HG I KJ F

CHAPTER TWO

2.1 (a) 3 24 3600

1 18144 10⁹

wk 7 d h s 1000 ms

1 wk 1 d h 1 s = . × ms

(b) 38 3600

25 98 26 0

.1 ft / s 0.0006214 mi s . .

3.2808 ft 1 h= mi / h⇒ mi / h

(c) 554 1 1

1000 g 3

m 1 d h kg 10 cm

d kg 24 h 60 min 1 m 85 10 cm g

4 8 4

4

4 4

⋅ = . × / min⋅

2.2 (a) 760 mi

3600 340 1 m 1 h

h 0.0006214 mi s= m / s (b) 921 kg

35.3145 ft 57 5 2.20462 lb 1 m

m 1 kg^m lb / ft

3

3 3 m

= . 3

(c) 5 37 10 1000 J 1

1 119 93 120

. 3

× × .

= ⇒

kJ 1 min .34 10 hp

min 60 s 1 kJ J / s hp hp

-3

2.3 Assume that a golf ball occupies the space equivalent to a 2in in in ×2 ×2 cube. For a classroom with dimensions 40 ft×40ft×15ft:

n_balls

3 3

ft (12) in 1 ball 6

ft in 10 5 million balls

=40×40×15 = × ≈

2 518

3

3 3 3 .

The estimate could vary by an order of magnitude or more, depending on the assumptions made.

2.4 4 3 24 3600 s

1 0 0006214

.

.

light yr 365 d h 1.86 10 mi 3.2808 ft 1 step 1 yr 1 d h 1 s mi 2 ft

7 10 steps

5 16

× = ×

2.5 Distance from the earth to the moon = 238857 miles

238857 mi 1

4 10¹¹ 1 m report

0.0006214 mi 0.001 m= × reports 2.6

19 0 0006214 1000

264 17 44 7

500 25 1

14 500 0 04464

700 25 1

21 700 0 02796

km 1000 m mi L

1 L 1 km 1 m gal mi / gal Calculate the total cost to travel miles.

Total Cost gal (mi)

gal 28 mi

Total Cost gal (mi)

gal 44.7 mi Equate the two costs 4.3 10 miles

American

European

5

.

. .

$14, $1.

, .

$21, $1.

, .

=

= + = +

= + = +

⇒ = ×

x

x x

x x

x

2.7

6 3

3

5

5320 imp. gal 14 h 365 d 10 cm 0.965 g 1 kg 1 tonne plane h 1 d 1 yr 220.83 imp. gal 1 cm 1000 g 1000 kg

tonne kerosene 1.188 10

plane yr

⋅

= ×

⋅

9

5

4.02 10 tonne crude oil 1 tonne kerosene plane yr yr 7 tonne crude oil 1.188 10 tonne kerosene 4834 planes 5000 planes

× ⋅

×

= ⇒

2.8 (a) 25 0 . 25 0

lb 32.1714 ft / s 1 lb .

32.1714 lb ft / s lb

m

2

f m

2 f

⋅ =

(b)

2 2

25 N 1 1 kg m/s

2.5493 kg 2.5 kg 9.8066 m/s 1 N

⋅ = ⇒

(c) 10 1000 g 980.66 cm 1

9 10⁹

ton 1 lb / s dyne

5 10 ton^m 2.20462 lb 1 g cm / s dynes

2 -4

m

× ⋅ 2 = ×

2.9 50 15 2 85 3 32 174

1 4 5 10⁶

× ×

⋅ = ×

m 35.3145 ft lb ft 1 lb

1 m 1 ft s 32.174 lb ft s lb

3 3

m f

3 3 2

m

2 f

. .

/ .

2.10 500 lb

5 10 1 2

1

10 25

2 m

3

m

1 kg 1 m 3

2.20462 lb 11.5 kg ≈ ×

F

m

HG I KJ F

HG I KJ

^≈

2.11 (a)

m m V V h r H r

h H

f f c c f c

c f

displaced fluid cylinder

3

cm cm g / cm 3

30 cm g / cm

= ⇒ = ⇒ =

= = −

=

ρ ρ ρ π ρ π

ρ ρ

2 2

30 14 1 1 00

( . )( . ) 0 53

. (b) ρ ρ

f

cH

= h = ( )( . )=

30 0 53 . cm g / cm 1 71

(30 cm - 20.7 cm) g / cm

3

3

H

h ρ_f ρ_c

2.12

V R H

V R H r h R

H r

h r R

Hh

V R H h Rh

H

R H h

H

V V R

H h H

R H

H H h

H

H

H h h

H

s f

f

f f s s f s

f s s s

= = − = ⇒ =

⇒ = −

F

HG I

KJ

⁼

F

⁻

HG I

KJ

= ⇒

F

−

HG I

KJ

⁼

⇒ =

−

= − =

−

F HG I

KJ

π π π

π π π

ρ ρ ρ π ρ π

ρ ρ ρ ρ

2 2 2

2 2 2 3

2

2 3

2

2

3 2

3

3 3 3

3 3 3

3 3 3

3 3

1 1

; ;

ρ_f ρ_s

R r h H

2.13

Say

h m( ) =

depth of liquid

A

( )

m ² h

1 m

⇒

y

x y = 1

y = 1 – h x =

1

– y ²

d A

( )

( ) ^{( ) ( ) ( )}

2

2

1 1

2 2 2

1 1

1 2

2 2 1 1

1

Table of integrals or trigonometric substitution

2 1 2 1

m 1 sin 1 1 1 sin 1

2

π

− − +

− − −

− − −

−

= ⋅ = − ⇒ = −

= − + ⎤⎥⎦ = − − − + − +

⇓

∫

^y

∫

^h

y

h

dA dy dx y dy A m y dy

A y y y h h h

W N

A

A

A

g g

b g

^{= × = ×}

E

4 0 879 1

1 10 3 45 10

2

3 4

0

m m g 10 cm kg 9.81 N

cm m g kg

Substitute for

2 6

3 3

( ) .

N .

W

b g

N ^{= ×}

L b g b g

h^{− −} h^{− +}

b g

h^{− +}

NM O

−

QP

3 45 10 1 1 1 1

2

4 2 1

. sin π

2.14 1 1 32 174 1

1 1

32 174

lb slug ft / s lb ft / s slug = 32.174 lb poundal = 1 lb ft / s lb

f

2

m 2

m

m 2

f

= ⋅ = ⋅ ⇒

⋅ =

. . (a) (i) On the earth:

M W

= =

= ⋅ = ×

175 lb 1

5 44

175 1

1

m

m m

2

m 2

3

slug

32.174 lb slugs lb 32.174 ft poundal

s lb ft / s 5.63 10 poundals .

(ii) On the moon M

W

= =

= ⋅ =

175 lb 1

5 44

175 1

1

m

m m

2

m 2

slug

32.174 lb slugs lb 32.174 ft poundal

6 s lb ft / s 938 poundals .

( )b F=ma⇒ =a F m/ = ⋅

=

355 poundals lb1 ft / s 1 slug 1 m 25.0 slugs 1 poundal 32.174 lb 3.2808 ft 0.135 m / s

m 2

m 2

y= –1

y= –1+h x

dA

2.15 (a) F =ma⇒

F HG I

KJ

^{= ⋅}

⇒ ⋅

1 1

6 5 3623 1

fern = (1 bung)(32.174 ft / s bung ft / s fern

5.3623 bung ft / s

2 2

2

) .

(b) On the moon: 3 bung 32.174 ft 1 fern

6 s 5.3623 bung ft / s fern

On the earth: = 18 fern

2 2

W W

= ⋅ =

=

3 3 32 174 5 3623

( )( . ) / .

2.16 (a) ≈ =

= ( )( ) ( . )( . )

3 9 27 2 7 8 632 23

(b)

4

5

4 6

4.0 10 40 1 10

(3.600 10 ) / 45 8.0 10

− −

− −

≈ × ≈ ×

× = ×

(c) ≈ + =

+ =

2 125 127 2 365 125 2. . 127 5.

(d) ≈ × − × ≈ × ≈ ×

× − × = ×

50 10 1 10 49 10 5 10

4 753 10 9 10 5 10

3 3 3 4

4 2 4

.

2.17

1 5 4

2 3

6

3

(7 10 )(3 10 )(6)(5 10 )

42 10 4 10 (3)(5 10 )

3812.5 3810 3.81 10

exact

R R

×

−

× ×

≈ ≈ × ≈ ×

×

= ⇒ ⇒ ×

(Any digit in range 2-6 is acceptable)

2.18 (a)

A: C

C

C

o

o

o

R X

s

= − =

= + + + + =

= − + − + − + − + −

−

=

731 72 4 0 7

72 4 731 72 6 72 8 73 0

5 72 8

72 4 72 8 731 72 8 72 6 72 8 72 8 72 8 73 0 72 8 5 1

0 3

2 2 2 2 2

. . .

. . . . .

.

( . . ) ( . . ) ( . . ) ( . . ) ( . . )

.

B: C

C

C

o

o

o

R X

s

= − =

= + + + +

=

= − + − + − + − + −

−

=

1031 97 3 58

97 3 1014 98 7 1031 100 4

5 100 2

97 3 100 2 1014 100 2 98 7 100 2 1031 100 2 100 4 100 2 5 1

2 3

2 2 2 2 2

. . .

. . . . .

.

( . . ) ( . . ) ( . . ) ( . . ) ( . . )

.

(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.

2.19 (a) X

X

s

X

X s

X s

i

i i

= = =

−

− =

= − = − =

= + = + =

= =

∑ ∑

1 12

2 1

12

12 73 5

73 5

12 1 1 2

2 73 5 2 1 2 711 2 73 5 2 1 2 75 9

.

( . )

.

. ( . ) .

. ( . ) .

C C

min=

max=

(b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter.

(c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness

2.20 (a), (b)

(c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12.

2.21 (a)

4 2 2 2

2

2.36 10 kg m 2.20462 lb 3.2808 ft 1 h ' h kg m 3600 s Q

× − ⋅

=

(b)

4

( 4 3) 6 2

approximate 3

6 2 2

exact

(2 10 )(2)(9)

' 12 10 1.2 10 lb ft / s

3 10

' =1.56 10 lb ft / s 0.00000156 lb ft / s Q

Q

− − − −

−

≈ × ≈ × ≈ × ⋅

×

× ⋅ = ⋅

(a) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

X 134 131 129 133 135 131 134 130 131 136 129 130 133 130 133 Mean(X) 131.9

Stdev(X) 2.2

Min 127.5

Max 136.4

(b) Run X Min Mean Max 1 128 127.5 131.9 136.4 2 131 127.5 131.9 136.4 3 133 127.5 131.9 136.4 4 130 127.5 131.9 136.4 5 133 127.5 131.9 136.4 6 129 127.5 131.9 136.4 7 133 127.5 131.9 136.4 8 135 127.5 131.9 136.4 9 137 127.5 131.9 136.4 10 133 127.5 131.9 136.4 11 136 127.5 131.9 136.4 12 138 127.5 131.9 136.4 13 135 127.5 131.9 136.4 14 139 127.5 131.9 136.4

126 128 130 132 134 136 138 140

0 5 10 15

2.22 N C k

C C N

p o

Pr o

Pr

. .

.

( )( )( )

( )( )( ) . . .

= = ⋅

⋅ ⋅

≈ × × ×

×⁻ ₋ × ≈ × ≈ × ×

μ _{0 583 1936 3 2808}

0 286

6 10 2 10 3 10

3 10 4 10 2

3 10

2 15 10 1 63 10

1 3 3

1 3

3

3 3

J / g lb 1 h ft 1000 g

W / m ft h 3600 s m 2.20462 lb The calculator solution is

m

m

2.23

Re . . .

.

Re ( )( )( )( )

( )( )( )( )

(

= =

× ⋅

≈ × ×

× × ≈ × ≈ × ⇒

−

− −

−

− −

Duρ μ

0 48 2 067 0 805

0 43 10

5 10 2 8 10 10

3 4 10 10 4 10

5 10

3 2 10

3

1 1 6

3 4

1 3)

4

ft 1 m in 1 m g 1 kg 10 cm s 3.2808 ft kg / m s 39.37 in cm 1000 g 1 m

the flow is turbulent

6 3

3 3

2.24

1/ 3 1/ 2

1/ 3 1/ 2

5 2 3

3 5 2 5 2

(a) 2.00 0.600

1.00 10 N s/m (0.00500 m)(10.0 m/s)(1.00 kg/m ) 2.00 0.600

(1.00 kg/m )(1.00 10 m / s) (1.00 10 N s/m ) (0.00500

44.426

μ ρ

ρ μ

−

− −

⎛ ⎞

⎛ ⎞

= + ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎡ × ⋅ ⎤ ⎡ ⎤

= + ⎢⎣ × ⎥ ⎢⎦ ⎣ × ⋅ ⎥⎦

= ⇒

g p p

g

k d y d u

D D

k

5 2

m)(0.100)

44.426 0.888 m / s

1.00 10 m / s⁻ = ⇒ =

× k^g

(b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error.

(c)

d_p (m) y D (m²/s) μ (N-s/m²) ρ (kg/m³) u (m/s) k_g 0.005 0.1 1.00E-05 1.00E-05 1 10 0.889 0.010 0.1 1.00E-05 1.00E-05 1 10 0.620 0.005 0.1 2.00E-05 1.00E-05 1 10 1.427 0.005 0.1 1.00E-05 2.00E-05 1 10 0.796 0.005 0.1 1.00E-05 1.00E-05 1 20 1.240 2.25 (a) 200 crystals / min mm; 10 crystals / min mm⋅ ⋅ ²

(b) r =

⋅ −

⋅

= ⇒ =

200 10

4 0 crystals 0.050 in 25.4 mm

min mm in

crystals 0.050 in (25.4) mm

min mm in

238 crystals / min 238 crystals 1 min

60 s crystals / s

2 2 2 2

2 2

min .

(c) D D

D

mm in mm

b g

₌ ^′

b g

^{25 4}1 in^. _{=25 4. ′; r} ^crystals _{r r} min

crystals 60 s s 1 min

F HG I

KJ

^{= ′ =60 ′}

⇒60r′ =200 25 4

b

. D′ −

g b

10 25 4. D′ ⇒ ′ =

g

² r 84 7. D′ −108

b g

D′ ²

2.26 (a) 70 5. lb_m /ft 8.27³; ×10^-7 in / lb² _f

(b)

7 2 6 2

3 f

m 2 5 2

f

3 3

m

3 3 6 3

m

8.27 10 in 9 10 N 14.696 lb / in (70.5 lb / ft ) exp

lb m 1.01325 10 N/m 70.57 lb 35.3145 ft 1 m 1000 g

1.13 g ft m 10 cm 2.20462 lb

ρ

⎡ × − × ⎤

⎢ ⎥

= ⎢⎣ × ⎥⎦

= = /cm³

(c) ρ ^lb ρ ρ

ft

g lb cm

cm g 1 ft

m 3

m

3

3 3

F HG I

KJ

^{= ′} 453 593^{1 28 317}, ⁼62 43 ^′

. .

P lb P P

in

N .2248 lb m

m N 39.37 in

f 2

f

2

2 2 2

F HG I

KJ

^{= ' 0} 1 ^{1 =}1 45 10^{. × − '}

2

4

⇒62 43. ρ′ =70 5. exp

d

8 27. ×10⁻⁷

id

1 45 10. × ^−4P'

i

⇒ ′ =ρ 113. exp .

d

1 20×10^−10P'

i

P'=9 00. ×10⁶ N / m² ⇒ =ρ' 113. exp[( .1 20×10⁻¹⁰)( .9 00×10⁶)]=113. g / cm³

2.27 (a) V V

V

cm in 28,317 cm

in

3

3 3

d i d i

^{= '} 3 ^{= . '}

1728 16 39 ; t

b g

s ⁼³⁶⁰⁰t^′

b g

hr

⇒16 39. V'=exp

b

3600t′ ⇒ =

g

V' 0 06102. exp

b

3600t′

g

(b) The t in the exponent has a coefficient of s^-1. 2.28 (a) 3 00. mol / L, 2.00 min^-1

(b) t C

C

= ⇒ =

⇒ =

0 3 00

3 00 .

.

exp[(-2.00)(0)] = 3.00 mol / L t = 1 exp[(-2.00)(1)] = 0.406 mol / L For t=0.6 min: C

C

int

. .

( . ) . .

.

= −

− − + =

=

0 406 3 00

1 0 0 6 0 3 00 1 4 3 00

mol / L exp[(-2.00)(0.6)] = 0.9 mol / L

exact

For C=0.10 mol/L: t t

int

exact

min

= - 1 2.00ln C

3.00= -1 2ln0.10

3.00= 1.70 min

= −

− − + =

1 0

0 406 3 0 10 3 00 0 112

. ( . . ) .

(c)

0 0.5 1 1.5 2 2.5 3 3.5

0 1 2

t (m in)

C (mol/L)

(t=0.6, C =1.4)

(t=1.12, C =0.10) Cexact vs. t

2.29 (a) p*

. . ( . )

= −

− − + =

60 20

199 8 166 2 185 166 2 20 42 mm Hg (b) c MAIN PROGRAM FOR PROBLEM 2.29

IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6

READ (5, *) TD(I), PD(I)

1 CONTINUE

WRITE (5, 902)

902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X,

* ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5

T = 100 + I

CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P

903 FORMAT (10X, F5.1, 10X, F5.1)

2 CONTINUE

END

SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6)

I = 1

1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I = I + 1

IF (I.EQ.6) STOP GO TO 1

2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN

END

DATA OUTPUT

98.5 1.0 TEMPERATURE VAPOR PRESSURE

131.8 5.0 (C) (MM HG)

# # 100.0 1.2

215.5 100.0 105.0 1.8

# #

215.0 98.7

2.30 (b) ln ln

(ln ln ) / ( ) (ln ln ) / ( ) .

ln ln ln . ( ) . .

y a bx y ae

b y y x x

a y bx a y e

bx

x

= + ⇒ =

= − − = − − = −

= − = + ⇒ = ⇒ = ⁻

2 1 2 1

0.693

2 1 1 2 0 693

2 0 63 1 4 00 4 00

(c) ln ln ln

(ln ln ) / (ln ln ) (ln ln ) / (ln ln )

ln ln ln ln ( ) ln( ) /

y a b x y ax

b y y x x

a y b x a y x

= + ⇒ = b

= − − = − − = −

= − = − − ⇒ = ⇒ =

2 1 2 1 2 1 1 2 1

2 1 1 2 2

(d) ^{/ /}

2 1 2 1

3 /

ln( ) ln ( / ) ( / ) [can't get ( )]

[ln( ) ln( ) ]/[( / ) ( / ) ] (ln 807.0 ln 40.2) /(2.0 1.0) 3

ln ln( ) ( / ) ln 807.0 3ln(2.0) 2 2

[can't solve explicitly for

by x by x

y x

xy a b y x xy ae y a x e y f x

b xy xy y x y x

a xy b y x a xy e

= + ⇒ = ⇒ = =

= − − = − − =

= − = − ⇒ = ⇒ =

( )] y x

2.30 (cont’d)

(e) ln( / ) ln ln( ) / ( ) [ ( ) ]

[ln( / ) ln( / ) ] / [ln( ) ln( ) ] (ln . ln . ) / (ln . ln . ) .

ln ln( / ) ( ) ln . . ln( . ) .

/ . ( ) . ( )

/

.33 / .

y x a b x y x a x y ax x

b y x y x x x

a y x b x a

y x x y x x

b b

2 2 1 2

2 2

2

1 2 1

2

2 4 1 2 2 165

2 2 2

2 2

807 0 40 2 2 0 1 0 4 33

2 807 0 4 33 2 0 40 2

40 2 2 6 34 2

= + − ⇒ = − ⇒ = −

= − − − −

= − − =

= − − = − ⇒ =

⇒ = − ⇒ = −

2.31 (b) Plot y² vs. x³ on rectangular axes. Slope=m,Intcpt= −n

(c) 1 1 1 a 1

Plot vs. [rect. axes], slope = , intercept =

ln( 3) ln( 3) b b

a x x

y = +b b ⇒ y

− −

(d) 1

1 3 1

1 3

2

3

2

3

( ) ( )

( ) ( ) , ,

y a x

y x a

+ = − ⇒

+ −

Plot vs. [rect. axes] slope = intercept = 0

OR

2 1 3 3

1 3

2

ln( ) ln ln( )

ln( ) ln( )

ln

y a x

y x

a

+ = − − −

+ −

⇒ − −

Plot vs. [rect.] or (y + 1) vs. (x - 3) [log]

slope = 3

2, intercept =

(e) ln ln y a x b

y x y x

= +

Plot vs. [rect.] or vs. [semilog ], slope = a, intercept = b

(f)

Plot vs. [rect.] slope = a, intercept = b

log ( ) ( )

log ( ) ( )

10

2 2

10

2 2

xy a x y b

xy x y

= + +

+ ⇒

(g) Plot vs. [rect.] slope = , intercept =

OR b

Plot 1 vs. 1

[rect.] , slope = intercept = 1

1 1

2 2

2 2

y ax b x

x

y ax b x

y x a b

y ax b

x xy a

x xy x b a

= + ⇒ = + ⇒

= + ⇒ = + ⇒

,

,

2.32 (a) A plot of y vs. R is a line through ( R= 5 , y = 0 011. ) and ( R= 80 , y = 0169. ).

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18

0 20 40 60 80 100

R

y

y a R b a b

y R

= + = −

− = ×

= − × = ×

U V|

W|

^{⇒ = × + ×}

−

− −

− −

0 169 0 011

80 5 2 11 10

0 011 2 11 10 5 4 50 10

2 11 10 4 50 10

3

3 4

3 4

. .

.

. . .

. .

d ib g

(b) R=43⇒ =y

d

2 11 10. × ⁻³

ib g

43 +4 50. ×10⁻⁴ =0 092. kg H O kg₂ 1200 kg h 0 092 kg H O kg₂ 110 kg H O h ₂

b gb

^.

g

⁼

2.33 (a) ln ln ln

(ln ln ) / (ln ln ) (ln ln ) / (ln ln ) .

ln ln ln ln ( . ) ln( ) . . ^.

T a b T a

b T T

a T b a T

= + ⇒ = b

= − − = − − = −

= − = − − ⇒ = ⇒ = ⁻

φ φ

φ φ

φ φ

2 1 2 1

1 19

120 210 40 25 119

210 119 25 9677 6 9677 6

(b) T T

T C

T C

T C

= ⇒ =

= ⇒ = =

= ⇒ = =

= ⇒ = =

9677 6 − 9677 6

85 9677 6 85 53 5

175 9677 6 175 29 1

290 9677 6 290 19 0

1 19 0.8403

0.8403

0.8403

0.8403

. . /

. / .

. / .

. / .

φ . φ

φ φ

φ

b g

b g

b g

b g

o

o

o

L / s L / s

L / s

(c) The estimate for T=175°C is probably closest to the real value, because the value of

temperature is in the range of the data originally taken to fit the line. The value of T=290°C is probably the least likely to be correct, because it is farthest away from the date range.

2.34 (a) Yes, because when ln[(C_A−C_Ae) / (C_A0−C_Ae)]is plotted vs. t in rectangular coordinates, the plot is a straight line.

-2 -1.5 -1 -0.5 0

0 50 100 150 200

t (m in) ln ((CA-CAe)/(CA0-CAe))

Slope = -0.0093⇒k = 9.3 10 min× ^{-3 −1} (b)

3

0 0

(9.3 10 )(120) -2

-2

ln[( ) /( )] ( )

(0.1823 0.0495) 0.0495 9.300 10 g/L 9.300 10 g 30.5 gal 28.317 L

= / = 10.7 g

L 7.4805 gal

−

−

− ×

− − = − ⇒ = − +

= − + = ×

⇒ = × =

kt

A Ae A Ae A A Ae Ae

A

C C C C kt C C C e C

C e

C m V m CV 2.35 (a) ft and h , respectively^{3 -2}

(b) ln(V) vs. t² in rectangular coordinates, slope=2 and intercept= ln( .3 53 10× ⁻²); or V(logarithmic axis) vs. t² in semilog coordinates, slope=2, intercept= 3 53 10. × ⁻² (c) V (m³)=1 00 10. × ⁻³exp( .15 10× ⁻⁷t²)

2.36 PV^k = ⇒ =C P C V/ ^k ⇒lnP=lnC−klnV

lnP = -1.573(lnV ) + 12.736 6

6.5 7 7.5 8 8.5

2.5 3 3.5 4

lnV

lnP

slope (dimensionless)

Intercept = ln mm Hg cm^4.719

k

C C e

= − = − − =

= ⇒ = = × ⋅

( . ) .

. ^.736 .

1573 1573

12 736 ¹² 3 40 10⁵

2.37 (a) G G

G G K C

G G

G G K C G G

G G K m C

L L

m

L L

m

L

L

−

− = ⇒ −

− = ⇒ −

− = +

0

0 0

1 ln ln ln

ln (G0-G )/(G -GL)= 2 .4 8 3 5 ln C - 1 0 .0 4 5

-1 0 1 2 3

3 .5 4 4 .5 5 5 .5

ln C ln(G0-G)/(G-GL)

2.37 (cont’d) m

K_L K_L

= =

= − ⇒ = × ⁻

slope (dimensionless)

Intercept = ln ppm^-2.483

2 483

10 045 4 340 10 ⁵ .

. .

(b) C G

G G

= ⇒ − ×

× ₋ −⁻ = × ⁻ ⇒ = × ⁻

475 180 10

3 00 10 4 340 10 475 1806 10

3 3

5 2 3

.

. . ( ) ^.483 .

C=475 ppm is well beyond the range of the data.

2.38 (a) For runs 2, 3 and 4:

Z aV p Z a b V c p

a b c

a b c

a b c

b c

= ⇒ = + +

= + +

= + +

= + +

 ln ln ln  ln

ln( . ) ln ln( . ) ln( . ) ln( . ) ln ln( . ) ln( . ) ln( . ) ln ln( . ) ln( . )

3 5 1 02 9 1

2 58 1 02 11 2

3 72 1 75 11 2

b c

=

⇒ = −

⋅ 0 68

1 46 .

.

a = 86.7 volts kPa^1.46 / (L / s)^0.678

(b) When P is constant (runs 1 to 4), plot lnZ vs. lnV . Slope=b, Intercept= lna+clnp

lnZ = 0.5199lnV + 1.0035 0 0.5 1 1.5 2

-1 -0.5 0 0.5 1 1.5

lnV

lnZ

b

a c P

= =

+ =

slope Intercept = ln

0 52

1 0035 .

ln .

When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a+c Vln 

lnZ = -0.9972lnP + 3.4551 0

0.5 1 1.5 2

1.5 1.7 1.9 2.1 2.3

lnP

lnZ

c slope

a b V

= = − ⇒

+ =

0 997 1 0 3 4551

. .

ln  . Intercept = ln

Plot Z vs V P^{b c}. Slope=a (no intercept)

Z = 31.096V^bP^c 1 2 3 4 5 6 7

0.05 0.1 0.15 0.2

V^bP^c

Z

a=slope=311. volt kPa / (L / s)⋅ ^.52 The results in part (b) are more reliable, because more data were used to obtain them.

2.39 (a)

s n x y

s n x

s n x s

n y a s s s

s s

xy i i

i n

xx i

i n

x i

i n

y i

i n

xy x y

xx x

= = + + =

= = + + =

= = + + = = = + + =

= −

− = −

=

=

= =

∑

∑

∑ ∑

1 0 4 0 3 2 1 1 9 31 3 2 3 4 677

1 0 3 1 9 3 2 3 4 647

1 0 3 1 9 3 2 3 18 1

0 4 2 1 31 3 1867 4 677 18 1

1 2 1

2 2 2

1 1

2

[( . )( . ) ( . )( . ) ( . )( . )] / .

( . . . ) / .

( . . . ) / . ; ( . . . ) / .

. ( . )( .

b g

^{4 647 18867 0 936}

4 647 1867 4 677 18

4 647 18 0 182

0 936 0 182

2

2 2

)

. ( . ) .

( . )( . ) ( . )( . )

. ( . ) .

. .

− =

= −

− = −

− =

= +

b s s s s

s s

y x

xx y xy x

xx

b g

x

(b) a s

s^xy y x

xx

= = 4 677= ⇒ =

4 647. 1 0065 1 0065

. . .

y = 1.0065x y = 0.936x + 0.182

0 1 2 3 4

0 1 2 3 4

x

y

2.40 (a) 1/C vs. t. Slope= b, intercept=a

(b) b=slope = 0.477 L / g h⋅ ; a=Intercept = 0.082 L / g

1/C = 0.4771t + 0.0823 0

0.5 1 1.5 2 2.5 3

0 1 2 3 4 5 6

t

1/C

0 0.5 1 1.5 2

1 2 3 4 5

t

C

C C-fitted

(c) C a bt

t C a b

= + ⇒ + =

= − = − =

1 1 0 082 0 477 0 12 2

1 1 0 01 0 082 0 477 209 5

/ ( ) / [ . . ( )] .

( / ) / ( / . . ) / . .

g / L

h (d) t=0 and C=0.01 are out of the range of the experimental data.

(e) The concentration of the hazardous substance could be enough to cause damage to the biotic resources in the river; the treatment requires an extremely large period of time; some of the hazardous substances might remain in the tank instead of being converted; the decomposition products might not be harmless.

2.41 (a) and (c)

1 10

0.1 1 10 100

x

y

(b) y=ax^b⇒lny=lna+bln ; Slope = b, Intercept = ln x a

ln y = 0.1684ln x + 1.1258

0 0.5 1 1.5 2

-1 0 1 2 3 4 5

ln x

ln y

b

a a

= =

= ⇒ =

slope Intercept = ln

0 168

11258 3 08

.

. .

2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0 (b)

Lab 1 ln(1-Cp/Cao) = -0.0062t

-4 -3 -2 -1 0

0 200 400 600 800

t

ln(1-Cp/Cao)

^{Lab 2}

ln(1-Cp/Cao) = -0.0111t -6

-4 -2 0

0 100 200 300 400 500 600

t

ln(1-Cp/Cao)

k = 0 0062. s^-1 k = 0 0111. s^-1

^{Lab 3}

ln(1-Cp/Cao) = -0.0063t-6 -4

-2 0

0 200 400 600 800

t

ln(1-Cp/Cao)

Lab 4 ln(1-Cp/Cao)= -0.0064t-6

-4 -2 0

0 200 400 600 800

t

ln(1-Cp/Cao)

k = 0 0063. s^-1 k = 0 0064. s^-1

(c) Disregarding the value of k that is very different from the other three, k is estimated with the average of the calculated k’s. k= 0 0063. s^-1

(d) Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor.

2.43

y ax a d y ax d

da y ax x y x a x

a y x x

i i i

i n

i i

i n

i i

i n

i i i

i n

i i

n

i i i

n

i i

n

= ⇒ = = − ⇒ = = − ⇒ − =

⇒ =

= = = = =

= =

∑ ∑ ∑ ∑ ∑

∑ ∑

φ φ

( )

/

2 1

2

1 1 1

2 1

1

2 1

0 2 0

b g b g

2.44 DIMENSION X(100), Y(100) READ (5, 1) N

C N = NUMBER OF DATA POINTS

1FORMAT (I10)

READ (5, 2) (X(J), Y(J), J = 1, N 2FORMAT (8F 10.2)

SX = 0.0 SY = 0.0 SXX = 0.0 SXY = 0.0 DO 100J = 1, N SX = SX + X(J) SY = SY + Y(J)

SXX = SXX + X(J) ** 2 100SXY = SXY + X(J) * Y(J)

AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN

CALCULATE SLOPE AND INTERCEPT A = (SXY - SX * SY)/(SXX - SX ** 2) B = SY - A * SX

WRITE (6, 3)

3FORMAT (1H1, 20X 'PROBLEM 2-39'/) WRITE (6, 4) A, B

4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/) C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF

RESIDUALS SSQ = 0.0

DO 200J = 1, N YC = A * X(J) + B RES = Y(J) - YC

WRITE (6, 5) X(J), Y(J), YC, RES

5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X

* 'RESIDUALb =', F6.3) 200SSQ = SSQ + RES ** 2

WRITE (6, 6) SSQ

6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3) STOP

END

$DATA

5

1.0 2.35 1.5 5.53 2.0 8.92 2.5 12.15

3.0 15.38

SOLUTION: a=6 536. ,b= −4 206.

2.45 (a) E(cal/mol), D0 (cm²/s)

(b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0.

(c) Intercept = ln = -3.0151D₀ ⇒D₀= 0.05 cm / s² .

Slope =−E R/ = -3666 K⇒E= (3666 K)(1.987 cal / mol K) = 7284 cal / mol⋅

ln D = -3666(1/T) - 3.0151

-14.0 -13.0 -12.0 -11.0

-10.0 ^{2.0E-03 2.1E-03 2.2E-03 2.3E-03 2.4E-03 2.5E-03 2.6E-03 2.7E-03 2.8E-03 2.9E-03 3.0E-03}

1/T

ln D

(d) Spreadsheet

T D 1/T lnD (1/T)*(lnD) (1/T)**2 347 1.34E-06 2.88E-03 -13.5 -0.03897 8.31E-06 374.2 2.50E-06 2.67E-03 -12.9 -0.03447 7.14E-06 396.2 4.55E-06 2.52E-03 -12.3 -0.03105 6.37E-06 420.7 8.52E-06 2.38E-03 -11.7 -0.02775 5.65E-06 447.7 1.41E-05 2.23E-03 -11.2 -0.02495 4.99E-06 471.2 2.00E-05 2.12E-03 -10.8 -0.02296 4.50E-06

Sx 2.47E-03 Sy -12.1 Syx -3.00E-02 Sxx 6.16E-06 -E/R -3666 ln D₀ -3.0151

D₀ 7284

E 0.05

CHAPTER THREE

3.1 (a) m= × ×

≈ × ≈ ×

16 6 2 1000

2 10 5 2 10³ 2 10⁵

m kg

m kg

3

3

b gb gb gd i

(b) m= ≈ ×

× ≈ ×

8 10

2 32

4 10

3 10 10 1 10

6 6

3

oz 1 qt cm 1 g 2

s oz 1056.68 qt cm g / s

3

3

b gd i

(c) Weight of a boxer ≈ 220 lb_m W_max≥ ×

12 220 ≈

220 stones lb 1 stone

14 lb

m

m

(d) dictionary

V = D L=

≈ × × × × × × ×

× × ≈ ×

π ^{2 2}

2 3

7

4

314 4 5 4

3 4 5 8 10 5 10 7

4 4 10 1 10

. . ft 800 miles 5880 ft 7.4805 gal 1 barrel 1 mile 1 ft 42 gal

barrels

2

3

d i d i

(e) (i)V ≈ × ×

≈ × × ≈ ×

6 ft 1 ft 0.5 ft 28,317 cm 3 3 10⁴ 1 10⁵

1 ft cm

3 3

3

(ii)V ≈ ≈ × ×

150 28 317 150 3 10 ≈ ×

60 1 10

4

lb 1 ft cm 5

62.4 lb 1 ft cm

m

3 3

m

3

, 3

(f) SG≈ 105.

3.2 (a) (i) 995 1 0 028317

0 45359 1 62 12

kg lb m

m ^mkg ft lb / ft

3

3 3 m

. 3

. = .

(ii) 995 62 43

1000 62 12

kg / m lb / ft

kg / m lb / ft

3

m 3

3 m

. 3

= .

(b) ρ ρ= _{H O}×SG= × =

2 62 43. lb / ft_m ³ 5 7. 360 lb / ft_m ³ 3.3 (a) 50

10₃ 35 L 0.70 10 kg 1 m

m L kg

3 3

3

× =

(b) 1150

1 60 27

kg m 1000 L 1 min

0.7 1000 kg m s L s

3

min × 3 =

(c) 10 1 0 70 62 43

2 gal 7 481 ft 1 lb 29

min gal ft lb / min

3

m

3 m

. .

.

× ≅

3.3 (cont’d)

(d) Assuming that 1 cm³kerosene was mixed with V_g (cm³) gasoline V_g

d

cm gasoline³

i

^{⇒ 0 70.} V_g

d

g gasoline

i

1

d

cm kerosene³

i

^{⇒ .}0 82

d

g kerosene

i

SG V

V ^g V

g

= + g

+ = ⇒ = −

− =

0 70 0 82

1 0 78 0 82 0 78

0 78 0 70 0 5

. .

. . .

. . .

d id i

d

^{cm blend3 g blend}

i

^{0 cm3}

Volumetric ratio cm

cm cm gasoline / cm kerosene

gasoline kerosene

3 3

3 3

=

V

= =

V

0 50

1 0 50

. .

3.4 In France: 50 0 5

0 7 1 0 1 5 22 42

. $1

. . . $68.

kg L Fr

kg L Fr

× =

In U.S.: 50 0 1 20

0 70 1 0 3 7854 1 64

. $1.

. . . $22.

kg L gal

kg L gal

× =

3.5

(a) V = .

× =

700 lb ft 1319 h ^m 0.850 62.43 lb ft / h

3

m

3

 

. 

m V

B

= B ×

=

3

m

3 B

ft 0.879 62.43 lb

h

d i

ft V kg / h

b g

^{54 88}

b g

  . . . 

m_H =

d hb

V_H ^{0 659}×^{62 43}

g

=⁴¹¹⁴V_H ^{kg / h}

b g

  . /

V_B +V_H = 1319 ft h³

  .  . 

m_B +m_H =54 88V_B +4114V_H =700 lb_m

⇒ V_B =11 4 ft. ³/h ⇒ m_B =628 lb_m/h benzene

 . /  . /

V_H =1 74 ft³ h ⇒ m_H =71 6 lb_m h hexane (b) – No buildup of mass in unit.

– ρ_B and ρ_H at inlet stream conditions are equal to their tabulated values (which are strictly valid at 20^oC and 1 atm.)

– Volumes of benzene and hexane are additive.

– Densitometer gives correct reading.

 ( ) ,  ( )

VH f t / h³ mH lbm / h

 ( ),  ( )

V_B ft / h³ m_B lb_m / h

700 lb / h

_m

( ), .

V ft / h³ SG= 0850

3.6 (a) V =

× =

195 5 1

0 35 1 2563 1 000 445

.

. . .

kg H SO kg solution L

kg H SO kg L

2 4

2 4

(b)

V_ideal ²

2 4 2

2 4

kg H SO L kg

kg H SO kg H O L

kg H SO kg L

= ×

+ =

195 5

1 8255 1 00

195 5 0 65

0 35 1 000 470

. 4

. .

. .

. .

% error=470−445× = .

445 100% 5 6%

3.7 Buoyant force up

b g

⁼

E

Weight of block down

b g

Mass of oil displaced + Mass of water displaced = Mass of block

ρ_oil ρ_{H O} ρ_c

2

0 542. 1 0 542.

b g

^V+

b

⁻

g

^{V = V}

From Table B.1: ρ_c =2 26. g / cm , ³ ρ_w=1 00. g / cm ³ ⇒ ρ_oil =3 325. g / cm³ m_oil =ρ_oil× =V 3 325. g / cm³×35 3. cm³=117 4. g

moil + flask =117 4. g+124 8. g=242g

3.8 Buoyant force up = Weight of block down

b g b g

⇒Wdisplaced liquid=Wblock⇒(ρVg)disp. Liq =(ρVg)block

Expt. 1: ρ_w 15A g ρ_B 2A g ρ_B ρ_w 15

. 2.

b g

⁼

b g

^{⇒ = ×}

ρ^w ρ_B SG _B

=¹ =0 75 ⇒ =0 75

.00

. .

g/cm

3

3

g / cm

b g

Expt. 2: ρ_soln

b g

A g⁼ρ_B

b g

²A g^⇒ρ_soln⁼²ρ_B ^=15. g / cm^3⇒

b g

SG _soln^=15. 3.9

W + W hs

A B

hb hρ1

Before object is jettisoned 1

1

Let ρ_w= density of water. Note: ρ_A>ρ_w (object sinks) Volume displaced: V_d1= A h_{b si} = A h_b

d

_p1−h_b1

i

⁽¹⁾

Archimedes ⇒ ρ_{w d}V g₁ =W_A+W_B weight of displaced water



Subst. (1) for V_{d 1}, solve for h

d

_p1−h_b1

i

h h W W

p b p gA

A B

w b

1− 1= +

(2)

Volume of pond water: V_w A h_{p p} V_d V A h A h h

i

w p p b p b

= ₁− ₁⇒

b g

= ₁− ₁− ₁

d i

for subst. 2

b h

w p p

A B

w

p w

p

A B

w p

p b

V A h W W

p g h V

A

W W

1 1 p gA

1 1

−

= − + ⇒ = + +

b g

(3)

2 solve for subst. 3 for in

b g

b g b g

, h

h b

w p

A B

w p b

b p

h V

A

W W

p g A A

1 1

1

1 1

= + +

L

−

N MM O

Q PP

⁽⁴⁾