Fundamentals of momentum heat and mass transfer 6th edition welty solutions manual

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Chapter 9 End of Chapter Problem Solutions 9.1

~

=

r

zr

=

r

r

zr

+

r

-

r

+

zr

+

-

zr

z

~ Substituting into C.V. relationship & evaluating in limit as r (r r)+

r

=0

z go to 0

r

9.2 = x + =

+

* =

x

Note

*

* =

y

+

z

+ (

)+

*

(

y

=1 For j=i =0 For j i x

+

y

+

z

*

)+

z

(

*

)

9.3 4

3

4’

3’

1

2

1’

2’

For 2-Dimensional flow: Volume change= ( )(

)-(

y

x

= x = x+[

)(

)

= y x(x+

x,y)-

3 2 = y[ y(x+ x,y+ ( )( )= x y

x(x,y)]

)-

y(x+

x,y)]

( )( )= x y+[ y(x+ x,y+ [ y(x+ x,y+ )- y(x+ x,y)] 2

x,y)]

+[

Dividing by x y

& evaluating in limit as x y

->0

Volume change =

+

= *

But, from continuity * =0

)-

y(x+

x(x+

x,y)-

x(x,y)]

+

9.4

=

r

+

d =

dr+

d + dt

=

+

+

=

+

+

=

+

+

+

r

r

+

= cos + sin =- sin =

=

=0

= - sin

=

In similar fashion =0

=-

Giving: =

+

=(

- )

+(

For

to be

then

= +( r

+ r) =

&

= =

+

- )

r

+

r

+

+

)

9.5 Navier-Stokes Eqn.- Incompressible form: = -

+

(a) For small- all terms involving ( & ) are small relative to other terms. (b) For small & large the product cannot be considered small relative to other terms

9.6

y x

2L

Incompressible N.S. Eqn. in x-direction + With

=

+

x

=gx-

y

=

=

=

=

+

y=0

y+ c1

=

+ c1y+c2

B.C. c1=0

=0 @ y= c2=

=

(y2-L2)

L2

9.7 = =

(r )+

+

~ * =

(

)=

( =0

And continuity is satisfied

9.8 =

+

=-

At y=100,000 ft. =

=

=

=20,000 ft./s = 0.0096

s-1

9.9 P= ( - ) =400[(y/L)2 +(x/L)2 ] =

+

+

= 400(y/L)2800

Evaluated at (L,2L) we get P=

- [2g+

]

+400(x/L)2800

=

[x2y +xy2 ]

9.10 In x-direction: = gx= gx-

-

+

)+

(

[

( (

(

Similarly in y & zA total of 45 terms!

)+ ( + )+

+ (

)+ (

) [

+

(

)+

+

+

)] )+

(

)+

(

)

]

9.11 For = o+ r o- of coordinate origin r- relative to coordinate origin =

+

=

N.S. Eqn. reduces to =

- P and P= ( - )

9.12 Given that (r )+ a) For

=0

=0

(r )=0

and r ( )=F( or = F( /r b) For =0 =f(r)

=0

9.13 N.S. for Incompressible, laminar flow = -

+

For negligible a) Vector properties~ and are independent by themselves but in same relationship must lie in same plane. b) If viscous force are negligible = is determined by

& is positive in direction of decreasing pressure.

c) In similar fashion, any fluid- either moving or static-will move or tend to move in direction of decreasing P.

9.14 For 1-D Steady flow: = (x) = =0

(

=-

+ [

=-

+3

)=0

2

)+

3

)

]

9.15

Continuity:

+ (

Momentum: (

+

)=0 )=-

9.16 Taking z as positive down with

=

=0 &

=f(r)

Eqn. E.6. yields z direction 2

(

+

+

+

=

+ gz+ [

(r )+ 2

2

+

2

2

]

Realizing that: 2

=

=

= = =

2 2

=

2

=0

and since gz=-g = (r ) Proceed as was done in solutions to problems 8.17 & 8.18 9.17 For incompressible, steady flow, with r direction

= =0 Eqn. (E-4) has the form 2

2

(

+

+

+

=

+ gr+ [

Realizing that the following terms are zero: 2

=

= = =2

2

2 2

2

The initial equation becomes 2

(P+

2

)= gr

2

=

2

=0

)+ 2

2

2

- 2

2

+

2

]

9.18 Governing Eqns. Are r direction 2

2

(

+

+

+

=

+ gr+ [

)+ 2

2

2 - 2

2

+

2

]

direction 2

(

+

+

+

=

+

+ [

)+ 2

2

2 - 2

2

+

2

z direction 2

(

+

When

+

+

=

=f(r) and with

2

= = Q.E.D.

(r )+ 2

2 2 +

2

]

= =0 the only non-zero term on the left-hand side of all 2

component eqns. is

+ gz+ [

]

9.19 Eqn. (E-5) is simplified for this case as

direction:

(

+

2

+

+

+

=

Realizing that the following terms are zero: 2

=

=

=

=2

2

= 2=

2

=0

& in the absence of gravity we have =

[

(r )]

+ [

)+ 2

2

2 - 2

2

+

2

]

9.20 From Prob. 9.19 & steady flow =0 Giving

(r

=c1

Integrating again: r =c1lnr+c2 B.C.

(R1)=R1 (R2)=R2

= [R12

+

2 2 2 2

2 2

]

9.21

Fluid flow around center section

RO

RI

r

V0 Need to begin with z-direction Navier-Stokes equation in cylindrical coordinates: 2 2

Applying the assumptions given in the problem statement:

2

2

2

2

Boundary Conditions: 1.

2.

Apply Boundary Condition #1:

2 2

2

2 2

Apply Boundary Condition #2: 2 2

Solve for

,

2

2

2

2

Solve for 2 by plugging back into equation for Boundary Condition #1 (if you plug into the Boundary Condition #2 equation your result will be slightly different and no less correct), 2

2

2 2

2

2

2

2

Thus, 2

2

2

2

2

2

9.22 VB

Fluid 1

Fluid 2

L1

L2

VA

(I will show two different solutions, the second solution may be slightly easier. The only difference is the identification of the boundary conditions that are in red in the figures below.)

Need y-direction Navier-Stokes Equation 2

2 2

2 2

2

Using the conditions in the problem statement this equation becomes, 2 2

Integrate twice solving for velocity, 2 2

2

2

2

Next, we develop equations for the velocity of Fluid 1 and Fluid 2 using the above equation.

Fluid 1 VB

VA

Fluid 1

Fluid 2

L1

L2

x=0

x=L1

x=L1+L2 2 2

2 Boundary Conditions: BC#1: BC#2: 2 Apply BC#1: 2

2

Apply BC#2: 2

2 Solve for

, 2

2 Thus, plugging in the expression

,

2

2

2

2

Fluid 2 2 2

Boundary Conditions: BC#2: 2 BC#3: 2 Apply BC#2:

2

2

(this boundary condition applies to both fluids) 2

2 2

2 Apply BC#3: 2

2

2

2 Set

2,

2

2

2

2 ,

Solve for

2

2 2

2 2

2

2

Now solve for

2

2

2

2

2 2

2 Insert the equation for

,

2 2

2

2

2

2

2

2

2

2

2

2

2

2

Thus, 2 2

2

2

And, plugging 2 2

2

2

and

2 back into the original equation for

2 2

2

2

2

2

2

2

2

2, 2

2

2

2

2

2

2 2

2

2 2 2 This equation could be simplified, but that is not necessary.

2

2

2

Solution 2 (this may be a slightly easier solution, but both are correct).

Need y-direction Navier-Stokes Equation 2

2 2

Using the conditions in the problem statement this equation becomes, 2 2

Integrate twice solving for velocity, 2 2

2 2

2

2 2

2

Next, we develop equations for the velocity of Fluid 1 and Fluid 2 using the above equation.

Fluid 1 VB

Fluid 1

Fluid 2

L1

L2

VA x= -L1

x=0

x= +L2

2 2

2 Boundary Conditions: BC#1: BC#2: 2 2

2

2 Apply BC#2: 2

2 2

2

Apply BC#1 (and insert the result of BC#2): 2 2

2

2

2

2 Thus, plugging in the expression for

2 ,

2

2 Fluid 2

2

2

2

2

BC#2: BC#3:

(this boundary condition applies to both fluids)

2 2

2 2 2

2

2

Apply BC#2: 2 2

Apply BC#3, inserting the value for 2 2

2: 2

2 2

Plugging the values of

and 2

2

2

2

2 2 back into the original equation: 2 2 2

2

2

9.23

Thin viscous fluid with thickness = h

Belt moving up with a velocity=Vw

Air

Begin with the y-direction equation in rectangular coordinates: 2

2 2

2 2

2 2

Thus, 2 2

Boundary Conditions: BC#1: at the moving wall: BC#2: at the free surface: Rearrange, 2 2

Integrate,

at x = 0 at x = h

2 2

2

Integrate again,

2

2

From BC#2:

From BC#1: 2

Thus, 2

2

2

9.24

Vz

z-direction Navier-Stokes Equation in Cylindrical Coordinates 2 2

To obtain the desired equation we must make the following assumptions:

2 2 2 2

Rearrange,

Integrate, 2

2 Rearrange, 2 Substitute in Newton's Law of Viscosity

Boundary condition:

2 , so 2

,

2

2 2

2