/s)2

F p = k pV m2

0.0219

F

m

1

2

ρ⋅ A

m

⋅ C

D

⋅ V

m

2⋅= or F

m

k

m

V

m

2⋅=



V p (m/s) 75 100 125 150 175 200 225 250

F p (kN)(Trendline)

137 243 380 547 744 972 1231 1519

Lift vs Speed for an Airplane Model

y = 0.0219x1.9954

R 2

= 0.9999

0

10

20

30

40

50

60

0 10 20 30 40 50 60

V m (m/s)

F m ( N

)

Model

Power Curve Fit

Lift vs Speed for an

Airplane Prototype

0

200

400

600

800

1000

1200

1400

1600

0 50 100 150 200 250 300

V p (m/s)

F p

( k N )



Lift vs Speed for an Airplane Model

(Log-Log Plot)

y = 0.0219x1.9954

R 2 = 0.9999

1

10

100

10 100

V m (m/s)

F m ( N )

Model

Power Curve Fit

Lift vs Speed for an Airplane Prototype (Log-Log

Plot)

1

10

100

1000

10000

10 100 1000

V p (m/s)

F p (

k N )



Problem 7.63




Given: Data on centrifugal water pump

Find: Π groups; plot pressure head vs flow rate for range of speeds

Solution










REPEATING PARAMETERS: Choose , g , d

M L t

ρ 1 -3

ω -1

D 1

GROUPS:

M L t M L t

∆ p 1 -1 -2 Q 0 3 -1

Π1: a = -1 Π2: a = 0

b = -2 b = -1

c = -2 c = -3



The following Π groups from Example Problem 7.1 are not used:

M L t M L t

0 0 0 0 0 0

Π3: a = 0 Π4: a = 0

b = 0 b = 0

c = 0 c = 0

The data is

Q (m3/hr) 0 100 150 200 250 300 325 350

p (kPa) 361 349 328 293 230 145 114 59

ρ = 999 kg/m3

ω = 750 rpm

D = 1 m ( D is not given; use D = 1 m as a scale)

Q /( D3) 0.00000 0.000354 0.000531 0.000707 0.000884 0.00106 0.00115 0.00124

p /(2 D

2) 0.0586 0.0566 0.0532 0.0475 0.0373 0.0235 0.0185 0.00957

Centifugal Pump Data and Trendline

y = -42371x2

+ 13.399x + 0.0582

R 2

= 0.9981

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.0000 0.0002 0.0004 0.0006 0.0008 0.0010 0.0012 0.0014

Q /( D3)

p / (

2 D 2 )

Pump Data

Parabolic Fit

Hence and with Π1 = f (Π2).

Based on the plotted data, it looks like the relation between Π1 and Π2 may be parabolic

Hence

221 D p

ρω ∆=Π 32

DQ

ω =Π

2

3322

+

+=

∆

D

Qc

D

Qba

D

p

ω ω ρω



From the Trendline analysis

a = 0.0582

b = 13.4

c = -42371

and

Finally, data at 500 and 1000 rpm can be calculated and plotted

ω = 500 rpm

Q (m3/hr) 0 25 50 75 100 150 200 250

p (kPa) 159 162 161 156 146 115 68 4

ω = 1000 rpm

Q (m3/hr) 0 25 50 100 175 250 300 350

p (kPa) 638 645 649 644 606 531 460 374

+

+=∆

2

33

22

D

Qc

D

Qba D p

ω ω

ρω

Centifugal Pump Curves

0

100

200

300

400

500

600

700

0 50 100 150 200 250 300 350 400

Q (m3 /hr)

p (

k P a )

Pump Data at 750 rpm

Pump Curve at 500 rpm

Pump Curve at 1000 rpm



Problem 7.65



Problem 7.66



Problem 7.68



Problem 7.69



Problem 7.70



Problem 7.71



Problem 7.72

4800866 introduction fluid mechanics solution chapter 07

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