Problem Set 1 Answers

Diving Myths & Realities

Dive Physics

Solutions To

Problem Set 1

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

20. decrease; adiabatic

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

39. add

40. remove

41. positive; less

42. weight; volume

43. horizontal

44. four

45. positive; negative

46. work

47. radiant; chemical

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

62. radiation; conduction; convection; evaporation

63. radiation

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

Determine weight of equivalent volume of fresh water:

80 l x 1.000 kg = 80 kg

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

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 ^oR

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 ^oR

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

Top

Physics Problem Set 1

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About The Author:

Larry "Harris" Taylor, Ph.D. is a biochemist and Diving Safety Coordinator at the University of Michigan. He has authored more than 200 scuba related articles. His personal dive library (See Alert Diver, Mar/Apr, 1997, p. 54) is considered one of the best recreational sources of information In North America.

Use of these articles for personal or organizational profit is specifically denied.

These articles may be used for not-for-profit diving education