Dive Physics: Solutions to Problem Set 1
By
Larry “Harris” Taylor, Ph.D.
. This material is copyrighted and all rights retained by the author. This article is made
available as a service to the diving community by the author and the Occupational Safety
and Environmental Health Department of the University of Michigan. This article may be
distributed for any non-commercial or Not-For-Profit use.
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
2
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
3
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
4
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
5
Part C: Unit Conversions:
1.
16.5 ft
x
2.
10,000,000 m
x
1 m =
3.28 ft
x
3.28
5.03 meter
ft
m
x
3.
3 lbs
4.
56 l
5.
2400 l
x
1 ft3
28.3 l
6.
50 ft3
x
28.3 l
= 1415 l
ft3
7.
1 unit
min
8.
100 fsw
9.
78 oF + 460 = 538 oR
10.
100 oC + 273 = 373 K
x
1 kg x
2.2 lb
1000 g
kg
1.06 qt
l
x
x
x
=
60 min
hr
1 m
3.28 ft
1 gal
qt
1 mi
5280 ft
=
=
6212.12 miles
1,363.6 g
=
14.8 gal
84.8 ft3
x
=
24 hr = 1440 units / day
day
30.5 m
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 =
Density =
3 kg
16 l
0.19 kg/l
6
3.
ratio submerged = ratio density
ratio =
4.
50.6 lb/ft3
64 lb/ft3
= 0.79
=>
79 % submerged
fresh water density = 1.000 g/cc
density ratio = 0.678 g/cc = 0.678
1.000 g/cc
5.
=>
68 % submerged
Determine equivalent weight of sea water displaced:
2.8 ft3 x
64 lbs
ft3
=
179.2 lbs
Determine equivalent weight of fresh water displaced:
2.8 ft3
x
62.4 lbs
ft3
Apply "force arrows"
=
174.7 lbs
for sea water:
weight of displaced water:
weight of diver:
net force on diver:
179 ↑
157 ↓
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 /l
=
82 kg
Determine weight of equivalent volume of fresh water:
80 l
x
1.000 kg / l
=
80 kg
Apply "force arrows" for sea water:
weight of displaced water:
weight of diver:
net force:
82 kg ↑
70 kg ↓
12 kg ↑
Since net force is upward, the diver needs 12 kg to compensate.
7
Apply "force arrows" for fresh water:
weight of displaced water:
weight of diver:
net force on diver:
80 kg ↑
70 kg ↓
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
ft3
=
212.2 lbs
Determine net force using "force arrows"
weight of displaced water:
weight of diver:
Net Force:
212 lbs ↑
196 lbs ↓
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
l
=
77 kg
Apply "force arrows" to determine buoyancy:
weight of displaced water:
weight of diver:
net force:
77 kg ↑
73 kg ↓
4 kg ↑
Since net force is upward, the diver needs 4 kg to compensate.
8
9.
Hydrostatic =
1.39 atm
x
46
fsw x
1
33 fsw/atm
= 1.39 atm
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 m
x
12 msw
= 1.19 atm
10.1 msw/atm
1 bar
10 m
= 1.2 bar
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
T1
=
V2
T2
Substituting:
36 ft3
538 oR
=
V2
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
T1
=
V2
T2
9
Substituting:
2000 l
298 K
=
V2
291 K
Solving:
V2
13.
= 1953 l
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
T1
=
P2
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
T1
=
P2
T2
10
Substituting:
201.01 bar
298 K
=
P2
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
33 fsw/atm
= 1.5 atm
Determine absolute pressure at depth:
Absolute = hydrostatic + atmospheric pressure
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
10 msw
=
1.8 bar
Determine absolute pressure at depth:
Absolute = hydrostatic + atmospheric
1.8 bar + 1.01 bar = 2.81 bar
11
Using Boyle's Law:
P1 V1 = P2 V2
(1.01 bar) (2000 l) = (2.81 bar) V2
Solving:
V2 =
17.
718.9 l
Convert to absolute temperature:
T1 = 78 oF + 460 = 538 oR
T2 = 50 oF + 460 = 510 oR
Determine hydrostatic pressure at depth:
58 ffw
=
34 ffw/atm
1.7 atm
Absolute pressure at depth
1.7 atm + 1.0 atm = 2.7 ata
Using General Gas Law:
P1 V1 =
T1
P2 V2
T2
Substituting:
(1 ata) 71.2 ft3 =
538 R
(2.7 ata) V2
510 R
Solving:
V2 =
18.
25.0 ft3
Convert to absolute temperature:
T1 = 25 oC + 273 = 298 K
T2 = 5 oC + 273 = 278 K
Determine hydrostatic pressure at depth:
22 mfw
x
1 atm
10.3 mfw
x
1.01 bar
atm
=
2.16 bar
Determine absolute pressure at depth:
2.16 bar + 1.01 bar = 3.17 bar
12
Using General Gas Law:
P1 V1 =
T1
P2 V2
T2
Substituting:
(1.01 bar) (2400 l) =
298 K
(3.17 bar) V2
278 K
Solving:
V2 =
19.
713 l
Determine hydrostatic pressures:
84 fsw
33 fsw/atm
=
2.5 atm
18 fsw
33 fsw/atm
=
0.5 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
10.1 msw/atm
=
2.4 atm
9 msw
10.1 msw/atm
=
0.9 atm
13
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
34 ffw/atm
= 1.12 atm
Determine absolute pressure at depth:
1.12 atm + 1 atm = 2.12 ata
The absolute air consumption in terms of pressure:
125 psig
2.12 ata- 2 minutes
=
29.5 psig
ata-min
The SAC rate is at the surface (1 ata):
29.5 psig
ata-min
22.
x
1 ata
=
29.5
psig
min
Determine hydrostatic pressure at depth:
5 msw
x
1 atm
10.1 msw
= 0.5 atm
Determine absolute pressure at depth:
0.5 atm + 1 atm = 1.5 ata
The absolute air consumption:
60 bar
(1.5 ata) 3 min
= 13.33
bar
ata-min
The SAC rate:
13.33 bar
ata-min
x 1 ata
=
13.33 bar
min
14
23.
Determine hydrostatic pressure at depth:
45 fsw
33 fsw/atm
=
1.4 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
24.
x
ata-min
40 psig
x
1
=
2.4 ata
19.8 min
Determine air consumption allowed:
200 bar - 70 bar = 130 bar
Determine hydrostatic pressure at depth:
20 msw
10.1 msw/atm
=
1.98 atm
Determine absolute pressure:
1.98 atm + 1 atm = 2.98 ata
Determine duration using absolute air consumption factor:
130 bar
25.
x
x
1 =
2.98 ata
21.8 min
71.55 ft3 =
2475 psig
0.72
ft3
ata-min
Multiply absolute air consumption by cylinder parameters:
2.0 bar
ata-min
27.
ata-min
2.0 bar
Multiply absolute air consumption by cylinder parameters:
25 psig
ata-min
26.
x
x
2000 l
200 bar
=
20
l
ata-min
Determine hydrostatic pressure at depth:
90 fsw
33 fsw/atm
= 2.7 atm
15
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
ata-min
3.7 ata
x
25 min =
66.6 ft3
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
10.1 msw/atm
= 2.38 atm
Determine absolute pressure at depth:
2.38 atm + 1 atm = 3.38 ata
Use absolute volume factor to determine duration:
20
l
ata-min
x 3.38 ata x 45 min = 3042 l
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
34 ffw/atm
=
1.06 atm
Determine absolute pressure at depth:
1.06 atm + 1 atm = 2.06 ata
Use volume consumption factor to derive a duration:
ata - min
0.72 ft3
30.
x
40 ft3
x
1
=
2.06 ata
27 minutes
Determine hydrostatic pressure at depth:
18 mfw
10.3 mfw/atm
=
1.75 atm
Determine absolute pressure at depth:
1.75 atm + 1 = 2.75 ata
16
Use volume consumption factor to derive a duration:
ata-min
20 l
31.
x
1
2.75 ata
=
21.8 minutes
x
3000 psig
50.43 ft3
=
42.8 psig
ata-min
Use volume factor; multiply by cylinder characteristics:
20 l
ata-min
33.
1200 l
Use volume factor; multiply by cylinder characteristics:
0.72 ft3
ata-min
32.
x
x
200 bar
2000 l
=
2.0
bar
ata-min
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
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
100 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.
17