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### Transcript of Felder -solucionario

• 2- 1

CHAPTER TWO

2.1 (a) 3 24 3600

118144 109

wk 7 d h s 1000 ms

1 wk 1 d h 1 s ms= .

(b) 38 3600

2598 26 0.

. .1 ft / s 0.0006214 mi s

3.2808 ft 1 h mi / h mi / h=

(c) 554 1 11000 g

3 m 1 d h kg 10 cm

d kg 24 h 60 min 1 m85 10 cm g

4 8 4

44 4

= . / min

2.2 (a) 760 mi

3600340

1 m 1 h

h 0.0006214 mi s m / s=

(b) 921 kg35.3145 ft

57 52.20462 lb 1 m

m 1 kg lb / ftm

3

3 3 m3= .

(c) 537 10 1000 J 11

119 93 1203.

.

= 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 2 2 2 in in in cube. For a

classroom with dimensions 40 40 15 ft ft ft :

nballs3 3

6 ft (12) in 1 ballft in

10 5 million balls= = 40 40 152

5183

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 steps5 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 00006214 1000264 17

44 7

50025 1

14 500 0 04464

70025 1

21700 002796

km 1000 m mi L1 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

xx

xx

x

• 2- 2

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) 250 250.

. lb 32.1714 ft / s 1 lb

32.1714 lb ft / s lbm

2f

m2 f

=

(b) 25 2 55 2 6 N 1 1 kg m / s

9.8066 m / s 1 N kg kg

2

2

= . .

(c) 10 1000 g 980.66 cm 1 9 109 ton 1 lb / s dyne

5 10 ton 2.20462 lb 1 g cm / s dynesm

2

-4m

2 =

2.9 50 15 2 853 32 1741

45 106

=

m 35.3145 ft lb ft 1 lb 1 m 1 ft s 32.174 lb ft s

lb3 3

m f3 3 2

m2 f

. ./

.

2.10 500 lb 5 1012

110

252m3

m

3 1 kg 1 m2.20462 lb 11.5 kg

m FHG

IKJFHG

IKJ

2.11

(a) m m V V h r H r

h

H

f f c c f c

cf

displaced fluid cylinder

33 cm cm g / cm

30 cm g / cm

= = =

= =

=

2 2

30 14 1 100053

( . )( . ).

(b)

fc Hh

= = =( )( . )

.30 053

171 cm g / cm

(30 cm - 20.7 cm) g / cm

33

H

hf

c

• 2- 3

2.12 V

R HV

R H r h RH

rh

rRH

h

VR H h Rh

HR

HhH

V VR

HhH

R H

H

HhH

HH h h

H

s f

f

f f s s f s

f s s s

= = = =

= FHGIKJ =

FHG

IKJ

= FHG

IKJ =

=

=

=

FHGIKJ

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

; ;

fs

R

rh

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

dA dy dx y dy

A m y dy

A y y y h h h

y

y

h

h

= =

=

= + = + +

+

zz

E

1

12

2 2

1

1

2 1

1

12 1

2

2

2 1

2 1

1 1 1 1 12

d i

d i b g b g b g Table of integrals or trigonometric substitution

m 2 sin sin

W NA

A

A

g g

b g =

=

E

4 0879 11 10 345 10

2

3 4

0

m m g 10 cm kg 9.81 Ncm m g kg

Substitute for

2 6

3 3

( ) ..{

W h h hNb g b g b g b g= + +LNMOQP

345 10 1 1 1 12

4 2 1. sin

2.14 1 1 32 174 1

11

32174

lb slug ft / s lb ft / s slug = 32.174 lb

poundal =1 lb ft / s lb

f2

m2

m

m2

f

= =

=

.

.

y= 1

y= 1+h

dA

• 2- 4

2.14(contd)

(a) (i) On the earth:

M

W

= =

=

=

175 lb 1544

175 11

m

m

m2

m2

3

slug32.174 lb

slugs

lb 32.174 ft poundal s lb ft / s

5.63 10 poundals

.

(ii) On the moon

M

W

= =

=

=

175 lb 15 44

175 11

m

m

m2

m2

slug32.174 lb

slugs

lb 32.174 ft poundal 6 s lb ft / s

938 poundals

.

( ) /b F ma a F m= = =

=

355 pound 1 1als lb ft / s 1 slug m 25.0 slugs 1 poundal 32.174 lb 3.2808 ft

0.135 m / s

m2

m

2

2.15 (a) F ma= FHGIKJ =

116

53623

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 2W

W

=

=

=

3

3 32174 5 3623( )( . ) / .

2.16 (a) =

=

( )( )

( . )( . )

3 9 27

2 7 8 632 23

(b) 4

5

4 6

4.0 101 10

40(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 36

3

(7 10 )(3 10 )(6)(5 10 )42 10 4 10

(3)(5 10 )

3812.5 3810 3.81 10exact

R

R

=

(Any digit in range 2-6 is acceptable)

• 2- 5

2.18 (a) A: C

C

C

o

o

o

R

X

s

= =

= + + + + =

= + + + +

=

731 72 4 0 7

72 4 731 72 6 728 73 05

72 8

72 4 728 731 72 8 72 6 728 72 8 728 730 72 85 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 987 1031 100 45

100 2

973 1002 1014 1002 987 1002 1031 1002 1004 10025 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)

XX

sX

X s

X s

ii i= = =

=

= = =

= + = + =

= =

1

122

1

12

12735

735

12 112

2 735 2 12 711

2 735 2 12 759

.( . )

.

. ( . ) .

. ( . ) .

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- 6

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) Q'.

= 2 36 10 4 kg m 2.10462 lb 3.2808 ft 1 h h kg m 3600 s

2 2 2

2

(b) Q

Q

'( )( )( )

. /

' . / /

(approximate

2

exact2 2

lb ft s

= lb ft s 0.00000148 lb ft s

=

2 10 2 93 10

12 10 12 10

148 10

4

34 3) 6

6

2.22 NC

kCC

N

po

oPr

Pr

. ..

( )( )( )( )( )( )

. . .

= =

0583 1936 3 28080 286

6 10 2 10 3 103 10 4 10 2

3 102

15 10 163 101 3 3

1 3

33 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( )( )( )( )( )( )( )( )

(

= =

Du

048 2 067 0 8050 43 10

5 10 2 8 10 103 4 10 10 4 10

5 103

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

(a) R