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 

Go To Site Page:       Home     About "Harris"     Articles     War Stories     Biblios     Editorials     Links     Site Map     Fini

 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

                                  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

  Top

Physics Problem Set 1

Go To Site Page:       Home     About "Harris"     Articles     War Stories     Biblios     Editorials     Links     Site Map     Fini

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.

  Copyright 2001-2004 by Larry "Harris" Taylor

All rights reserved.

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

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