Chapter 1 End of Chapter Problem Solutions

1.1

n = 4 x 1020

molecules/in3

= = 1.32x104 in./s

A=

(10

-3 in)

2

NA =

n A=1.04x10

8 m/s

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1.2

Flow Properties: Velocity, Pressure Gradient, Stress

Fluid Properties: Pressure, Temperature, Density, Speed of Sound, Specific Heat

1.3

mass of solid= ρsvs

mass of fluid= ρfvf

mix=

=

=

1.4

Given

=

7

For P1= 1 atm

= 1.01

P= 3001(1.01)7 – 3000 = 217 atm

1.5

At Constant Temperature

= constant

= constant

For 10% increase in

P must also increase by 10 %

1.6

Since density varies as

& =nM (M=Molecular wt.)

n250,000= nS.L.

= 4 x 10

20

= 2.5 x 10

16

1.7

= x + y

= +

= x + y

= +

Q.E.D.

1.8

= + =

= =

Q.E.D.

1.9 Transformation from (x,y) to (r, )

=

+

=

+

r2 = x

2+y

2

=

so:

=

=

=

=

=

=

=

=

=

=

+

1.10

=

+

+

= (

) + (

+

) +

= ( + )

+

( + )

+

Thus: =

+

1.11

P=

+

P(a,b)= 2

+

(sin1cos1)

= 2{[

+ 2) +

(

)

1.12

T(x,y)= Toe-1/4

[

(cos

cosh

) +

(sin

sinh

)

T(a,b)= Toe-1/4

[

(cos cosh1) +

(sin sinh1)

= Toe-1/4

[

+

]

=

[

(1+e

-2) +

1.13

In problem 1.12 T(x,y) is dimensionally homogeneous (D.H.)

P(x,y) in Prob 1.11 will be D.H. if

~

LBf s

2/ ft

4

or using the conversion factor gc

1.14 = 3x2y+4y

2

A scalar field is given by the function:

(a) Find at the point (3,5)

For the value of at the point (3,5)

(b) Find the component of that makes a -60 angle with the axis at the point (3,5)

Let the unit vector be represented by

At the point (3,5) this becomes:

1.15

For an ideal gas

P=

From Prob 1.3:

1.16

= Arsin (1-

)

a) =

+

= Asin (1-

)

b) = A

1/2

max is given by d = 0 or

dr+

d = 0

Requiring

=

=0

For

= 0: - (1+

) + (1-

) = 0 (1)

And for

=0:

(2)

From Eq. 2: 4a2/r

2=0

If a 0, r then for which = 0,

(3)

Subst. into Eq. 1 =0, 1-a2/r

2=0

Giving a = r

For =

1+ a

2/r

2 =0 impossible

Thus conditions for max are r = a

1.17

P=Po+

2

=

2

= +

2

= +

2

P =

2

1.18

Vertical cylinder d=10m, h=6m

V=

(10m)

2(6m)= 471.2 m

3

@ 20 w=998.2 kg/m3

m= wV= (998.2)(471.2)= 470350 kg

@ 80 w=971.8 kg/m3

m=(971.8)(471.2)= 457910 kg

12440 kg

1.19

Liquid V= 1200 cm3 @ 1.25 MPa

V=1188 cm3 @ 2.5 MPa

= -V

T -V

V= 1194 cm3= 1.194 x 10

-3 m

3

∆V= -12 cm3= -1.2 x 10

-7 m

3

= -1.194 x 10-3

= +12440 MPa

1.20

= -V

T -V

V=0.25 m3

V= -0.005 m3

P= 10 mPa

= -0.25

= 500 MPa

1.21

For H20: =2.205 GPa

= -0.0075

-V

or ∆P =

∆P= (2.205 GPa)(0.0075) = 0.0165 GPa= 16.5 MPa

1.22

For H20: P1=100kPa P2=120 MPa 2.205

= -V

or

=

=

=

= 0.999 x 10

-3 = 0.0999 percent

1.23

H20@ 68 (341 K)

0.123 [1-0.00139(341)] = 0.0647 N/m

In a clean tube- =0

h =

=

= 9.37 x 10

-3m = 9.37 mm

1.24

Parallel Glass Plates

Gap=1.625 mm

= 0.0735 N/m

For a unit depth:

Surface Tension Force = 2(1) cos

Weight of H20 = (1)(1.625x10-3

)

For clean glass cos =1

Equating Forces:

2(1) = (1)(1.625x10-3

)

h =

= 0.00922m= 9.22 mm

1.25

Glass Tube: di=0.25 mm

do=0.35 mm

=130

Surface Tension Force-

Inside: 2 i cos

Outside: 2 o cos

Total Upward Force

F= 2 cos (ri+ro) = 2 (0.44)(cos130 )(

x 10

-3) = 5.33x 10

-4 N

1.26

H20-Air-Glass Interface @40

Tube Radius= 1 mm

h=

cos

= 0.123[1-0.0139(313)]=0.0695 N/m

h =

= 0.0143m (1.43 cm)

1.27

Soap Bubble- T=20 d=4mm

= 0.025 N/m (Table 1.2)

Force Balance for Bubble:

r2 r

so

=

= 25 N/m

2 = 25 Pa

1.28

Glass Tube in Hg (S.G.= 13.6)

For Hg/ Glass: =0.44 N/m

h=

r= 3mm

=

= -0.0277 m

= 2.77 cm Depression

1.29

@60 C = 0.0662 N/m

= 0.44 N/m

Tube Diameter= 0.55 mm

h=

For H20:

= 0.0499m (4.99 cm Rise)

For Hg:

= -0.0157 m (1.57 cm Depression)

1.30

H20/ Glass Interface

T=30 C

= 0.123[1-0.0139(303)] = 0.0712 N/m

= 996 kg/m3

h 1 mm

h =

r =

=

= 0.0146 m (1.457 cm)

d = 2r = 2.915 cm

1.31

Bubble Diameter = 0.25 cm = 0.0025 m, and so Radius = 0.00125 m

Capillary Tube: Diameter = 0.2cm = 0.002 m, and so Radius = 0.001 m

Beginning with:

Rearrange and remember the unit conversion Pa=kg/ms2,

Next, we can calculate the height of the fluid in the tube:

1.32

First, calculate the surface tension of water using the temperature of the water:

Next, using the equation for the height of a fluid in a capillary,

Rearranging and solving for the radius:

Diameter is 2r so D=1.50 mm

1.32

First, calculate the surface tension of water using the equation for the height:

Rearrange, solving for ,

Now use the surface tension to calculate the temperature:

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