8-Practice Problems.pdf

Chapter 8

Dimensional Analysis and

Similitude

Problem 8.1

The discharge, !" of an ideal ß uid (no viscous e ! ects) through an ori Þ ce depends

on the ori Þ ce diameter, # , the pressure drop across the ori Þ ce, !$ , and the ß uid

density. Find a nondimensional relationship for the discharge.

Solution

The functional relationship is

! = %(#"!$"&)

Based on the Buckingham " theorem, there should be 4!3 = 1 '! groups. Use

the step-by-step method, as shown in the following table.

Variable [ ] Variable [ ] Variable [ ] Variable [ ]

! 3

$ 3

$ 2

# (

! 3

&# 3 )

!" 2 !$#

" 2

%$ 2

" 2

As shown in the table, the length dimension is Þ rst eliminated with # , then the

mass dimension is eliminated with &# 3 " and Þ nally the time dimension is eliminated

with !$*&# 2 + Thus

# 2

!$ = ,

CHAPTER 8. DIMENSIONAL ANALYSIS AND SIMILITUDE

where , is a constant. This may be expressed as

! = ,# 2

Problem 8.2

The terminal velocity of a sphere (maximum drop velocity) depends on the sphere

diameter, sphere density, ß uid density, ß uid viscosity, and acceleration due to grav-

ity.

- ( = %(."& ) "& * "/"0)

Find a nondimensional form for the terminal velocity.

Solution

Based on the Buckingham " theorem, there should be 6!3 = 3 ' -groups. Use the

step-by-step method as shown in the following table.

Variable [ ] Variable [ ] Variable [ ] Variable [ ]

- (

+ !

+ ! % " ,

& )

! 3

& ) . 3 )

% #

% "

& *

! 3

& * . 3 )

% " , 2

. (

" 2

.% " , 3

- 2

First, length is eliminated with ." then mass is eliminated with & * . 3 " and Þ nally

time is eliminated with /*& * . 2 +

The nondimensional grouping becomes

- ( & * .

= %( & )

& * " 0& * . 3

/ 2

)

which can also be written as

& * . %( & )

0.& * .

- ( =

& * "

)

0.%( & )

0.& * .

- ( =

& * "

)

Problem 8.3

The pressure drop in a smooth horizontal pipe in a turbulent, incompressible ß ow

dependsonthepipediameter, pipe length, ß uidvelocity, ß uiddensity, andviscosity.

!$ = %(."1"-"&"/)

Find a nondimensional relationship for the pressure drop.

Solution

By the Buckingham " theorem, the number of dimensionless ' -groups is 6!3 = 3 .

The exponent method will be used. First, express the equation as

!$ = . / 1 0 - 1 & 2 / 3

Substitute the dimensions of each variable

)

(2 2 = ( / ( 0

(

)

( 3

)

Equate the powers of each dimension

) : 1 = 3 +4

( :

!1 = 5+6 +7 !33 !4

2 :

!2 = !7 !4

Solving for 5" 7 and 3 in terms of 6 and 4

3 = 1!4

7 = 2!4

5 = !6 !4

Substituting back into the equation for pressure

!$ = . !0!3 1 0 - 2!3 & 1!3 / 3

-.&

= &- 2

This relation can be expressed as

&- 2 = %

. " &-.

CHAPTER 8. DIMENSIONAL ANALYSIS AND SIMILITUDE

Problem 8.4

A 25 scale model of an airship is tested in water at 20 o C. If the airship trav-

els 5 m/s in air at atmospheric pressure and 20 o C, Þ nd the velocity for the model to

achieve similitude. Also, Þ nd the ratio of the drag force on the prototype to that

on the model. The densities of water and air at these conditions are 1000 kg/m 3

and 1.2 kg/m 3 + The corresponding dynamic viscosities of water and air are 10 !3

N · s/m 2 and 1.81 ×10 !5 N · s/m 2 .

Solution

The signi Þ cant nondimensional number for this problem is the Reynolds number.

Thus, for similitude

RE model = RE prototype

- 4 ( 4 & 4

/ 4

= - & ( & & &

/ &

- 4 = - & ( &

( 4

& &

& 4

/ 4

/ &

= 5×25×

1+2

1000

10 !3

1+81×10 !5

= 8+29 m/s

Dimensional analysis for the force yields

8 = &- 2 ( 2 %(RE)

Thus, for the ratio of forces

8 4 = & &

- &

- 4

( &

( 2 4

%(RE & )

%(RE 4 )

& 4

Since the Reynolds numbers are the same

8 4 = & &

- &

- 4

( &

( 2 4

& 4

The force ratio is

8 4 = 1+2

5 2

8+29 2 25 2

1000

= 0+273

8 &

Problem 8.5

A scale model of a pumping system is to be tested to determine the head losses

in the actual system. Air with a speci Þ c weight of 0.085 lbf/ft 3 and a viscosity of

3.74 ×10 !7 lbf · s/ft 2 is to be used in the model. Water with a speci Þ c weight of 62.4

lbf/ft 3 and a viscosity of 2.36 ×10 !5 lbf-s/ft 2 is used in the prototype. The velocity

in the prototype is 2 ft/s. A practical upper limit for the air velocity in the model

to avoid compressibility e ! ects is 100 ft/s. Find the scale ratio for the model and

the ratio of the pressure losses in the prototype to those in the model.

Solution

In this problem, the Reynolds number is the important scaling parameter so

RE model = RE prototype

- 4 ( 4 & 4

/ 4

= - & ( & & &

/ &

Therefore

( & = - & & & / 4

- 4 & 4 / &

= 2

100

(62+4*32+2)

(0+085*32+2)

3+74×10 !7

2+36×10 !5

= 0+233

or about a 4 scale model. Note that the speci Þ c weight is changed to mass density

by dividing by 32.2.

Since the Reynolds numbers are the same, the pressure coe " cients are also the

same.

9 &54 = 9 &5&

&- 2

!$ 4 = & &

- &

- 4

& 4

which gives

!$ 4 = (62+4*32+2)

2 2

100 2

(0+085*32+2)

= 0+294

( 4

!$ &

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