SOLUTIONS MANUAL FOR Multiphase Flows with Droplets and Particles Second Edition
by
Clayton Crowe
SOLUTIONS MANUAL F...

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SOLUTIONS MANUAL FOR Multiphase Flows with Droplets and Particles Second Edition

by

Clayton Crowe

SOLUTIONS MANUAL FOR Multiphase Flows with Droplets and Particles Second Edition

by Clayton Crowe

Boca Raton London New

York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2011 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20150127 International Standard Book Number-13: 978-1-4398-7320-5 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Solution Manual Table of contents

There are no exercises for Chapters 1, 8 and 9.

Chapter 2

1

Chapter 3

6

Chapter 4

16

Chapter 5

38

Chapter 6

49

Chapter 7

59

Chapter 10

69

Chapter 2 Problem 2.1 The equation for droplet spacing in a lattice configuration is =

µ

6

¶13

=

µ

6 × 04

¶13

= 109

Problem 2.2 The data given are ¯ = 05 kg/m3 , = 100, ¯ = 12 kg/m3 and = 3 1000 kg/m Number density: =

¯ 6¯ = = 95 × 108 /m3 3

Disperse phase volume fraction: = ¯ = 05103 = 05 × 10−3 Continuous phase bulk density: 3

¯ = (1 − ) = 12 kg/m Concentration: = ¯ ¯ = 0416 Problem 2.3

Consider a cube in which particles each corner and they are all touching a particle at the cube center. The particle locations are illustrated in the diagram. The total distance across the diagonal is 2 The length of a side, would be related to the distance across the diagonal by 32 = (2)2

1

4 = ( )12 3 There are two whole particles within the volume so the solids volume fraction is 33 3 = ( )32 = 0680 3 3 4 The maximum solids volume fraction for the lattice configuration is = and gives =

=

= 0524 6

Problem 2.4 Droplets released in hot stream will cool the carrier gas, increase the density and reduce the velocity to satisfy the continuity equation. The reduced gas velocity will lead to less acceleration of the droplets as they proceed toward velocity equilibrium with the gas.

Tc

one-way coupling two-way coupling

c u v x

Problem 2.5

p1

p2

Control volume

2

Consider the duct shown in the figure. The control volume is designated by the dashed lines. The momentum equation states that the rate of change of momentum in the control volume plus the net eﬄux of momentum through the control surface is equal to the forces acting on the control surface. When the particle bounce from the wall they enter the control volume and when they impinge on the wall the exit the control volume. The pressure force is given by (1 − 2 ) where is the cross-sectional area. The force due to skin friction is − ∆ 12 2 The net eﬄux of momentum due to the particles is ˙ ( − 0 ) where 0 is the rebound velocity and ˙ is the particle flow rate per unit area into the control volume through the control surface adjacent to the wall. The flow is steady and there is no momentum change of the continuous phase. Thus the momentum equation is 1 (1 − 2 ) − 2 = ˙ ∆( − 0 ) 2 The value for 0 is 05 so the momentum equation can be written as ˙ ∆ 2 ∆ = 2 + 05 ∆ ∆ The Darcy-Weisbach friction factor is defined as 1 ∆ = 2 ∆ 2 so the equation for pressure drop can be expressed as ˙ ∆ 1 2 2 = 2 + 05 2 ∆ But 10% of the particle mass flow impinges on the wall per diameter length of duct so ∆ ˙ ∆ = 01˙ so the equation for the friction factor is = 4 + 01

˙

For ∼ 1 one has = 4 + 01

3

Problem 2.6

p single phase low coupling high coupling x

With small momentum coupling, the pressure variation varies little from the single phase. However, with a large momentum coupling there is a greater pressure drop since the gas has to accelerate the particles. Also the minimum pressure drop occurs after the throat because the particles are still being accelerated beyond the throat. Problem 2.7 The data provided are ˙ = 01 kg/s ˙ = 001 kg/s, = 30 m/s, = 100m, = 5 cm and = 002 cm2 s. For standard conditions, = 12 kg/m3 Evaporation time is 2 = 0005 s. Carrier phase velocity is = ˙ = 424 m/s. The mass transfer Stokes number is =

= 424

The concentration is =

= 0141

The mass coupling parameter is Π = = 0141424 = 003

4

The latent heat coupling parameter is Π = Π

= 0278

In this case the mass coupling is negligible but the heat transfer due to eﬀect evaporation should be considered although the eﬀect may not be significant. Problem 2.8

one-way coupling two-way coupling

Tc u p

Td x

The temperature of the ice particle does not change because of the change in phase. With two-way coupling the gas phase temperature drops due to heat transfer to the ice particles. The density increases due to lower temperature so the velocity increases to satisfy continuity. The pressure decreases more due to the acceleration of the gas.

5

Chapter 3 Problem 3.1 The given data are = 60m and = 03 1. The number median diameter is 2

= −3 = 60131 = 458m 2. The Sauter mean diameter is 5

2

32 = 2 = 458 × 125 = 574m 3. The mass mean diameter is 2

= 2 = 60 × 1046 = 628m 4. The mass mode is the value for which () is a maximum.

() = =

∙ ¸ 1 1 2 1 √ exp − 2 (ln − ln ) 2 2 ∙ ¸ 1 1 2 √ exp − 2 (ln − ln ) − ln 2 2

Taking the derivative with respect to ln and setting to zero gives −

1 (ln − ln ) − 1 = 0 2 2

2

Therefore ln mmod e = − = 60−03 = 548m

Problem 3.2 Rosin-Rammler distribution with = 18 and = 120 m. 1. The parameter is = 0693118 = 1471 m 2. The mass median diameter is R = () = Γ( 1 + 1) = 1471 × Γ(1555) = 1307 m The variance is R 2 = 2 () − 2 = 2 Γ( 2 + 1) − 2 = 14712 Γ(2111) − 13072 = 5607 m2 The corresponding standard deviation is 74.9 m. 3. The Sauter mean diameter is 6

3 () = 2 () Γ(1) 1471 Γ(444) = 738

32 =

Γ(1) Γ(1− 1 )

=

m

Problem 3.3 The number frequency distribution is = 1100 = 001 m−1 for 0 ≤ ≤ 100m and = 0 everywhere else. 1. The number mean is Z 100 = 001 = 50m 0

The variance is

2 =

Z

100

0012 − 2 = 833m2

0

and the standard deviation is = 289m. The Sauter mean diameter is R 100 0013 3 32 = R0100 = × 100 = 75 m 2 4 001 0 2. The number cumulative distribution is () = The number median diameter is

Z

001 = 001 0

05 = 001 = 50m 3. The mass frequency distribution is

() = =

3 R 100 3 0

0013 = 4 × 10−8 3 25 × 105

The mass mode is 100m. The mass mean diameter is Z 100 = 4 × 10−8 4 = 80m 0

The mass variance is

7

2 =

Z

100

4 × 10−8 5 − 2 = 66667 − 6400 = 2667m2

0

The mass standard deviation is = 163m. 4. The mass cumulative distribution is () =

Z

0

The mass median diameter is

4 × 10−8 3 = 10−8 4

4 05 = 10−8 = 841m

Particle diameter, microns

Problem 3.4 1. Plot on log-probability coordiantes

40 30 20

10

10

30

50

70

90

% less than

From the plot, = 22 m and 84 = 37 m. The value for is = ln

84 = 052 50

2. The number median diameter: 2

= −3 = 98 m The Sauter mean diameter:

8

2

= 5 2 = 192 m The variance: 2

2 2 2

2

2 = (2 − ) = 196 m2 = 14 m = 39 m2 = 624 m

The ratio of the standard deviation to the median diameter is 0.636 so the distribution is not monodisperse.

Problem 3.5 The mass median diameter is 20 m and = 2. 1. From Equation 3.30 =

20 = = 24 m 1 0693 069312

1. The gamma function is defined by Z ∞ − −1 Γ() = 0

For the integral Z

∞ 0

∙ µ ¶ ¸ µ ¶−1 µ ¶ exp −

−1

let = () . Then = () Z

0

∞

() and the integral becomes

∙ µ ¶ ¸ µ ¶−1 µ ¶ Z ∞ µ ¶ exp − − = 0

But

= 1 µ ¶ =

so

and

Z

0

∞µ

¶

−

=

Z

∞

−

0

Comparing with the definition of the gamma function =−1

so = 9

+1

giving Z

∞ 0

∙ µ ¶ ¸ µ ¶−1 µ ¶ exp − = Γ( + 1)

2. The mass variance is

2

∙ ¸ 1 2 2 () − 2 = 2 Γ( + 1) − Γ2 ( + 1) 0 ¤ £ 2 2 2 2 = 24 Γ(2) − Γ (15) = 24 × (1 − 0886 )

=

Z

max

= 124 m2

3. The number distribution is obtained from 3

() = R ∞ 0

() = ()3

The Sauter mean diameter is

h ¡ ¢ i ¡ ¢−4 exp − ¡ ¢ 3 Γ 1−

R∞ 3 () Γ(1) = R0∞ 2 = Γ(1 − 1 ) () 0 Problem 3.6 The mass associated with each size is 30 60

= 06total = 04total

0430 = 0660 so 3 3 0430 30 = 0660 60

Thus 30 06 60 3 = ( ) = 12 60 04 30 The Sauter mean diameter is P 30 303 + 30 603 12 3 = = 375 m = P 2 30 302 + 30 302 12

10

Problem 3.7 Since the distribution is linear, the mode is equal to max The number distribution is represented by () =

2 2 max

1. The number mean is =

Z

max

0

The number variance is 2

=

Z

max

0

2 22 = max = 67 m 2 max 3

23 4 2 1 2 − max = = 555 m2 2 max 9 18 max

The Sauter mean diameter is R max

= R0max 0

2 24 max 2 23 max

=

4 max = 80 m 5

2. The mass frequecy distribution is 3 ()

() = R max 0

3 ()

=

3 2 2

max 5 2max

5

=5

4 5 max

The mass mean is

=

Z

max 0

55 5 = max = 833 m 5 max 6

The mass variance and standard deviation are

2

Z

∞

6 5 5 25 5 2 2 − ( max )2 = max ( − )= max = 198 m 5 6 7 36 252 0 max = 141 m =

5

3.The number median is obtained by solving 05 = yielding

Z

0

2 2 max

1 = √ max = 707 m 2 The mass median diameter is obtained by solving

11

05 = The result is

Z

5 0

4 5 = ( ) 5 max max

1 max = 87 m 215

=

Problem 3.8 Expressing as exp( ln ) and incorporating it into the exponential function gives

√ 1 2

∙ ³ ´2 ¸ 1 ln −ln exp − 2 ln −∞

R∞

√ 1 2

=

h ³ 2 ´i 2 2 1 − ln +ln −2 ln ln +ln exp − ln 2 2 −∞

R∞

= exp( ln ) exp( exp( =

z

=1

1 2

2 2 √ 2 )

Z

}| { ∙ ¸ 1 [ln − (ln + 2 )]2 exp − ln 2 2 −∞ ∞

2 2 2 )

Problem 3.9 From Equation 3.21 2 2 2

=

Z

∞

()

(1)

0

and 3 = 3 ln

(2)

The equation of the mass frequency distribution can be written as h ¢i ¡ 2 2 1 2 1 exp − 22 ln − 2 ln ln − ln − 6 ln 1 () = √ 3 exp( 92 2 ) 2 Completing the square in the argument of the exponential term gives n o £ ¡ ¢¤ 1 9 2 2 2 exp − + 3 + 3 ln + ln − ln 2 2 2 1 1 () = √ 3 exp( 92 2 ) 2 12

or ½ ¾ ¡ ¢¤2 1 1 1 £ () = √ exp − 2 ln − ln + 3 2 2 2

Thus the mass median diameter is ln = ln + 3 2 or 2 = 3

(3)

Problem 3.10 Find the diameter from the cross-sectional area, the discrete number distribution is ˜ (m) 0.2 60.3 0.3 100 0.5 120 1. The number average diameter is X ¯ = ˜ = 102 m 2. The mass distribution is obtained from

3 ˜ ˜ = P 3 ˜

The distribution is m ˜ 0.037 60.3 0.248 100 0.715 120 The mass average diameter is X ¯ = ˜ = 1128 m

3. The Sauter mean diameter is P 3˜ 32 = P = 1105 m 2 ˜

Problem 3.11 The table is reduced for the cumulative number distribution in terms of the logarithm of the diameter.

13

1.75 2.21 2.78 3.51 4.42 5.75 7.02 8.58 11.15 14.0 17.7 22.3 28.1

No. 0.0 3797 3119 2522 1878 1396 850 523 297 179 93 18 2

F () 0.0 0.258 0.471 0.643 0.771 0.866 0.924 0.960 0.980 0.992 0.998 0.999 1

ln 0.55 0.792 1.022 1.25 1.486 1.75 1.95 2.15 2.41 2.62 2.87 3.10 3.33

Particle diameter, microns

100 70 40 20 10 7 4 2 1 30

50

70

90 % less than

99

99.9

1. From the plot of cumulative distribution, = 3 m and 84 = 6 = ln(63) = 0693 2. The mass median diameter is 2

= 3 = 3 × 144 = 127 m 2. The Sauter mean diameter is 2

= 5 2 = 3 × 120 = 997 m

14

Problem 3.12 Using the data provided in the problem, the following table is generated. D 10 0.0766 0.0766 0.0012 0.0012 20 0.1916 0.2682 0.0248 0.0260 30 0.2729 0.5479 0.1224 0.1484 40 0.2375 0.7854 0.2466 0.3950 50 0.1264 0.9118 0.2564 0.6514 60 0.0690 0.9808 0.2418 0.8932 70 0.0192 1.0000 0.1068 1.0000 ¯ = P = 343m = 283m ¯ = P = 488m...... = 441m Problem 3.13 The frequency function is n o £ ¤12 = exp − 2 + ( − )2 + ( − )

Taking the derivative with respect to and setting equal to zero gives £ ¤−12 − 2 + ( − )2 ( − ) + = 0 ¤ £ 2 −12 ( − ) = + ( − )2

This equation can be squared and written as

( − )2 2 = 2 2 + ( − )2 Solving for ( − ) gives

2 − 2

so ln mod e = + √

−= p 2 − 2

The value of the frequency function at this condition is

"

µ = exp − 2 +

22 2 − 2 h ¡ ¢12 i = exp − 2 − 2

15

¶12

2

+p 2 − 2

#

Chapter 4 Problem 4.1

wf

wb

U

The thrust is given by Equation A.32 Z =− 0

0

where = ˙ + Take the gas density on the surface as the carrier phase density. Let be the mass flux velocity on the front hemisphere and on the rear hemisphere. Thus = 11 The mass flux from the droplet is ˙ = ( + 11 ) = 105 2 ³ ³ ´ ´ ˙ So = 105 ˙ and = 1048 ˙ The velocity or = 0952 with respect to the center of the droplet is 0 = + ˙ = −

= (1 − )

The thrust becomes R 0 ³ = − ´ = − 0 − 0 in direction of ´2 ¡ ³ ¢ = − (1 − ) ˙ 09522 − 10482 ´2 ³ = 0192 (1 − ) ˙

Thus

³

˙

´2 = 0192(1 − 16

)

in the direction of Problem 4.2 From the Reynolds transport theorem Z Z + = If the temperature is constant in the droplet ( = ) and the mass flux and entropy of the gases at the surface is uniform, then ˙ = ( ) − From the second law of thermodynamics ˙ ˙ > ( ) − where ˙ is the heat transfer rate to the droplet. This equation can also be written as

˙ + ˙ ( − ) >

Problem 4.3 The momentum equation for the rotating droplet is

=

Z ( + ) Z ( + + 0 ) +

This equation can rewritten as R ( ) + = ( )R R + + 0 + R

where is the droplet radius. By definition of the center of mass Z = 0

and Z

= 0

Thus the equation reduces to

17

=

˙ = ( ) −

Problem 4.4

y

x

z

1. The specific kinetic energy can be expressed as 2

2

| + | | | 2 = + + 2 2 2 The kinetic energy associated with the droplet motion is =

Z

2 = 2

Z

+

Z

1 + 2

Z

2 2 sin2

The first integral becomes 2 2 The second integral is zero

Z Z

=

2 2

= 0

since the velocity is axisymmetric about the origin (() = −( + )) The last integral is 1 2

Z

2 2

2

sin

=

= =

Z Z 2 Z 2 1 2 4 sin3 2 0 0 0 Z 5 1 2 sin3 2 2 160 0 5 4 2 2 1 1 2 2 = 2 160 3 2 10

The momentum of inertia of a sphere is 2 10 so = 12 18

2. The work term becomes ˙

= −

Z

( + )( − ) Z Z ( − ) − = − Z = − −

since the dot product is equal to zero. Problem 4.5 The continuity equation from Appendix A (Equation a.19) is Z =− Since there is no mass transfer at the surface ( = 0) the continuiy equation reduces to =0 The momentum equation for the droplet reduces to Z = + The velocity is the velocity of the fluid at any point in the droplet. The droplet moves with a velocity which also corresponds to the velocity at the center of the "Hill’s vortex". Assume the velocity of fluid inside the droplet is given by = + where is the deviation of the velocity with respect to the center of the Hill vortex. Then Z Z = ( + ) Z = ( ) + Consider the diﬀerential volume shown in the figure. The integration of the velocities in the -direction for a ring with radius and with is Z

2 =

Z

2

−2

19

Z

()2

2 0

where ()2 is radius of the droplet at -plane. From continuity Z

()2

2 = 0

0

For the velolcity in the -dierction Z Z 2 =

From continuity

Z

2

2 −2

Z

()2

0

()2

= 0

0

Thus

Z

= 0

␦vr

dV

␦vz r ␦vz

z

and ( ) = + = + since the droplet mass is constant. The velocity of the fluid at the surface will reduce the shear stress and the drag force on the droplet. This eﬀect is included in the drag coeﬃcient given in Equation 4.51. Problem 4.6 The momentum equation is

=

Z

+

Z

·

The rate of particle mass change due to flux from the particle surface is ˙ = −7 20

where is the surface area of one side. Along the surfaces parallel to the flow, the velocity with respect to an inertial reference rate is On the windward face, the velocity at the control surface with respect to the inertial reference frame is − and on the leeward face, + 2 The momentum equation becomes =

() + 4 + ( − ) + 2( + 2) =

+ ˙ + 7 + 3

=

3 − ˙ 7

Another approach using

= +

Z

·

In this case, on the sides of the particle parallel to the flow the component of in the x-direction is zero so there is no contribution from these four faces. On the other two faces Z £ ¤ · = (2)2 − 2 = 3 2 so the equation for particle motion becomes

3 − ˙ 7 Since ˙ 0 the thrust is in the −direction. =

Problem 4.7 1. The energy equation (Equation A.67) is Z + () =

This term can be written as

Z

+

+ =

Z

˙ + 2 +

Including this term in the droplet energy equation results in

= ˙ + ˙ + ( − ) −

or ( +

) = ˙ + ˙ + ( − ) 21

2. The internal circulation would lead to an additional term in the kinetic energy. However there would also have to be a work term to account for the eﬀect of the work done to overcome friction forces. Ultimately when the mechanical energy equation is subtracted from the total energy equation there will be a term that reflects a dissipation term due to friction which will be responsible for droplet heating. 3. If the particle were porous with not mass flux, the energy equation would be the same with an adjusted density and energy density. If the particle is porous and there is an internal reaction to produce a mass transfer, then the energy associated with the change of phase has to be included. This will not be controlled by conditions on the surface corresponding to evaporation or condensation. Of course, the enthalpy flux from the surface has to be included. Problem 4.8 1. The equation of motion for the evaporating droplet is 18 =− =− 2 (1 − )

where is based on the initial droplet diameter. Integrating this equation gives ln =

ln(1 − ) +

Applying the initial condition (0) = gives = (1 − ) The distance the droplet travels is obtained by integrating the equation for velocity = −

+1 (1 − ) + +

Applying the initial condition (0) = 0 gives ∙ ¸ +1 = ) 1 − (1 − + 2. For = and À 1 = and for ¿ 1, = . These are the relations one would expect. Problem 4.9 1. The equation of motion for the particle is = + The equation for the particle velocity will be of the form 22

= Substituting into the equation of motion gives ( + 1) = Solving for one has −1

− tan 1 − = = 2 12 1 + 2 (1 + 2 2 ) The equation for particle velocity is −1

=

(−tan

) 12

(1 + 2 2 )

The amplitude ratio is || 1 = 12 || (1 + 2 2 ) and the phase shift is = tan−1 The product is equivalent to the Stokes number. For small values of one expects the amplitude ratio to approach unity and the phase shift to zero. For large values of the amplitude ratio is small and the phase approaches 2 2. The velocity response time for a 10 micron particle is =

2 2500 × 10−10 = = 77 × 10−4 sec 18 18 × 18 × 10−5

The product is 0.077. The amplitude ratio is 0.997 and the phase shift is 4.40 Problem 4.10 Part 1: Assume that the humidity level is such that the vapor mass fraction of the surrounding air is small and less than the vapor mass fraction at the droplet surface. In the beginning the second term on the right hand side is small and the conduction term controls the droplet temperature. As the droplet heats up, the vapor mass fraction at the surface increases and the second term becomes most important. Finally a state is reached when the second and first term are numerically equal and lead to no further change in droplet temperature. At this point the droplet has reached the wet-bulb temperature. Part 2: The vapor pressure at 20 is 0.02339 bars so the partial pressure is 0.02339. The mass fraction of water vapor in the free stream is

23

18 × 06 = 0008711 29

∞ = 002339 × The equation for temperature becomes

293 − − 2855( − 0008711) = Taking values for and setting up a table for an iterative solution 280 0009912 000615 2031 285 001388 000862 8260 290 001919 001191 −6133 Interpolating for = 0 gives = 2879 K or 14.9 C. Part 3: The equations are written as = 0 + 0 = + 0 The equation for droplet temperature during evaporation is = +

( ∞ − )

which can be expressed by = + ( ∞ − ) where is taken as a constant and evaluated at the equilibrium condition. The mass fraction for water vapor can be expressed as =

18 29

where is the vapor pressure and is the total pressure. In the region near the equilibrium condition, the vapor pressure can be assumed to vary linearly with temperature so = 0 + ( − ) and the vapor mass fraction is = 0 +

18 ( − ) = 0 + Γ0 29

Substituting into the energy equation gives = (0 + 0 − − 0 ) + ( ∞ − 0 − Γ0 ) Cancelling out the equilibrium wet bulb condition, one has

0 = (0 − 0 ) − Γ0 24

Problem 4.11 The wetness is the ratio of the liquid to the bone-dry solid. =

The wetness varies with time as = − so = exp(−) The mass of the slurry droplet is = + = [1 + exp(−)] The equation of motion assuming the droplet diameter does not vary is = −3 assuming Stokes drag coeﬃcient. The equation of motion can be written as

−3 = [1 + exp(−)] Integrating this equation gives ∙ ¸ 3 exp(−) ln = ln + 1 + exp(−) Using the initial condition = at = 0 gives ∙ ¸ 3 (1 + ) exp(−) = 1 + exp(−) Problem 4.12 Equation of motion normal to wall. = + Solving for with (0) = 0 gives ¸ ∙ = 1 − exp(− ) Integrating again for distance with the initial condition (0) = 0 yields ¸ ∙ = − 2 1 − exp(− ) Setting = and solving for one has

25

2

µ

¶−1

=

− 1 + exp(− )

The equation of motion in the axial direction is =0 + Integrating once with (0) = one has = exp(−

)

Integrating again with (0) = 0 the result is ¸ ∙ = 1 − exp(− ) Setting = one has = 1 − exp(− ) and solving for gives = − ln(1 − ) Substituting in the equation above for one has µ ¸ ¶∙ 2 )− = − ln(1 − This is an implicit equation for as a function of is the “stopping distance” for a particle. It could never go any farther than this distance. Therefore ≤ 1 Problem 4.13 The volume and surface equivalent diameters (from example in text) for a 1-mm cubical particle are

= 124 mm = 113 mm

and the circularity is = 0887 Substituting into the equation for the drag factor gives £ ¤ = 0911 1 + 0149Re 0687 +

00145Re ¤ 0942 1 + 473 × 104 Re −116

26

£

The density ratio is much less than unity so the equation for terminal velocity is = The velocity response time for the particle is =

2 1400 × (124 × 10−3 )2 = 661 s = 18 18 × 181 × 10−5

so =

648 m/s

The Reynolds number is calculated from Re =

124 × 10−3 = 821 = 151 × 10−5

The equations have to be solved iteratively for The answer is = 448 m/s Problem 4.14 The velocity response time of the 200 micron particle is =

1500 × (2 × 10−4 )2 = 0185 s 18 × 18 × 10−5

The terminal velocity for Stokes flow is 1.82 m/s. This gives a Re value for 24.3 and a corresponding Re of 13 and terminal velocity of 0.975 m/s. The relative turbulent intensity is √ √ 10 02 02 = = = 005 × = 051 | − | | − | 0975 The critical Reynolds number corresponding to this relative turbulence intensity is (Eqn. 4.122) log10 Re = 3371 − 175 = 248 so the critical Reynolds number is 300. Thus the particle is in the subcritical regime so one would expect the drag coeﬃcient to be about the same or, perhaps, somewhat larger than the steady state value for a sphere. Problem 4.15 The equation of motion for the particle reduces to Z 3 2 √ √ = −3 − − 2 2 − 0 27

This equation can be rewritten as ³ =− 1 +

2

´ −

s

18 1 1 ³ ´ 1 + 2

Z

0

√ −

The integral can be approximated by Z h i X √ ˙ ( − )12 − ( − − ∆)12 ' − 0 Substituting the values into the equation gives

and

h i X 12 ˙ ( − ) − ( − − ∆)12 = −600 − 338 = 1 + 015Re067 = 1 + 355 067

This can be integrated numerically using a two-step Euler method. Problem 4.16 The equation of motion for a droplet is

= 3 ( − ) +

Dividing by the droplet mass and using the 2 law gives 18 ( − ) = + 2 (1 − )

For no gravitational force and constant free stream velocity of the equation reduces to

or

( − ) = (1 − ) + = (1 − ) (1 − )

Integrating this equation using the integrating factor and employing the initial condition (0) = 0 gives ∙ ¸ = 1 − (1 − ) Integrating one more time to get the distance with the initial condition (0) = 0 yields

28

=

½ +

1 +

¸¾ ∙ 1+ ) −1 (1 −

At time = the distance travelled is ( ) =

2 +

Problem 4.17 The equation of motion is

= 3 ( − ) +

which can be written as 18 ( − ) + = 2 The buoyancy has been neglected because À Using the 2 law 2 = 2 − the equation becomes for = 0 =− + (1 − )

where is the velocity response time based on the initial diameter and is the evaporation time. This is a first order ordinary diﬀerential equation which can be integrated using the integrating factor. ∙ ¸ Z Z exp( ) = exp( ) (1 − ) (1 − ) Integrating this equation using the initial condition = 0 at = 0 gives # "µ ¶ = − (1 − ) 1− 1 −

The maximum velocity occurs where =0 Taking the derivative yields

=1−

µ

¶1−

Substituting into the equation for velocity gives

29

max = (

− )

Problem 4.18 The equation of motion is written as 1 2 = − 2 2 4 Using the expression for drag coeﬃcient gives

3 2 =− − 8 For convenience let

8 3

= so the equation becomes = − 2 +

Integrating with the initial condition (0) = gives ) = exp(− 1+ Solving for =

exp(− ) = 1 − exp(− )

Integrating with respect to time and setting (0) = 0 gives ∙ ¸ 1 − = − ln 1 − ( ) exp(− )

As time approaches infinity, the distance is

= − ln(1 − )

For a 100 micron particle with a material density of 2000 kg/m3 in air at standard conditions, the velocity response time is 0.062 s. The nondimensional length is 0.444 m. With an initial velocity of 10 m/s, the constant is 4.18 m/s. Thus the distance travelled is = −0444 ln(1 − 0062 × 4180444) = 04 m After one velocity response time, the velocity will be =

0368 × 418 = 196 m/s 1 − 0368 × 0062 × 4180444

Using Stokes law the velocity would be

30

= 0368 × 10 = 368 m/s This velocity is larger because Stokes law corresponds to a smaller drag coeﬃcient. For Stokes flow the stopping distance would be = = 10 × 0062 = 062 m Problem 4.19 The equation of motion for the wet particle is 3 =− + Using the rate equation for the water mass, the above equation can be written as 3 = = The solution of this equation is =

µ

¶

where is the initial velocity and is the initial water mass in the particle. The equation for distance can be written as =

= =

or (− ) = +

µ

µ

¶

¶

This equation can be re-expressed as =− ( + ) Integrating this equation results in µ ¶ +1 =− + + +1 Evaluating the constant using ( ) = 0 one has ( " " ¶ # ¶+1 #) µ µ = + 1− 1− +1 31

The distance travelled before water removal is complete ( = 0) is µ ¶ = + +1 or ¶ µ 3 + = 3 3 + Problem 4.20 The expression for terminal velocity is =

neglecting the buoyant force. The response time for the 0.1 micron particle with a material density of 800 kg/m3 is air at standard conditions is =

2 800 × 10−14 = = 247 × 10−7 18 18 × 18 × 10−5

The terminal velocity based on Stokes drag is

= 981 × 247 × 10−7 = 242 × 10−6 m/s

The mean free path can be estimated from1

= 3 where is the molecular mean velocity given by r 8 = where is the gas constant for the specific gas. For air at standard conditions, = 463 m/s so the mean free path is = 0097×10−6 m. The Knudsen number is = 097 Substituting into the Cunningham correct factor gives =

= 1355 Thus, the terminal velocity is = 242 × 10−6 × 355 = 86 × 10−6 m/s 1 This is Maxwell’s equation which was developed in 1860. More accurate coeﬃcients are available from kinetic theory but the form of the equation is the same.

32

Problem 4.21 The equation for rotational motion can be written as 3 = − The response time is =

5 2 = = 3 3 60 60

compared to =

2 18

so the rotational response time is about 1/3 of the velocity response time. The rotating sphere will reach equilibrium faster than a translating particle. Problem 4.22 Boundary layer eﬀects play a minimum role in the drag of a prismatic particle because the separation (and form drag) is established by the flow separation at the corners. Therefore the Basset term would be unimportant for prismatic particles. Problem 4.23 For creeping flows the momentum equation at the particle surface is µ ¶ 1 = 2 2 = Using the velocity distribution for Stokes flow one has 3 = sin 2 Integrating this equation one has 3 cos 2 where is the pressure at the forward stagnation point ( = 0). The form force is = −

= −

Z

0

cos = −

= 2 = which is 1/3 of the Stokes drag.

33

Z

0

cos 2 sin

Problem 4.24 The equation for particle velocity in the y-direction is =0 + The velocity is = exp(− ) and the penetration distance = [1 − exp(− )] The maximum penetration is max = The equation of motion in the x-direction is = = [1 − exp(− )] + Using the integrating factor the equation can be written as [ exp( )] = [1 − exp(− )] exp( ) Integrating this equation yields exp( ) = [exp( ) − 1] − With the initial condition (0) = 0 the equation becomes = [1 − exp(− )] − exp(− ) Integrating with respect to time and using the initial condition (0) = 0 gives ª © = [1 + exp(− )] + 2 2 [exp(− ) − 1]

After a long time

= max Problem 4.25 Setting the lift force equal to the weight = −12 |( − ) | 3 = 1612 ( )12 |ω | 6

34

Solving for =

6 161 ( )12 |ω |−12 |( − ) |

where is given by Equation 4.150 in text. The equation will have to be solved iteratively because is not a linear function of The rotation of the fluid in the laminar sublayer is =

=

The shear stress is given by 2 8 where is the Darcy-Weisbach friction factor and is duct velocity Thus =

=

2 8

The Reynolds number of the flow in the duct is Re =

005 m × 10 m/s = = 33 × 104 15 × 10−5 m2 /s

The Dracy-Weisbach friction factor can be determined from =£

log10

For a smooth pipe, = 0 so =h

log10

³

¡

025 37

025 574 (33×104 )09

The rotational rate of the fluid is =

+

574 Re09

¢¤2

´i2 = 0023

0023 102 = 192 × 103 rad/s 8 15 × 10−5

The values for the factors in the above equation for are 6 ( )

12

6 m2 s2 = 78 × 10−5 × 2500 × 981 kg ¡ ¢ 12 = 18 × 10−5 × 12 = 465 × 10−3 =

|( − ) | = 10 × 192 × 103 = 192 × 104

35

m s2

kg m2 s12

Solving for with = 1 gives =1

¡ ¢12 = 161 × 78 × 10−5 × 465 × 10−3 × 192 × 104 192 × 103 = 255 × 10−4 m

Now = =1 Assume, to start with, that Re 40 and use equation 4.150 for = (1 − 03314 12 ) exp(− Re 10) + 03314 12 with Re

=

=

10 = 667 × 105 15 × 10−5 192 × 103 = 96 2 × 10

Taking values for , evaluating Re and then calculating using =1 and plotting =1 − yields the graph shown in the figure. The value of for which =1 − = 0 is 33 m. The corresponding relative Reynolds number is 22 which is less than 40 required for the correlation.

3x10-4

CSD0-D

2x10-4

1x10-4 5x10-5 0 -5x10-5 0

-6

2x10 3x10-6 1x10-5-6 Particle diameter, m

4x10-6-5 4x10

Problem 4.26 The equation of motion can be written as ¶ µ ¶ µ − − + =− + It is noted that as → 0 36

µ

−

¶

→

1

µ ¶ − +

We will use the method of small perturbations with this limit as the zeroth order solution. µ ¶ µ ¶ − 1 = = − + The function is expanded in powers of = + 1 + 2 2 + Substituting into the diﬀerential equation one has

1 2 + 2 + 3 + + 1 + 2 2 = − +

From the zeroth order solution = −

+

Collecting terms with first powers of + 1 = 0 Thus 1 =

1 2 2 2

The equation for becomes ¶ µ 1 1 2 − =− + + 2 2 + ( 2 ) = Solving for =+ − +

37

µ

¶2

2 + ( 3 ) 2

Chapter 5 Problem 5.1 Thew coeﬃcient of restitution for a sphere from Equation 5.67b is (2)

=

(1)

=

8 = 08 10

The magnitude of the approach velocity is p || = 102 + 142 = 172 cm/s

so

(0)

10 =− = −058 || 172

From Table 5.4 the value of the parameter −

2 2 =− = −0794 7 ( + 1) 7 × 01 × 18

(0)

Because || −0794, use second column in Table 5.4 with = 1. Thus

(0)

(0)

= + ( + 1) = 14 − 02 × 18 × 10 = 104 cm/s

Also

5 (0) ( − 1) 2 5 = 0+ × 02 × 18 × (−10 × 10−2 m/s) 2 × 10−4 m = −1800 rad/s (0)

= +

Problem 5.2 The component of velocity normal to the surface is = −162 m/s The magnitude of velocity is p |v| = 1132 + 1622 + 0482 = 203 m/s

Thus

2 = −0798 − = −0577 |v| 7 (1 + )

Therefore the particle is not sliding (it is rolling) before leaving surface. From first column in Table 5.4 the post-collisional velocities are 38

=

5 2 ( − ) = 0806 m/s 7 5

= − = −(065)(−162) = 1053 m/s 5 2 ( + ) = −0337 m/s 7 5 The post-collision rotations are =

=

= −178 rad/s

= −002 rad/s = −

= −424 rad/s

Problem 5.3 The velocities and rotations of the two spheres are v1 = (433 −25 0) v2 = (−10 0 0) ω 1 = (0010) ω 2 = (0 0 20) The vector G is G = v1 − v2 = (1433 −25 0) At collision, is equal to (1 0 0)The tangential component of at contact is G = G − (G · n)n + (ω 1 + ω 2 ) × n = (1433 −25 0) − (1433 00) + (3 × 10−4 )(0 0 30) × (1 0 0) = (0 −2491 0) For aluminum, = 033 and = 68 × 109 Pa. The value for becomes = 623 × 108 N/m32 From Figure 5.8, = 02 for a coeﬃcient of restitution of 0.8. Thus the damping coeﬃcient is √ = 14 √ 1 = 02 × 305 · 10−7 · 623 · ×108 × (25 · 10−6 ) 4 = 0195 Ns m Take = The normal force is

39

= (− 32 − G · n) = [−623 · 10 × (25 · 10−6 )32 ] − 0195 · (1433) = −751N 8

The tangential force is = − − = −0195 · −294 = 0486N However | | | |. Therefore there is no slipping. Problem 5.4 1. Conservation of linear momentum: 1 1 + 2 2 = 1 1 + 2 2 By definition of the coeﬃcient of restitution, 2 − 1 = (1 − 2 ) Substituting into the first equation gives (1 + 2 ) 1 = (1 − 2 )1 + 2 (1 + )2 Solving for 1 gives 1 = 1 +

2 (1 + )(2 − 1 ) 1 + 2

In the same manner one finds 2 = 1 +

1 (1 + )(1 − 2 ) 1 + 2

2. From Equation 5.14a in text with t = 0 one has 1 = 1 −

2 (1 + )(1 − 2 ) 1 + 2

which is the same as the above equation. The same procedure for Equation 5.14b yields the same result for 2 . 3. For 1 = 5 m/s and 2 = 10 m/s and 1 =

(10−5 )3 × 2500 = 131 × 10−12 kg 6 2 = 163 × 10−13 kg

Substituting into the above equations for velocity gives 1 = 605 m/s 2 = 155 m/s

40

Problem 5.5 The force is given by 3˙ = − 2

Z

0

with = 0 +

2 2

= −3

2 ˙ 20

and 0 Show

Evaluating the integral Z 0

3 3

3 = 3

Z

3 ¡ ¢ 2 3 0 + 2

0

Change variables with = 2 2 to have Z Z 2 3 2 = 2 ¡ 3 2 ¢3 ( 0 0 0 + ) 0 + 2

By particle fractions

+

3

2

=

(0 + ) (0 + ) 1 3 ( + + ) = (0 + )

(0 + )3 (0 + )

3

Therefore = 1, The integral becomes Z 2 22 3 ( 0 0 + )

+ = 0 so = −0

2

= 2

Z

2

"

−

+

#

(0 + )2 " #2 1 1 2 0 − = 2 2 (0 + )2 (0 + ) 0 ∙ 1 0 2 = 2 − − 2(0 + 2)2 (0 + 2) ∙ 0 1 = 22 − + 2(0 + 2)2 (0 + 2) 0

(0 + )3

1

Since 2 0 , the integral reduces to Z 2 2 22 3 ' 0 (0 + ) 0 41

1 1 + 20 0 ¸ 1 20

¸

The force is then = −

3˙ 2

Z

0

32 ˙ 3 = − 3 20

Problem 5.6 The magnitude of the approach velocity is p || = 52 + 32 + 12 = 592 m/s

The velocity in the − plane is p || = 52 + 32 = 583 m/s

The angle of the velocity vector with respect to the vertical is tan−1 (53) = 59o

The angle with respect to the normal direction to the wall is = 59 − 30 = 29o 26.3

o

3.7

o

o

29

59

30

o

o

Pre and post collisional velocities. The velocity components tangential and normal to the wall are 0(0)

= 583 sin 29o = 283 m/s

0(0)

= −583 cos 29o = −510 m/s

0(0)

and

= 1 m/s 0(0)

The value of the parameter

= −0861 ||

2 2 = = 0397 7 ( + 1) 7 × 4 × 18 42

Since -0.397-0.861, use column 1 of Table 5.4 0

0 0

5 0(0) 5 = × 283 = 202 m/s 7 7 0(0) = − = −08 × −51 = 408 m/s 5 0(0) 5 = = × 1 = 0714 m/s 7 7 =

The magnitude of the vector in the 0 − 0 plane is p |0 | = 4082 + 2022 = 455 m/s

The angle of the velocity with respect to the normal from the inclined plane is tan−1 (202408) = 263o The angle with respect to the original coordinate system is = 263 − 30 = −37o The post-collisional velocities with respect to original coordinate system are

= 455 × sin(−37o ) = −029 m/s = 455 × cos(−37o ) = 454 m/s = 0714 m/s

Problem 5.7 The only way to have a collision with no sliding with no initial particle (0) rotation is to have the tangential component of the relative velocity, G = 0Thus (0) G = G(0) − (G(0) · n)n = 0

(0) This condition can only¯happen ¯ for a "head on" collision in which G is parallel (0) (0) ¯ ¯ for n so that G = n and ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ G(0) − (G(0) · n)n = n ¯(0) ¯ − n ¯(0) ¯ = 0

The angular velocity after collision will be zero ω1 = 0

ω2 = 0

The translational velocities will be h ³ ´i 2 (0) (0) (0) v1 = v1 − (1 + ) v1 − v2 1 + 2 ´i h ³ 1 (0) (0) (0) 2 = 2 + (1 + ) v1 − v2 1 + 2

43

Problem 5.8 The mass of the particle is 3 = × 920 × 10−9 = 482 × 10−7 kg 6 6 The force in the normal direction is given by =

= 32 where √ 2 = 3(1 − 2 )

For the properties given

= 42 × 108 N/m32 so = 42 × 108 × (10−6 )32 = 042 N The spring constant for a 1 m displacement is = 12 = 42 × 105 N/m The damping factor in the normal direction (using critical damping) is p p = 2 = 2 482 × 10−7 × 42 × 105 = 090 Ns/m In the tangential direction

where

√ 2 2 12 = 2 − =

2(1 + )

Substituting in the given values gives =

30 × 109 = 1010 Pa 2(1 + 05)

The spring constant in the tangential direction is √ 8 2 12 = 42 × 105 N/m = 2 − The relative velocity between the two particles is G = v − v = 10i − 1559i − 9j = −559i − 9j 44

The direction n is −j. The value for the tangential velocity diﬀerence is G = G − (G · n)n + (ω + ω ) × n G = −559i − 9j−9(−j) + 00005(−200k) × (−j) G = −559i + 01i = −558i The force in the normal direction is F = (−042 − 090 × 9)(−j) = 852j N The force in the tangential direction is F = 42 × 105 × 10−6 i − 090 × (−558i) = 544i N Check to see if there is a slip contact. If |F | |F | then slippage occurs. With = 01 slippage does occur so F − |F | t = −01 × 852(−i) = 852i N Problem 5.9 For this problem, (0)

= −1 ||

and −

(4)

2 = −0793 7 ( + 1)

Therefore

(0)

−

2 7 ( + 1) ||

50so use column 1 of table 5.4. The ratio of the post-collisional and is (0)

(0) − 2 5 2 2 = 5 7 (0) = = (−05) = −0178 (0) 7 7 × 08 −

The angle after bouncing is = tan−1 (−0178) = −101o

45

Problem 5.10 The coordinate system for this problem is

n y 5 m/s x

o

50 10 m/s

The unit normal vector for collision is n=j and = 09. The pre-collisional translational and rotational velocities are (0)

v1

= 10j m/s

(0) v2

= 5 cos 50o j − 5 sin 50o i = 321j − 383i m/s

(0)

ω1

(0) ω2

= −10 000k rad/s = 5 000k rad/s

The pre-collisional relative velocity is (0)

(0)

G(0) = v1 − v2 = 383i+679j The relative velocity at contact is G(0)

= 383i+679j + (25 × 10−6 × −104 k) × j + (25 × 10−6 × 5 × 103 k) × j = 383i+679j+025i−0125i = 396i+679j

Use Equations 5.21a through 5.21d. The tangential component at contact is (0)

G

(0) = G(0) · n)n − (G = 396i+679j − (679)j = 396i

46

The velocities are µ ¶ 2 1 = 10j − 19 × 679j + × 396i × = 355j − 057i 7 2 = 321j − 383i + 645j + 057i =966j−326i m/s 5 1 = −10 000k − 396(j × i) × 7 × 25 × 10−6 2 = 103 × 105 k rad/s 5 1 = 5000 − 396(j × i) × 7 × 25 × 10−6 2 = 118 × 105 k rad/s

v1 v2 ω1

2

m/s

The impulse force (Equation 5.12) is = −

1 (1 + )( · (0) ) = − (19 × 679) = −645 1 + 2 2

The particle mass is = 3 = 2500 × × (50 × 10−6 )3 = 164 × 10−10 kg 6 6 Thus

= −645 × 164 × 10−10 = −106 × 10−9 N·s 2 1 ¯¯ (0) ¯¯ = − ¯G ¯ = − × 396 = −928 × 10−11 N · s 7 1 + 2 7

where the tangential impulse component is from Equation 5.20. The impulse force is J = −106 × 10−9 j − 928 × 10−11 i Problem 5.11 The Hamaker constant for copper and alumina is p p 12 = copp er alumina = 155 × 10−20 × 284 × 10−20 = 21 × 10−20 J

The force between two spheres is (Equation 5.14) is =

12 2

The force is equal to the weight of the alumina sphere (1) µ ¶ 1 2 1 13 = 1 = 12 6 1 + 2 12( + )2 where = 04 × 10−9 m and = (1 + 2 )2 47

Putting in values

205 × 104 12

3 2 4000 kg/m × 981 m/s × 13 6 µ ¶ −2 × 10 m2 3 × 1 −20 = 21 × 10 J× 1 + 3 × 10−2 m ∙ ¸2 1 1 × 12 04 × 10−9 + (10−6 + 0011 )2 m ∙ ¸2 525 × 10−22 1 = 1 + 3 × 10−2 04 × 10−9 + (10−6 + 0011 )2 m

This equation must be solved by interpolation. The result is 1 = 182 × 10−6 m = 182 m Problem 5.12 Stokes drag based on approach velocity 32 ˙ Stokes = −6˙ = − 20 Solving for 0 gives 0 1 = 4

48

Chapter 6 Problem 6.1 From Figure 6.5, the number of particles necessary for a variation of 5% or less at the 99% confidence level is 104 . Thus the dimension of the volume necessary for this condition is 13 = 10 00013 = 21 The interparticle spacing is =

µ

6

¶13

=8

so = 160 m and the dimension of the volume is 3.4 mm. Problem 6.2 a) The particle number flow rate is ˙ = ˜ so = ˙ hi since the particles are in kinetic equilibrium with the fluid. The mass source term is written as ˙ =− = − or

hi hi 3 ˙ =− ×− hi hi 2

=

3 hi 2 hi

where is the initial mass concentration, the continuous phase volume fraction is taken and the carrier phase density is assumed constant. b) The continuity equation is 3 hi ∆ ( hi) = = ∆ 2 hi

or

hi

3 ∆ hi 3 ∆ hi = hi = = hi ∆ 2 2

µ ¶32 1−

Integrating with the initial condition hi = hi at time = 0 yields hi 32 = (1 + ) − (1 − ) hi 49

When =

hi = (1 + ) hi

Problem 6.3 Solution: The mass source term can be written as = ¯ =

˙

˙

From the D2 -law, µ ¶32 2 = 1 − 6

Taking the derivative with respect to time and dividing by the mass gives ˙ 3 1 = 2 Thus, the source term becomes

µ ¶−1 1−

¯ 3 = 2

µ ¶−1 1−

Problem 6.4 Applying Equation B.20 in Appendix B for the volume average of a spatial derivative Z 1 = ( h i) − For the interior particle Z

=

Z

= 0

For the boundary particles 1

Z

= −

( ˜ )

Thus = ( h i) + ( ˜ ) = 0 From Equation 6.45 with no mass coupling and constant particle material density ( ˜ ) = − = 50

so

( h i) = 0 +

Problem 6.5 Applying Equation B.20 in Appendix B for the volume average of a spatial derivative Z 1 = ( h i) − [ + (˙ + ) ] For the interior particle Z − [ + (˙ + ) ]

= −

Z

−

˙ ˙ +

˙ ˙ = − +

For all the interior and boundary particles Z X ˙ X ˙ 1 [ + (˙ + ) ] = + ( ˜ ) − + −

so X ˙ X ˙ ( h i) + ( ˜ ) − + " # X X 1 ˙ ( h i) + ( ˜ ) − ˙ + X ˙ ( h i) − +

= 0

= 0 = 0

X ( h i) = − ˙ = mass ( ) +

Problem 6.6 The mass source term is X X X = − ˙ = − ˙ = − ˙ ˜ ( )

In terms of a continuous number distribution Z = ()() ˙

For evaporation

() ˙ = ( − ∞ ) 51

which gives a mass source term of = ( − ∞ ) For a Rosin-Rammler distribution () = () The source term becomes = ( − ∞ )

Z

()

()−−2 Γ(1 − 3)

+2 Γ(1 − 3)

Z

()−−1

Using the Equation 3.34. The source term becomes = ( − ∞ )

Γ(−1) Γ(1 − 3)

or in terms of and the mass median diameter = ( − ∞ )

Γ(−1) 06931 Γ(1 − 3)

Problem 6.7 The bulk density of the coal particles is ¯ = 001 × 1300 = 13 kg/m3 The relative Reynolds number is Re = 1 × 10−4 × 1218 × 10−5 = 67 The drag factor from Equation 4.111 is = 0 + 1 Re where 0 = (1 −

)2

"

# p µ ¶ 1 + 3 2 + 211 ln + 179 + 10 1 + 0681 − 1102 + 1543

and 1 = 04673 + 001833 and = [−10(04− ) ] Evaluating = exp(−10

039 ) = exp(−390) = 0 001 52

Evaluating 0

2

= 099

"

= 1286

# √ 1 + 3 0005 + 211 × 001 × ln 001 + 179 × 001 1 + 0681 × 001 − 110 × 0012 + 154 × 0013

The value for 1 is = 0467 × 0993 × 01 + 00183 × 0993 = 00223 so = 1286 + 00223 × 67 = 143 The velocity response time is =

2 = 004 s 18

Finally, evaluating =

¯ = 465 kg/m3 s

Problem 6.8 The gas is stagnant and there is no mass transfer so the momentum equation, Equation 6.69 reduces to X − ∆ hi + 3 = 0

since the flow is horizontal. The positive direction for is in the direction of particle motion. Summing over all the particles in the volume and dividing by ∆ gives ¯ ∆ hi = = 3 = ∆ The particle volume fraction is =

2 = 00008 2500

so ' 1. The velocity response time is 3

=

2 2500 kg/m × (2 × 10−4 m)2 = = 031 s 18 18 × 181 × 10−5 Pa · s

The Reynolds number is Re =

30 m/s × 2 × 10−4 m = = 397 151 × 10−5 m2 /s 53

The drag factor is = 1 + 015 Re0687 = 102 The pressure gradient is 2 kg/m3 × 102 Pa ∆ hi = × 30 m/s = 1970 ∆ 031 s m Note that the pressure increases in the direction of particle motion which is the force which balances the particle drag force. Problem 6.9 The force that appears in the coupling term in Equation 6.72 is the hydrodynamic forces acting on the particle. When the equation of motion is used to evaluate this force = + the already includes the buoyant force since that contributes to the hydrodynamic force on the particle (through the pressure gradient). Thus the force coupling term should be ∙ ¸ 1 X 1 X − = −

Thus the formulation as stated in the problem is incorrect. Problem 6.10 Applying Equations 6.64 and 6.66 for the carrier phase in a vertical duct with uniform and steady flow and no mass transfer gives 0 = −∆ hi − ∆ −

X

+ ∆

Take the positive -direction as positive upward. The particles are not accelerating so ∆ hi ∆

− 4 = − −

= −

Problem 6.11 The fluid dynamic force acting on each particle is

= +

Since there is no particle acceleration = − 54

Applying Equation 6.64 ∆ hi = − = ³´ ∆ hi = = ¯ ∆

Notice that the pressure gradient is positive opposing the force due to the Coulomb force on the particles. Problem 6.12 The ratio of the kinetic energy to the enthalpy is 2 −1 = 2 2

µ

¶2

=

−1 2 2

(5)

The kinetic energy can be neglected with respect to the enthalpy if the Mach number is small. Generally, the kinetic energy can be neglected for 03. Problem 6.13 Non-dimensionalizing the conduction term gives µ ¶ µ ¶ h i ¯ eﬀ eﬀ = eﬀ 2 ¯ ¯ Comparing the conduction and convection term [eﬀ ( h i )] [ h i]

∼ ∼

eﬀ 2 1 eﬀ 1 ∼ Pr Re

for ∼ 1. The product of the Reynolds number and Prandtl number is also called the Peclet number. The Prandtl number is usually the order of unity so for large Reynolds numbers the conduction term can be neglected. Problem 6.14 The thermal dissipation is

Φ=

∙µ

¶µ

+

¶¸

= 2

µ

+

¶µ

+

Non dimensionalizing the velocity with and the length with µ ¶µ ¶ 2 + Φ= 2 55

¶

The convection term can be non-dimensionalized to ®¢ ¡ ( h i h i) = h i

where is the representative temperature. Then

1 2 2 Φ ∼ = × × Re ( h i h i)

The ratio 2 is much less than unity because the thermal energies are much larger than mechanical energies. For a gas this would be proportional to Mach number squared which is generally small. Also the Reynolds number would be significantly larger than unity. Thus the thermal dissipation term is generally negligible. Problem 6.15 The energy equation is 2

2

hi hi )] + ∆[ hi ( + )] 2 2 ∙ ¸ X 2 ) − ˙ ( + ) = −∆ ( hi ˜ 2 ∆ [ ( +

+ hi −

X

−

X

˙ − ˙ ∆ − ∆ (,eﬀ )

Steady horizontal flow with no mass transfer and insulated wall reduces the equation to hi2 ∆[ hi ( + )] X2 X = −∆ ( hi ˜ ˙ − ∆ (,eﬀ ) ) − −

The particle volume fraction is the order of 10−2 so ∼ 1 and the flow work due to particle motion can be neglected. Also the flow velocity is much less than sonic so the kinetic energy can be neglected. Also the heat conduction term can be neglected since the Reynolds number is large. Thus the equation reduces to X X ∆[ hi ] = − − ˙

Also, from the example in the text, the energy associated with the work due to fluid dynamic drag is much less than the heat transfer so X ˙ ∆[ hi ] = −

56

The radiative heat transfer to the particles will be transferred to the gas. If ˙ is the rate of heat transfer to particle then the final form of the equation is X ∆[ hi ] = ˙

The gas will heat up with distance along the channel. The velocity will also increase as the gas density decreases to satisfy continuity. There will also be a pressure gradient associated with the velocity change. Problem 6.16 The energy equation is 2

2

hi hi )] + ∆[ hi ( + )] 2 2 ∙ ¸ X 2 ) − ˙ ( + ) = −∆ ( hi ˜ 2 ∆ [ ( +

+ hi −

X

−

X

˙ − ˙ ∆ − ∆ (,eﬀ )

Eliminating the terms for steady flow in horizontal channel with no heat transfer yields hi2 )] 2 ∙ ¸ X 2 ) − ˙ ( + ) = −∆ ( hi ˜ 2 X X − − ˙ − ∆ (,eﬀ ) ∆[ hi ( +

The velocities are much less than sonic so the kinetic energies can be neglected with respect to thermal energies. Also small particle volume fraction allows ∼ 1 and the flow work due to particle motion is insignificant. Also the magnitude of the Reynolds number permits neglecting the conduction term. Thus X X X ∆[ hi ] = − ˙ − − ˙

Finally the work due to particle drag is much smaller than the particle heat transfer so X X ∆[ hi ] = − ˙ − ˙

For each droplet, the heat transfer to the droplet can be related to the change in enthalpy ˙ = − ˙ ( − ) 57

so the energy equation reduces to ∆[ hi ] =

X

˙

The enthalpy of the carrier fluid will decrease with distance (will cool) because ˙ 0 Mass transfer will continue until the temperature of the carrier gas becomes equal to the droplet temperature. Problem 6.17 The entropy equation is written as ( ) = − ( ) +

µ

˙

¶

+

where is the irreversibility term and is always positive. Applying Equations B.20 and B.28 to the volume average of the left side gives ( ) = ( h i) () + ( hi) + The volume average of the irreversibility term is ¯ = hi The volume average of the entropy flux term is

µ

˙

¶

=

µ ¿ À¶ Z ˙ ˙ −

The integral over the particle surfaces will be the sum of entropy transfer from the particle surfaces and the entropy transfer through the boundary particles.

58

Chapter 7 Problem 7.1 For the assumptions listed in the problem, the turbulent kinetic energy equation reduces to µ ¶ h i 0 = − 3 − + |h i − ˆ |2 Subtracting oﬀ the single phase velocity shows µ ¶ ¡ ¢ ¡ ¢ ¡ ¢ h i − − + − − + 3 2 = |h i − ˆ | where is the total eﬀect and is the single phase eﬀect. From the above equation, the right hand side is solely responsible for the eﬀects of the augmentation or attenuation relative to the single phase eﬀects. Here we see that 2 3 ˆ | is always positive, therefore augmentation must occur for |h i − both cases. The dissipation equation is µ µ ¶ ¶ 0 h i + + = − 1 0

−2 or

2 2 2 P | − | + 3

h i 0 = −1 + 0

0

−2 + 3

µµ ¶ ¶ +

2 2 P | − |

Using the same method to get the single phase volume averaged equation and then subtract it shows 3

2 2 P | − |

¡ ¢ h i = 1 − µµ ¶ ¶ ¢ ¡ + − − £ ¤ 0 +2 −

The same conclusion is drawn for the dissipation. 59

Consider the center of the channel where the gradients are zero. This reduces the turbulent energy and dissipation equations to ¢ 3 ¡ |h i h i − 2 h i ˆ + ˆ ˆ | − =

and the dissipation equation reduces to

£ ¤ 3 2 − = 0 |h i h i + 2 − 2 h i ˆ + ˆ ˆ | 2

where the redistribution terms are neglected. Substituting shows

3 |h i h i + 2 − 2 h i ˆ + ˆ ˆ | = 3 |h i h i − 2 h i ˆ + ˆ ˆ | 0 2 If h i h i − 2 h i ˆ + ˆ ˆ 2, then the presence of particles alters the turbulence time scale by 3 2 = Re The turbulence time scale is related to the turbulence viscosity2 . Although it is diﬃcult to speculate, this simplified case shows that small particles and moderate particle Reynolds numbers may cause attenuation by increasing the turbulent viscosity, whereas large particles and moderate particle Reynolds numbers may cause augmentation by increasing the turbulent viscosity. However, at low particle Reynolds number, turbulence is not produced and the particle interactions may play a role. Problem 7.2 The dissipation equation is

0 h i + = − 1 0

−2

µ µ ¶ ¶ +

2 2 2 P | − | + 3

for a steady stream of particles, uniformly dispersed, falling into fluid with zero mean velocity, the above equation is reduced to 3

¯ 0 2 P 1 ¯¯ 2 2 − + 2 ¯ = 2

decomposing the relative velocities and assuming that h i shows ¯ 0 2 P 1 ¯¯ 2 2¯ 3 ¯2 − h i + h i ¯ = 2

2 As the particle Reynolds number approachs zero, either −→ 0 or | − | −→ 0. However, if the mean velocities are equal, then there is still an eﬀect of the particles within a turbulent eddy. Thus it may be diﬃcult for the local relative fluctuations | − | to be zero. Thus it is hypothesized that on the average 2 −→ 0 before | − | −→ 0.

60

3

0 6 2 2 2 h i = 2 4

(6)

where = 6 3 . The turbulent kinetic energy equation is ( h i ) ( ) + µ ¶ = h i − +3 |h i − ˆ |2 ³ ´ \ \ +3 − −

Applying the above conditions shows 2

3 |ˆ | = or 18 h i2

1 = 2

(7)

Substituting Equation 7 into 6 shows 3 6 2 2 h i = 0 4 2

µ ¶2 µ ¶2 2 4 2 1 18 h i = 18 4 h i 2 2 2

and

h i2 1 − 006 Re142 ³√ ´ Determine the turbulence intensity if the particle Reynolds number is = 54

300, the particle volume fraction is 5×10−4 . According to Equation 4.51, the drag factor is estimated as = 1 + 015 Re0687 = 854

The turbulence intensity is then !12 Ã √ 854 5 · 10−4 ≈ 006 = 54 h i 1 − 5 · 10−4 006 (300)142 Problem 7.3

61

For fully developed pipe flow, the single phase momentum equation is hi h 1 i 1 − + = The continuous phase momentum is hi 1 h 1 i − + − ( − ) = Assuming the wall shear is the same for both cases, then the above equation can be rewritten as µ µ ¶ ¶ µ ¶ h 1 i h 1 i hi 1 1 hi = − − + − (−) − where signifies the single phase. The diﬀerence in pressure gradient is attributed to the diﬀerence in the Reynolds stress gradient and the drag; that is, ¸ µ ¶ ∙ µ ¶ hi 1 1 hi − + ( − ) + = If the pressure gradient is equal to the single phase pressure gradient (i.e. ghost particles), then the eﬀect of the particles on the Reynolds stress gradient must balance with the drag. h i particle ³ ´ In order to reduce the pressure gradient, 1 1 + ( − ) . Problem 7.4 Contracting Equation 7.50 by setting = shows (recall that = 3) ( h i ) ( ) + h i = −2 − (1 + 1 ) ( − ) µ ¶ h i h i − 2 +2 2 µ ¶ + −2 3 X + [2 (h i − ) ( − )] h i´ ³ 2 [(h i − ˜ ) (h i − ˜ )] − |h i − ˜ | −3

62

The above equation is now a scalar equation. Knowing that = 2, then ( h i ) ( ) + 2 h i = −2 µ ¶ +2 −2 3 X [(h i − ) ( − )] +2 2

Assuming isotropic diﬀusion, then = 23 within the diﬀusion term. Dividing out the factor of 2 and using the condition = 32 , then ( h i ) ( ) + h i = −2 µ ¶ 3 2 + 2 3 −2 3 X [(h i − ) ( − )] +2

and finally the turbulent energy equation is: ( h i ) ( ) + h i = −2 µ ¶ 2 + −2 3 X [(h i − ) ( − )] +2

Problem 7.5 a) For a steady stream of falling particles and the turbulence energy equation reduces to o ¯ ¢ n 0 0 ¯¯¡ = + ( − 1)¯ − 0 0 63

¯ ¯¡ ¢ and therefore 0 0 + ¯ ( − 1)¯ 2 to ensure that the turbulence dissipation is always positive. The dissipation equation reduces to ¯ ¯ 2 + 3 ¯ ( − 1)¯ µ ¶ = 0 0 2 − 1

and therefore if 0

0 0

=

0 0 ,

then the turbulence energy goes to infinity,

0

thus 1 so that the turbulent energy remains positive. b) For the case where = 0 and 0 = 0, the turbulence energy equation is =−

2

and the dissipation equation reduces to 2 = −

2

both of which are unrealizable. The equations are suspect from the beginning because the change in turbulent energy (and ¡ ¢ dissipation) should not depend on the sign of the relative velocity − . Problem 7.6 For fully developed turbulent wall flows, Equation 7.25 reduces to 0 = −

3 hi 2 |h i − ˆ | −+

with the redistribution terms neglected. For wall flows, = − hi . Assuming that the diﬀusion of kinetic energy and dissipation near the wall in a fully developed channel flow with particles is negligible, the volume averaged turbulent kinetic energy equation is reduced to µ ¶2 hi 3 0 = −+ |h i − ˆ |2 Assuming that the law of the wall applies, the dissipation near the wall is found to be 3 2 = + |h i − ˆ | where is described in Equation 6.48. The kinetic energy is also found by assuming a constant shear layer near the wall. The shear at the wall is related to the friction velocity by 2 hi 2 = 64

Again, assuming the law of the wall is valid and substituting in the dissipation near the wall shows a relation for the particle laden turbulent kinetic energy near the wall of the form µ ¶12 2 2 =p |h i − ˆ | 1+ 3

with the redistribution terms neglected. The above equations for and assure that production of turbulent kinetic energy due to particles and mean velocity gradients is balanced by dissipation near the wall and the single phase boundary conditions are obtained in the limit of vanishing particles. Problem 7.7 For steady, homogeneous flow Equation 7.9 reduces to: ´ ³ 0 0 = − 0 0 ´ ¡ ¢ 1 ³ 0 0 0 + − 0 0 0 + − 0 0

Since particles are stationary, the fluctuating particle velocity is zero. Also, since the particles are fixed in space, 0 = 0. This special case reduces to 0 0 = −2

= −

which is not possible. Problem 7.8 For single phase, constant density, fully developed channel flow near the center of the channel, the equation for 12 is h1 i − 1) 2 − (1 ) (12 ) 0 = (2 µ ¶ 12 + If diﬀusion is negligible, then 12 =

(2 − 1) h1 i 22 1 2

The equation for 22 is 0 = − 22

=

9 22 8 + 5 15

8 27 65

Substituting 22 into the equation for 12 shows 12 = −

16 2 h1 i 243 2

This is the Boussinesq approximation for turbulence shear near the wall. Problem 7.9 Equation 7.50 reduces to ¶ 1 2 − − 3 3 µ ¶ i h 1 \ \ +2 − ˜ − ˜ ˜ ˜ ˜ ˜ + − 3 3

0 = − (1 + 1 )

µ

a) In the direction of falling particles ==3 33

=

+ 1 − 1) 2 (1 (1 + 1 ) 3 i h 2 \ \ − ˜3 ˜3 + + 3 3 3 3 (1 + 1 ) µ ¶ 1 3 ˜3 ˜3 − ˜3 ˜3 − (1 + 1 ) 3

b) In the direction transverse to the falling particles ==2 22

=

11

=

+ 1 − 1) 2 (1 (1 + 1 ) 3 i h \ 2 \ 2 2 − + 2 2 (1 + 1 ) 3 + ˜3 ˜3 3 (1 + 1 )

(1 + 1 − 1) 2 (1 + 1 ) 3 i h \ 2 \ 1 1 − + 1 1 (1 + 1 ) 3 + ˜3 ˜3 3 (1 + 1 )

22 and 11 diﬀer only by the redistribution terms. c) For =1, =2, + 1 ) (1

i h \ \ ˜1 ˜2 + (12 ) = 2 1 2 − 2 1 66

with ˜1 = ˜2 = 0, then 12 =

(1

i 2 h \ \ 1 2 − 2 1 + 1 )

For =2, =1 21 =

i 2 h \ \ − 2 1 1 2 (1 + 1 )

For =2, =3 23 =

(1

i 2 h \ \ 2 3 − 3 2 + 1 )

For =3, =2 32 =

(1

i h \ 2 \ 3 2 − 2 3 + 1 )

For =1, =3 13 =

i h \ 2 \ − 1 3 3 1 (1 + 1 )

For =3, =1 31 =

(1

i h \ 2 \ 3 1 − 1 3 + 1 )

The above oﬀ-diagonal components show symmetry and are dependent on the redistribution terms. The turbulent energy is defined as =

1 (11 + 22 + 33 ) 2

Substituting in the Reynolds stress terms and solving for the dissipation shows: ⎛

⎞ +1 −1) (1 2 ⎜ " ( 1 +1 ) # ⎟ ⎟ 1⎜ \ \ \ ⎜ ⎟ ˜3 ˜3 + 1 1 − 1 1 + 3 3 = ⎜ + 2 ⎟ ⎟ 2 ⎜ (1 +1 ) \ \ \ − 3 3 + 2 2 − 2 2 ⎝ ⎠ 3 3 (˜ ˜ ) + (˜ ˜ ) − + 3 3 3 3 ( 1 1 ) (1 +1 ) ⎡ ⎤ \ \ ˜3 ˜3 + 1 1 − 1 1 1 ⎢ ⎥ \ \ ¶ µ = ⎣ ⎦ + 3 3 − 3 3 +1 −1) (1 \ \ (1 + 1 ) 1 − + +2 2 − 2 2 ( 1 ) 1

67

⎡

⎤ \ \ ˜3 ˜3 + 1 1 − 1 1 1 ⎢ ⎥ \ \ = ⎣ ⎦ + 3 3 − 3 3 ((1 + 1 ) − (1 + 1 − 1)) \ \ + − 2 2 2 2 " # \ \ \ ˜3 ˜3 + 1 1 − 1 1 + 2 2 = \ \ \ − 2 2 + 3 3 − 3 3 From Equation 7.25, the turbulent kinetic energy equation is reduced to ! Ã \ \ \ 2 6 ˆ3 ˆ3 + 1 1 − 1 1 + 2 2 = 3 2 \ \ \ 6 − 2 2 + 3 3 − 3 3 and

=

=

Ã ! \ \ \ 18 ˆ3 ˆ3 + 1 1 − 1 1 + 2 2 − 3 2 \ \ \ 6 2 2 + 3 3 − 3 3 Ã ! \ \ \ ˆ3 ˆ3 + − + 1 1 1 1 2 2 \ \ \ − 2 2 + 3 3 − 3 3

The two equations are the same.

68

Chapter 10 Problem 10.1 The bulk density of the powder in the sampling probe is ∆ 60 gm = = ∆ 60 s = × 20 m/s × × (001 m)2 4 kg/s = × 157 × 10−3 m3 s ˙

10−3

=

= 064 kg/m3

The relationship between the mean velocity and the centerline velocity is ¯ 22 2 × 72 = = = 0816 ( + 1) (2 + 1) 8 × 15 Thus =

20 = 245 m/s 0816

Velocity at 50 mm is (50 mm) =

µ

50 100

¶12

× 245 = 222 m/s

Evaluation of parameter at 50 mm and centerline 2 18Prob e 2500 × (2 × 10−4 )2 × 222 (50 mm) = = 790 18 × (184 × 10−5 118) × 001 (center) = 872

=

From Fig. 10.22, the value for is infinity. For 50 mm, = 2022 = 09 0 =1.05 so 064 3 0 = = 061 kg/m 105 For the centerline, = 20245 = 0816 so 0 = 1.1. The bulk density in the free stream is 064 0 = = 058 kg/m3 11 Problem 10.2 Determine required dimensions for probe volume for coincidence error of 5%. The number density is = 69

The particle mass is 3 = × 2500 × 3 6 6 (15 m) = 44 × 10−12 kg (60 m) = 28 × 10−10 kg

=

The number density is (15 m) = (60 m) =

2 × 12 3 = 545 × 109 (1/m ) 44 × 10−12 2 × 12 = 857 × 1011 (1/m3 ) 28 × 10−10

Probe volume 01 max (15 m) = 183 × 10−13 m3 (605 m) = 112 × 10−11 m3

=

= 0057 mm = 0226 mm

Problem 10.3 The beam angle is given by = sin−1

µ

2

¶

For a frequency of 50 MHz, velocity of 500 m/s and a wave length of 514.5 nm, µ ¶ 50 × 106 × 5145 × 10−9 = sin−1 = 15o 2 × 500 The number of fringes is =

4 ∆

so spacing is

∆ = = 10 × × 2 mm = 157 cm 4 4 Chose a spacing of 20 mm to give 12 fringes. Also

so

∆ = 2b tan

∆ 20 mm = 318 mm = 2 tan 2 × tan 15o Use 500 mm focal length lens This will give an angle of µ ¶ µ ¶ ∆ 20 mm = tan−1 = tan−1 = 114o 2 × 500 mm 2b b =

70

This gives a Doppler frequency for 500 m/s particle of =

2 sin 2 × 500 m/s × sin 114o = = 387 MHz 5145 nm

The diameter of the probe volume is =

4b 4 × 500 mm × 5145 nm = = 163 m × 2 mm

The final configuration of the LDA system is

beam spacing focal length beam half angle probe volume diameter Number of fringes

20 mm 500 1.14o 163 m 12

Problem 10.4 The probe volume waist diameter for the 1000 mm focal length is =

4b 4 × 1000 mm × 05145 m = = 484 m 0 × 135 mm

and for the 500 mm focal length is 282 m. To reduce Gaussian beam eﬀects, the 1000 mm would be better but one would have to consider the number flow rated to insure that coincidence error is not a problem. Taking an initial beam spacing of 40 mm and the 1000 mm focal length lens would give a beam angle of tan() = 201000 = 002 Assume that refraction is responsible for the scattering, then equation 10.77 can be used for ΦThe value for Φ can be obtained from Equation 10.79 with ∆ = 2 05145 m Φ= = = 193 × 10−3 133 × 200 m 0.006

0.005

⌽

0.004

tan( tan( tan( tan()=0.04

0.003

0.002

0.001

0.000

0.00

0.01

0.02

tan()

71

0.03

0.04

A plot of Φ versus the tangent of the half angle between the transmitting beams is shown in the figure. The limiting line of Φ = 193 × 10−3 is shown on the figure. A slit spacing of 10 mm and a focal length of the receiving lens of 200 mm would give a tan of 0.025 and a Φ of 0.0015. The equation relating particle size and phase shift would be = 41∆ (m) with a phase shift of 2 corresponding to 256 m. This would give an adequate range with acceptable accuracy. The final configuration would be Beam spacing Transmitting lens focal length Receiving lens focal length slit spacing

40 mm 1000 mm 200 10 mm

If coincidence error is an issue, the smaller focal length transmitting lens could be used with other options for the slit-receiving lens combination. Problem 10.5 a) The diﬀraction limited spot diameter is = 244( + 1) # = 244 × 2 × 8 × 05106 = 199 m

Image diameter

¡ 2 2 ¢12 + 2 + 2 ¡ ¢12 = 802 + 1992 + 202 = 848 m =

b) For image separation of 5 ∆ = 5 = 424 m For a velocity of 10 m/s ∆ =

∆ 424 m = = 42 × 10−5 s 10 × 106 m/s

The frequency would be 1 = 238 kHz ∆ c) The blur is the pulse time times the velocity =

∆blur = 30 × 10−9 × 10 = 03 m 72

The relation to image diameter 03 ∆blur = = 00035 848 d) Uncertainty in velocity measurement max

∆ ∆ + max ∆ ∆ 02 × 848 2 × 10−9 = + 6 −5 10 × 10 × 42 × 10 42 × 10−5 −4 = 0040 + 0476 × 10 = 004 =

73

K13392

an informa business www. taylorandfrancisgroup.co m

6000 Broken Sound Parkwa y, NW Suite 300, Boca Raton, FL 33487 270 Madison A venue New Y ork, NY 10016 2 Park Square, Milton Park Abingdon, Oxon OX14 4RN, UK

9 781439 873205

by

Clayton Crowe

SOLUTIONS MANUAL FOR Multiphase Flows with Droplets and Particles Second Edition

by Clayton Crowe

Boca Raton London New

York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2011 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20150127 International Standard Book Number-13: 978-1-4398-7320-5 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Solution Manual Table of contents

There are no exercises for Chapters 1, 8 and 9.

Chapter 2

1

Chapter 3

6

Chapter 4

16

Chapter 5

38

Chapter 6

49

Chapter 7

59

Chapter 10

69

Chapter 2 Problem 2.1 The equation for droplet spacing in a lattice configuration is =

µ

6

¶13

=

µ

6 × 04

¶13

= 109

Problem 2.2 The data given are ¯ = 05 kg/m3 , = 100, ¯ = 12 kg/m3 and = 3 1000 kg/m Number density: =

¯ 6¯ = = 95 × 108 /m3 3

Disperse phase volume fraction: = ¯ = 05103 = 05 × 10−3 Continuous phase bulk density: 3

¯ = (1 − ) = 12 kg/m Concentration: = ¯ ¯ = 0416 Problem 2.3

Consider a cube in which particles each corner and they are all touching a particle at the cube center. The particle locations are illustrated in the diagram. The total distance across the diagonal is 2 The length of a side, would be related to the distance across the diagonal by 32 = (2)2

1

4 = ( )12 3 There are two whole particles within the volume so the solids volume fraction is 33 3 = ( )32 = 0680 3 3 4 The maximum solids volume fraction for the lattice configuration is = and gives =

=

= 0524 6

Problem 2.4 Droplets released in hot stream will cool the carrier gas, increase the density and reduce the velocity to satisfy the continuity equation. The reduced gas velocity will lead to less acceleration of the droplets as they proceed toward velocity equilibrium with the gas.

Tc

one-way coupling two-way coupling

c u v x

Problem 2.5

p1

p2

Control volume

2

Consider the duct shown in the figure. The control volume is designated by the dashed lines. The momentum equation states that the rate of change of momentum in the control volume plus the net eﬄux of momentum through the control surface is equal to the forces acting on the control surface. When the particle bounce from the wall they enter the control volume and when they impinge on the wall the exit the control volume. The pressure force is given by (1 − 2 ) where is the cross-sectional area. The force due to skin friction is − ∆ 12 2 The net eﬄux of momentum due to the particles is ˙ ( − 0 ) where 0 is the rebound velocity and ˙ is the particle flow rate per unit area into the control volume through the control surface adjacent to the wall. The flow is steady and there is no momentum change of the continuous phase. Thus the momentum equation is 1 (1 − 2 ) − 2 = ˙ ∆( − 0 ) 2 The value for 0 is 05 so the momentum equation can be written as ˙ ∆ 2 ∆ = 2 + 05 ∆ ∆ The Darcy-Weisbach friction factor is defined as 1 ∆ = 2 ∆ 2 so the equation for pressure drop can be expressed as ˙ ∆ 1 2 2 = 2 + 05 2 ∆ But 10% of the particle mass flow impinges on the wall per diameter length of duct so ∆ ˙ ∆ = 01˙ so the equation for the friction factor is = 4 + 01

˙

For ∼ 1 one has = 4 + 01

3

Problem 2.6

p single phase low coupling high coupling x

With small momentum coupling, the pressure variation varies little from the single phase. However, with a large momentum coupling there is a greater pressure drop since the gas has to accelerate the particles. Also the minimum pressure drop occurs after the throat because the particles are still being accelerated beyond the throat. Problem 2.7 The data provided are ˙ = 01 kg/s ˙ = 001 kg/s, = 30 m/s, = 100m, = 5 cm and = 002 cm2 s. For standard conditions, = 12 kg/m3 Evaporation time is 2 = 0005 s. Carrier phase velocity is = ˙ = 424 m/s. The mass transfer Stokes number is =

= 424

The concentration is =

= 0141

The mass coupling parameter is Π = = 0141424 = 003

4

The latent heat coupling parameter is Π = Π

= 0278

In this case the mass coupling is negligible but the heat transfer due to eﬀect evaporation should be considered although the eﬀect may not be significant. Problem 2.8

one-way coupling two-way coupling

Tc u p

Td x

The temperature of the ice particle does not change because of the change in phase. With two-way coupling the gas phase temperature drops due to heat transfer to the ice particles. The density increases due to lower temperature so the velocity increases to satisfy continuity. The pressure decreases more due to the acceleration of the gas.

5

Chapter 3 Problem 3.1 The given data are = 60m and = 03 1. The number median diameter is 2

= −3 = 60131 = 458m 2. The Sauter mean diameter is 5

2

32 = 2 = 458 × 125 = 574m 3. The mass mean diameter is 2

= 2 = 60 × 1046 = 628m 4. The mass mode is the value for which () is a maximum.

() = =

∙ ¸ 1 1 2 1 √ exp − 2 (ln − ln ) 2 2 ∙ ¸ 1 1 2 √ exp − 2 (ln − ln ) − ln 2 2

Taking the derivative with respect to ln and setting to zero gives −

1 (ln − ln ) − 1 = 0 2 2

2

Therefore ln mmod e = − = 60−03 = 548m

Problem 3.2 Rosin-Rammler distribution with = 18 and = 120 m. 1. The parameter is = 0693118 = 1471 m 2. The mass median diameter is R = () = Γ( 1 + 1) = 1471 × Γ(1555) = 1307 m The variance is R 2 = 2 () − 2 = 2 Γ( 2 + 1) − 2 = 14712 Γ(2111) − 13072 = 5607 m2 The corresponding standard deviation is 74.9 m. 3. The Sauter mean diameter is 6

3 () = 2 () Γ(1) 1471 Γ(444) = 738

32 =

Γ(1) Γ(1− 1 )

=

m

Problem 3.3 The number frequency distribution is = 1100 = 001 m−1 for 0 ≤ ≤ 100m and = 0 everywhere else. 1. The number mean is Z 100 = 001 = 50m 0

The variance is

2 =

Z

100

0012 − 2 = 833m2

0

and the standard deviation is = 289m. The Sauter mean diameter is R 100 0013 3 32 = R0100 = × 100 = 75 m 2 4 001 0 2. The number cumulative distribution is () = The number median diameter is

Z

001 = 001 0

05 = 001 = 50m 3. The mass frequency distribution is

() = =

3 R 100 3 0

0013 = 4 × 10−8 3 25 × 105

The mass mode is 100m. The mass mean diameter is Z 100 = 4 × 10−8 4 = 80m 0

The mass variance is

7

2 =

Z

100

4 × 10−8 5 − 2 = 66667 − 6400 = 2667m2

0

The mass standard deviation is = 163m. 4. The mass cumulative distribution is () =

Z

0

The mass median diameter is

4 × 10−8 3 = 10−8 4

4 05 = 10−8 = 841m

Particle diameter, microns

Problem 3.4 1. Plot on log-probability coordiantes

40 30 20

10

10

30

50

70

90

% less than

From the plot, = 22 m and 84 = 37 m. The value for is = ln

84 = 052 50

2. The number median diameter: 2

= −3 = 98 m The Sauter mean diameter:

8

2

= 5 2 = 192 m The variance: 2

2 2 2

2

2 = (2 − ) = 196 m2 = 14 m = 39 m2 = 624 m

The ratio of the standard deviation to the median diameter is 0.636 so the distribution is not monodisperse.

Problem 3.5 The mass median diameter is 20 m and = 2. 1. From Equation 3.30 =

20 = = 24 m 1 0693 069312

1. The gamma function is defined by Z ∞ − −1 Γ() = 0

For the integral Z

∞ 0

∙ µ ¶ ¸ µ ¶−1 µ ¶ exp −

−1

let = () . Then = () Z

0

∞

() and the integral becomes

∙ µ ¶ ¸ µ ¶−1 µ ¶ Z ∞ µ ¶ exp − − = 0

But

= 1 µ ¶ =

so

and

Z

0

∞µ

¶

−

=

Z

∞

−

0

Comparing with the definition of the gamma function =−1

so = 9

+1

giving Z

∞ 0

∙ µ ¶ ¸ µ ¶−1 µ ¶ exp − = Γ( + 1)

2. The mass variance is

2

∙ ¸ 1 2 2 () − 2 = 2 Γ( + 1) − Γ2 ( + 1) 0 ¤ £ 2 2 2 2 = 24 Γ(2) − Γ (15) = 24 × (1 − 0886 )

=

Z

max

= 124 m2

3. The number distribution is obtained from 3

() = R ∞ 0

() = ()3

The Sauter mean diameter is

h ¡ ¢ i ¡ ¢−4 exp − ¡ ¢ 3 Γ 1−

R∞ 3 () Γ(1) = R0∞ 2 = Γ(1 − 1 ) () 0 Problem 3.6 The mass associated with each size is 30 60

= 06total = 04total

0430 = 0660 so 3 3 0430 30 = 0660 60

Thus 30 06 60 3 = ( ) = 12 60 04 30 The Sauter mean diameter is P 30 303 + 30 603 12 3 = = 375 m = P 2 30 302 + 30 302 12

10

Problem 3.7 Since the distribution is linear, the mode is equal to max The number distribution is represented by () =

2 2 max

1. The number mean is =

Z

max

0

The number variance is 2

=

Z

max

0

2 22 = max = 67 m 2 max 3

23 4 2 1 2 − max = = 555 m2 2 max 9 18 max

The Sauter mean diameter is R max

= R0max 0

2 24 max 2 23 max

=

4 max = 80 m 5

2. The mass frequecy distribution is 3 ()

() = R max 0

3 ()

=

3 2 2

max 5 2max

5

=5

4 5 max

The mass mean is

=

Z

max 0

55 5 = max = 833 m 5 max 6

The mass variance and standard deviation are

2

Z

∞

6 5 5 25 5 2 2 − ( max )2 = max ( − )= max = 198 m 5 6 7 36 252 0 max = 141 m =

5

3.The number median is obtained by solving 05 = yielding

Z

0

2 2 max

1 = √ max = 707 m 2 The mass median diameter is obtained by solving

11

05 = The result is

Z

5 0

4 5 = ( ) 5 max max

1 max = 87 m 215

=

Problem 3.8 Expressing as exp( ln ) and incorporating it into the exponential function gives

√ 1 2

∙ ³ ´2 ¸ 1 ln −ln exp − 2 ln −∞

R∞

√ 1 2

=

h ³ 2 ´i 2 2 1 − ln +ln −2 ln ln +ln exp − ln 2 2 −∞

R∞

= exp( ln ) exp( exp( =

z

=1

1 2

2 2 √ 2 )

Z

}| { ∙ ¸ 1 [ln − (ln + 2 )]2 exp − ln 2 2 −∞ ∞

2 2 2 )

Problem 3.9 From Equation 3.21 2 2 2

=

Z

∞

()

(1)

0

and 3 = 3 ln

(2)

The equation of the mass frequency distribution can be written as h ¢i ¡ 2 2 1 2 1 exp − 22 ln − 2 ln ln − ln − 6 ln 1 () = √ 3 exp( 92 2 ) 2 Completing the square in the argument of the exponential term gives n o £ ¡ ¢¤ 1 9 2 2 2 exp − + 3 + 3 ln + ln − ln 2 2 2 1 1 () = √ 3 exp( 92 2 ) 2 12

or ½ ¾ ¡ ¢¤2 1 1 1 £ () = √ exp − 2 ln − ln + 3 2 2 2

Thus the mass median diameter is ln = ln + 3 2 or 2 = 3

(3)

Problem 3.10 Find the diameter from the cross-sectional area, the discrete number distribution is ˜ (m) 0.2 60.3 0.3 100 0.5 120 1. The number average diameter is X ¯ = ˜ = 102 m 2. The mass distribution is obtained from

3 ˜ ˜ = P 3 ˜

The distribution is m ˜ 0.037 60.3 0.248 100 0.715 120 The mass average diameter is X ¯ = ˜ = 1128 m

3. The Sauter mean diameter is P 3˜ 32 = P = 1105 m 2 ˜

Problem 3.11 The table is reduced for the cumulative number distribution in terms of the logarithm of the diameter.

13

1.75 2.21 2.78 3.51 4.42 5.75 7.02 8.58 11.15 14.0 17.7 22.3 28.1

No. 0.0 3797 3119 2522 1878 1396 850 523 297 179 93 18 2

F () 0.0 0.258 0.471 0.643 0.771 0.866 0.924 0.960 0.980 0.992 0.998 0.999 1

ln 0.55 0.792 1.022 1.25 1.486 1.75 1.95 2.15 2.41 2.62 2.87 3.10 3.33

Particle diameter, microns

100 70 40 20 10 7 4 2 1 30

50

70

90 % less than

99

99.9

1. From the plot of cumulative distribution, = 3 m and 84 = 6 = ln(63) = 0693 2. The mass median diameter is 2

= 3 = 3 × 144 = 127 m 2. The Sauter mean diameter is 2

= 5 2 = 3 × 120 = 997 m

14

Problem 3.12 Using the data provided in the problem, the following table is generated. D 10 0.0766 0.0766 0.0012 0.0012 20 0.1916 0.2682 0.0248 0.0260 30 0.2729 0.5479 0.1224 0.1484 40 0.2375 0.7854 0.2466 0.3950 50 0.1264 0.9118 0.2564 0.6514 60 0.0690 0.9808 0.2418 0.8932 70 0.0192 1.0000 0.1068 1.0000 ¯ = P = 343m = 283m ¯ = P = 488m...... = 441m Problem 3.13 The frequency function is n o £ ¤12 = exp − 2 + ( − )2 + ( − )

Taking the derivative with respect to and setting equal to zero gives £ ¤−12 − 2 + ( − )2 ( − ) + = 0 ¤ £ 2 −12 ( − ) = + ( − )2

This equation can be squared and written as

( − )2 2 = 2 2 + ( − )2 Solving for ( − ) gives

2 − 2

so ln mod e = + √

−= p 2 − 2

The value of the frequency function at this condition is

"

µ = exp − 2 +

22 2 − 2 h ¡ ¢12 i = exp − 2 − 2

15

¶12

2

+p 2 − 2

#

Chapter 4 Problem 4.1

wf

wb

U

The thrust is given by Equation A.32 Z =− 0

0

where = ˙ + Take the gas density on the surface as the carrier phase density. Let be the mass flux velocity on the front hemisphere and on the rear hemisphere. Thus = 11 The mass flux from the droplet is ˙ = ( + 11 ) = 105 2 ³ ³ ´ ´ ˙ So = 105 ˙ and = 1048 ˙ The velocity or = 0952 with respect to the center of the droplet is 0 = + ˙ = −

= (1 − )

The thrust becomes R 0 ³ = − ´ = − 0 − 0 in direction of ´2 ¡ ³ ¢ = − (1 − ) ˙ 09522 − 10482 ´2 ³ = 0192 (1 − ) ˙

Thus

³

˙

´2 = 0192(1 − 16

)

in the direction of Problem 4.2 From the Reynolds transport theorem Z Z + = If the temperature is constant in the droplet ( = ) and the mass flux and entropy of the gases at the surface is uniform, then ˙ = ( ) − From the second law of thermodynamics ˙ ˙ > ( ) − where ˙ is the heat transfer rate to the droplet. This equation can also be written as

˙ + ˙ ( − ) >

Problem 4.3 The momentum equation for the rotating droplet is

=

Z ( + ) Z ( + + 0 ) +

This equation can rewritten as R ( ) + = ( )R R + + 0 + R

where is the droplet radius. By definition of the center of mass Z = 0

and Z

= 0

Thus the equation reduces to

17

=

˙ = ( ) −

Problem 4.4

y

x

z

1. The specific kinetic energy can be expressed as 2

2

| + | | | 2 = + + 2 2 2 The kinetic energy associated with the droplet motion is =

Z

2 = 2

Z

+

Z

1 + 2

Z

2 2 sin2

The first integral becomes 2 2 The second integral is zero

Z Z

=

2 2

= 0

since the velocity is axisymmetric about the origin (() = −( + )) The last integral is 1 2

Z

2 2

2

sin

=

= =

Z Z 2 Z 2 1 2 4 sin3 2 0 0 0 Z 5 1 2 sin3 2 2 160 0 5 4 2 2 1 1 2 2 = 2 160 3 2 10

The momentum of inertia of a sphere is 2 10 so = 12 18

2. The work term becomes ˙

= −

Z

( + )( − ) Z Z ( − ) − = − Z = − −

since the dot product is equal to zero. Problem 4.5 The continuity equation from Appendix A (Equation a.19) is Z =− Since there is no mass transfer at the surface ( = 0) the continuiy equation reduces to =0 The momentum equation for the droplet reduces to Z = + The velocity is the velocity of the fluid at any point in the droplet. The droplet moves with a velocity which also corresponds to the velocity at the center of the "Hill’s vortex". Assume the velocity of fluid inside the droplet is given by = + where is the deviation of the velocity with respect to the center of the Hill vortex. Then Z Z = ( + ) Z = ( ) + Consider the diﬀerential volume shown in the figure. The integration of the velocities in the -direction for a ring with radius and with is Z

2 =

Z

2

−2

19

Z

()2

2 0

where ()2 is radius of the droplet at -plane. From continuity Z

()2

2 = 0

0

For the velolcity in the -dierction Z Z 2 =

From continuity

Z

2

2 −2

Z

()2

0

()2

= 0

0

Thus

Z

= 0

␦vr

dV

␦vz r ␦vz

z

and ( ) = + = + since the droplet mass is constant. The velocity of the fluid at the surface will reduce the shear stress and the drag force on the droplet. This eﬀect is included in the drag coeﬃcient given in Equation 4.51. Problem 4.6 The momentum equation is

=

Z

+

Z

·

The rate of particle mass change due to flux from the particle surface is ˙ = −7 20

where is the surface area of one side. Along the surfaces parallel to the flow, the velocity with respect to an inertial reference rate is On the windward face, the velocity at the control surface with respect to the inertial reference frame is − and on the leeward face, + 2 The momentum equation becomes =

() + 4 + ( − ) + 2( + 2) =

+ ˙ + 7 + 3

=

3 − ˙ 7

Another approach using

= +

Z

·

In this case, on the sides of the particle parallel to the flow the component of in the x-direction is zero so there is no contribution from these four faces. On the other two faces Z £ ¤ · = (2)2 − 2 = 3 2 so the equation for particle motion becomes

3 − ˙ 7 Since ˙ 0 the thrust is in the −direction. =

Problem 4.7 1. The energy equation (Equation A.67) is Z + () =

This term can be written as

Z

+

+ =

Z

˙ + 2 +

Including this term in the droplet energy equation results in

= ˙ + ˙ + ( − ) −

or ( +

) = ˙ + ˙ + ( − ) 21

2. The internal circulation would lead to an additional term in the kinetic energy. However there would also have to be a work term to account for the eﬀect of the work done to overcome friction forces. Ultimately when the mechanical energy equation is subtracted from the total energy equation there will be a term that reflects a dissipation term due to friction which will be responsible for droplet heating. 3. If the particle were porous with not mass flux, the energy equation would be the same with an adjusted density and energy density. If the particle is porous and there is an internal reaction to produce a mass transfer, then the energy associated with the change of phase has to be included. This will not be controlled by conditions on the surface corresponding to evaporation or condensation. Of course, the enthalpy flux from the surface has to be included. Problem 4.8 1. The equation of motion for the evaporating droplet is 18 =− =− 2 (1 − )

where is based on the initial droplet diameter. Integrating this equation gives ln =

ln(1 − ) +

Applying the initial condition (0) = gives = (1 − ) The distance the droplet travels is obtained by integrating the equation for velocity = −

+1 (1 − ) + +

Applying the initial condition (0) = 0 gives ∙ ¸ +1 = ) 1 − (1 − + 2. For = and À 1 = and for ¿ 1, = . These are the relations one would expect. Problem 4.9 1. The equation of motion for the particle is = + The equation for the particle velocity will be of the form 22

= Substituting into the equation of motion gives ( + 1) = Solving for one has −1

− tan 1 − = = 2 12 1 + 2 (1 + 2 2 ) The equation for particle velocity is −1

=

(−tan

) 12

(1 + 2 2 )

The amplitude ratio is || 1 = 12 || (1 + 2 2 ) and the phase shift is = tan−1 The product is equivalent to the Stokes number. For small values of one expects the amplitude ratio to approach unity and the phase shift to zero. For large values of the amplitude ratio is small and the phase approaches 2 2. The velocity response time for a 10 micron particle is =

2 2500 × 10−10 = = 77 × 10−4 sec 18 18 × 18 × 10−5

The product is 0.077. The amplitude ratio is 0.997 and the phase shift is 4.40 Problem 4.10 Part 1: Assume that the humidity level is such that the vapor mass fraction of the surrounding air is small and less than the vapor mass fraction at the droplet surface. In the beginning the second term on the right hand side is small and the conduction term controls the droplet temperature. As the droplet heats up, the vapor mass fraction at the surface increases and the second term becomes most important. Finally a state is reached when the second and first term are numerically equal and lead to no further change in droplet temperature. At this point the droplet has reached the wet-bulb temperature. Part 2: The vapor pressure at 20 is 0.02339 bars so the partial pressure is 0.02339. The mass fraction of water vapor in the free stream is

23

18 × 06 = 0008711 29

∞ = 002339 × The equation for temperature becomes

293 − − 2855( − 0008711) = Taking values for and setting up a table for an iterative solution 280 0009912 000615 2031 285 001388 000862 8260 290 001919 001191 −6133 Interpolating for = 0 gives = 2879 K or 14.9 C. Part 3: The equations are written as = 0 + 0 = + 0 The equation for droplet temperature during evaporation is = +

( ∞ − )

which can be expressed by = + ( ∞ − ) where is taken as a constant and evaluated at the equilibrium condition. The mass fraction for water vapor can be expressed as =

18 29

where is the vapor pressure and is the total pressure. In the region near the equilibrium condition, the vapor pressure can be assumed to vary linearly with temperature so = 0 + ( − ) and the vapor mass fraction is = 0 +

18 ( − ) = 0 + Γ0 29

Substituting into the energy equation gives = (0 + 0 − − 0 ) + ( ∞ − 0 − Γ0 ) Cancelling out the equilibrium wet bulb condition, one has

0 = (0 − 0 ) − Γ0 24

Problem 4.11 The wetness is the ratio of the liquid to the bone-dry solid. =

The wetness varies with time as = − so = exp(−) The mass of the slurry droplet is = + = [1 + exp(−)] The equation of motion assuming the droplet diameter does not vary is = −3 assuming Stokes drag coeﬃcient. The equation of motion can be written as

−3 = [1 + exp(−)] Integrating this equation gives ∙ ¸ 3 exp(−) ln = ln + 1 + exp(−) Using the initial condition = at = 0 gives ∙ ¸ 3 (1 + ) exp(−) = 1 + exp(−) Problem 4.12 Equation of motion normal to wall. = + Solving for with (0) = 0 gives ¸ ∙ = 1 − exp(− ) Integrating again for distance with the initial condition (0) = 0 yields ¸ ∙ = − 2 1 − exp(− ) Setting = and solving for one has

25

2

µ

¶−1

=

− 1 + exp(− )

The equation of motion in the axial direction is =0 + Integrating once with (0) = one has = exp(−

)

Integrating again with (0) = 0 the result is ¸ ∙ = 1 − exp(− ) Setting = one has = 1 − exp(− ) and solving for gives = − ln(1 − ) Substituting in the equation above for one has µ ¸ ¶∙ 2 )− = − ln(1 − This is an implicit equation for as a function of is the “stopping distance” for a particle. It could never go any farther than this distance. Therefore ≤ 1 Problem 4.13 The volume and surface equivalent diameters (from example in text) for a 1-mm cubical particle are

= 124 mm = 113 mm

and the circularity is = 0887 Substituting into the equation for the drag factor gives £ ¤ = 0911 1 + 0149Re 0687 +

00145Re ¤ 0942 1 + 473 × 104 Re −116

26

£

The density ratio is much less than unity so the equation for terminal velocity is = The velocity response time for the particle is =

2 1400 × (124 × 10−3 )2 = 661 s = 18 18 × 181 × 10−5

so =

648 m/s

The Reynolds number is calculated from Re =

124 × 10−3 = 821 = 151 × 10−5

The equations have to be solved iteratively for The answer is = 448 m/s Problem 4.14 The velocity response time of the 200 micron particle is =

1500 × (2 × 10−4 )2 = 0185 s 18 × 18 × 10−5

The terminal velocity for Stokes flow is 1.82 m/s. This gives a Re value for 24.3 and a corresponding Re of 13 and terminal velocity of 0.975 m/s. The relative turbulent intensity is √ √ 10 02 02 = = = 005 × = 051 | − | | − | 0975 The critical Reynolds number corresponding to this relative turbulence intensity is (Eqn. 4.122) log10 Re = 3371 − 175 = 248 so the critical Reynolds number is 300. Thus the particle is in the subcritical regime so one would expect the drag coeﬃcient to be about the same or, perhaps, somewhat larger than the steady state value for a sphere. Problem 4.15 The equation of motion for the particle reduces to Z 3 2 √ √ = −3 − − 2 2 − 0 27

This equation can be rewritten as ³ =− 1 +

2

´ −

s

18 1 1 ³ ´ 1 + 2

Z

0

√ −

The integral can be approximated by Z h i X √ ˙ ( − )12 − ( − − ∆)12 ' − 0 Substituting the values into the equation gives

and

h i X 12 ˙ ( − ) − ( − − ∆)12 = −600 − 338 = 1 + 015Re067 = 1 + 355 067

This can be integrated numerically using a two-step Euler method. Problem 4.16 The equation of motion for a droplet is

= 3 ( − ) +

Dividing by the droplet mass and using the 2 law gives 18 ( − ) = + 2 (1 − )

For no gravitational force and constant free stream velocity of the equation reduces to

or

( − ) = (1 − ) + = (1 − ) (1 − )

Integrating this equation using the integrating factor and employing the initial condition (0) = 0 gives ∙ ¸ = 1 − (1 − ) Integrating one more time to get the distance with the initial condition (0) = 0 yields

28

=

½ +

1 +

¸¾ ∙ 1+ ) −1 (1 −

At time = the distance travelled is ( ) =

2 +

Problem 4.17 The equation of motion is

= 3 ( − ) +

which can be written as 18 ( − ) + = 2 The buoyancy has been neglected because À Using the 2 law 2 = 2 − the equation becomes for = 0 =− + (1 − )

where is the velocity response time based on the initial diameter and is the evaporation time. This is a first order ordinary diﬀerential equation which can be integrated using the integrating factor. ∙ ¸ Z Z exp( ) = exp( ) (1 − ) (1 − ) Integrating this equation using the initial condition = 0 at = 0 gives # "µ ¶ = − (1 − ) 1− 1 −

The maximum velocity occurs where =0 Taking the derivative yields

=1−

µ

¶1−

Substituting into the equation for velocity gives

29

max = (

− )

Problem 4.18 The equation of motion is written as 1 2 = − 2 2 4 Using the expression for drag coeﬃcient gives

3 2 =− − 8 For convenience let

8 3

= so the equation becomes = − 2 +

Integrating with the initial condition (0) = gives ) = exp(− 1+ Solving for =

exp(− ) = 1 − exp(− )

Integrating with respect to time and setting (0) = 0 gives ∙ ¸ 1 − = − ln 1 − ( ) exp(− )

As time approaches infinity, the distance is

= − ln(1 − )

For a 100 micron particle with a material density of 2000 kg/m3 in air at standard conditions, the velocity response time is 0.062 s. The nondimensional length is 0.444 m. With an initial velocity of 10 m/s, the constant is 4.18 m/s. Thus the distance travelled is = −0444 ln(1 − 0062 × 4180444) = 04 m After one velocity response time, the velocity will be =

0368 × 418 = 196 m/s 1 − 0368 × 0062 × 4180444

Using Stokes law the velocity would be

30

= 0368 × 10 = 368 m/s This velocity is larger because Stokes law corresponds to a smaller drag coeﬃcient. For Stokes flow the stopping distance would be = = 10 × 0062 = 062 m Problem 4.19 The equation of motion for the wet particle is 3 =− + Using the rate equation for the water mass, the above equation can be written as 3 = = The solution of this equation is =

µ

¶

where is the initial velocity and is the initial water mass in the particle. The equation for distance can be written as =

= =

or (− ) = +

µ

µ

¶

¶

This equation can be re-expressed as =− ( + ) Integrating this equation results in µ ¶ +1 =− + + +1 Evaluating the constant using ( ) = 0 one has ( " " ¶ # ¶+1 #) µ µ = + 1− 1− +1 31

The distance travelled before water removal is complete ( = 0) is µ ¶ = + +1 or ¶ µ 3 + = 3 3 + Problem 4.20 The expression for terminal velocity is =

neglecting the buoyant force. The response time for the 0.1 micron particle with a material density of 800 kg/m3 is air at standard conditions is =

2 800 × 10−14 = = 247 × 10−7 18 18 × 18 × 10−5

The terminal velocity based on Stokes drag is

= 981 × 247 × 10−7 = 242 × 10−6 m/s

The mean free path can be estimated from1

= 3 where is the molecular mean velocity given by r 8 = where is the gas constant for the specific gas. For air at standard conditions, = 463 m/s so the mean free path is = 0097×10−6 m. The Knudsen number is = 097 Substituting into the Cunningham correct factor gives =

= 1355 Thus, the terminal velocity is = 242 × 10−6 × 355 = 86 × 10−6 m/s 1 This is Maxwell’s equation which was developed in 1860. More accurate coeﬃcients are available from kinetic theory but the form of the equation is the same.

32

Problem 4.21 The equation for rotational motion can be written as 3 = − The response time is =

5 2 = = 3 3 60 60

compared to =

2 18

so the rotational response time is about 1/3 of the velocity response time. The rotating sphere will reach equilibrium faster than a translating particle. Problem 4.22 Boundary layer eﬀects play a minimum role in the drag of a prismatic particle because the separation (and form drag) is established by the flow separation at the corners. Therefore the Basset term would be unimportant for prismatic particles. Problem 4.23 For creeping flows the momentum equation at the particle surface is µ ¶ 1 = 2 2 = Using the velocity distribution for Stokes flow one has 3 = sin 2 Integrating this equation one has 3 cos 2 where is the pressure at the forward stagnation point ( = 0). The form force is = −

= −

Z

0

cos = −

= 2 = which is 1/3 of the Stokes drag.

33

Z

0

cos 2 sin

Problem 4.24 The equation for particle velocity in the y-direction is =0 + The velocity is = exp(− ) and the penetration distance = [1 − exp(− )] The maximum penetration is max = The equation of motion in the x-direction is = = [1 − exp(− )] + Using the integrating factor the equation can be written as [ exp( )] = [1 − exp(− )] exp( ) Integrating this equation yields exp( ) = [exp( ) − 1] − With the initial condition (0) = 0 the equation becomes = [1 − exp(− )] − exp(− ) Integrating with respect to time and using the initial condition (0) = 0 gives ª © = [1 + exp(− )] + 2 2 [exp(− ) − 1]

After a long time

= max Problem 4.25 Setting the lift force equal to the weight = −12 |( − ) | 3 = 1612 ( )12 |ω | 6

34

Solving for =

6 161 ( )12 |ω |−12 |( − ) |

where is given by Equation 4.150 in text. The equation will have to be solved iteratively because is not a linear function of The rotation of the fluid in the laminar sublayer is =

=

The shear stress is given by 2 8 where is the Darcy-Weisbach friction factor and is duct velocity Thus =

=

2 8

The Reynolds number of the flow in the duct is Re =

005 m × 10 m/s = = 33 × 104 15 × 10−5 m2 /s

The Dracy-Weisbach friction factor can be determined from =£

log10

For a smooth pipe, = 0 so =h

log10

³

¡

025 37

025 574 (33×104 )09

The rotational rate of the fluid is =

+

574 Re09

¢¤2

´i2 = 0023

0023 102 = 192 × 103 rad/s 8 15 × 10−5

The values for the factors in the above equation for are 6 ( )

12

6 m2 s2 = 78 × 10−5 × 2500 × 981 kg ¡ ¢ 12 = 18 × 10−5 × 12 = 465 × 10−3 =

|( − ) | = 10 × 192 × 103 = 192 × 104

35

m s2

kg m2 s12

Solving for with = 1 gives =1

¡ ¢12 = 161 × 78 × 10−5 × 465 × 10−3 × 192 × 104 192 × 103 = 255 × 10−4 m

Now = =1 Assume, to start with, that Re 40 and use equation 4.150 for = (1 − 03314 12 ) exp(− Re 10) + 03314 12 with Re

=

=

10 = 667 × 105 15 × 10−5 192 × 103 = 96 2 × 10

Taking values for , evaluating Re and then calculating using =1 and plotting =1 − yields the graph shown in the figure. The value of for which =1 − = 0 is 33 m. The corresponding relative Reynolds number is 22 which is less than 40 required for the correlation.

3x10-4

CSD0-D

2x10-4

1x10-4 5x10-5 0 -5x10-5 0

-6

2x10 3x10-6 1x10-5-6 Particle diameter, m

4x10-6-5 4x10

Problem 4.26 The equation of motion can be written as ¶ µ ¶ µ − − + =− + It is noted that as → 0 36

µ

−

¶

→

1

µ ¶ − +

We will use the method of small perturbations with this limit as the zeroth order solution. µ ¶ µ ¶ − 1 = = − + The function is expanded in powers of = + 1 + 2 2 + Substituting into the diﬀerential equation one has

1 2 + 2 + 3 + + 1 + 2 2 = − +

From the zeroth order solution = −

+

Collecting terms with first powers of + 1 = 0 Thus 1 =

1 2 2 2

The equation for becomes ¶ µ 1 1 2 − =− + + 2 2 + ( 2 ) = Solving for =+ − +

37

µ

¶2

2 + ( 3 ) 2

Chapter 5 Problem 5.1 Thew coeﬃcient of restitution for a sphere from Equation 5.67b is (2)

=

(1)

=

8 = 08 10

The magnitude of the approach velocity is p || = 102 + 142 = 172 cm/s

so

(0)

10 =− = −058 || 172

From Table 5.4 the value of the parameter −

2 2 =− = −0794 7 ( + 1) 7 × 01 × 18

(0)

Because || −0794, use second column in Table 5.4 with = 1. Thus

(0)

(0)

= + ( + 1) = 14 − 02 × 18 × 10 = 104 cm/s

Also

5 (0) ( − 1) 2 5 = 0+ × 02 × 18 × (−10 × 10−2 m/s) 2 × 10−4 m = −1800 rad/s (0)

= +

Problem 5.2 The component of velocity normal to the surface is = −162 m/s The magnitude of velocity is p |v| = 1132 + 1622 + 0482 = 203 m/s

Thus

2 = −0798 − = −0577 |v| 7 (1 + )

Therefore the particle is not sliding (it is rolling) before leaving surface. From first column in Table 5.4 the post-collisional velocities are 38

=

5 2 ( − ) = 0806 m/s 7 5

= − = −(065)(−162) = 1053 m/s 5 2 ( + ) = −0337 m/s 7 5 The post-collision rotations are =

=

= −178 rad/s

= −002 rad/s = −

= −424 rad/s

Problem 5.3 The velocities and rotations of the two spheres are v1 = (433 −25 0) v2 = (−10 0 0) ω 1 = (0010) ω 2 = (0 0 20) The vector G is G = v1 − v2 = (1433 −25 0) At collision, is equal to (1 0 0)The tangential component of at contact is G = G − (G · n)n + (ω 1 + ω 2 ) × n = (1433 −25 0) − (1433 00) + (3 × 10−4 )(0 0 30) × (1 0 0) = (0 −2491 0) For aluminum, = 033 and = 68 × 109 Pa. The value for becomes = 623 × 108 N/m32 From Figure 5.8, = 02 for a coeﬃcient of restitution of 0.8. Thus the damping coeﬃcient is √ = 14 √ 1 = 02 × 305 · 10−7 · 623 · ×108 × (25 · 10−6 ) 4 = 0195 Ns m Take = The normal force is

39

= (− 32 − G · n) = [−623 · 10 × (25 · 10−6 )32 ] − 0195 · (1433) = −751N 8

The tangential force is = − − = −0195 · −294 = 0486N However | | | |. Therefore there is no slipping. Problem 5.4 1. Conservation of linear momentum: 1 1 + 2 2 = 1 1 + 2 2 By definition of the coeﬃcient of restitution, 2 − 1 = (1 − 2 ) Substituting into the first equation gives (1 + 2 ) 1 = (1 − 2 )1 + 2 (1 + )2 Solving for 1 gives 1 = 1 +

2 (1 + )(2 − 1 ) 1 + 2

In the same manner one finds 2 = 1 +

1 (1 + )(1 − 2 ) 1 + 2

2. From Equation 5.14a in text with t = 0 one has 1 = 1 −

2 (1 + )(1 − 2 ) 1 + 2

which is the same as the above equation. The same procedure for Equation 5.14b yields the same result for 2 . 3. For 1 = 5 m/s and 2 = 10 m/s and 1 =

(10−5 )3 × 2500 = 131 × 10−12 kg 6 2 = 163 × 10−13 kg

Substituting into the above equations for velocity gives 1 = 605 m/s 2 = 155 m/s

40

Problem 5.5 The force is given by 3˙ = − 2

Z

0

with = 0 +

2 2

= −3

2 ˙ 20

and 0 Show

Evaluating the integral Z 0

3 3

3 = 3

Z

3 ¡ ¢ 2 3 0 + 2

0

Change variables with = 2 2 to have Z Z 2 3 2 = 2 ¡ 3 2 ¢3 ( 0 0 0 + ) 0 + 2

By particle fractions

+

3

2

=

(0 + ) (0 + ) 1 3 ( + + ) = (0 + )

(0 + )3 (0 + )

3

Therefore = 1, The integral becomes Z 2 22 3 ( 0 0 + )

+ = 0 so = −0

2

= 2

Z

2

"

−

+

#

(0 + )2 " #2 1 1 2 0 − = 2 2 (0 + )2 (0 + ) 0 ∙ 1 0 2 = 2 − − 2(0 + 2)2 (0 + 2) ∙ 0 1 = 22 − + 2(0 + 2)2 (0 + 2) 0

(0 + )3

1

Since 2 0 , the integral reduces to Z 2 2 22 3 ' 0 (0 + ) 0 41

1 1 + 20 0 ¸ 1 20

¸

The force is then = −

3˙ 2

Z

0

32 ˙ 3 = − 3 20

Problem 5.6 The magnitude of the approach velocity is p || = 52 + 32 + 12 = 592 m/s

The velocity in the − plane is p || = 52 + 32 = 583 m/s

The angle of the velocity vector with respect to the vertical is tan−1 (53) = 59o

The angle with respect to the normal direction to the wall is = 59 − 30 = 29o 26.3

o

3.7

o

o

29

59

30

o

o

Pre and post collisional velocities. The velocity components tangential and normal to the wall are 0(0)

= 583 sin 29o = 283 m/s

0(0)

= −583 cos 29o = −510 m/s

0(0)

and

= 1 m/s 0(0)

The value of the parameter

= −0861 ||

2 2 = = 0397 7 ( + 1) 7 × 4 × 18 42

Since -0.397-0.861, use column 1 of Table 5.4 0

0 0

5 0(0) 5 = × 283 = 202 m/s 7 7 0(0) = − = −08 × −51 = 408 m/s 5 0(0) 5 = = × 1 = 0714 m/s 7 7 =

The magnitude of the vector in the 0 − 0 plane is p |0 | = 4082 + 2022 = 455 m/s

The angle of the velocity with respect to the normal from the inclined plane is tan−1 (202408) = 263o The angle with respect to the original coordinate system is = 263 − 30 = −37o The post-collisional velocities with respect to original coordinate system are

= 455 × sin(−37o ) = −029 m/s = 455 × cos(−37o ) = 454 m/s = 0714 m/s

Problem 5.7 The only way to have a collision with no sliding with no initial particle (0) rotation is to have the tangential component of the relative velocity, G = 0Thus (0) G = G(0) − (G(0) · n)n = 0

(0) This condition can only¯happen ¯ for a "head on" collision in which G is parallel (0) (0) ¯ ¯ for n so that G = n and ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ G(0) − (G(0) · n)n = n ¯(0) ¯ − n ¯(0) ¯ = 0

The angular velocity after collision will be zero ω1 = 0

ω2 = 0

The translational velocities will be h ³ ´i 2 (0) (0) (0) v1 = v1 − (1 + ) v1 − v2 1 + 2 ´i h ³ 1 (0) (0) (0) 2 = 2 + (1 + ) v1 − v2 1 + 2

43

Problem 5.8 The mass of the particle is 3 = × 920 × 10−9 = 482 × 10−7 kg 6 6 The force in the normal direction is given by =

= 32 where √ 2 = 3(1 − 2 )

For the properties given

= 42 × 108 N/m32 so = 42 × 108 × (10−6 )32 = 042 N The spring constant for a 1 m displacement is = 12 = 42 × 105 N/m The damping factor in the normal direction (using critical damping) is p p = 2 = 2 482 × 10−7 × 42 × 105 = 090 Ns/m In the tangential direction

where

√ 2 2 12 = 2 − =

2(1 + )

Substituting in the given values gives =

30 × 109 = 1010 Pa 2(1 + 05)

The spring constant in the tangential direction is √ 8 2 12 = 42 × 105 N/m = 2 − The relative velocity between the two particles is G = v − v = 10i − 1559i − 9j = −559i − 9j 44

The direction n is −j. The value for the tangential velocity diﬀerence is G = G − (G · n)n + (ω + ω ) × n G = −559i − 9j−9(−j) + 00005(−200k) × (−j) G = −559i + 01i = −558i The force in the normal direction is F = (−042 − 090 × 9)(−j) = 852j N The force in the tangential direction is F = 42 × 105 × 10−6 i − 090 × (−558i) = 544i N Check to see if there is a slip contact. If |F | |F | then slippage occurs. With = 01 slippage does occur so F − |F | t = −01 × 852(−i) = 852i N Problem 5.9 For this problem, (0)

= −1 ||

and −

(4)

2 = −0793 7 ( + 1)

Therefore

(0)

−

2 7 ( + 1) ||

50so use column 1 of table 5.4. The ratio of the post-collisional and is (0)

(0) − 2 5 2 2 = 5 7 (0) = = (−05) = −0178 (0) 7 7 × 08 −

The angle after bouncing is = tan−1 (−0178) = −101o

45

Problem 5.10 The coordinate system for this problem is

n y 5 m/s x

o

50 10 m/s

The unit normal vector for collision is n=j and = 09. The pre-collisional translational and rotational velocities are (0)

v1

= 10j m/s

(0) v2

= 5 cos 50o j − 5 sin 50o i = 321j − 383i m/s

(0)

ω1

(0) ω2

= −10 000k rad/s = 5 000k rad/s

The pre-collisional relative velocity is (0)

(0)

G(0) = v1 − v2 = 383i+679j The relative velocity at contact is G(0)

= 383i+679j + (25 × 10−6 × −104 k) × j + (25 × 10−6 × 5 × 103 k) × j = 383i+679j+025i−0125i = 396i+679j

Use Equations 5.21a through 5.21d. The tangential component at contact is (0)

G

(0) = G(0) · n)n − (G = 396i+679j − (679)j = 396i

46

The velocities are µ ¶ 2 1 = 10j − 19 × 679j + × 396i × = 355j − 057i 7 2 = 321j − 383i + 645j + 057i =966j−326i m/s 5 1 = −10 000k − 396(j × i) × 7 × 25 × 10−6 2 = 103 × 105 k rad/s 5 1 = 5000 − 396(j × i) × 7 × 25 × 10−6 2 = 118 × 105 k rad/s

v1 v2 ω1

2

m/s

The impulse force (Equation 5.12) is = −

1 (1 + )( · (0) ) = − (19 × 679) = −645 1 + 2 2

The particle mass is = 3 = 2500 × × (50 × 10−6 )3 = 164 × 10−10 kg 6 6 Thus

= −645 × 164 × 10−10 = −106 × 10−9 N·s 2 1 ¯¯ (0) ¯¯ = − ¯G ¯ = − × 396 = −928 × 10−11 N · s 7 1 + 2 7

where the tangential impulse component is from Equation 5.20. The impulse force is J = −106 × 10−9 j − 928 × 10−11 i Problem 5.11 The Hamaker constant for copper and alumina is p p 12 = copp er alumina = 155 × 10−20 × 284 × 10−20 = 21 × 10−20 J

The force between two spheres is (Equation 5.14) is =

12 2

The force is equal to the weight of the alumina sphere (1) µ ¶ 1 2 1 13 = 1 = 12 6 1 + 2 12( + )2 where = 04 × 10−9 m and = (1 + 2 )2 47

Putting in values

205 × 104 12

3 2 4000 kg/m × 981 m/s × 13 6 µ ¶ −2 × 10 m2 3 × 1 −20 = 21 × 10 J× 1 + 3 × 10−2 m ∙ ¸2 1 1 × 12 04 × 10−9 + (10−6 + 0011 )2 m ∙ ¸2 525 × 10−22 1 = 1 + 3 × 10−2 04 × 10−9 + (10−6 + 0011 )2 m

This equation must be solved by interpolation. The result is 1 = 182 × 10−6 m = 182 m Problem 5.12 Stokes drag based on approach velocity 32 ˙ Stokes = −6˙ = − 20 Solving for 0 gives 0 1 = 4

48

Chapter 6 Problem 6.1 From Figure 6.5, the number of particles necessary for a variation of 5% or less at the 99% confidence level is 104 . Thus the dimension of the volume necessary for this condition is 13 = 10 00013 = 21 The interparticle spacing is =

µ

6

¶13

=8

so = 160 m and the dimension of the volume is 3.4 mm. Problem 6.2 a) The particle number flow rate is ˙ = ˜ so = ˙ hi since the particles are in kinetic equilibrium with the fluid. The mass source term is written as ˙ =− = − or

hi hi 3 ˙ =− ×− hi hi 2

=

3 hi 2 hi

where is the initial mass concentration, the continuous phase volume fraction is taken and the carrier phase density is assumed constant. b) The continuity equation is 3 hi ∆ ( hi) = = ∆ 2 hi

or

hi

3 ∆ hi 3 ∆ hi = hi = = hi ∆ 2 2

µ ¶32 1−

Integrating with the initial condition hi = hi at time = 0 yields hi 32 = (1 + ) − (1 − ) hi 49

When =

hi = (1 + ) hi

Problem 6.3 Solution: The mass source term can be written as = ¯ =

˙

˙

From the D2 -law, µ ¶32 2 = 1 − 6

Taking the derivative with respect to time and dividing by the mass gives ˙ 3 1 = 2 Thus, the source term becomes

µ ¶−1 1−

¯ 3 = 2

µ ¶−1 1−

Problem 6.4 Applying Equation B.20 in Appendix B for the volume average of a spatial derivative Z 1 = ( h i) − For the interior particle Z

=

Z

= 0

For the boundary particles 1

Z

= −

( ˜ )

Thus = ( h i) + ( ˜ ) = 0 From Equation 6.45 with no mass coupling and constant particle material density ( ˜ ) = − = 50

so

( h i) = 0 +

Problem 6.5 Applying Equation B.20 in Appendix B for the volume average of a spatial derivative Z 1 = ( h i) − [ + (˙ + ) ] For the interior particle Z − [ + (˙ + ) ]

= −

Z

−

˙ ˙ +

˙ ˙ = − +

For all the interior and boundary particles Z X ˙ X ˙ 1 [ + (˙ + ) ] = + ( ˜ ) − + −

so X ˙ X ˙ ( h i) + ( ˜ ) − + " # X X 1 ˙ ( h i) + ( ˜ ) − ˙ + X ˙ ( h i) − +

= 0

= 0 = 0

X ( h i) = − ˙ = mass ( ) +

Problem 6.6 The mass source term is X X X = − ˙ = − ˙ = − ˙ ˜ ( )

In terms of a continuous number distribution Z = ()() ˙

For evaporation

() ˙ = ( − ∞ ) 51

which gives a mass source term of = ( − ∞ ) For a Rosin-Rammler distribution () = () The source term becomes = ( − ∞ )

Z

()

()−−2 Γ(1 − 3)

+2 Γ(1 − 3)

Z

()−−1

Using the Equation 3.34. The source term becomes = ( − ∞ )

Γ(−1) Γ(1 − 3)

or in terms of and the mass median diameter = ( − ∞ )

Γ(−1) 06931 Γ(1 − 3)

Problem 6.7 The bulk density of the coal particles is ¯ = 001 × 1300 = 13 kg/m3 The relative Reynolds number is Re = 1 × 10−4 × 1218 × 10−5 = 67 The drag factor from Equation 4.111 is = 0 + 1 Re where 0 = (1 −

)2

"

# p µ ¶ 1 + 3 2 + 211 ln + 179 + 10 1 + 0681 − 1102 + 1543

and 1 = 04673 + 001833 and = [−10(04− ) ] Evaluating = exp(−10

039 ) = exp(−390) = 0 001 52

Evaluating 0

2

= 099

"

= 1286

# √ 1 + 3 0005 + 211 × 001 × ln 001 + 179 × 001 1 + 0681 × 001 − 110 × 0012 + 154 × 0013

The value for 1 is = 0467 × 0993 × 01 + 00183 × 0993 = 00223 so = 1286 + 00223 × 67 = 143 The velocity response time is =

2 = 004 s 18

Finally, evaluating =

¯ = 465 kg/m3 s

Problem 6.8 The gas is stagnant and there is no mass transfer so the momentum equation, Equation 6.69 reduces to X − ∆ hi + 3 = 0

since the flow is horizontal. The positive direction for is in the direction of particle motion. Summing over all the particles in the volume and dividing by ∆ gives ¯ ∆ hi = = 3 = ∆ The particle volume fraction is =

2 = 00008 2500

so ' 1. The velocity response time is 3

=

2 2500 kg/m × (2 × 10−4 m)2 = = 031 s 18 18 × 181 × 10−5 Pa · s

The Reynolds number is Re =

30 m/s × 2 × 10−4 m = = 397 151 × 10−5 m2 /s 53

The drag factor is = 1 + 015 Re0687 = 102 The pressure gradient is 2 kg/m3 × 102 Pa ∆ hi = × 30 m/s = 1970 ∆ 031 s m Note that the pressure increases in the direction of particle motion which is the force which balances the particle drag force. Problem 6.9 The force that appears in the coupling term in Equation 6.72 is the hydrodynamic forces acting on the particle. When the equation of motion is used to evaluate this force = + the already includes the buoyant force since that contributes to the hydrodynamic force on the particle (through the pressure gradient). Thus the force coupling term should be ∙ ¸ 1 X 1 X − = −

Thus the formulation as stated in the problem is incorrect. Problem 6.10 Applying Equations 6.64 and 6.66 for the carrier phase in a vertical duct with uniform and steady flow and no mass transfer gives 0 = −∆ hi − ∆ −

X

+ ∆

Take the positive -direction as positive upward. The particles are not accelerating so ∆ hi ∆

− 4 = − −

= −

Problem 6.11 The fluid dynamic force acting on each particle is

= +

Since there is no particle acceleration = − 54

Applying Equation 6.64 ∆ hi = − = ³´ ∆ hi = = ¯ ∆

Notice that the pressure gradient is positive opposing the force due to the Coulomb force on the particles. Problem 6.12 The ratio of the kinetic energy to the enthalpy is 2 −1 = 2 2

µ

¶2

=

−1 2 2

(5)

The kinetic energy can be neglected with respect to the enthalpy if the Mach number is small. Generally, the kinetic energy can be neglected for 03. Problem 6.13 Non-dimensionalizing the conduction term gives µ ¶ µ ¶ h i ¯ eﬀ eﬀ = eﬀ 2 ¯ ¯ Comparing the conduction and convection term [eﬀ ( h i )] [ h i]

∼ ∼

eﬀ 2 1 eﬀ 1 ∼ Pr Re

for ∼ 1. The product of the Reynolds number and Prandtl number is also called the Peclet number. The Prandtl number is usually the order of unity so for large Reynolds numbers the conduction term can be neglected. Problem 6.14 The thermal dissipation is

Φ=

∙µ

¶µ

+

¶¸

= 2

µ

+

¶µ

+

Non dimensionalizing the velocity with and the length with µ ¶µ ¶ 2 + Φ= 2 55

¶

The convection term can be non-dimensionalized to ®¢ ¡ ( h i h i) = h i

where is the representative temperature. Then

1 2 2 Φ ∼ = × × Re ( h i h i)

The ratio 2 is much less than unity because the thermal energies are much larger than mechanical energies. For a gas this would be proportional to Mach number squared which is generally small. Also the Reynolds number would be significantly larger than unity. Thus the thermal dissipation term is generally negligible. Problem 6.15 The energy equation is 2

2

hi hi )] + ∆[ hi ( + )] 2 2 ∙ ¸ X 2 ) − ˙ ( + ) = −∆ ( hi ˜ 2 ∆ [ ( +

+ hi −

X

−

X

˙ − ˙ ∆ − ∆ (,eﬀ )

Steady horizontal flow with no mass transfer and insulated wall reduces the equation to hi2 ∆[ hi ( + )] X2 X = −∆ ( hi ˜ ˙ − ∆ (,eﬀ ) ) − −

The particle volume fraction is the order of 10−2 so ∼ 1 and the flow work due to particle motion can be neglected. Also the flow velocity is much less than sonic so the kinetic energy can be neglected. Also the heat conduction term can be neglected since the Reynolds number is large. Thus the equation reduces to X X ∆[ hi ] = − − ˙

Also, from the example in the text, the energy associated with the work due to fluid dynamic drag is much less than the heat transfer so X ˙ ∆[ hi ] = −

56

The radiative heat transfer to the particles will be transferred to the gas. If ˙ is the rate of heat transfer to particle then the final form of the equation is X ∆[ hi ] = ˙

The gas will heat up with distance along the channel. The velocity will also increase as the gas density decreases to satisfy continuity. There will also be a pressure gradient associated with the velocity change. Problem 6.16 The energy equation is 2

2

hi hi )] + ∆[ hi ( + )] 2 2 ∙ ¸ X 2 ) − ˙ ( + ) = −∆ ( hi ˜ 2 ∆ [ ( +

+ hi −

X

−

X

˙ − ˙ ∆ − ∆ (,eﬀ )

Eliminating the terms for steady flow in horizontal channel with no heat transfer yields hi2 )] 2 ∙ ¸ X 2 ) − ˙ ( + ) = −∆ ( hi ˜ 2 X X − − ˙ − ∆ (,eﬀ ) ∆[ hi ( +

The velocities are much less than sonic so the kinetic energies can be neglected with respect to thermal energies. Also small particle volume fraction allows ∼ 1 and the flow work due to particle motion is insignificant. Also the magnitude of the Reynolds number permits neglecting the conduction term. Thus X X X ∆[ hi ] = − ˙ − − ˙

Finally the work due to particle drag is much smaller than the particle heat transfer so X X ∆[ hi ] = − ˙ − ˙

For each droplet, the heat transfer to the droplet can be related to the change in enthalpy ˙ = − ˙ ( − ) 57

so the energy equation reduces to ∆[ hi ] =

X

˙

The enthalpy of the carrier fluid will decrease with distance (will cool) because ˙ 0 Mass transfer will continue until the temperature of the carrier gas becomes equal to the droplet temperature. Problem 6.17 The entropy equation is written as ( ) = − ( ) +

µ

˙

¶

+

where is the irreversibility term and is always positive. Applying Equations B.20 and B.28 to the volume average of the left side gives ( ) = ( h i) () + ( hi) + The volume average of the irreversibility term is ¯ = hi The volume average of the entropy flux term is

µ

˙

¶

=

µ ¿ À¶ Z ˙ ˙ −

The integral over the particle surfaces will be the sum of entropy transfer from the particle surfaces and the entropy transfer through the boundary particles.

58

Chapter 7 Problem 7.1 For the assumptions listed in the problem, the turbulent kinetic energy equation reduces to µ ¶ h i 0 = − 3 − + |h i − ˆ |2 Subtracting oﬀ the single phase velocity shows µ ¶ ¡ ¢ ¡ ¢ ¡ ¢ h i − − + − − + 3 2 = |h i − ˆ | where is the total eﬀect and is the single phase eﬀect. From the above equation, the right hand side is solely responsible for the eﬀects of the augmentation or attenuation relative to the single phase eﬀects. Here we see that 2 3 ˆ | is always positive, therefore augmentation must occur for |h i − both cases. The dissipation equation is µ µ ¶ ¶ 0 h i + + = − 1 0

−2 or

2 2 2 P | − | + 3

h i 0 = −1 + 0

0

−2 + 3

µµ ¶ ¶ +

2 2 P | − |

Using the same method to get the single phase volume averaged equation and then subtract it shows 3

2 2 P | − |

¡ ¢ h i = 1 − µµ ¶ ¶ ¢ ¡ + − − £ ¤ 0 +2 −

The same conclusion is drawn for the dissipation. 59

Consider the center of the channel where the gradients are zero. This reduces the turbulent energy and dissipation equations to ¢ 3 ¡ |h i h i − 2 h i ˆ + ˆ ˆ | − =

and the dissipation equation reduces to

£ ¤ 3 2 − = 0 |h i h i + 2 − 2 h i ˆ + ˆ ˆ | 2

where the redistribution terms are neglected. Substituting shows

3 |h i h i + 2 − 2 h i ˆ + ˆ ˆ | = 3 |h i h i − 2 h i ˆ + ˆ ˆ | 0 2 If h i h i − 2 h i ˆ + ˆ ˆ 2, then the presence of particles alters the turbulence time scale by 3 2 = Re The turbulence time scale is related to the turbulence viscosity2 . Although it is diﬃcult to speculate, this simplified case shows that small particles and moderate particle Reynolds numbers may cause attenuation by increasing the turbulent viscosity, whereas large particles and moderate particle Reynolds numbers may cause augmentation by increasing the turbulent viscosity. However, at low particle Reynolds number, turbulence is not produced and the particle interactions may play a role. Problem 7.2 The dissipation equation is

0 h i + = − 1 0

−2

µ µ ¶ ¶ +

2 2 2 P | − | + 3

for a steady stream of particles, uniformly dispersed, falling into fluid with zero mean velocity, the above equation is reduced to 3

¯ 0 2 P 1 ¯¯ 2 2 − + 2 ¯ = 2

decomposing the relative velocities and assuming that h i shows ¯ 0 2 P 1 ¯¯ 2 2¯ 3 ¯2 − h i + h i ¯ = 2

2 As the particle Reynolds number approachs zero, either −→ 0 or | − | −→ 0. However, if the mean velocities are equal, then there is still an eﬀect of the particles within a turbulent eddy. Thus it may be diﬃcult for the local relative fluctuations | − | to be zero. Thus it is hypothesized that on the average 2 −→ 0 before | − | −→ 0.

60

3

0 6 2 2 2 h i = 2 4

(6)

where = 6 3 . The turbulent kinetic energy equation is ( h i ) ( ) + µ ¶ = h i − +3 |h i − ˆ |2 ³ ´ \ \ +3 − −

Applying the above conditions shows 2

3 |ˆ | = or 18 h i2

1 = 2

(7)

Substituting Equation 7 into 6 shows 3 6 2 2 h i = 0 4 2

µ ¶2 µ ¶2 2 4 2 1 18 h i = 18 4 h i 2 2 2

and

h i2 1 − 006 Re142 ³√ ´ Determine the turbulence intensity if the particle Reynolds number is = 54

300, the particle volume fraction is 5×10−4 . According to Equation 4.51, the drag factor is estimated as = 1 + 015 Re0687 = 854

The turbulence intensity is then !12 Ã √ 854 5 · 10−4 ≈ 006 = 54 h i 1 − 5 · 10−4 006 (300)142 Problem 7.3

61

For fully developed pipe flow, the single phase momentum equation is hi h 1 i 1 − + = The continuous phase momentum is hi 1 h 1 i − + − ( − ) = Assuming the wall shear is the same for both cases, then the above equation can be rewritten as µ µ ¶ ¶ µ ¶ h 1 i h 1 i hi 1 1 hi = − − + − (−) − where signifies the single phase. The diﬀerence in pressure gradient is attributed to the diﬀerence in the Reynolds stress gradient and the drag; that is, ¸ µ ¶ ∙ µ ¶ hi 1 1 hi − + ( − ) + = If the pressure gradient is equal to the single phase pressure gradient (i.e. ghost particles), then the eﬀect of the particles on the Reynolds stress gradient must balance with the drag. h i particle ³ ´ In order to reduce the pressure gradient, 1 1 + ( − ) . Problem 7.4 Contracting Equation 7.50 by setting = shows (recall that = 3) ( h i ) ( ) + h i = −2 − (1 + 1 ) ( − ) µ ¶ h i h i − 2 +2 2 µ ¶ + −2 3 X + [2 (h i − ) ( − )] h i´ ³ 2 [(h i − ˜ ) (h i − ˜ )] − |h i − ˜ | −3

62

The above equation is now a scalar equation. Knowing that = 2, then ( h i ) ( ) + 2 h i = −2 µ ¶ +2 −2 3 X [(h i − ) ( − )] +2 2

Assuming isotropic diﬀusion, then = 23 within the diﬀusion term. Dividing out the factor of 2 and using the condition = 32 , then ( h i ) ( ) + h i = −2 µ ¶ 3 2 + 2 3 −2 3 X [(h i − ) ( − )] +2

and finally the turbulent energy equation is: ( h i ) ( ) + h i = −2 µ ¶ 2 + −2 3 X [(h i − ) ( − )] +2

Problem 7.5 a) For a steady stream of falling particles and the turbulence energy equation reduces to o ¯ ¢ n 0 0 ¯¯¡ = + ( − 1)¯ − 0 0 63

¯ ¯¡ ¢ and therefore 0 0 + ¯ ( − 1)¯ 2 to ensure that the turbulence dissipation is always positive. The dissipation equation reduces to ¯ ¯ 2 + 3 ¯ ( − 1)¯ µ ¶ = 0 0 2 − 1

and therefore if 0

0 0

=

0 0 ,

then the turbulence energy goes to infinity,

0

thus 1 so that the turbulent energy remains positive. b) For the case where = 0 and 0 = 0, the turbulence energy equation is =−

2

and the dissipation equation reduces to 2 = −

2

both of which are unrealizable. The equations are suspect from the beginning because the change in turbulent energy (and ¡ ¢ dissipation) should not depend on the sign of the relative velocity − . Problem 7.6 For fully developed turbulent wall flows, Equation 7.25 reduces to 0 = −

3 hi 2 |h i − ˆ | −+

with the redistribution terms neglected. For wall flows, = − hi . Assuming that the diﬀusion of kinetic energy and dissipation near the wall in a fully developed channel flow with particles is negligible, the volume averaged turbulent kinetic energy equation is reduced to µ ¶2 hi 3 0 = −+ |h i − ˆ |2 Assuming that the law of the wall applies, the dissipation near the wall is found to be 3 2 = + |h i − ˆ | where is described in Equation 6.48. The kinetic energy is also found by assuming a constant shear layer near the wall. The shear at the wall is related to the friction velocity by 2 hi 2 = 64

Again, assuming the law of the wall is valid and substituting in the dissipation near the wall shows a relation for the particle laden turbulent kinetic energy near the wall of the form µ ¶12 2 2 =p |h i − ˆ | 1+ 3

with the redistribution terms neglected. The above equations for and assure that production of turbulent kinetic energy due to particles and mean velocity gradients is balanced by dissipation near the wall and the single phase boundary conditions are obtained in the limit of vanishing particles. Problem 7.7 For steady, homogeneous flow Equation 7.9 reduces to: ´ ³ 0 0 = − 0 0 ´ ¡ ¢ 1 ³ 0 0 0 + − 0 0 0 + − 0 0

Since particles are stationary, the fluctuating particle velocity is zero. Also, since the particles are fixed in space, 0 = 0. This special case reduces to 0 0 = −2

= −

which is not possible. Problem 7.8 For single phase, constant density, fully developed channel flow near the center of the channel, the equation for 12 is h1 i − 1) 2 − (1 ) (12 ) 0 = (2 µ ¶ 12 + If diﬀusion is negligible, then 12 =

(2 − 1) h1 i 22 1 2

The equation for 22 is 0 = − 22

=

9 22 8 + 5 15

8 27 65

Substituting 22 into the equation for 12 shows 12 = −

16 2 h1 i 243 2

This is the Boussinesq approximation for turbulence shear near the wall. Problem 7.9 Equation 7.50 reduces to ¶ 1 2 − − 3 3 µ ¶ i h 1 \ \ +2 − ˜ − ˜ ˜ ˜ ˜ ˜ + − 3 3

0 = − (1 + 1 )

µ

a) In the direction of falling particles ==3 33

=

+ 1 − 1) 2 (1 (1 + 1 ) 3 i h 2 \ \ − ˜3 ˜3 + + 3 3 3 3 (1 + 1 ) µ ¶ 1 3 ˜3 ˜3 − ˜3 ˜3 − (1 + 1 ) 3

b) In the direction transverse to the falling particles ==2 22

=

11

=

+ 1 − 1) 2 (1 (1 + 1 ) 3 i h \ 2 \ 2 2 − + 2 2 (1 + 1 ) 3 + ˜3 ˜3 3 (1 + 1 )

(1 + 1 − 1) 2 (1 + 1 ) 3 i h \ 2 \ 1 1 − + 1 1 (1 + 1 ) 3 + ˜3 ˜3 3 (1 + 1 )

22 and 11 diﬀer only by the redistribution terms. c) For =1, =2, + 1 ) (1

i h \ \ ˜1 ˜2 + (12 ) = 2 1 2 − 2 1 66

with ˜1 = ˜2 = 0, then 12 =

(1

i 2 h \ \ 1 2 − 2 1 + 1 )

For =2, =1 21 =

i 2 h \ \ − 2 1 1 2 (1 + 1 )

For =2, =3 23 =

(1

i 2 h \ \ 2 3 − 3 2 + 1 )

For =3, =2 32 =

(1

i h \ 2 \ 3 2 − 2 3 + 1 )

For =1, =3 13 =

i h \ 2 \ − 1 3 3 1 (1 + 1 )

For =3, =1 31 =

(1

i h \ 2 \ 3 1 − 1 3 + 1 )

The above oﬀ-diagonal components show symmetry and are dependent on the redistribution terms. The turbulent energy is defined as =

1 (11 + 22 + 33 ) 2

Substituting in the Reynolds stress terms and solving for the dissipation shows: ⎛

⎞ +1 −1) (1 2 ⎜ " ( 1 +1 ) # ⎟ ⎟ 1⎜ \ \ \ ⎜ ⎟ ˜3 ˜3 + 1 1 − 1 1 + 3 3 = ⎜ + 2 ⎟ ⎟ 2 ⎜ (1 +1 ) \ \ \ − 3 3 + 2 2 − 2 2 ⎝ ⎠ 3 3 (˜ ˜ ) + (˜ ˜ ) − + 3 3 3 3 ( 1 1 ) (1 +1 ) ⎡ ⎤ \ \ ˜3 ˜3 + 1 1 − 1 1 1 ⎢ ⎥ \ \ ¶ µ = ⎣ ⎦ + 3 3 − 3 3 +1 −1) (1 \ \ (1 + 1 ) 1 − + +2 2 − 2 2 ( 1 ) 1

67

⎡

⎤ \ \ ˜3 ˜3 + 1 1 − 1 1 1 ⎢ ⎥ \ \ = ⎣ ⎦ + 3 3 − 3 3 ((1 + 1 ) − (1 + 1 − 1)) \ \ + − 2 2 2 2 " # \ \ \ ˜3 ˜3 + 1 1 − 1 1 + 2 2 = \ \ \ − 2 2 + 3 3 − 3 3 From Equation 7.25, the turbulent kinetic energy equation is reduced to ! Ã \ \ \ 2 6 ˆ3 ˆ3 + 1 1 − 1 1 + 2 2 = 3 2 \ \ \ 6 − 2 2 + 3 3 − 3 3 and

=

=

Ã ! \ \ \ 18 ˆ3 ˆ3 + 1 1 − 1 1 + 2 2 − 3 2 \ \ \ 6 2 2 + 3 3 − 3 3 Ã ! \ \ \ ˆ3 ˆ3 + − + 1 1 1 1 2 2 \ \ \ − 2 2 + 3 3 − 3 3

The two equations are the same.

68

Chapter 10 Problem 10.1 The bulk density of the powder in the sampling probe is ∆ 60 gm = = ∆ 60 s = × 20 m/s × × (001 m)2 4 kg/s = × 157 × 10−3 m3 s ˙

10−3

=

= 064 kg/m3

The relationship between the mean velocity and the centerline velocity is ¯ 22 2 × 72 = = = 0816 ( + 1) (2 + 1) 8 × 15 Thus =

20 = 245 m/s 0816

Velocity at 50 mm is (50 mm) =

µ

50 100

¶12

× 245 = 222 m/s

Evaluation of parameter at 50 mm and centerline 2 18Prob e 2500 × (2 × 10−4 )2 × 222 (50 mm) = = 790 18 × (184 × 10−5 118) × 001 (center) = 872

=

From Fig. 10.22, the value for is infinity. For 50 mm, = 2022 = 09 0 =1.05 so 064 3 0 = = 061 kg/m 105 For the centerline, = 20245 = 0816 so 0 = 1.1. The bulk density in the free stream is 064 0 = = 058 kg/m3 11 Problem 10.2 Determine required dimensions for probe volume for coincidence error of 5%. The number density is = 69

The particle mass is 3 = × 2500 × 3 6 6 (15 m) = 44 × 10−12 kg (60 m) = 28 × 10−10 kg

=

The number density is (15 m) = (60 m) =

2 × 12 3 = 545 × 109 (1/m ) 44 × 10−12 2 × 12 = 857 × 1011 (1/m3 ) 28 × 10−10

Probe volume 01 max (15 m) = 183 × 10−13 m3 (605 m) = 112 × 10−11 m3

=

= 0057 mm = 0226 mm

Problem 10.3 The beam angle is given by = sin−1

µ

2

¶

For a frequency of 50 MHz, velocity of 500 m/s and a wave length of 514.5 nm, µ ¶ 50 × 106 × 5145 × 10−9 = sin−1 = 15o 2 × 500 The number of fringes is =

4 ∆

so spacing is

∆ = = 10 × × 2 mm = 157 cm 4 4 Chose a spacing of 20 mm to give 12 fringes. Also

so

∆ = 2b tan

∆ 20 mm = 318 mm = 2 tan 2 × tan 15o Use 500 mm focal length lens This will give an angle of µ ¶ µ ¶ ∆ 20 mm = tan−1 = tan−1 = 114o 2 × 500 mm 2b b =

70

This gives a Doppler frequency for 500 m/s particle of =

2 sin 2 × 500 m/s × sin 114o = = 387 MHz 5145 nm

The diameter of the probe volume is =

4b 4 × 500 mm × 5145 nm = = 163 m × 2 mm

The final configuration of the LDA system is

beam spacing focal length beam half angle probe volume diameter Number of fringes

20 mm 500 1.14o 163 m 12

Problem 10.4 The probe volume waist diameter for the 1000 mm focal length is =

4b 4 × 1000 mm × 05145 m = = 484 m 0 × 135 mm

and for the 500 mm focal length is 282 m. To reduce Gaussian beam eﬀects, the 1000 mm would be better but one would have to consider the number flow rated to insure that coincidence error is not a problem. Taking an initial beam spacing of 40 mm and the 1000 mm focal length lens would give a beam angle of tan() = 201000 = 002 Assume that refraction is responsible for the scattering, then equation 10.77 can be used for ΦThe value for Φ can be obtained from Equation 10.79 with ∆ = 2 05145 m Φ= = = 193 × 10−3 133 × 200 m 0.006

0.005

⌽

0.004

tan( tan( tan( tan()=0.04

0.003

0.002

0.001

0.000

0.00

0.01

0.02

tan()

71

0.03

0.04

A plot of Φ versus the tangent of the half angle between the transmitting beams is shown in the figure. The limiting line of Φ = 193 × 10−3 is shown on the figure. A slit spacing of 10 mm and a focal length of the receiving lens of 200 mm would give a tan of 0.025 and a Φ of 0.0015. The equation relating particle size and phase shift would be = 41∆ (m) with a phase shift of 2 corresponding to 256 m. This would give an adequate range with acceptable accuracy. The final configuration would be Beam spacing Transmitting lens focal length Receiving lens focal length slit spacing

40 mm 1000 mm 200 10 mm

If coincidence error is an issue, the smaller focal length transmitting lens could be used with other options for the slit-receiving lens combination. Problem 10.5 a) The diﬀraction limited spot diameter is = 244( + 1) # = 244 × 2 × 8 × 05106 = 199 m

Image diameter

¡ 2 2 ¢12 + 2 + 2 ¡ ¢12 = 802 + 1992 + 202 = 848 m =

b) For image separation of 5 ∆ = 5 = 424 m For a velocity of 10 m/s ∆ =

∆ 424 m = = 42 × 10−5 s 10 × 106 m/s

The frequency would be 1 = 238 kHz ∆ c) The blur is the pulse time times the velocity =

∆blur = 30 × 10−9 × 10 = 03 m 72

The relation to image diameter 03 ∆blur = = 00035 848 d) Uncertainty in velocity measurement max

∆ ∆ + max ∆ ∆ 02 × 848 2 × 10−9 = + 6 −5 10 × 10 × 42 × 10 42 × 10−5 −4 = 0040 + 0476 × 10 = 004 =

73

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