Diving Myths & Realities Dive Physics   Solutions To Problem Set 1 by Larry "Harris" Taylor. Ph.D.   Diving Safety Coordinator University of Michigan Ann Arbor, Michigan

Physics Problem Set 1

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Part A: Concepts

1. air breathing; brain

2. fun

3. space; mass

4. solid; liquid; gas

5. solid; gas; liquid

6. atoms

7. element

8. compounds

9. mixtures

10. lightest; not used

11. deep; ease; less; hypothermia; HPNS

12. physiologically; narcosis; decompression sickness

13. beyond

14. very; increases

15. hypoxia; hyperoxia

16. Hyperoxia; CNS; grand-mal-like; not

17. waste product; breathing

18. CO2

19. must

21. not

22. 78; 21; 1

23. length; time; mass; force; energy

24. easier

25. metric; English

26. foot; meter

27. second

28. pounds; kilograms

29. inertia; force

30. will not; vary

31. mass; volume

32. units

33. units; in error

34. 1.00; 62.4

35. 1.0256; 64

36. force arrows

37. displacement; weight; volume

38. work

40. remove

41. positive; less

42. weight; volume

43. horizontal

44. four

45. positive; negative

46. work

48. potential; kinetic

49. push; pull; magnitude; direction

50. mass; distance; no

51. thermal; kinetic

52. 1; 1

53. 1; 1

54. store

55. low

56. thermal conductivity

57. hypothermia

58. warmer; colder

59. cannot

60. air

61. 25

64. Conduction; convection

65. is

66. 4; 39.2

67. lower

68. ROY G BIV

69. diffusion

70. Turbidity

71. refraction

72. reflected

73. is not

74. force; area

75. barometer

76. not; all of the atmosphere

77. shorter

78. 29.27; 760; 33; 34; 10.1; 10.3; 14.7; 1.01; 10

79. hydrostatic

80. absolute; ambient

81. pressure; depth

82. greater; greater; slower

83. increases

84. increases; increases

85. decreases

86. increases

87. decreases; increases

88. increases

89. near the surface

90. constant

91. 1; 28.3

92. increases

93. doubles

94. do not; must

95. zero; are not; mixing

96. sum

97. 2 ata; 3 ata

98. increases

99. rate

100. fun

Part B: Names

1. Archimedes

2. Joule-Thompson

3. Daniel Fahrenheit

4. Anders Celsius

5. Rankine

6. Kelvin

7. Issac Newton

8. Aristole

9. Evangesta Torricelli

10. Jacques Charles

11. Guillaume Amontons

12. Joseph Guy-Lussac

13. Sir Robert Boyle

14. John Dalton

15. William Henry

16. Lou Fead, The Easy Diver

Part C: Unit Conversions:

USE COURIER (or Other Non-proportional)  FONT TO MAINTAIN EQUATION ALIGNMENT

1.    16.5 ft  x     1 m  =  5.03 meter

3.28 ft

2.    10,000,000 m  x   3.28  ft  x      1 mi   =  6212.12 miles

m       5280 ft

3.     3 lbs  x   1 kg  x  1000 g    =  1,363.6 g

2.2 lb          kg

4.     56 l  x   1.06 qt   x   1 gal   =  14.8 gal

l         4 qt

5.     2400 l  x   1 ft3   =  84.8 ft3

28.3 l

6.     50 ft3   x  28.3 l   = 1415 l

ft3

7.     1 unit   x    60 min  x  24 hr  = 1440 units / day

min            hr         day

8.     100 fsw   x     1 m    =  30.5 m

3.28 ft

9.     78 oF + 460 = 538 oR

10.    100 oC + 273 = 373 K

Part D: Numerical Problems:

1. Density =    mass

volume

Substituting:

Density =    6 lbs

9 ft3

Density = 0.67 lbs/ft3

2. Density =   mass

volume

Substituting:

Density =     3 kg

16 l

Density =  0.19 kg/l

3. ratio submerged = ratio density

ratio =  50.6 lb/ft3  = 0.79  =>  79 % submerged

64  lb/ft3

4. fresh water density = 1.000 g/cc

density ratio = 0.678 g/cc = 0.678  =>  68 % submerged

1.000 g/cc

5. Determine equivalent weight of sea water displaced:

2.8 ft3  x    64 lbs   =  179.2 lbs

ft3

Determine equivalent weight of fresh water displaced:

2.8 ft3  x   62.4 lbs  =  174.7 lbs

ft3

Apply "force arrows"  for sea water:

weight of displaced water:   179

weight of diver:             157

net force on diver:           22

Since the net force is upward, the diver needs 22 lbs to compensate.

Apply "force arrows" for fresh water:

weight of water displaced:  175

weight of dive:             157

net Force:                   18

Since net force is upward, the diver needs 18 lbs to compensate.

6. Determine weight of equivalent volume of sea water:

80 l  x  1.0256 kg   =   82 kg

l

Determine weight of equivalent volume of fresh water:

80 l  x   1.000 kg   =   80 kg

l

Apply "force arrows" for sea water:

weight of displaced water:  82 kg

weight of diver:            70 kg

net force:                  12 kg

Since net force is upward, the diver needs 12 kg to compensate.

Apply "force arrows"  for fresh water:

weight of displaced water:  80 kg

weight of diver:            70 kg

net force on diver:         10 kg

Since net force is upward, the diver needs 10 kg to compensate

7. Determine volume based on weight needed to be neutral in sea water:

Total weight: 196 lbs + 22 lbs = 218 lbs

Converting to sea water volume:

218 lbs  x    ft3  = 3.4 ft3

64 lbs

Buoyant force from this volume of fresh water:

3.4 ft3   x     62.4 lbs  =  212.2 lbs

ft3

Determine net force using "force arrows"

weight of displaced water:  212 lbs

weight of diver:            196 lbs

Net Force:                   16 lbs

Since net force is upward, the diver needs 16 lbs of lead.

8. Downward force of diver:

73 kg + 6 kg = 79 kg

Convert this weight into volume of displaced sea water:

79 kg  x   1 l     = 77 l

1.0256 kg

Determine buoyant force of this volume of fresh water:

77 l  x  1.0 kg  =   77 kg

l

Apply "force arrows" to determine buoyancy:

weight of displaced water:  77 kg

weight of diver:            73 kg

net force:                   4 kg

Since net force is upward, the diver needs 4 kg to compensate.

9. Hydrostatic =    46  fsw  x    1        = 1.39 atm

33 fsw/atm

1.39 atm  x  14.7 psi = 20.43 psi

atm

Absolute = ambient + 1

1.39 atm + 1 atm =  2.39 ata

20.43 psi + 14.7 psi = 35.13 psia

10. Hydrostatic =     12 msw       = 1.19 atm

10.1 msw/atm

12 m  x    1 bar  = 1.2 bar

10 m

Absolute = ambient + 1

1.19 atm + 1 atm    = 2.19 atm

1.2 bar + 1.01 bar = 2.21 bar

11. Converting temperature to absolute:

T1 = 78 oF + 460 = 538 oR

T2 = 40 oF + 460 = 500 o

Using Charles' Law:

V1   V2

T1      T2

Substituting:

36 ft3   =     V2

538 oR        500 oR

Solving:

V2 =   33.4 ft3

12. Converting temperature to absolute:

T1 = 25 oC + 273 = 298 K

T2 = 18 oC + 273 = 291 K

Using Charles' Law:

V1   V2

T1      T2

Substituting:

2000 l  =   V2

298 K      291 K

Solving:

V2  = 1953 l

13. Convert to absolute temperature:

T1 =  75 oF + 460 = 535 oR

T2 = 126 oF + 460 = 586 o

Determine absolute pressure:

P1 = 3000 psig + 14.7 psi = 3014.7 psia

Using Guy-Lussac's Law:

P1   =   P2

T1       T2

Substituting:

3014.7 psia   =   P2

535 oR         586 oR

Solving:

P2 = 3302.1 psia

Converting to gauge pressure:

3302.1 psia - 14.7 psi = 3287.4 psig

14. Converting to absolute temperature:

T1 = 25 oC + 273 = 298 K

T2 = 42 oC + 273 = 315 K

Determine absolute pressure of cylinder:

P1 = 200 bar + 1.01 bar = 201.01 bar

Using Guy-Lussac's Law:

P1   =   P2

T1       T2

Substituting:

201.01 bar   =    P2

298 K         315 K

Solving:

P2 = 212.48 bar

Converting to gauge:

212.48 bar - 1.01 = 211.47 bar

15. Determine hydrostatic pressure at depth:

48 fsw        =    1.5 atm

33 fsw/atm

Determine absolute pressure at depth:

Absolute = hydrostatic + atmospheric = 1.5 atm + 1 atm = 2.5 ata

Using Boyle's Law:

P1 V1 = P2 V2

Substituting:

(1 ata) (80 ft3) = (2.5 ata) V2

Solving:

V2 = 32 ft3

16. Determine hydrostatic pressure at depth:

18 msw x   1 bar   =  1.8 bar

10 msw

Determine absolute pressure at depth:

Absolute = hydrostatic + atmospheric

1.8 bar + 1.01 bar = 2.81 bar

Using Boyle's Law:

P1 V1 = P2 V2

(1.01 bar) (2000 l) = (2.81 bar) V2

Solving:

V2 =  718.9 l

17. Convert to absolute temperature:

T1 = 78 oF + 460 = 538 oR

T2 = 50 oF + 460 = 510 oR

Determine hydrostatic pressure at depth:

58 ffw     =  1.7 atm

34 ffw/atm

Absolute pressure at depth =  1.7 atm + 1.0 atm = 2.7 ata

Using General Gas Law:

P1 V1 =  P2 V2

T1        T2

Substituting:

(1 ata) 71.2 ft3 =  (2.7 ata) V2

538 R                510 R

Solving:

V2 =  25.0 ft3

18. Convert to absolute temperature:

T1 = 25 oC + 273 = 298 K

T2 =  5 oC + 273 = 278 K

Determine hydrostatic pressure at depth:

22 mfw  x      atm  x  1.01 bar  =  2.16 bar

10.3 mfw          atm

Determine absolute pressure at depth:

2.16 bar + 1.01 bar = 3.17 bar

Using General Gas Law:

P1 V1 =  P2 V2

T1        T2

Substituting:

(1.01 bar) (2400 l) =  (3.17 bar) V2

298 K                278 K

Solving:

V2 =  713 l

19. Determine hydrostatic pressures:

84 fsw      =  2.5 atm

33 fsw/atm

18 fsw      =  0.5 atm

33 fsw/atm

Determine absolute pressures:

P1 = 2.5 atm + 1 atm = 3.5 ata

P2 = 0.5 atm + 1 atm = 1.5 ata

Using a "form" of Boyle's Law:  ("Duration" can be viewed as a "volume")

(Duration 1) (Pressure 1) = (Duration 2) (Pressure 2)

Substituting:

(22 min) (3.5 ata) = (Duration 2) (1.5 ata)

Solving:

Duration 2 = 51.3 min

20. Determine hydrostatic pressure at depth:

24 msw             =  2.4 atm

10.1 msw/atm

9 msw             =  0.9 atm

10.1 msw/atm

Determine absolute pressure at depth:

P1 = 2.4 atm + 1 atm = 3.4 ata

P2 = 0.9 atm + 1 atm = 1.9 ata

Using a "form" of Boyle's Law:

(Duration 1) (Pressure 1) = (Duration 2) (Pressure 2)

(30 min) (3.4 ata) = (Duration 2) (1.9 ata)

Solving:

Duration 2 = 53.7 min

21. Determine absolute pressure at depth:

38 ffw        = 1.12 atm

34 ffw/atm

Determine absolute pressure at depth:

1.12 atm + 1 atm = 2.12 ata

The absolute air consumption in terms of pressure:

125 psig               =   29.5 psig

2.12 ata- 2 minutes             ata-min

The SAC rate is at the surface (1 ata):

29.5 psig   x  1 ata   =   29.5  psig

ata-min                         min

22. Determine hydrostatic pressure at depth:

5 msw  x    1 atm       = 0.5 atm

10.1 msw

Determine absolute pressure at depth:

0.5 atm + 1 atm = 1.5 ata

The absolute air consumption:

60 bar         =  13.33   bar

(1.5 ata) 3 min             ata-min

The SAC rate:

13.33 bar   x 1 ata  =     13.33 bar

ata-min                         min

23. Determine hydrostatic pressure at depth:

45 fsw      =  1.4 atm

33 fsw/atm

Determine absolute pressure at depth:

1.4 atm + 1 atm =  2.4 ata

Determine allowed gauge pressure consumption:

2900 psig - 1000 psig = 1900 psig

Determine duration based on absolute air consumption:

1900 psig    x   ata-min    x     1       =    19.8 min

40 psig        2.4 ata

24. Determine air consumption allowed:

200 bar - 70 bar = 130 bar

Determine hydrostatic pressure at depth:

20 msw         =  1.98 atm

10.1 msw/atm

Determine absolute pressure:

1.98 atm + 1 atm = 2.98 ata

Determine duration using absolute air consumption factor:

130 bar  x    ata-min  x     1                =   21.8 min

2.0 bar        2.98 ata

25. Multiply absolute air consumption by cylinder parameters:

25 psig  x    71.55 ft3  =   0.72      ft3

ata-min     2475 psig            ata-min

26. Multiply absolute air consumption by cylinder parameters:

2.0 bar    x    2000 l    =   20 l

ata-min         200 bar           ata-min

27. Determine hydrostatic pressure at depth:

90 fsw       = 2.7 atm

33 fsw/atm

Determine absolute pressure at depth:

2.7 atm + 1 atm = 3.7 ata

Use volume consumption factor to determine air consumption needs:

0.72 ft3  x  3.7 ata  x  25 min  =  66.6 ft3

ata-min

Yes! Since the 80 ft3 cylinder contains more than 66.6 ft3 with ~30 % of air remaining to begin ascent, the dive can be made.

28. Determine hydrostatic pressure at depth.

24 msw         = 2.38 atm

10.1 msw/atm

Determine absolute pressure at depth:

2.38 atm + 1 atm = 3.38 ata

Use absolute volume factor to determine duration:

20 l        x 3.38 ata x 45 min = 3042 l

ata-min

No! Since consumption of the entire cylinder will deliver only 2000 l, a 45 minute dive with this air supply is not possible.

29. Determine hydrostatic pressure at depth:

36  ffw      =  1.06 atm

34 ffw/atm

Determine absolute pressure at depth:

1.06 atm + 1 atm = 2.06 ata

Use volume consumption factor to derive a duration:

ata - min    x  40 ft3  x      1       =   27 minutes

0.72 ft3                       2.06 ata

30. Determine hydrostatic pressure at depth:

18 mfw         =  1.75 atm

10.3 mfw/atm

Determine absolute pressure at depth:

1.75 atm + 1 = 2.75 ata

Use volume consumption factor to derive a duration:

ata-min     x   1200 l  x    1           =  21.8 minutes

20 l                       2.75 ata

31. Use volume factor; multiply by cylinder characteristics:

0.72 ft3    x   3000 psig    =     42.8 psig

ata-min        50.43 ft3                ata-min

32. Use volume factor; multiply by cylinder characteristics:

20 l       x   200 bar    =     2.0  bar

ata-min         2000 l                ata-min

33. The percentage of O2 at depth is 21%.

The partial pressure changes with depth,

the percentage of composition does not vary!

Using Dalton's Law:

p(component) = p(total) x fraction of component

pO2 =  4 ata x 0.21

pO2 =  0.84 ata

pN2 =  4 ata x 0.78

pN2 =  3.12 ata

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Physics Problem Set 1

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