 Diving Myths & Realities Dive Physics   Solutions To Problem Set 2 Larry "Harris" Taylor. Ph.D. Diving Safety Coordinator University of Michigan Ann Arbor, Michigan Divegeek Content Page:   Home     About "Harris"     Articles      Slides     War Stories     Editorials        Fini

Physics Problem Set  2

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1.  Rather than memorize hundreds of conversions, it is best to memorize only a few and make all conversions as a long string of conversions to change from given to final:

a.     100 m    x    60 sec     x    100 cm    x        1 in        x        1 ft   x         1 mi      =  23.3 mi/hr

9.6 sec             min                  m             2.54 cm         12 in            5,280 ft

b.      1  mi      x   60 min     =    15 mi/hr

4 min             hr

c.     400 yd    x   60 min    x   3 ft    x          1 mi        =  1.1 mi/hr

12 min         hr                yd             5280 ft

2.       100 ft    x    12  in    x   2.54 cm     x        1 m        =  0.33 m/sec

92 sec          1 ft                1 in            100 cm

100 ft    x   60 sec     x   60 min     x    1 mi          =  0.74 mi/hr

92 sec      min              hr                5,280 ft

Measuring time to travel a short distance is a common method for determining current velocity.

3. A camera focus is set at the apparent distance (actual distance altered by the refraction of light). The refraction of light underwater results in the water acting as a lens that makes everything appear 75  % closer. So, the focus is set at:

0.75   x   6 ft   = 4.5 ft

4.  First, convert units to area to make the comparison

Jox   =  30 ft   x   60 ft = 1800 ft2

Field  =   132 ft  x  300 ft = 34,600 ft

Next, determine ratio  ... looking for Jox per football field

Jox         x    34,600  ft2      =  22 Jox / field

1800 ft2                 field

This is a bad pun. In the United States, the slang term for an athlete is "jock," which, in English, sounds like "jox."  In addition, a US football team is 11 players, so a game of American football has 2 teams of 11 players each, or 22 jocks on the field.

5. First, determine the funds needed

5 ibs   x   12  oz     x    \$35.00     x      8   =   \$16,800.00

lbs              oz

Next, determine amount of time need to earn this amount of money

\$16,800.00    x    100 cents      x                    hr        =  112,000 hrs

dollar                        15 cents

Convert to years

112,00 hrs     x        1 day     x       1 year        =   12.79 years

24 hrs          365 days

6. Dew point is the temperature at which the saturated water vapor in the air must condense. Dew point is commonly reported in weather reports as an indicator of the temperature at which fog will form. Since the 52 F is below the reported dew point of 56 Fo, the warm, saturated air from your breath will be visible as soon as it cools to below dew point.

7. a. The temperature difference between surface and altitude:

70o F  -  50o F  20o  F

So, the temperature difference corresponds to:

20o  F  x   1000 ft    =   4,500 ft

4.5o  F

b. For 2 degree drop every hour, the time involved for a 10 degree 68o F - 58o F)  drop:

1 hr     x   10o F   =  5 hr

2o F

Five hours from 9 pm:

9 +  5 = 14 ==>   2 am on the 12 hour clock; 02 on the 24 hour clock

Since 2 am is later than midnight, the eclipse should be visible.

8.    100 g   x   1.0  cal      x 10o C    =  1000 cal   (or 1 kcal)

g oC

9.  The meal calories total to

605 kcal + 214 kcal + 332 kcal = 1151 kcal

So, the time needed to "burn" 1151 kcals:

1151 kcal      x    1 min       =   127.9 min

9 kcal

10. The temperature difference needed to overcome the hypothermia state and restore normal body temperature:

98.6  oF   -   90.0  oF  =    8.6  oF

The heat content this represents:

60 kcal     x  8.6  oF   =  516 kcal

oF

The amount of water at 110 oF that holds this amount of heat:

516 kcal   x   1 liter     = 30.4 liters

17 kcal

Drinking approximately 30 liters to raise the temperature suggests that drinking warm water is not a viable hypothermia treatment.

11.  a. Heat capacity (Cp) is a measure of the amount of heat a particular substance can absorb at constant pressure. The less the heat capacity, the less the tendency to absorb heat from the surroundings and the better the substance will serve as an insulator    Assuming 100 g of gas is raised 10 degree Celsius.

For argon:

100 g   x     0.1252 cal     x   10 oC     =   125.2 cal

g oC

For Air:

100 g   x     0.3439 cal     x   10 oC     =   243.9 cal

g oC

For Helium

100 g   x     4.9680 cal     x   10 oC     =   4,968 cal

g oC

b. Carbon dioxide reacts with water from perspiration to form a skin irritant, carbonic acid. The skin irritation from the formed acid suggests that using carbon dioxide as heat insulator inside a dry suit is not a particularly pleasant technique. The chemical reaction involved is

CO  +   H2O   =   H2CO3

12. Density = Mass / Volume

Since the mass of a gas stays constant and the volume (Boyle's Law) decreases with depth, the density must increase.

Using Boyle's law with density (D) substituted for Volume:

P1D  =   P2D2

Gas laws require absolute pressure:

Pabs = Pambient  + P atmosphere

For the surface:

Pabs  = 0 + 1 ata = 1 ata

At depth  (1 atm corresponds to a pressure equivalent of 33 feet of sea water):

Pabs  = ( 78 fsw  x     1 atm  )   + 1 atm = 2.36  atm + 1  atm = 3.36 ata

33 fsw

So, the Boyle's Law expression becomes:

(1 ata) (Dsurface)  = (3.36 ata)(Dat depth)

Thus, gas is 3.36 times as dense at 78 fsw as it is on the surface.

Gas density contributes to breathing resistance and the work associated with breathing. This is a good reason to dive with top-of-the line regulators.

13.   The height of a column of a substance is inversely proportional to its density.

(Remembering that 1 atm is defined as 760 mm Hg)

For an air barometer:

0.0012 g/cm     =        760 mm

13,6 g/cm            height for air

height of air =   8.61 x 106 mm

Converting units

8.61 x 106 mm   x            1 m    x     100 cm    x             1 in     x         1 ft     x              1 mi     =    5.35 miles

1000 mm            1 m                2.54 cm            12 in                5,280 ft

So, most of the weight of the earth's atmosphere is in the volume between the surface and 5.35 miles above the surface.

14.  Depth gauges measure pressure, so depth readings must first be converted to a pressure.

For

seawater:     1 atm = 33 fsw

fresh water:  1 atm = 34 ffw

So, ambient pressure of 100 fsw:

100 fsw x       1 atm      =  3.03 atm

33 fsw

Thus, the fresh water gauge will read a water pressure equivalent to 3.03 ata

3.03 atm   x        34 ffw       =  103 ffw

1 atm

Fresh water has no dissolved salts, so it will take a greater depth to reach an equivalent salt water pressure. Since diver depth gauges can be calibrated in either ffw or fsw, it is wise to check your gauges and dive the depth gauge calibrated to your diving environment.

15.  Although there are faster ways to solve this problem (e.g. memorizing a specific formula), I prefer to take the solution in several logical steps so that,  regardless of a particular situation, understanding of what is known, what is unknown and  how  I convert the units of known to the unknown will direct my solution.

1. Water depth for atmospheric pressure at this altitude:

Depth of a 1 atmosphere column of water will be inversely proportional to specific gravity: Converting to feet fresh water The water pressure at 93 feet of fresh water This corresponds as an absolute pressure of:

3.81 atm + 1 atm = 4.81 atm total pressure (ata) at altitude

The pressure in mm Hg at this altitude would be: This hydrostatic pressure at sea level  (where 1 atm is 760 mm Hg) would be:

2626.3 mm Hg  -   760 mm Hg  =  1866.3 mm Hg

Converting to a sea level pressure: Converting sea level atm to salt water depth:

2.46 atm x 33 fsw / atm = 81.2 fsw

1. The ocean equivalent depth would be:

3.82 atm  x  33 fsw / atm  = 126.1 fsw

1. The US Navy entry would be:

The 126.1 depth is rounded to the next greater 10 foot entry, so as divemaster, you recommend that Indy's dive be controlled by the 130 fsw entry allowing him 10 minutes to find and recover the headpiece.

Thus, Indy will have to dive to 93 feet of fresh water at an altitude near 9000 feet above sea level, his depth gauge will read 81 fsw and he will treat the dive as if his decompression schedule obligation was determined by a sea level dive of 130 fsw.

16. a.     On Earth

67 fsw x    1 atm     =  2.03 atm   ===>   3.03 ata

33 fsw

On Xandor

Determine equivalent water depth to equal 1 atm (680 mm Hg compared to 760 mm Hg on Earth)

680 mm Hg   =      1.47 g/cc

x mm H2O          13.47  g/cc

x = 6,231 mm

Converting to feet

6,231 mm   x       1 cm      x          1 in      x   1    foot      =  20.44 fsw

10 mm            2.54 cm            12 in

Determine absolute pressure of Xandor at 67 feet

67 fsw   x            1 atm     =  3.27 atm ===>  2.27 ata

20.44 fsw

In terms of mm Hg for Xandor

3.27 ata x 680 mm Hg      =   2,223.6 mmHg

ata

Converting to Earth Equivalent for absolute pressure

2,223.6 mm Hg    x      1      ata          =  2.92 ata   (which is 1.92 atm of water depth)

760 mm Hg

So, earth gauge reading will be

1.92 atm   x   33  fsw/atm =  63.5 ==> 64 fsw

c. Round to next highest 10 foot increment = 70 fsw schedule

d. on the bottom at 2700 psig and wishing to return at 1000 psig leaves 1700 psig for the dive.

1700 psig   x        ata-min      x                           =       14.97  min  ===> 15 minutes

50 psig                2.27 ata

e. The force diagram for earth

weight =  225 lbs buoyancy =     ? Net force = 12 lbs so, buoyant force on earth was 225 lbs - 12 lbs = 213 lbs This corresponds to a volume of sea water at

213 lbs   x          1 ft  =  3.33 ft

64 lbs

Determine buoyant force associated immersed Xandor volume of 3.33ft

3.33 ft x    (12 in)  x  (2.54 cm)3    x   1.47   g           1 lbs       =  305  lbs

ft                 1 in3                 cc                454 g

Set up forces diagram for Xandor

weight =      225 lbs buoyancy = 305 lbs Net force =   70 lbs So, to dive in Xandor,  the diver would need 70 lbs of lead to compensate for the buoyancy associated with diving in a dense sea

17.  Depth (in ata)  = Pressure (ata) / Fraction oxygen

For 1.6 ata

On Oxygen

D = 1.6 ata / 1

D = 1.6 ata

D (in water) = 0 .6 x 33 fsw/atm = 19.8 fsw

On  Air

D = (1.6 ata) / 0.21

D = 7.62 ata

D (in water)  =  6.62 atm  x  33 fsw/atm  = 218 fsw

For 1.4 ata

On Oxygen

D = (1.4 ata) / 1

D = 0.4 ata

D (in water)  =  0.4 atm  x  33 fsw/atm  = 13.2 fsw

On  Air

D = (1.4 ata) / 0.21

D = 6.67 ata

D (in water)  =  5.67 atm  x  33 fsw/atm  = 187 fsw

18. Need both absolute pressure and temperature for gas law related problems

Absolute Temps:

45 oF = 45 oF + 460 oF = 505 oR

78 oF = 78 oF + 460 oF = 538 oR

Absolute Pressure = gauge plus atmospheric; in psi:

Pabsolute  = 3000 psig + 14.7 psi = 3014  psi

Using law of Charles for determining pressure-volume at constant temp

P1 / T1 = P2 / T2

Substituting

3014 psi     =       P2

505 oR             538 oR

P2 = 3210.95 psi  (This is absolute pressure)

Converting to gauge pressure:

3210.95 psi -  14.7 psi = 3196.35

19. Using a Boyle's Law approach

Converting pressure to absolute:

99 fsw   x   1 atm    =  3 atm  ambient  ==> 4 ata

33 fsw

132 fsw x   1 atm     =  4 atm  ambient  ==> 5 ata

33 fsw

Substituting into Boyle's Law

(4 ata) (25 min)  =  (5 ata) )x min)

x = 20 minutes

20. a. Volume of cylinder is independent of gas content. So, the cylinder will hold the same amount of nitrox as air; 80 cubic feet at 3000 psig.

b. Working with absolute pressures:

Pabsolute1 = 3000 psi + 14.7 psi = 3014.7 psi:

Pabsolute2 = 2400 psi + 14.7 psi = 2414.7 psi

Now, determine volume available at 2400 psi on the gauge:

80 ft3   =         x

3014.7            2414.7

X = 64.1ft3

The volume of gas available from a fixed volume cylinder will be directly proportional to the absolute pressure.  Converting gauge pressure to absolute:

c. Convert gauge pressure to absolute:

1800 psig + 14.7 psig  = 1814.7 psia

3000 psig + 14.7 psig =  3014.7 psia

2475 psig + 14.7 psig =  2489.7 psia

For 80 cubic foot cylinder:

80 ft3   =    Volume

3014.7 psia          1814.7 psia

Volume = 48.15 ft3

For 71.6 cubic foot cylinder:

71.6 ft3    =    Volume

2489.7 psia          1814.7 psia

Volume = 52.18  ft3

21. a. The buoyant force of the object is equal to volume of sea water displaced = 4 cubic foot

4 ft3   x  64 lbs      =   256 lbs

ft3

Since the object weighs less than its buoyant force, it will float.

Density of the object = mass/volume:

Density =  96 lbs  /  4 ft3  =  24 lbs / ft3

Amount submerged is the ratio of the density of the object and the density of the fluid:

Submerged  =   24 lbs / ft3   =   37.5 %

64 lbs/ ft3

b. This requires balancing weight (downward force) with buoyancy (upwards force).

The total of weight (diver equipment plus lead) needed to hover in salt water is equivalent to the buoyant force = 198  lbs + 18 lbs = 216 lbs

To hover, the volume of water displaced by the diver must exert an upward buoyant force equal the total weight of the diver plus gear (downward force). This is the upward buoyant force exerted by the displaced volume of seawater (density = 64 lb./cubic foot) that weighs 216 pounds.

Determine Volume of diver: (Divers use weight and mass as equivalent terms)

Density =  Weight / Volume

Rearranging:

Volume =  Weight / Density

Volume =    216 lbs        =  3.38  ft

64 lb/ft3

The upward buoyant force from fresh water (density 62.4 lb./ cubic foot) that the diver would displace:

3.38  ft  x    62.4 lb     =   210.9  lbs   ===> 211 lbs

ft3

Applying force arrows:

Fresh water buoyant force =  211 lbs Diver weight                     =  198 lbs Net =                                    13 lbs So, the diver that was comfortable with eighteen pounds of lead on the weight belt in seawater must remove 5 pounds (for a total of 13 lb.) from the weight belt to dive in fresh water. The difference in density between fresh and seawater is the reason why different amounts of weight must be used when diving in different environments. When moving from fresh to seawater (with the same equipment configuration), divers must add weight. When moving from seawater to less dense fresh water, divers should remove weight.

Thirteen pounds cannot be worn symmetrically on the weight belt, so in this case, the diver would choose 14 pounds of lead to be a tiny bit heavy, but balanced.

22. a. Determine buoyancy force (weight of displaced water):

2.0  ft  x    64 lb     =    128  lbs

ft3

Determine amount of lift needed to raise the anchor

Weight of anchor = 524 lbs Buoyancy  =          128 lbs Net  Force =          396 lbs To lift the anchor, 396 pounds of lift must be supplied. This is equivalent to the a water volume of:

396 lbs  x      1  ft3    =   6.19 ft3

64 lbs

Determine absolute pressure at depth

66 fsw x     1 atm      = 2 atm   ===> 3 ata

33 fsw

Use Boyle's Law to determine surface equivalent

(1 ata) (Vsurface) =  (6.19  ft3 )(3 ata)

Vsurface  = 18.57  ft3

b.     Determine Volume of concrete block using definition of density:

150 lbs/ ft3 =  2460 lbs / Volume

Volume = 16.4 ft3

Determine Weight of displaced water (buoyant force):

16.4 ft3    x    64 lbs     =  1,049.6  ==>  1050  lbs

ft3

Forces diagram:

Weight of the block = 2460 lbs Buoyant force  =        1050 lbs Net Force =           1410 lbs So, need 1410 pounds of lift to raise the object

If each device can only supply 75% of capacity, then need to supply

1410 lbs    =  1880 lbs of lift

0.75

The number of 55 gal drums needed is:

1880 lbs       =   4.31  ==>   5 drums

436 lbs/drum

The volume of water needed to supply lifting force:

1880  lbs    x          ft3       =   29.4 ft3

64 lbs

The absolute pressure at depth

86 fsw   x       1  atm    =  2.6  atm  ===>  3.6 ata

33 fsw

The volume of at the surface needed to supply lifting volume

(1 atm) Vsurface)  =  (3.6 ata) (29.4 ft3)

Vsurface   = 105.8 ft3

The number of 80's needed

105.8 ft3        = 1.32  tanks  ==>  2 tanks

80 ft3 /tank

23. Since salt water is more dense than fresh water, an equal volume of sea water will weigh more (have more buoyant force). So, for the same immersed object, less lift will be needed in salt water than in fresh water. So, equipment that can manage a lift in fresh water should be able to accomplish same lift in sea water.

Volume  of the object:

V = pi r2 l

V  = (3.14)  (2.5 ft)2 (50 ft)

V = 981.8  ft3

Weight of the object using definition of density:

150 lbs/  ft3    =     weight /  981.8  ft3

weight = 147,270  lbs

Weight of water displaced in fresh water:

981.8  ft3   x   62.4  lbs/  ft3  = 61,264.32 lbs

Weight of water displaced in salt water:

981.8  ft3   x   64  lbs/  ft3  = 62,835.2 lbs

Force Needed to lift in fresh water:

Weight of object  = 147,270  lbs Buoyant force  =       61,264  lbs Net Force =      86,006 lbs ==>  Need at least 86,006 pounds of lift to overcome in-water weight

Force Needed to in salt water:

Weight of object  = 147,270  lbs Buoyant force  =       62,835  lbs Net Force =      84,435 lbs ==>  Need at least 84,435 pounds of lift to overcome in-water weight

24. a.   A mole (the molecular mass of a substance expressed in grams) of any gas occupies 22.4 liters at STP (standard temperature and pressure: 0 oC (273 K); 1 ata).  Since one mole of dry gas at STP occupies 22.4 liters, the density of each pure substance:

Density O2 = 31.998 g/mole x 1 mole/22.4 L = 1.428 g/L

Density N2 = 28.014 g/mole x 1 mole/22.4 L = 1.251 g/L

Density He =    4.00 g/mole x 1 mole/22.4 L = 0.178 g/L

The density of a mix is the sum of the components; at STP:

O    0.21  x 1.428 g/L  =  0.300 g/L

N2       0.29  x  1.251 g/L  = 0.363 g/L

He     0.50  x 0.178  g/L = 0.089 g/L

Density Mix = 0.752 g/L

b. Absolute pressure at 225 fsw

225 fsw  x    1 atm     =  6.81 atm  ==>  7.81 ata

33 fsw

Since density is directly proportional to pressure:

Trimix     Ddepth = 0.752 g/L  x 7.81 =  5.87 g/L

Oxygen   Ddepth = 1.296 g/L  x 7.81 = 10.12 g/L

25. a. Each gas is treated separately:

Final volume is  200 l +  300 L = 500 l

Using Boyle's Law

For Oxygen:

(200 l) (200 mm)  =  (500 l ) (P)

P = 80 mm

For nitrogen:

(300 l) (100 mm) = (500 l) (P)

P = 60 mm

b. Using Henry's Law

Total pressure = 80 mm  + 60 mm =  140 mm Hg

26. For an ideal gas:

PV =nRT    ==> P = nRT/V

P = (80 moles) (0.0821 l-atm/deg mole) (298 K) / (11.3 l)

P = 173.21 ata

P psi = 173.21 ata x 14.7  psi /ata = 2546.2 psi

Pgauge = 2546. 2  psi  - 14.7 psi   = 2531.5 psig

For a real gas (using Van derWaal's)

(P + an2/V2) (V -nb) = nRT

Rearranging:

P =   nRT     -     n2a

(V-nb)         V2

P = (80 moles) (0,0821 l-atm/deg-mole) (298 K)       -        (80 moles)2 (0.03412  atm/mole2)

(11.3 l - 80(0.0237 l/mole))                                      (11.3 l)2

P = 208.12 atm - 1.71 atm  =  206.4 atm

P psi = 206.4  atm   x   14.7 psi/atm = 3,034.2 psi

P gauge = 3024.2 psi - 14.7psi  =  3,019,5 psi

Since there is a significant difference in calculated pressures, it would be unwise to use ideal equations for helium containing mixes.

Physics Problem Set 2

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