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Bell Diving Supervisors QUESTIONS > Mastering Sat Physics Questions > Flashcards

Mastering Sat Physics Questions Flashcards

1
Q

In a chamber at 40 msw, the Oxygen percentage reading is 8.5%. What is the ppO2 in the chamber?

A

5 x 8.5% = 425mb or 0.425bar

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2
Q

In a chamber at 120 msw, the Oxygen percentage reading is 6.5% . What is the ppO2 in the chamber?

A

13 x 6.5% = 0.845bar or 845mb

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3
Q

A diver at 120msw is breathing a 3.5% mix. What is his ppO2?

A

13 x 3.5% = 0.455bar or 455mb

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4
Q

During a saturation dive at 260fsw, divers require a ppO2 between 0.5 and 0.7 ata. What is a suitable mix?

A

260fsw = 8.88ata

  1. 5/ 8.8 = 5.6%
  2. 7/ 8.8 = 7.9%
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5
Q

During a saturation dive at 160msw, divers require a ppO2 between 0.5 and 0.7 bar. What is a suitable mix?

A

160msw = 17bar

  1. 5/17 = 2.9%
  2. 7/17 = 4.1%

Between 2.9% and 4.1%

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6
Q

If the ppO2 must lie between 1.2 and 1.5 bar, What is the deepest depth (in msw) at which a 12% mix could be used?

A
  1. 2/12% = 10 -1 = 9 x 10 = 90m
  2. 5/12% = 12.5 - 1 = 11.5 x10 = 115m

Deepest is 115msw

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7
Q

Prepare to make 200 bar of 8%, using 4% and 23%. What pressure of each gas do you need?

A

Pmix1 = Fp x (%Fmix - %mix2) / (%mix1 - %mix2)

200 x (8 - 4) / (23 - 4) = 42 bar of mix 1 (23%)

PS: The formula always give the results of mix1 (richest mix 23%) deduct from the total 200 - 42 = 158 bar of mix 2 (4%)

or

200/19 = 10.52 comum factor
(19 is difference between 23 - 4 on the triangle method)

4 x 10.52 = 42 bar of 23%
15 x 10.52 = 158 bar of 4%

PS: The 4 and 15 came out of the difference between the gases on the triangle method.
just don’t loose on the concept of directions of the triangle.

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8
Q

You require 190 bar of 10% mixes available are 4% and 14% O2. How much 4% required?

A

Pmix1 = Fp x (%Fmix - %mix2) / (%mix1 - %mix2)

190 x (10 - 4) / (14 - 4) = 114bar of mix 1 (14%)

PS: The formula always give the results of mix1 (richest mix 14%) deduct from the total 190 - 114 = 76 bar of mix 2 (4%)

or

190/10 = 19 comum factor
(10 came out of 14 - 4)

4 x 19 = 76bar of 4%
( 4 came out of the difference between 14 and 10 )

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9
Q

You require 190 bar of 8% mixes available are 4% and 10%. How much 4% required?

A

Pmix1 = Fp x (%Fmix - %mix2) / (%mix1 - %mix2)

190 x (8 - 4) / (10 - 4) = 126 bar of mix 1 ( 10%)

PS: The formula always give the results of mix1 (richest mix 10%) deduct from the total 190 - 127= 63 bar of mix 2 (4%)

or

190/6 = 31.6 comum factor
2 x 31.6 = 63 bar of 4%

2 is the difference between 10 and 8
6 is the difference between 10 and 4

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10
Q

You have 60 bar of 5% and you want to turn it into 7% by pumping in 10%. What will be the final pressure of the mixture?

A

Fp = Pmix1 x (% mix1 - %mix2) / (%Fmix - %mix2)

60 (5 - 10) / (7 - 10) = 100 bar final pressure

or

60/3 = 20 comum factor
20 x 2 = 40 bar of 10%

PS: If you have 60 bar of 5%, you just need adding 40 bar of 10% to turn in 100 bar total of 7%

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11
Q

You have 90 bar of 6% and you want to turn it into 10% by pumping in 12%. What will be the final pressure of the mixture?

A

Fp = Pmix1 x (% mix1 - %mix2) / (%Fmix - %mix2)

90 x (6 - 12) / (10 - 12) = 270 bar final pressure

or

90/2 = 45 comum factor ( 2 is difference between 12 and 10)
4 x 45 = 180 bar (4 is difference between 10 and 6)

If you have 90 bar of 6%, you just need adding 180 bar of 12% to turn in 270 bar total of 10%

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12
Q

You require 2900psi of 10%. Mixes available are 4% and 14% O2. How much 4% required?

A

Fp = Pmix1 x (% mix1 - %mix2) / (%Fmix - %mix2)

2900 x (10 - 4) / (14 - 4) = 1740 psi of mix 1 ( 14%)

PS: The formula always give the results of mix1 (richest mix 14%) deduct from the total 2900 - 1740 = 1160psi of mix 2 (4%)

or

2900/10 = 290 comum factor ( 10 is the difference between 14 and 4)
4 x 290 = 1160 psi ( 4 is the difference between 4 and 14)

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13
Q

You require 3000psi of 11%. Mixes available are 4% and 14% O2. How much 14% required?

A

Fp = Pmix1 x (% mix1 - %mix2) / (%Fmix - %mix2)

3000 x (11 - 4) / (14 - 4) = 2100 psi of mix 1 ( 14%)

PS: The formula always give the results of mix1 (richest mix 14%)

or

3000/10 = 300 comum factor ( 10 is the difference between 4 and 14)
7 x 300 = 2100 psi of 14% (7 is the difference between 11 and 4)

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14
Q

A chamber is to be pressurised to 90 msw, using 12% and 2%. The final ppO2 must be 600 mb. What depth of 12% should be added to start the pressurisation?

A

Depth rich mix = (PPo2reqmb - PPo2 presmb) - (Bottom depth x weak mix) / (rich mix - weak mix)

(600 - 210) - (90 x 2) / (12 - 2) = 21m

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15
Q

A chamber is to be pressurised to 70 msw, using 14% and 2%. The final ppO2 must be 500 mb. What depth of 14% should be added to start the pressurisation?

A

Depth rich mix = (PPo2reqmb - PPo2 presmb) - (Bottom depth x weak mix) / (rich mix - weak mix)

(500 - 210) - (70 x 2) / (14 - 2) = 12.5m

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16
Q

A chamber is to be pressurised to 58msw, using 20% and 2%. The final ppO2 must be 400 mb. What depth of 20% should be added to start the pressurisation?

A

Depth rich mix = (PPo2reqmb - PPo2 presmb) - (Bottom depth x weak mix) / (rich mix - weak mix)

(400 - 210) - (58 x 2) / (20 - 2) = 4.1m

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17
Q

A chamber is to be pressurised to 200 fsw, using 20% and 2%. The final ppO2 must be 0.5 ata. What depth of 20% should be added to start the pressurisation?

A

Depth rich mix = (3300 (PPo2req ata - PPo2 pres ata)) - (Bottom depth x weak mix) / (rich mix - weak mix)

3300 (0.5 - 0.21)) - (200 x 2) / (20 - 2) = 31 fsw

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18
Q

A chamber is to be pressurised to 280fsw using 10% and 2%. The final ppO2 must be 0.6 ata. What depth of 10% should be added to start the pressurisation?

A

Depth rich mix = (3300 (PPo2req ata - PPo2 pres ata)) - (Bottom depth x weak mix) / (rich mix - weak mix)

3300 (0.6 - 0.21)) - (280 x 2) / (10 - 2) = 91 fsw

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19
Q

A chamber is to be pressurised to 390 fsw, using 18% and 2%. The final ppO2 must be 0.5 ata. What depth of 18% should be added to start the pressurisation?

A

Depth rich mix = (3300 (PPo2req ata - PPo2 pres ata)) - (Bottom depth x weak mix) / (rich mix - weak mix)

3300 (0.5 - 0.21)) - (390 x 2) / (18 - 2) = 11 fsw

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20
Q

A chamber system has a volume of 30 m3. What volume of gas would it take to pressurise it to 120 msw?

A

30 x 12 = 360m3

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21
Q

A chamber system has a volume of 1500 ft3. What volume of gas would it take to pressurise it to 480 fsw?

A

14.54 x 1500 = 21.818 ft3

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22
Q

A chamber system has a volume of 20 m3. It is to be pressurised to 70 msw, with 12 msw of 14% and 58 msw of 2%. What volume of each gas would be needed?

A

7 x 20 = 140m3

12 x 140 / 70 = 24m3 of 14%
58 x 140 / 70 = 116m3 of 2%

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23
Q

A medical lock is 0.6 metres long and 0.3 metres in diameter. The chamber is at 140msw. How much gas is used when the lock is operated?

A

Volume of cylinder = Pi x radius² x length

  1. 14 x 0.15² x 0.6 = 0.042m³
    than:
  2. 042 x 14 = 0.588m³ of gas for each run.
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24
Q

An equipment lock is 0.8 metres long and 1.3 metres in diameter. The chamber is at 80 msw. How much gas is used when the equipment lock is operated?

A

Volume of cylinder = Pi x radius² x length

3.14 x 0.65² x 0.8 = 1.06m³

than

1.06 x 8 = 8.48m³ each run

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25
Q

An equipment lock is 2 feet 8 inches long and 3 feet in diameter. The chamber is at 265 fsw. How much gas is used when the equipment lock is operated?

A

Volume of cylinder = Pi x radius² x length

2ft and 8inches (8 divide by 12, convert in ft) = 2.66

3.14 x 1.5² x 2.66 = 18.79 ft³

than

265fsw = 8.03ata
18.79 x 8.03 = 150.8ft³ each run

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26
Q

An equipment lock is 2 feet 4 inches long and 5 feet in diameter. The chamber is at 165 fsw. How much gas is used when the equipment lock is operated?

A

Volume of cylinder = Pi x radius² x length

2ft and 4 inches (4 divide by 12, convert in ft) = 2.33

3.14 x 2.5² x 2.33 = 45.72ft³

than

165fsw = 5ata
5 x 45.72 = 229ft³ each run

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27
Q

What is IMCA recommended ppO2 range for living in a chamber?

a 0.4 to 0.5 ata/bar or 400 to 500mb
b 0.21 to 0.5 ata/bar or 210 to 500mb
c 0.35 to 0.6 ata/bar or 350 to 600mb
d 0.35 to 0.5 ata/bar or 350 to 500mb

A

d

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28
Q

What is the IMCA recommended ppO2 range for saturation diving when out of the bell?

a 0.5 to 0.9 ata/bar or 500 to 900mb
b 0.6 to 0.9 ata/bar or 600 to 900mb
c 0.35 to 0.5 ata/bar or 350 to 500mb
d 0.35 to 0.6 ata/bar or 350 to 600mb

A

b

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29
Q

What is the formula for working out the volume of gas required to carry out chamber pressurization from one depth to another using metric requirements?

a. FGV = Absolute Depth x Chamber Volume/10
b. FGV = Depth added msw x Chamber Volume/10
c. FGV = Absolute Pressure x Chamber Volume/10
d. FGV = Depth added fsw x Chamber Volume/10

A

a

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30
Q

12 divers are in saturation for 10 days. How much oxygen will the use metabolically in the chamber? (Answer in m3).

A

O2 consumption

12 x 0.72 x 10 = 86.4m³

PS: Metabolic O2 consumption is unaffected by the depth

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31
Q

6 divers are in saturation for 5.5 days. How much oxygen will they use metabolically in the chamber? (Answer in m3).

A

6 x 0.72 x 5.5 = 23.7m³

PS: Metabolic O2 consumption is unaffected by the depth

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32
Q

A DDC is decompressed from 150msw to surface. The ppO2 is 500mb. The DDC has a floodable volume of 12m3. How much O2 is used in the decompression only?

A

O2 used = (Ln initial abs pressure x Po2 bar x volume chamber
Ln of 16 = 2.77

2.77 x 0.5 x 12 = 16.62m³

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33
Q

A DDC is decompressed from 88msw to surface. The ppO2 is 550mb. The DDC has a floodable volume of 15m3. How much O2 is used in the decompression only?

A

O2 used = (Ln initial abs pressure x Po2 bar x volume chamber
Ln of 16 = 2.28

2.28 x 0.55 x 15 = 18.81m³

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34
Q

A decompression from 95 msw takes 2 days, with a ppO2 of 600 mb. There are two divers in the chamber, and the chamber volume is 10m3. How much oxygen is used?

A

2 x 2 x 0.72 = 2.88m³ (metabolic consumption)

O2 used = (Ln initial abs pressure x Po2 bar x volume chamber
Ln of 10.5 = 2.35

2.35 x 0.6 x 10 = 14.1m³

than

14.1 + 2.88 = 16.98m³

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35
Q

A decompression from 65m takes 3 days, with a ppO2 of 500 mb. There are 4 divers in the chamber and the chamber volume is 12 m3. How much oxygen is used?

A

3 x 4 x 0.72 = 8.64m³ (metabolic consumption)

O2 used = (Ln initial abs pressure x Po2 bar x volume chamber
Ln of 7.5 = 2.01

  1. 01 x 0.5 x 12 = 12.06
  2. 64 + 12.06 = 20.7m³
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36
Q

Six divers are in saturation for a total of 28 days including decompression at 180 msw, with a ppO2 of 400 mb. Before starting the decompression, the ppO2 is raised to 550 mb. The chamber volume is 18 m3. How much oxygen is used?

A

28 x 6 x 0.72 = 120.96m³ (metabolic consumption)

than

O2 used = (Ln initial abs pressure x Po2 bar x volume chamber
Ln of 19 = 2.94

2.94 x 0.55 x 18 = 29.1m³

than

  1. 15 x 18 = 2.7m³
  2. 96 + 29.1 + 2.7 = 152.76m³
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37
Q

At the start of the shift, a chamber is at 145msw and 32°C. During the shift, the temperature fell to 28°C. If the life support crew had not added gas to maintain depth, by how much would the depth have decreased?

A

145 x 301/305 = 143.09
Depth decrease by 2m

⁰C + 273
⁰F + 460
⁰C x 1.8 + 32 = ⁰F

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38
Q

At the start of the shift, a chamber is at 180msw and 30°C. During the shift, the temperature fell to 25°C. If the life support crew had not added gas to maintain depth, by how much would the depth have decreased?

A

180 x 298/303 = 177m
depth decrease 3m

PS:
⁰C + 273
⁰F + 460
⁰C x 1.8 + 32 = ⁰F

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39
Q

At the start of the shift, a chamber is at 290fsw and 60°F. During the shift, the temperature fell to 50°F. If the life support crew had not added gas to maintain depth, by how much would the depth have decreased?

A

290 x 510/520 = 284fsw

⁰C + 273
⁰F + 460
⁰C x 1.8 + 32 = ⁰F

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40
Q

At the start of the shift, a chamber is at 310fsw and 72°F. During the shift, the temperature fell to 40°F. If the life support crew had not added gas to maintain depth, by how much would the depth have decreased?

A

310 x 500/ 532 = 291fsw

⁰C + 273
⁰F + 460
⁰C x 1.8 + 32 = ⁰F

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41
Q

A chamber is at 88msw with a ppO2 of 400mb. It is blown down to 120 msw using 4%. What is the ppO2 at 120 msw?

A

120 - 88 = 32 or 3.2bar
3.2 x 4% = 0.128 or 128mb

400 + 128 = 528mb or 0.528bar

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42
Q

Chamber 1 is at 30msw with a ppO2 of 400mb. It is blown down to 100msw using 2.5%. What is the ppO2 at 100msw?

A

100 - 30 = 70 = 7 bar
7 x 2.5% = 0.175
175 + 400 = 575mb at 100msw

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43
Q

A chamber is at 190 fsw, with a ppO2 of 0.45 ata. It is blown down to 240 fsw, using 2%. What is the ppO2 at 240 fsw?

A

50/33 = 1.51ata x 2% = 0.030ata

0.45 + 0.030 = 0.480ata at 240fsw

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44
Q

A chamber and bell are at 150msw with a ppO2 of 400mb. For the dive, the bell is separated from the chamber and blown down on 6% to a working depth of 170msw. What is the ppO2 in the bell at working depth?

A

170 - 150 = 20 or 2bar
2 x 6% = 0.12bar or 120mb
400 + 120 = 520mb at 170m

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45
Q

A chamber is at 250 fsw, with a ppO2 of 0.4 ata, percentage 4.66%. It is bled to 70 fsw. What is the ppO2 at 70 fsw?

A

70/33 = 2.12ata + 1

3.12 x 4.66% = 0.145ata ppo2 at 70fsw

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46
Q

Chamber 1 is at 102 msw, with a PO2 of 0.42 bar, percentage 3.75%. It is bled to 40 msw. What is the ppO2 at 40 msw?

A

40m = 5bar

5 x 3.75% = 0.187bar or 187mb at 40msw

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47
Q

After equalisation, Chamber 1 has a volume of 12m³ and a ppO2 of 480 mb. Chamber 2 has a volume of 8m³ and a ppO2 of 400 mb. What is the final ppO2 when the atmospheres are completely mixed?

A

Chamber 1 = 12 x 0.48 = 5.76
Chamber 2 = 8 x 0.4 = 3.2

than

3.2 + 5.76 = 8.96/20 = 0.448mb

PS: 20 came from the sum of the 2 chambers

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48
Q

After equalisation, Chamber 1 has a volume of 15m³ and a ppO2 of 550 mb. Chamber 2 has a volume of 10m³ and a ppO2 of 420 mb. What is the final ppO2 when the atmospheres are completely mixed?

A

Chamber 1 = 15 x 0.55 = 8.25
Chamber 2 = 10 x 0.42 = 4.2
8.25 + 4.2 = 12.45/25 = 0.498mb final ppo2

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49
Q

During a saturation 9 divers will live at 78msw for 28 days including decompression. How much soda lime would they use?

A

9 x 28 x 6 = 1512Kg or 1.512 ton

Regardless de depth 6kg per day per diver
1kg = 4hours per diver
6kg = 24hours per diver

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50
Q

The bellman uses a chemical sampling tube to take a CO2 reading in the bell. The scale reading is 0.47%. If the bell is at 138msw, give the true percentage of CO2, the ppCO2 and the SEP?

A

True Perc = 0.47/14.8 = 0.031%
SEP = 0.47%
PPCO2 = 0.47/100 = 0.0047bar

PS:

PPM to % = divide by 10.000
PPM to PP = divide by 1.000.000
% to PP = divide by 100

True percentage = scale reading/absolute depth
SEP = reading on the surface analyser
PPCO2 = scale reading/100

CO2 max limit
Bell = 20.000ppm or 2% or 0.02pp/bar
DDC = 5000ppm or 5% or 0.05pp/bar

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51
Q

The bellman uses a chemical sampling tube to take a CO2 reading in the bell. The scale reading is 1.6% if the bell is at 380 fsw , give the true percentage of CO2, the ppCO2 and the SEP.

A

SEP = 1.6%
True Perc = 1.6/12.51 = 0.127%
PPCO2 = 1.6/100 = 0.016ata

PS:

PPM to % = divide by 10.000
PPM to PP = divide by 1.000.000
% to PP = divide by 100

True percentage = scale reading/absolute depth
SEP = reading on the surface analyser
PPCO2 = scale reading/100

CO2 max limit
Bell = 20.000ppm or 2% or 0.02pp/bar
DDC = 5000ppm or 5% or 0.05pp/bar

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52
Q

Surface CO2 Read Out is 500ppm. What is the true CO2 % and pp CO2 in the bell at a depth of 150m?

A

True perc = 500/10000 = 0.05%

PPCO2 = 0.05 x 16 = 0.8/100 = 0.008bar or 8mb

PS:

PPM to % = divide by 10.000
PPM to PP = divide by 1.000.000
% to PP = divide by 100

True percentage = scale reading/absolute depth
SEP = reading on the surface analyser
PPCO2 = scale reading/100

CO2 max limit
Bell = 20.000ppm or 2% or 0.02pp/bar
DDC = 5000ppm or 5% or 0.05pp/bar

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53
Q

Surface CO2 Read Out is 700ppm. What is the true CO2 % and ppCO2 in the chamber at a depth of 170m?

A

True Perc = 700/10000 = 0.07%
PPCO2 = 0.07 x 18 = 1.26/100 = 0.0126bar or 12mb

PS:

PPM to % = divide by 10.000
PPM to PP = divide by 1.000.000
% to PP = divide by 100

True percentage = scale reading/absolute depth
SEP = reading on the surface analyser
PPCO2 = scale reading/100

CO2 max limit
Bell = 20.000ppm or 2% or 0.02pp/bar
DDC = 5000ppm or 5% or 0.05pp/bar

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54
Q

The bell is at 180m with a CO2 of 0.48% taken inside the bell at depth. What will be the surface read out in PPM?

A
  1. 48/19 = 0.0252%
  2. 0252 x 10000 = 252ppm

PS:

PPM to % = divide by 10.000
PPM to PP = divide by 1.000.000
% to PP = divide by 100

True percentage = scale reading/absolute depth
SEP = reading on the surface analyser
PPCO2 = scale reading/100

CO2 max limit
Bell = 20.000ppm or 2% or 0.02pp/bar
DDC = 5000ppm or 5% or 0.05pp/bar

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55
Q

The bell is at 60m with a CO2 of 0.6% taken inside the bell at depth. What will be the surface read out in PPM?

A
  1. 6/7 = 0.085%
  2. 085 x 10000 = 857ppm

PS:

PPM to % = divide by 10.000
PPM to PP = divide by 1.000.000
% to PP = divide by 100

True percentage = scale reading/absolute depth
SEP = reading on the surface analyser
PPCO2 = scale reading/100

CO2 max limit
Bell = 20.000ppm or 2% or 0.02pp/bar
DDC = 5000ppm or 5% or 0.05pp/bar

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56
Q

How long approximately will 1 kg of soda sorb will last 1 diver?

a. 2 hours
b. 4 hours
c. 6 hours
d. 24 hours

A

b

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57
Q

A chemical tube reading taken in the bell at depth is referred to as a surface equivalent?

a. True
b. False

A

True

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58
Q

1.Using a treatment mix with a PPO2 of 2.5 at 450fsw, what % O2 do we require?

(a) 13%
(b) 15.5%
(c) 17%
(d) 18.5%

A

c

450/33 +1 = 14.63
2.5/14.63 = 17%

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59
Q
  1. You have to pressurise a DDC with a FV of 15m3 to 122msw. How much gas is required?

(a) 55.39m3
(b) 74.4m3
(c) 183m3
(d) 198m3

A

c

12.2 x 15 = 183m3

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60
Q
  1. What is the ppO2 of a DDC at 250fsw if the O2 % is 4%?

(a) 0.34 ata
(b) 1.0 ata
(c) 0.004 ata
(d) 0.45 ata

A

a

250/33 + 1 = 8.57 x 4% =

0.34ata

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61
Q
  1. Which gas would be required to perform a dive to 262fsw using a ppO2 at 1.2 ata?

(a) 12%
(b) 12.5%
(c) 13.4%
(d) 14%

A

c

262/33 + 1 = 8.93

1.2/8.93 = 13.4%

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62
Q
  1. A DDC loses 12m3 per day from leaks, how many m3 per hour does that relate to?

(a) 0.25m3/hr
(b) 0.5m3/hr
(c) 2.88m3/hr
(d) 3.0m3/hr

A

b

12/24 = 0.5m3/h

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63
Q
  1. You are to pressurise a DDC from surface using a 5% O2 mix to establish a ppO2 of 400mb in a DDC. To what depth would you pressurise on 5%?

(a) 38 msw
(b) 3.8 msw
(c) 40 msw
(d) 80 msw

A

a

  1. 4 - 0.21 = 0.19/5% = 3.8
  2. 8 x 10 = 38m

PS: removed the 0.21 bar of the chamber.

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64
Q
  1. You require 190 bar of 10%. Mixes available are 4% O2 and 14%. How much 4% is required?

(a) 7.6 bar
(b) 26.6 bar
(c) 19 bar
(d) 76 bar

A

d

190/10 = 19 comum fator
(10 is the difference between 14 and 4)

4 x 14 = 76bar
(4 is the difference between 14 and 10)

Triangle method

Formula

Pmix1 = Fp x (%Fmix - %mix2) / (%mix1 - %mix2)

190 x (10 - 4) / (14 - 4) = 114bar of mix 1 (14%)

PS: The formula always give the results of mix1 (richest mix 14%) deduct from the total 190 - 114 = 76 bar of mix 2 (4%)

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65
Q
  1. Your company’s ppO2 during decompression is 0.6 bar. At what depth would the O2 reach 21%?

(a) 18.5 msw
(b) 28.5 msw
(c) 2.85 msw
(d) 30.0 msw

A

a

  1. 6/21% = 2.85 -1 =
  2. 85 x 10 = 18.5msw
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66
Q
  1. You vent a chamber from 75msw to surface. If the O2% was 4% at 75 msw, what will be the O2 % and ppO2 at the surface?

(a) 4% and 340mb
(b) 4% and 300mb
(c) 4% and 40mb
(d) 21% and 210mb

A

c

Percentage of O2 always the same in venting questions. 4%

1 x 4% = 0.04 x 1000 = 40mb

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67
Q
  1. You are to pressurise a DDC from the surface to 55msw using 2% O2 and 8% O2. ppO2 required is 400mb. How deep will you take the DDC on 8% O2?

(a) 10msw
(b) 13.3msw
(c) 40msw
(d) 48.3msw

A

b

(400 - 210) - (55 x 2)/ (8 -2)=

13.3msw

Depth rich mix = (PPo2reqmb - PPo2 presmb) - (Bottom depth x weak mix) / (rich mix - weak mix)

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68
Q
  1. You pressure a DDC from surface to 95msw using 3/97 mix. What is the helium % at 95msw?

(a) 0.315%
(b) 97%
(c) 87.7%
(d) 95%

A

C

9.5 x 97% = 9.215/10.5 = 87.7%

Same calculation as PPN2 in air calculations

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69
Q
  1. Using the following mixes, and using Mix 2 as your base, how much 2% will be required to make 200 bar of 4%?

Mix 1 = 185 bar of 10%
Mix 2 = 165 bar of 3%
Mix 3 = 175 bar of 2%

(a) 6 bar
(b) 12 bar
(c) 18bar
(d) 24 bar

A

A

200 x (4 - 3)/(10 - 3) = 29bar of 10%

So

165 bar of 3%
29 bar of 10%
6 bar of 2%

or

200 x 4% = 8bar
165 x 3% = 4.95
deducting
200 - 165 = 35
8 - 4.95 = 3.05

than

3.05/35 = 8.7%

make 35bar of 8.7% using 10% and 2% (triangle)

35/8 = 4.37
1.3 x 4.37 = 5.6 bar of 2%

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70
Q
  1. A chamber is brought to the surface from 120msw where O2 was 4.4%. What is the O2% and the partial pressure?

(a) 4.4% & 0.044 pp
(b) 4.4% & 0.0044 pp
(c) 4.4% & 4.4 pp
(d) 4.4% & 0.088 pp

A

a

4.4% Always the same percentage in venting questions

1 x 4.4% = 0.044pp

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71
Q
  1. If a diver consumes 0.5 litre/minute of oxygen metabolically at the surface, how much oxygen will he use per minute at 90 msw?

(a) 0.5 litres
(b) 5.0 litres
(c) 50.0 litres
(d) 500.0 litres

A

a

Same, Metabolic consumption do not changes by depth

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72
Q
  1. A 10 litre bailout is charged to 200bar. A dive is planned to 100msw. How much umbilical can be deployed? (Remember the regulator drive pressure)

(a) 20 metres
(b) 30 metres
(c) 40 metres
(d) 50 metres

A

c

200 - 11 - 10 = 179 x 10 = 1790 litres

11 x 40 = 440l/m

1790/440 = 4 min

For every minute of bailout is allowed 10m of umbilical.

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73
Q
  1. A 2% mix was inadvertently supplied to the diver at 55 msw. What would the diver be suffering from?

(a) Hypoxia
(b) Hypercapnia
(c) Anoxia
(d) Hypocapnia

A

a

6.5 x 2% = 130mb or 0.13bar

anything bellow 160mb is hypoxia

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74
Q
  1. A Chamber is pressurized from the surface to 20 msw using 6%. What is the ppO2 in the chamber at 20 msw?

(a) 210 mb
(b) 120 mb
(c) 330 mb
(d) 420 mb

  1. The same chamber is then pressurized from 20 msw to 57 msw using 4% He O2. What will be the ppO2 at 57 msw?

(a) 210 mb
(b) 120 mb
(c) 478 mb
(d) 358 mb

  1. The same chamber is then pressurized from 57 msw to 92 msw using 2.5 He O2. What will be the ppO2 at 92 msw?

(a) 87.5 mb
(b) 565.5 mb
(c) 297.5 mb
(d) 396 mb

  1. What is the % O2 and ppO2 at 92 msw of this same chamber?

(a) 565 mb / 5.5%
(b) 565 mb / 3%
(c) 210 mb / 21%
(d) 565 mb / 2%

A

17 - C
2 x 6% = 120mb + 210 = 330mb

18 - C
3.7 x 4% = 148mb + 330 = 478mb

19 - B
3.5 x 2.5% = 87.5mb + 478 = 565.5mb

20 - A
0.565/10.2 = 5.5%

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75
Q
  1. A DDC is pressurized from the surface to 267 msw using 2% He O2. What is the % He at 267 msw? (exam question)

(a) 94.5%
(b) 98%
(c) 92.5%
(d) 97%

A

a

26.7 x 98% = 26.16/27.7 = 94.5%

Exame question!!!!

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76
Q
  1. A typical bell CO2 scrubber canister with a capacity of 6 kg of soda sorb has an average duration of?

(a) 0 - 10 man hours
(b) 10 - 20 man hours
(c) 20 - 30 man hours
(d) 30 - 40 man hours

A

(c) 20 - 30 man hours

Every 1kg = 4h

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77
Q
  1. What is the total metabolic O2 consumption for 4 divers for 15 hours at a depth of 75msw?

(a) 255litres
(b) 1020 litres
(c) 1800 litres
(d) 15300 litres

A

c

4 x 15 x 30 = 1800

Not affected by the depth.

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78
Q
  1. Soda Lime works best when?

(a) Warm & moist
(b) Warm & dry
(c) Cold & wet
(d) Cold & moist

A

a

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79
Q
  1. A diver at 250 fsw is breathing a 15% mix. What is the PPO2 in his mix?
A

250/33 + 1 = 8.57 x 15% = 1.28ata

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80
Q
  1. In a chamber at 80msw, the oxygen percentage reading is 4.5%. What is the PPO2 in the chamber?
A

9 x 4.5% = 405mb or 0.405bar

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81
Q
  1. A diver at 125 msw is breathing a 4% mix. What is his PPO2?
A

13.5 x 4% = 0.54bar or 540mb

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82
Q
  1. A diver at 165 msw is breathing a 4% mix. What is the PPO2 in this mix?
A

17.5 x 4% = 0.7bar or 700mb

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83
Q
  1. A diver at 340 fsw is breathing a 6% mix. What is the PPO2 in his mix?
A

340/33 + 1 = 11.30 x 6% = 0.678ata

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84
Q
  1. A diver at 60 msw is breathing an 18% mix. What is the PPO2 in his mix?
A

7 x 18% = 1.26bar

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85
Q
  1. What is the PPO2 in a chamber at 50 fsw if the oxygen percentage is 23%?
A

50/33 + 1 = 2.51 x 23% = 0.578ata

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86
Q
  1. The PPO2 in the chamber at 108msw is 400mb. What is the oxygen percentage?
A

0.4/11.8 = 3.38%

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87
Q
  1. The PPO2 in the chamber at 327 fsw is 0.42 ATA. What is the percentage of oxygen in the chamber?
A 
327fsw = 10.9ata
0.42/10.90 = 3.85%
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88
Q
  1. During a saturation dive at 600 fsw, the divers require a PPO2 between 0.5 and 0.8 ATA. What is a suitable mix?
A

600fsw = 19.18ata

  1. 5/19.18 = 2.6%
  2. 8/19.18 = 4.1%
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89
Q
  1. If the PPO2 must lie between 1.2 and 1.6 bar, what is the greatest depth (in msw) at which you could use a 15% mix?
A
  1. 2/15% = 8 - 1 x 10 = 70msw

1. 6/15% = 10.6 - 1 x 10 = 96.6msw

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90
Q
  1. Assuming that air contains 21% oxygen and 79% nitrogen, what is the PPO2 and PPN2 in air at 165 fsw?
A

165/33 + 1 = 6 x 21%= 1.26ppo2 and 6 x 79% = 4.74ppn2

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91
Q
  1. A hydrox mix contains 1% oxygen. How deep could a diver go without exceeding a PPO2 of 750 mb?
A

0.75/1% = 75 - 1 =74 x 10 = 740m

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92
Q
  1. You want to make 200 bar of 9%, using 2% and 12%. What pressure of each gas do you need?
A

Pmix1 = Fp x (%Fmix - %mix2)/(%mix1 - %mix2) = 200 x (9 - 2)/(12 - 2) =
140 of mix1 (richest mix 12%).
200 - 140 = 60bar of 2%

or

200/10 = 20 comum factor
3 x 20 = 60bar of 2%
7 x 20 = 140bar of 12%

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93
Q
  1. You want to make 180 bar of 6%, using 2% and 18%. What pressure of each gas do you need?
A

Pmix1 = Fp x (%Fmix - %mix2)/(%mix1 - %mix2) = 180 x (6 - 2)/(18 - 2) =
45bar of 18%
180 - 45 = 135bar of 2%

or

180/16 = 11.25 factor
12 x 11.25 = 135bar of 2%
4 x 11.25 = 45bar of 18%

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94
Q
  1. You want to make 3000psi of 23%, using 12% and 50%. What pressure of each gas do you need?
A

Pmix1 = Fp x (%Fmix - %mix2)/(%mix1 - %mix2) = 3000 x (23 - 12)/(50 - 12) =
868psi of 50%
3000 - 868 = 2132psi of 12%

or

3000/38 = 78.94 comum factor
27 x 78.94 = 2.132psi of 12%
11 x 78.94 = 868psi of 50%

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95
Q
  1. You want to make 2500 PSI of 12%, using 2% and 18%. What pressure of each gas do you need?
A

Pmix1 = Fp x (%Fmix - %mix2)/(%mix1 - %mix2) = 2500 x (12 - 2)/(18 - 2) =
1562psi of 18%
2500 - 1562 = 938 of 2%

or

2500/16 = 156.25 comum factor
6 x 156.25 = 938psi of 2%
10 x 156.2 = 1562psi of 18%

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96
Q
  1. You have 100 bar of 4%, and you want to turn it into 10%, by pumping in 20%. What will be the final pressure of the mixture?
A

Fp = Pmix1 x (% mix1 - %mix2)/(%Fmix - %mix2) = 100 x (4 - 20)/(10 - 20) = 160bar

or

100/10 = 10 comum factor
6 x 10 = 60bar of 20%
endup with 160bar final pressure

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97
Q
  1. You have 45 bar of 2% and you want to turn it into 10%, by pumping in 18%. What will be the final pressure of the mixture?
A

Fp = Pmix1 x (% mix1 - %mix2)/(%Fmix - %mix2) = 45 x (2 - 18)/(10 - 18) = 90bar

or

45/8 = 5.62 comum factor
8 x 5.62 = 45bar of 18%
endup with 90bar final pressure

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98
Q
  1. You have 1800psi of 1.5% and you want to turn it into 4%, by pumping in 16%. What will be the final pressure of the mixture?
A

Fp = Pmix1 x (% mix1 - %mix2)/(%Fmix - %mix2) = 1800 x (1.5 - 16)/(4 - 16) = 2175psi

or

1800/12 = 150comum factor
2.5 x 150 = 375psi of 16%
endup with 2175psi final pressure

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99
Q
  1. You have 600 psi of 2% and you want to turn it into 6%, by pumping in 18%. What will be the final pressure be?
A

Fp = Pmix1 x (% mix1 - %mix2)/(%Fmix - %mix2) = 600 x (2 - 18)/(6 - 18) = 800psi

or

600/12 = 50 comum factor
4 x 50 = 200psi of 18%
endup with 800psi final pressure

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100
Q
  1. You have 50bar of 2%, 40bar of 4% and you want to mix them together and add 23% to make the mix up to 8%. What will the final pressure be?
A

50 x 2% = 1
40 x 4% = 1.6

1+1.6 = 2.6/90 = 2.9% of 90bar

than

90/15 = 6 comum factor
5.1 x 6 = 30bar of 23%

SO…. 90 + 30 = 120bar final pressure

or

Fp = Pmix1 x (% mix1 - %mix2)/(%Fmix - %mix2) = 90 x (2.9 - 23)/(8 - 23) = 120bar

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101
Q
  1. You have 500 psi of 6%, 400 psi of 4% and you want to mix them together and add 16% to make the mix up to 12%. What will the final pressure be?
A 
500 = 15.15 x 6% = 0.909ata
400 = 12.12 x 4% = 0.484ata
  1. 909 + 0.484 = 1.393
  2. 15 + 12.12 = 27.27

1.393/27.27 = 5.1% of 900bar

than

900/4 = 225 comum factor
6.9 x 225 = 1552psi of 16%
SO…. 1552 + 900 = 2452psi final pressure

or

Fp = Pmix1 x (% mix1 - %mix2)/(%Fmix - %mix2) = 900 x (5.1 - 16)/(12 - 16) = 2452psi

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102
Q
  1. You want to pressurise a chamber to 90msw, using 12% and 2%. The final PPO2 must be 600 mb. What depth of 12% should you add to start the pressurisation?
A

Depth rich mix = (PPo2reqmb - PPo2 presmb) - (Bottom depth x weak mix) / (rich mix - weak mix)

(600 - 210) - (90 x 2)/(12 - 2) = 21m

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103
Q
  1. You want to pressurise a chamber to 250 fsw, using 16% and 2%. The final PPO2 must be 0.5 ATA. What depth of 16% should you add to start the pressurisation?
A

Depth rich mix = 3300 x (PPo2reqmb - PPo2 presmb) - (Bottom depth x weak mix) / (rich mix - weak mix)

3300 x (0.5 - 0.21) - (250 x 2) / (16 - 2) = 32.6fsw

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104
Q
  1. You want to pressurise a chamber to 500 fsw, using 18% and 1%. The final PPO2 must be 0.6 ATA. What depth of 18% should you add to start the pressurisation?
A

Depth rich mix = 3300 x (PPo2reqmb - PPo2 presmb) - (Bottom depth x weak mix) / (rich mix - weak mix)

3300 x (0.6 - 0.21) - (500 x 1) / (18 - 1) = 46.3fsw

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105
Q
  1. You want to pressurise a chamber to 120 msw, using 20% and 1.5%. The final PPO2must not exceed 650 mb. What depth of 20% should you add to start the pressurisation?
A

Depth rich mix = (PPo2reqmb - PPo2 presmb) - (Bottom depth x weak mix) / (rich mix - weak mix)

(650 - 210) - (120 x 1.5) / (20 - 1.5) = 14m

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106
Q
  1. A chamber system has a volume of 40 m3. What volume of gas would be required to pressurize it to 150 msw?
A

15 x 40 = 600m3

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107
Q
  1. A chamber system has a volume of 1200 ft3. What volume of gas would it take to pressurise it to 500 fsw?
A

500/33 = 15.15 x 1200 = 18182ft3

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108
Q
  1. A chamber system has a volume of 38 m3. What volume of gas would be required to pressurise it to 212 msw?
A

21.2 x 38 = 805.6m3

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109
Q
  1. A chamber system has a volume of 875 ft3. What volume of gas would be required to pressurise it to 355 fsw?
A

355/33 = 10.75 x 875 = 9412ft3

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110
Q
  1. A chamber system has a volume of 30m3. It is to be pressurised to 90 msw, with 21 msw of 12% and 69 msw of 2%. What volume of each gas would be needed?
A

9 x 30 = 270m3

So….21 x 270 / 90 = 63m3 of 12% and 207m3 of 2%

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111
Q
  1. A chamber system has a volume of 1100 ft3. It is to be pressurised to 620 fsw, with 39 fsw of 18%, and 581 fsw of 1 %. What volume of each gas would be needed?
A

620/33 = 18.78 x 1100 = 20666ft3

So 39 x 20666 / 620 = 1300ft3 of 18% and 19366ft3 of 1%

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112
Q
  1. A chamber system has a volume of 45 m3. It is to be pressurised to 197 msw, with 7msw of 16%, and 190 msw of 1.5%. What volume of each gas would be needed?
A

19.7 x 45 = 886.5m3

so… 7 x 886.5/197 = 31m3 of 16% and 855m3 of 1.5%

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113
Q
  1. A medical lock is 0.8 meters long and 0.3 meters in diameter. The chamber is at 160msw. How much gas is used when the lock is operated?
A

V = Pi x R² x L

V = 3.14 x 0.15² x 0.8 = 0.056 x 16 = 0.896m³ every run

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114
Q
  1. An equipment lock is 4 ft long and 2 ft 6 inches in diameter. The chamber is at 510 fsw. How much gas is used when the lock is operated?
A

2ft and 6 inches (6/12) = 2.5ft diameter
510fsw = 15.45ata
V = Pi x R² x L
V = 3.14 x 1.25² x 4 = 19625 x 15.45 = 303ft³

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115
Q
  1. 9 divers are in saturation for 5.5 days. How much oxygen will they use in the chamber (answer in m3)?
A

9 x 5.5 x 0.72 = 35.64m3

PS: Metabolic O2 consumption is unaffected by the depth

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116
Q
  1. 6 divers are in saturation for 12 days. How much oxygen will they use in the chamber (answer in ft3)?
A

6 x 12 x 25 = 1800ft3

PS: Metabolic O2 consumption is unaffected by the depth

Metric Daily: 0.72m3
Imperial Daily: 25ft3

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117
Q
  1. A decompression from 95 msw takes 2 days, with a PPO2 of 600 mb. There are two divers in the chamber, and the chamber volume is 10m3. How much oxygen is used?
A

2 x 2 x 0.72 = 2.88m3

Ln (abs pressure) x PPO2 bar x Volume chamber
Ln of 10.5bar = 2.35
so….
2.35 x 0.6 x 10 = 14.10m3 + 2.88 = 16.98m3 of O2 in total

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118
Q
  1. A decompression from 180 msw takes 4 days, with a PPO2 of 600 mb. There are two divers in the chamber, and the chamber volume is 10m3. How much oxygen is used?
A

2 x 4 x 0.72 = 5.76m3

Ln (abs pressure) x PPO2 bar x Volume chamber
Ln of 19bar = 2.94
so…
2.94 x 0.6 x 10 = 17.66m3 + 5.76 = 23.42m3 of O2 in total

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119
Q
  1. 6 divers are in saturation for 11 days including the decompression at 110 msw, with a PPO2 of 400 mb. Before starting the decompression, the PPO2 is raised to 600 mb. The decompression takes 3 days. The chamber volume is 15m3. How much oxygen is used?
A

6 x 11 x 0.72 = 47.52m3

than

0.2 x 15 = 3m3

Ln (abs pressure) x PPO2 bar x Volume chamber
Ln of 12bar = 2.48
2.48 x 0.6 x 15 = 22.32m3

So…
22.32 + 3 + 47.52 = 72.84m3 of O2 in total

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120
Q
  1. During saturation the PPO2 in the chamber is maintained at 400 mb. Before starting the decompression the level is raised to 600 mb. If the chamber volume is 17m3, what volume of oxygen is required to raise the PPO2?
A

0.2 x 17 = 3.4m3

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121
Q
  1. 6 divers are in saturation at 275 fsw for 12 days including the decompression with a PPO2 of 0.4 ATA. The PPO2 is then raised to 0.6 ATA for a decompression which lasts 2 days. The chamber volume is 500 ft3. How much oxygen is used altogether?
A

6 x 12 x 25 = 1800ft3

than

0.2 x 500 = 100ft3

275fsw = 9.33ata
Ln (abs pressure) x PPO2 bar x Volume chamber
Ln of 9.33ata = 2.23
2.23 x 0.6 x 500 = 670ft3

So…
670 + 100 + 1800 = 2570ft3 of O2 in total

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122
Q
  1. A chamber is at 145 msw and 28⁰C. At the start of the shift, the temperature was 32⁰C.If the life support crew had not added gas to maintain depth, by how much would the depth have decreased?
A

145 x 301 / 305 = 143

Decrease by 2m

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123
Q
  1. A chamber is at 345 fsw and 31⁰C. The temperature drops to 26⁰C. If the life support crew took no action, what would the depth decrease by?
A

345 x 299 / 304 = 339fsw

Decrease 6ft

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124
Q
  1. Chamber 1 is at 97msw, with a PPO2 of 400mb. It is blown down to 130 msw using 2%. What is the PPO2 at 130msw?
A

130 - 97 = 33 = 3.3 x 2% = 0.066bar or 66mb + 400 = 466mb

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125
Q
  1. A chamber is blown down from 180 fsw to 210 fsw using 4%. If the PPO2 was 0.4 ATA at 180 fsw, what is it at 210 fsw?
A

210 - 180 = 30ft/33 = 0.9ata X 4% = 0.03ata + 0.4 = 0.433ata at 210fsw

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126
Q
  1. After a serious leak, a chamber is at 50 msw with a PPO2 of 180 mb. The chamber is blown back to the living depth of 125 msw using 5%.What is the PPO2 at living depth?
A

7.5 x 5% = 0.375bar or 375mb + 180mb = 555mb

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127
Q
  1. Chamber 1 is at 320 fsw, with a PPO2 of 0.4 ATA, Percentage 3.74%. It is bled to 100 fsw. What is the PPO2 at 100 fsw?
A

100/33 + 1 = 4.03 x 3.74% = 0.15ata

PS: Bleeding or Venting questions percentage still the same. Matters at what depth you are at the moment.

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128
Q
  1. A chamber is at 186 msw, with an oxygen percentage of 2.1%. It is bled to 143 msw. What is the PPO2 at 143 msw?
A

15.3 x 2.1% = 0.321bar or 321mb

PS: Bleeding or Venting questions percentage still the same. Matters at what depth you are at the moment.

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129
Q
  1. After equalisation, chamber 1 has a volume of 12m3 and a PPO2 of 480mb. Chamber 2 has a volume of 8m3 and a PPO2 of 400mb. What is the final PPO2 when the atmospheres are completely mixed?
A

Chamber 1 = 12 x 0.48 = 5.76
Chamber 2 = 8 x 0.4 = 3.2

12 + 8 = 20 and 5.76 + 3.2 = 8.96/20 = 0.448bar

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130
Q
  1. Chamber 1 has a volume of 400ft3 and the PPO2 is 0.45 ATA.
    Chamber 2 has the same volume, but the PPO2 is 0.38 ATA. What is the final PPO2 when the chamber atmospheres are fully mixed?
A

Chamber 1 = 0.45 x 400 = 180
Chamber 2 = 0.38 x 400 = 152

400 + 400 = 800 and 180 + 152 = 332/800 = 0.415ata

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131
Q
  1. Chamber 1 has a volume of 18m3 and the PPO2 is 420mb
    Chamber 2 has a volume of 12m3 and the PPO2 is 400mb
    Chamber 3 has a volume of 15m3 and the PPO2 is 600mb
    What is the final PPO2 when all the chamber atmospheres are fully mixed?
A

Chamber 1 = 18 x 0.42 = 7.56
Chamber 2 = 12 x 0.4 = 4.8
Chamber 3 = 15 x 0.6 = 9

So all the chambers = 45m3 and all ppo2 = 21.36/45 = 0.475bar

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132
Q
  1. A chamber and bell are at 185 msw, with a PPO2 of 400mb. For the dive, the bell is separated from the chamber and blown down on 4% to a working depth of 200 msw. What is the PPO2 in the bell at working depth?
A

200 - 185 = 15m or 1.5abs x 4% = 0.06bar or 60mb + 400 = 460mb

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133
Q
  1. Divers are being pressured in a chamber using 2% straight from the surface. The pressurisation is aborted at 70 fsw when the oxygen percentage is 8%. The chamber is bled straight back to the surface. What is the PPO2 on the surface? How do the divers feel about this?
A

1 x 8% = 0.08bar or 80mb

Divers feel shit or dont feel anything because they pass out!!!
Hypoxia

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134
Q
  1. The bell man uses a chemical sampling tube to take a CO2 reading in the bell. The scale reading is 1.4%. If the bell is at 400 fsw, give the true percentage of CO2, the PPCO2 and the SEP?
A

400fsw = 13.12ata

True percentage = 1.4/13.12 = 0.106%
PPCO2 = 1.4/100 = 0.014ata
SEP = 1.4%

PS:

PPM to % = divide by 10.000
PPM to PP = divide by 1.000.000
% to PP = divide by 100

True percentage = scale reading/absolute depth
SEP = reading on the surface analyser
PPCO2 = scale reading/100

CO2 max limit
Bell = 20.000ppm or 2% or 0.02pp/bar
DDC = 5000ppm or 5% or 0.05pp/bar

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135
Q
  1. The bell man uses a chemical sampling tube to take a CO2 reading in the bell. The scale reading is 2%. If the bell is at 145msw, give the true % of CO2, the PPCO2 and the SEP?
A

True Perc = 2/15.5 = 0.129%
PPCO2 = 2/100 = 0.02bar
SEP = 2%

PS:

PPM to % = divide by 10.000
PPM to PP = divide by 1.000.000
% to PP = divide by 100

True percentage = scale reading/absolute depth
SEP = reading on the surface analyser
PPCO2 = scale reading/100

CO2 max limit
Bell = 20.000ppm or 2% or 0.02pp/bar
DDC = 5000ppm or 5% or 0.05pp/bar

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136
Q
  1. A diving bell displaces 180ft3 of seawater and weighs 5.2 imperial tons (1 ton = 2240lbs). Is the bell positively buoyant?
A

180 x 64.38 = 11588/2240 = 5.17ton

No, is negative

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137
Q
  1. A diving bell displaces 180ft3 of seawater and weighs 5.2 US tons (1 US ton = 2000lbs). Is the bell positively buoyant?
A

180 x 64.38 = 11588/2000 = 5.79t

Yes, is positive.

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138
Q
  1. A block of concrete 1m by 1m by 1m is lying on the sea bed. The density of concrete is 2400kg/m3. How much force is required to lift the block clear of the seabed? (Ignore any suction).
A

2400/1000 = 2.4t
1 x 1 x 1 = 1 x 2.4t
1 x 1 x 1 = 1 x 1.03

2.4 - 1.03 = 1.37t

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139
Q
  1. A welding habitat weighs 25 tonnes in the air. It displaces 12m3. What does it weigh in the seawater?
A

12 x 1.03 = 12.36

25 - 12.36 = 12.64t

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140
Q
  1. A closed length of pipe weighs 12 imperial tons (2240lbs) and displaces 140ft3. How many 1 ton lifting bags would you need to lift it clear of the seabed? (ignore suction)?
A

140 x 64.38 = 9013.2/2240 = 4.02t
12 - 4.02 = 7.97t

8 lifting bags

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141
Q

A saturation diving operation is planned for 8 days of continuous work with 3 man bell runs. The 3 DDC’s including entry locks each have a floodable volume of 18 m3. The bell FV is 6 m3. HRL FV is 12 m3. The storage depth will be 120 msw. How much heliox will be required to press the whole system to the storage depth? Include any reserve gas required for blowdown reserve once at depth as according to DMCA D050.

A

1728m3

3 x 18 = 64 + 6 + 12 = 72m3

72 x 12 = 864m3

According to IMCA D050. double the amount as reserve.
So… total of 1728m3

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142
Q

A diver suffering from DCI is to be given a total of 6 cycles of 30 minutes on BIBS. Theconsumption rate is 20 litres/min at surface, the chamber is at 70 msw How much treatment mix will be used?

a. 2.8800 L
b. 28.800 L
c. 288.00 L
d. 2880.0 L

A

B

8 x 20 x 6 x 30 = 28800 Litres or 28.8m3

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143
Q

A DDC is at a depth of 380fsw. The analyser at the surface shows the CO2 to be 420ppm. What is the surface equivalent %?

A

0.525% SEP

380 fsw = 12.51ata

420ppm x 12.51 ata = 5250ppm / 10000 = 0.525% SEP

PS: shoud check this question… analyser at surface.

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144
Q

What is the total metabolic O2 consumption for 6 divers for 17 hours at a depth of 85msw?

A

3060 L

6 x 17 x 30 = 3060 L

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145
Q

Which gas would be required to perform a dive to 250fsw using a ppO2 at 1.3ata?

A

15.1%

250 fsw = 8.57 ata

1.3 / 8.57 = 15.1%

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146
Q

You are to use 6% O2 mix to establish a ppO2of 400mb in a DDC. To what depth would you pressurise on 6%?

A
  1. 6m
  2. 4 - 0.21 = 0.19 bar
  3. 19 / 6% = 3.16 x 10 = 31.6m
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147
Q

You require 195 bar of 8%. Mixes available are 5% O2 and 15%. How much 5% is required?

A

136 bar of 5%

195 / 10 = 19.5 factor

7 x 19.5 = 136 bar of 5%

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148
Q

Your company’s ppO2 during decompression is 0.5 bar. At what depth (MSW) would the O2 reach 21%?

A
  1. 8m

0. 5 / 21% = 2.38 - 1 x 10 = 13.8m

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149
Q

You vent a chamber from 60msw to surface. If the O2% was 5% at 60msw, what is theO2% and ppO2 at the surface?

A

Percentage of O2 dont change still 5%

PPO2 = 1 x 5% = 0.05 bar or 50mb

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150
Q

You are to pressurise a DDC from the surface to 75msw using 2% O2 and 8% 02. PPO2 required is 400mb, How deep will you take the DDC on 8% O2?

A

6.66m

(400 - 210) - (75 x 2) / (8 - 2) = 6.66m

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151
Q

If 300 psi of pure O2 is added to 1200 psi of 3%, what will be the final % of the mix?

A

22.4%

300 x 100% = 300
1200 x 3% = 36

300 + 36 = 336 / 1500 = 22.4%

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152
Q

You have a DDC at 220msw, the O2% is 2%. If the DDC is vented to 130msw, whatis the ppO2 and O2% at 130msw?

A

Percentage 2%

14 x 2% = 0.28bar or 280mb

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153
Q

A Drager tube reading taken by the bellman is showing 0.6%. Bell depth is 80msw. Howmany PPM CO2 is that equivalent to? (As read on analyser in control room)

A

666 ppm

0.6% / 9 = 0.0666% x 10000 = 666 ppm

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154
Q

A chamber is pressurized from the surface to 10 msw using 6%. What is the ppO2 in the chamber at 10 msw?

The same chamber is then pressurized from 10 msw to 70 msw using 4% He02.What will be the ppO2 at 70 msw?

What is the % O2 and ppO2 at 120 msw of this same chambers?

A

0.27 bar and 0.51 bar and 3.9%

1 x 6% = 0.06 + 0.21 = 0.27 bar

6 x 4% = 0.24 bar + 0.27 = 0.51 bar

0.51 / 13 = 3.9%
13 x 3.9% = 510mb or 0.51bar

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155
Q

A DDC is pressurized from the surface to 210 msw using 2% HeO2. What is the % He at 210 msw?

A

93.5%

21 x 98% = 20.58 / 22 = 93.5%

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156
Q

A 1O litre bailout is charged to 200bar. A dive is planned to 110msw. How much umbilical can be deployed? (Remember the regulator drive pressure)

A

30m of umbilical

200 - 12 - 10 = 178 x 10 = 1780 / 12 x 40 = 3.70 min

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157
Q

You have unlimited pure helium & oxygen onboard. You want to make 195 bar of 6%.How much 02 is required?

A

11.7 bar

195 / 100 = 1.95 x 6 = 11.7 bar

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158
Q

Using the following mixes & using mix 2 as your base, how much 2% will be required to make 200 bar of 8%.?

Mix I : 200 bar of 10%
Mix 2 : 50 bar of 3%
Mix 3 : 200 bar of 2%

A

7.5 bar of 2%

50 x 3% = 1.5
200 x 8% = 16

16 - 1.5 = 14.5
200 - 50 = 150

so….
14.5/150 = 9.6%

SO… We need to make 150 bar of 9.6% using 10% and 2%

150 / 8 = 18.75 factor
0.4 x 18.75 = 7.5 bar of 2%

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159
Q

During a saturation dive, 12 divers will live at 105msw for 8 days, including decompression. How much absorbent would they use?

A

12 x 8 x 6 = 576Kg

Regardless de depth 6kg per day per diver
1kg = 4hours per diver
6kg = 24hours per diver

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160
Q

Do you have 6 divers in a chamber at 25m. What is the minimum amount of gas required for the divers if they go on Bibs?

A

Minimum required Bibs gas in the chambers is 4h per diver.
Bibs gas consumption 20l/min or 0.7ft3/min

6 x 240 x 20 x 3.5 = 100.8m3

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161
Q

What is the gas consumption on bibs in a Chamber? (answer in metric)

a. 1.25ft/min
b. 5l/min
c. 20l/min
d. 0.7ft/min

A

c. 20l/min

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162
Q

What is the gas consumption on bibs in a Chamber? (answer in imperial)

a. 1.25ft/min
b. 5l/min
c. 20l/min
d. 0.7ft/min

A

d. 0.7ft/min

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163
Q

If a diver in saturation is using reclaim, what would be his gas consumption rate ?

a. 1.25ft/min or 35l/min
b. 0.18ft/min or 5l/min
c. 0.7ft/min or 20l/min
d. 1.5ft/min or 40l/min

A

b

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164
Q

Two divers are working at 600 fsw. How much gas would they use in 6 hours? How much gas would they use in 6 hours if they were using gas reclaim?

A

Normal Consumption: 17262 ft3
600fsw: 19.18ata
6h: 360min
360 x 19.18 x 1.25 x 2 = 17262 ft3

Reclaim: 2486 ft3
360 x 19.18 x 0.18 x 2 = 2486 ft3

PS: Remember reclaim consumption 0.18ft3/min or 5l/min

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165
Q

A diver is working at 75 msw. What is his gas consumption using gas reclaim?

A
  1. 0425 m3/min
  2. 5 x 5 = 42.5/1000 = 0.0425m3/min

PS: Remember reclaim consumption 0.18ft3/min or 5l/min

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166
Q

On a saturation job, there are expected to be three dives each day, with two divers out of the bell for 6 hours on each dive. If the working depth is 300 fsw, what is the expected daily gas use for the divers, using reclaim?

A

3923 ft3

300fsw = 10.09 ata
10.09 x 360 x 0.18 x 6 = 3923 ft3

PS: Remember reclaim consumption 0.18ft3/min or 5l/min

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167
Q

A body tissue can absorb 0.5 litres of helium when the PPHe is 2 bar. How much can it absorb if the PPHe is 5 bar?

A

1.25 litres

5 x 0.5 / 2 = 1.25 litres

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168
Q

A diver at 300 fsw is breathing a 5% mix. What is the PP02?

A

0.505 ata

300 = 10.09 ata x 5% = 0.505 ata

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169
Q

A body tissue can absorb 0.27 litres of nitrogen when the PPN2 is 0.79 ATA. How much can it absorb when the PPN2 is 1.4 ATA?

A
  1. 48 litres

1. 4 x 0.27 / 0.79 = 0.48 litres

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170
Q

A diver at 150 fsw is breathing a 12% mix. What is the PP02 in his mix?

A

0.665 at

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171
Q

A diver at 150 msw is breathing a 4% mix. What is his PP02?

A

640 mb

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172
Q

What is the diver’s PP02 if he is breathing a 7% mix at 85 msw?

A

665 mb

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173
Q

A diver is at 200 fsw. If he is breathing a 15% mix, what is his PP02?

A

1.06 at

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174
Q

A diver is breathing an 18% mix at 60 msw. What is his PP02?

A

1.26 bar

175
Q

A chamber is at a pressure of 70 msw and the oxygen percentage 5.2%. What is the PP02 in the chamber?

A

416 mb

176
Q

What is the PP02 in a chamber at 60 fsw if the oxygen percentage is 21%?

A

0.592 at

177
Q

The PP02 in a chamber at 165 msw is 400mb. What is the oxygen percentage?

A

2.29%

178
Q

The PP02 in a chamber at 400 fsw is 0.42 AT. What is the percentage of oxygen in the chamber?

A

3.20%

179
Q

In a chamber at 85 msw, the oxygen percentage reading is 4.5%. What is the PP02 in the chamber?

A

428 mb

180
Q

The C02 reading in a chamber at 90 msw is 250 ppm. What is the PPC02?

A

2.5mb

250/10.000 = 0.025%

For Partial Pressure is Scale reading/100 = 0.025/100 = 0.00025 x 10 (abs pressure) = 0.0025 bar or 2.5mb

or

250 x 10 / 1.000.000 = 0.0025 bar or 2.5mb

181
Q

The C02 reading in a chamber at 300 fsw is 150 ppm. What is the PPC02?

A

0.0015 ata

150/10000 = 0.015/100 x 10.09 = 0.0015 ata

PP = %/100

or

150 x 10.09 / 1.000.000 = 0.0015 ata

182
Q

During a saturation dive at 450 fsw, the divers require a PP02 between 0.5 and 0.7 AT. What is a suitable mix?

A

Anything between 3.4% and 4.7%

183
Q

If the PP02 must lie between 1.2 and 1.4 bar, what is the greatest depth at which you could use a 12% mix?

A

106.6 msw

184
Q

The divers require a PP02 between 0.5 and 0.7 AT during a saturation dive at 320 fsw. What is a suitable mix?

A

Anything between 4.7% and 6.5%

185
Q

During a bounce dive to 78 msw, the divers require a PP02 between 1.2 and 1.6 bar. What is a suitable mix?

A

Anything between 13.8% and 18.2%

186
Q

During a saturation dive at 145 msw, the divers require a PP02 between 550 and 750mb. What is a suitable mix?

A

Anything between 3.5% and 4.8%

187
Q

What is the PP02 and PPN2 in air at 150 fsw? (Assume that air contains 21% oxygen and 79% nitrogen).

A

PPO2: 1.16 At
PPN2: 4.38 At

188
Q

A diver is using a hydrox mix containing 1% oxygen. How deep could he go without exceeding a PP02 of 700mb?

A

690 msw

189
Q

The bellman uses a chemical sampling tube to take a C02 reading in the bell. The scale reading is 1.0 %. The bell is at 80 msw. What is the PPC02? What is the true percentage? What is the SEP?

A

0.01 bar or 10 mb

For Partial pressure is Scale reading/100 = 1/100 = 0.01bar or 10mb

For True Percentage = scale reading / absolute pressure = 1/9 = 0.11%

SEP = 1.0%

190
Q

The bellman uses a chemical sampling tube to take a C02 reading in the bell. The scale reading is 0.6%. The bell is at 400 fsw. What is the true percentage of C02? What is the PPC02?

A

True Percentage = %/absolute pressure = 0.6%/ 13.12 = 0.05%

PPCO2 = Scale reading/100 = 0.6/100 = 0.006 at

191
Q

A diver in the chamber uses a chemical sampling tube to check for hydrogen sulphide (H2S). The scale reading is 1.1ppm. The chamber is at 65 msw. What is the true ppm? What is the PPH2S?

A

True Ppm = 1.1ppm / 7.5 = 0.15 ppm
Convert ppm to bar = divide by 1.000.000

0.0000011bar or 0.001mb

{This doesn’t sound much, but you’d be able to smell it and might get the first H2S poisoning symptoms)

192
Q

An LST uses a chemical sampling tube to take a chamber C02 reading on the surface. The scale reading is 0.04%. The chamber is at 77 msw. What is the true percentage of C02? What is the PPC02?

A

0.04% because was taking on the surface. True percentage.

PPCO2 = 8.7 x 0.04 = 3.48mb or 0.00348bar

PS: Reading was taken on the surface!!!!!

193
Q

You want to make 200 bar of 6%, using 2% and 12%. What pressure of each gas do you need?

A

You need 80 bar of 12% and 120 bar of 2%.

194
Q

You have a quad containing 100 bar of 4%. You want to make 10% by adding 20% to the quad. What pressure of each gas do you need?

A

Add 60 bar of 20% to the 100 bar of 4%.

So will be total of 160 bar of 10%.

195
Q

You want to make 200 bar of 4%, using 2% and 12%. What pressure of each gas do you need?

A

You need 40 bar of 12% and 160 bar of 2%.

196
Q

You want to make 3000 psi of 18%, using 10% and 50%. What pressure of each gas do you need?

A

You need 600 psi of 50% and 2400 psi of 10%.

3000 / 40 = 75 factor
8 x 75 = 600 psi of 50%
32 x 75 = 2400 psi of 10%

197
Q

You want to make 2500 psi of 8%, using 2% and 18%. What pressure of each gas do you need?

A

You need 938 psi of 18% and 1562 psi of 2%.

2500 / 16 = 156.25 factor

6 x 156.25 = 937.5 psi of 18%
10 x 156.25 = 1562.5 psi of 2%

198
Q

You have a quad containing 70 bar of 4%. You plan to make 6% by pumping 10% into the quad. What will the final pressure of the mixture be?

A

105 bar

70 / 4 = 17.5 factor
2 x 17.5 = 35 bar of 10%

70 + 35 = 105 bar

199
Q

You wish to make 10% by adding 18% to a quad of 2%. If there’s 45 bar of 2% in the quad, what will the final pressure of the 10% be?

A

90 bar

(Short cut: since 10% is exactly halfway between 2% and 18%, you’ll need the same pressures of 2% and 18%.}

200
Q

You have a quad containing 2000 psi of 1.5%. You plan to make 4%, by pumping 16% into the quad. What pressure of 16% must you add?

A

417 psi

2000 / 12 = 166.66 factor

2.5 x 166.66 = 417 psi

201
Q

You pump 100 bar of 10% into a quad containing 60 bar of 2%. What is the percentage of the final mix?

A

7%
100 x 10% = 10
60 x 2% = 1.2

10 + 1.2 = 11.2/160 = 0.07 = 7%

202
Q

You have 30 bar of 2% and 60 bar of 5% and you want to mix them together and add 23% to make the mix up to 10%. What will the final pressure be?

{This is for amusement only. You wouldn’t expect this is an exam!)

A

132 bar

(Do this in two stages. Work out what you get by mixing the 2% and 6%, and then use the triangle.)

30 x 2% = 0.6
60 x 5% = 3
3 + 0.6 = 3.6/90 = 4%

Now I have 90 bar of 4%

90/13 = 6.92 (factor)
6.92 x 6 = 41.5 bar
90 + 41.5 = 132 bar

203
Q

You have 250 psi of 4%, 300 psi of 10% and you want to mix them together and add 18% to make the mix up to 12%. What will the final pressure be?

(This is for amusement only. You wouldn’t expect this is an exam!)

A

983 psi

Do this in two stages.

250/33 = 7.57 x 4% = 0.30300/33 = 9.09 x 10% = 0.900.3 + 0.9 = 1.21/16.66 = 7.2%

550 psi of 7.2%

550/6 = 91.66 factor
91.66 x 4.8 = 440 psi
550 + 440 = 990 psi of 12%

204
Q

You want to pressurise a chamber to 100 msw with a PP02 of 450 mb. You plan to start the pressurisation with a 16% mix and then use 2%. What depth of 16% do you need?

A

2.86msw

(450 - 210) - (100 x 2) / 16 -2 = 2.86msw

205
Q

You want to pressurise a chamber to 150 msw with a PP02 of 450 mb. You plan to start the pressurisation with a 16% mix and then use 2%. What depth of 16% do you need?

A
  • 4.29 msw

{It’s a negative answer, so it can’t be done- it would have to be 2% all the way.)

(450 - 210) - (150 x 2)/ 16 -2 = - 4.29 msw

206
Q

You want to pressurise a chamber to 300 fsw with a PP02 of 0.48 AT. You plan to start the pressurisation with a 18% mix and then use 2%. What depth of 18% do you need?

A

18.19 fsw

3300(0.48 - 0.21) - (300 x 2 ) / 18 - 2 = 18.19 fsw

207
Q

You want to pressurise a chamber to 100 msw, using 12% and 2%. The final PP02 must be 600 mb. What depth of 12% should you add to start the pressurisation?

A

19 msw

(600 - 210) - (2 x 100) / (12 - 2) = 19 msw

208
Q

Chamber depth will be 200 fsw and the PP02 0.5AT. Pressurisation will be carried out using 18% and 2%. What depth of 18% is needed to start the pressurisation?

A

34.8 fsw

209
Q

You want to pressurise a chamber to 300 fsw, using 18% and 2%. The final PP02 must be 0.6 AT. What depth of 18% should you add to start the pressurisation?

A

42.9 fsw

210
Q

You want to pressurise a chamber to 110 msw, using 20% and 2%. The final PP02 must not exceed 650 mb. What depth of 20% should you add to start the pressurisation?

A

12.2 msw

211
Q

You want to pressurise a 45 m3 chamber system to 130 msw. You are planning to use 8 msw of 16% and 122msw of 2%. What volume of each gas do you need?

A

36 m3 of 16% and 549 m3 of 2%

212
Q

A chamber system has a volume of 45m3. It is to be pressurised to 90 msw, with 21msw of 12% and 69 msw of 2%. What volume of each gas would be needed?

A

94.5 m3 of 12% and 310.5 m3 of 2%

213
Q

A chamber system has a volume of 1200 ft3. It is to be pressurised to 620 fsw, with 39 fsw of 18% and 581fsw of 1%. What volume of each gas would be needed?

A

1418.2 ft3 of 18% and 21127.3 ft3 of 1%

214
Q

A chamber system has a volume of 30m3. It is to be pressurised to 197 msw, with 7 msw of 16% and 190 msw of 1.5%. What volume of each gas would be needed?

A

21 m3 of 16% and 570m3 of 1.5%

215
Q

During a saturation, the PP02 in the chamber is maintained at 400mb. Before starting decompression, the level is raised to 500mb. If the chamber volume is 17m3 what volume of oxygen is required to raise the PP02?

A
  1. 7m3

0. 1 x 17 = 1.7m3

216
Q

You have to pressurise a chamber all the way on 2%. At a depth of 20 msw, the pressurisation is aborted. What Will the PP02 be in the chamber if it is bled straight back to surface?

A

8.3% and 83mb

2 x 2% = 0.04 + 0.21 = 0.25 / 3 = 8.3%

250mb at depth of 30m
83mb at surface

217
Q

A diver medic goes into a chamber at 70 msw, to treat an injured diver. The PP02 is 400mb. He expects to spend about 25 minutes in the chamber before starting decompression. You have bounce dive tables for 78 msw using an 18% mix, 80 msw using 16% and 82 msw using 16%. Which bounce dive table could be used to decompress the diver medic?

A

You could use 82 msw 16% table

At 70 msw, the pressure is 8 bar.
The PP02 is 400mb, or 0.4 bar.
The PPHe must be (8 - 0.4) bar ; 7.6 bar.

Try the 78 msw table using 18%:
Partial pressure ; absolute pressure x decimal percentage
absolute pressure : 8.8 bar decimal percentage 0.82
(It’s the helium percentage you needs)
Partial pressure 8.8 x 0.82 bar 7.22 bar
This is less than the PPHe in the chamber. This table is no good

Try the 80 msw table using 16%: Partial pressure : absolute pressure x decimal percentage absolute pressure : 9.0 bar decimal percentage : 0.84 Partial pressure : 9.0 x 0.84 bar : 7.56 bar
Still not enough!

Lets try 82 msw table using 16%: Partial pressure : absolute pressure x decimal percentage absolute pressure 9.2 bar decimal percentage : 0.84 Partial pressure : 9.2 x 0.84 bar 7.73 bar

That’s it! The PPHe in the table is greater than that in the chamber. you could use the 82 msw, 16% table to decompress the diver medic.

218
Q

A saturation pressurisation is aborted at 100 msw, when the PP02 is 430 mb. You have bounce dive tables from 100 msw to 112 msw, at 3 msw intervals, all using 12%. The Choose a suitable bounce table to decompress the divers.

A

112 msw table

(The PPHe in the chamber is 10.57 bar = 11 - 0.43).
The PPHe in the 112 msw table is 10.74.
( 12.2 - 1.46)

PS: The 109 msw table would only give a PPHe of 10.47 bar.
PPHe deco has to be higher of PPHe in the chamber

219
Q

An equipment lock is 2.0 metres long and 0.8 metres in diameter. The chamber is at 120 msw. How much gas is used when the lock is operated?

A

12.07m3

12 x 3.14 x 2 x 0.4² = 12.07m3

220
Q

An equipment lock is 3ft 6 inches long and 2ft 6 inches in diameter. The chamber is at 320 fsw. What volume of gas is used when the lock is operated?

A

167 ft

33.5 x 1.25² x 3.14 x 9.69 = 166.39 ft3

PS: Remember to convert inches in foot. Divide by 12

221
Q

A transfer chamber is at 90 msw with 4% oxygen. It is bled to 50msw. What is the PP02 at 50 msw?

A

240mb

PS: Remember the important here is where you are at the moment

222
Q

A chamber is at 340 fsw with 4% oxygen. It is bled to 300fsw. What is the PP02 at 300 fsw?

A

0.404 AT

PS: Remember the important here is where you are at the moment

223
Q

Chamber 1 is at 110 msw, with a PP02 of 390 mb. It is blowndown to 140 msw, using 2%. What is the PP02 at 130 msw?

A

If the chamber is at 140m the ppo2 is 450mb

but if

you stop the chamber at 130 msw is 430mb2 x 2% = 0.04390 + 40 = 430mb

224
Q

A chamber is blown down from 150 fsw to 230 fsw using 2%. If the PP02 was 0.4 AT at 150 fsw, what is it at 230 fsw?

A

0.448 AT

225
Q

After a serious leak, a chamber is at 40 msw with a PP02 of 150mb. The chamber is blown back to the living depth of 125 msw using 4%. What will the PP02 be at living depth?

A

490mb

226
Q

Chamber 1 is at 380 fsw, with a PP02 of 0.4 AT. It is bled to 265 fsw without adding oxygen. What is the PP02 at 265 fsw?

A

0.289 AT

227
Q

A chamber is at 155 msw, with an oxygen percentage of 2.5%. It is bled to 123 msw without adding oxygen. What is the PP02 at 123 msw?

A

333mb

228
Q

A chamber and bell are at 180 msw, with a PP02 of 400 mb. For the dive, the bell is separated from the chamber and blown down on 4% to a working depth of 195 msw. What is the PP02 in the bell at working depth?

A

460mb

229
Q

A chamber at 130 fsw, PP02 0.38AT is bled back to surface empty for maintenance. What is the PP02 on the surface?

A

0.077 AT

Not enough oxygen!

230
Q

Chamber 1 has a volume of 12m3 and Chamber 2 has a volume of 8m3. After equalisation, Chamber 1 has a PP02 of 480mb and Chamber 2 a PP02 of 400mb. What is the final PP02 when the atmospheres are completely mixed?

A

448mb

  1. 48 x 12 = 5.760.40 x 8 = 3.25.76 + 3.2 = 8.96 / 20 =
  2. 448 bar or 448 mb
231
Q

Chamber 1 has a volume of 350 ft3 and the PP02 is 0.45 AT. Chamber 2 has the same volume, but the PP02 is 0.4 AT. What is the final PP02 when the chamber atmospheres are fully mixed?

A

0.425 AT

(Short cut: The chamber volumes are the same so you would expect the final PP02 to be half way between the separate chamber PP02s.}

232
Q

Chamber 1 has a volume of 18 m3 and the PP02 is 420 mb. Chamber 2 has a volume of 12 m3 and the PP02 is 400 mb. Chamber 3 has a volume of 15 m3 and the PP02 is 600 mb. What is the final PP02 when all the chamber atmospheres are fully mixed?

A

475mb

  1. 42 x 18 = 7.56
  2. 40 x 12 = 4.8
  3. 60 x 15 = 9

7.56 + 4.8 + 9 = 21.36 / 45 (All chambers volume) = 0.475 bar or 475mb

233
Q

9 divers are in saturation at 80 msw for 24 days. How much oxygen will they use?

A

9 x 24 x 0.72 = 155.52 m3

Note: Metabolic consumption not affect by the depth.
Metric 0.5 litres/minor 30 litres/houror 0.72m3/day per diver
Imperial 0.018 ft3/min or 25 ft3/day per diver

234
Q

6 divers are in saturation at 300 fsw for 28 days. How much oxygen will they use?

A

6 x 28 x 25 = 4200ft3

Note: Metabolic consumption not affect by the depth.

Metric
0.5 litres/minor 30 litres/houror 0.72m3/day per diver
Imperial
0.018 ft3/min or 25 ft3/day per diver

235
Q

6 divers are in saturation for 28 days. How much oxygen will they­ use in the chamber? (Answer in m3)

A

6 x 28 x 0.72 = 120.96m3

Note: Metabolic consumption not affect by the depth.
Metric
0.5 litres/minor 30 litres/houror 0.72m3/day per diver

Imperial
0.018 ft3/min or 25 ft3/day per diver

236
Q

6 divers are in saturation for 12 days. How much oxygen will they use in the chamber? (Answer in ft3.)

A

6 x 12 x 25 = 1800 ft3

Note: Metabolic consumption not affect by the depth.

Metric
0.5 litres/minor 30 litres/houror 0.72m3/day per diver

Imperial
0.018 ft3/min or 25 ft3/day per diver

237
Q

A decompression from 95 msw takes 4 days, with a PP02 of 500 mb. There are 4 divers in the chamber, and the chamber volume is 15m3 Before the decompression starts, the PP02 is 400mb. How much oxygen is used?

A

Metabolic 4 x 4 x 0.72 = 11.52m3

Decompression used 10.5bar

Ln(abs pres) x volume x pp (bar)2.35 x 15 x 0.5 = 17.62m3
Raise O2
0.1 x 15 = 1.5m3
17.62 + 11.52 + 1.5 = 30.64 m3

Note:
Metric
0.5 litres/minor 30 litres/hour or 0.72m3/day per diver
Imperial
0.018 ft3/min or 25 ft3/day per diver
238
Q

A decompression from 150 msw takes 5.5 days, with a PP02 of 480 mb. There are 4 divers in the chamber, and the chamber volume is 12 m3. How much oxygen is used?

A

Metabolic
5.5 x 4 x 0.72 = 15.84m3 Decompression use 16 bar

Ln(abs pres) x volume x pp (bar)2.77 x 12 x 0.48 = 15.95m3
Total = 15.84 + 15.95 = 31.79m3

Note:
Metric
0.5 litres/minor 30 litres/hour or 0.72m3/day per diver
Imperial
0.018 ft3/min or 25 ft3/day per diver
239
Q

A decompression from 80 msw takes 3.25 days, with a PP02 of 500 mb. There are two divers in the chamber, and the chamber volume is 10m3. How much oxygen is used?

A

Metabolic 3.25 x 2 x 0.72 = 4.68m3
Decompression use 9 bar

Ln(abs pres) x volume x pp (bar) 2.19 x 10 x 0.5 = 10.95 m3
Total = 10.95 + 4.68 = 15.63m3

Note:
Metric
0.5 litres/minor 30litres/hour or 0.72m3/day per diverImperial0.018 ft3/min or 25 ft3/day per diver

240
Q

6 divers are in saturation for 18 days at 110 msw, with a PP02 of 400 mb. Before starting the decompression, the PP02 is raised to 500 mb. The decompression takes 4 days. The chamber volume is 15m3. How much oxygen is used?

A

Metabolic use (22 days)

6 x 22 x 0.72 = 95.04m3
Raising 02 level 15 x 0.1 = 1.5m3
Decompression use 18.7 m3 2.48 x 15 x 0.5 = 18.6m3

Total use = 95.04 + 1.5 + 18.6 = 115.14m3

Note:
Metric
0.5 litres/minor 30 itres/hour or 0.72m3/day per diver

Imperial
0.018 ft3/min or 25 ft3/day per diver

241
Q

You have 1800 psi of 1.5%, and you want to turn it into 4%, by pumping in 16%. What will the final pressure of the mixture be ?

a 2366 psi
b 2438 psi
c 1998 psi
d 2175 psi

A

d

1800 / 12 = 150 (factor)
14,5 x 150 = 2175 psi

242
Q

You want to make 3000 psi of 23% using 12% and 50%. What pressure of each gas do you need ?

a 2000 psi of 12 % and 1000 psi of 50 %

b 1978 psi of 12 % and 1022 psi of 50 %

c 2466 psi of 12 % and 534 psi of 50 %

d 2132 psi of 12 % and 868 psi of 50 %

A

d

3000 / 38 = 78.94
27 x 78.79 = 2131.7
11 x 78.79 = 868

243
Q

You have a bank of 20/80 heliox and we require 180 bar of 12/88 heliox. This must be done using 20/80 heliox and pure He. How much of each gas must be added to create the new mix?

a 72 bar 20/80 and 108 bar He

b 90 bar 20/80 and 90 bar He

c 108 bar 20/80 and 72 bar He

d 45 bar He and 135 bar 20/80

A

c

180 / 20 = 9 (factor)

12 x 9 = 108 bar of 20%
8 x 9 = 72 bar of pure He

244
Q

You have 124 bar of 10/90 and require some 12/88. You have 16% available to mix. How much 16% is required?

a 62 bar
b 52 bar
c 48 bar
d 60 bar

A

a

124 / 4 = 31 (factor)2 x 31 = 62 bar of 16%

245
Q

A quad has 1764 psi of pure helium and you require some 6%. How much O2 needs to be added to the helium to achieve this?

a 103 psi
b 100 psi
c 114.7 psi
d 88.2 psi

A

c

1764 / 94 x 6 = 112.59

246
Q

You require 190 bar of 8%. Mixes available are 5% and 12%. How much 5% is required?

a 119 bar
b 109 bar
c 90 bar
d 100 bar

A

b

190 / 7 = 27.14 (factor) x 4 = 108.57 bar

247
Q

You require 190 bar of 10%. Mixes available are 4% and 14%. How much 4% is required?

a 90 bar
b 72 bar
c 86 bar
d 76 bar

A

d

190/10 = 19 (factor) x 4 = 76 bar

248
Q

You want to make 2940 psi of 8% using 4% and 23%. What pressure of each gas do you need?

a 588 psi of 23 % and 2352 psi of 4 %
b 764.4 psi of 23% and 2161 psi of 4 %
c 529.2 psi of 23 % and 2412 psi of 4 %
d 617.4 psi of 23 % and 2321.1 psi of 4 %

A

d

2940/19 = 154.74 (factor)
15 x 154.74 = 2321.1 psi of 4%
4 x 154.74 = 618.7 psi of 23%

249
Q

You have 70 bar of 4%, you require 6% by pumping in 10%. What will the final pressure be?

a 110 bar
b 105 bar
c 95 bar
d 115 bar

A

b

70/4 = 17.5 (factor)

6 x 17.5 = 105 bar

250
Q

A spare quad is filled with the following gases 40 bar of 15%, 15 bar of 3%, 50 bar of 2% and 60 bar of 10%. What will be the final O2% in the quad?

a 5%
b 10%
c 8.2%
d 18.2%

A

c

40% of 15 = 6 15% of 3 = 0.45 2% of 50 = 1

10% of 60 = 66+0.45+1+6= 13.45/165
(total volume bailout)= 0.0815 x 100 = 8.2%

251
Q

A 20m³ chamber at 75 msw with a PPO2 of 400mb is equalised with a 30m³ chamber at 35m with a PPO2 of 500mb. What will be the new depth and PPO2?

a 55m and 450mb
b 41m and 460mb
c 51m and 460mb
d 55m and 475mb

A

c

Depth
8.5 x 20 = 170m3
4.5 x 30 = 135m3
170 + 135 = 305 / 50 = 6.1 -1 x 10 = 51m

PPO2 0.4 x 20 = 8
0.5 x 30 = 15

8 + 15 = 23 / 50 = 0.46bar or 460mb

Note: 50 is the sum of both chambers

252
Q

6 divers are in saturation at 275 fsw for 10 days with a PO² of 0.4 atm. The PO² is then raised to 0.6 atm for a decompression which lasts 2 days. The chamber volume is 500 ft³. How much oxygen is used altogether ?

a 2750 ft³
b 3002 ft³
c 2438 ft³
d 2570 ft³

A

d

Raise PPO2 0.2 x 500 = 100 ft3

Metabolic consumption
12 x 6 x 25 = 1800 ft3

Decompression Ln (abs pres) x ppo2 x chamber volume 2.23 x 0.6 x 500 = 670

670 + 1800 + 100 = 2570 ft3

253
Q

6 divers are in saturation for 12 days. What is the metabolic oxygen daily consumption will they use each? (answer in ft³)

a 45ft³
b 25ft³
c 49ft³
d 35ft³

A

b

Metabolic consumption daily imperial 25ft3 or 0.018ft3/min per diver

254
Q

During a saturation dive to 350 fsw, the divers require a PO2 between 0.5 ats and 0.75 ats. What is a suitable mix?

a Between 4.5.% and 7.5 %
b Between 4.8 % and 7.1 %
c Between 4 % and 6.9 %
d Between 4.3% and 6.5 %

A

d

350+33/33=11.60

  1. 5/11.60= 0.043x100=4.3%
  2. 75/11.60=0.064x100=6.4%
255
Q

If you have a 20m³ chamber at a depth of 80 msw and a 28m³ chamber at a depth of 45m, at what depth will they equalise?

a 70m
b 55m
c 60m
d 65m

A

c

9 x 20 = 180 m3
5.5 x 28 = 154 m3

180 + 154 = 334 m3

334 / 48 = 6.95 - 1 x 10 = 60m

256
Q

A chamber is at a depth of 200 fsw. O2% is 6%. If the chamber is vented back to 60 fsw, what would the % O2 and PPO2 be?

a 6% 0.28 ats
b 21% 0.21 ats
c 13% 0.40 ats
d 6% 0.17 ats

A

d

Percentage always the same so…. 6%
Important where the chamber is at the moment so….

60 / 33 + 1 x 6% = 0.17ats

257
Q

What % O2 would be required for a diver if a PPO2 of 0.7 bar is required? Team 1 is diving at 50 msw. Team 2 is diving at 130 msw.

a Team 1 - 12.5% Team 2 - 3.2%
b Team 1 - 14% Team 2 - 5.4%
c Team 1 - 10% Team 2 - 4.3%
d Team 1 - 11.7% Team 2 - 5%

A

d

Team 1 = 0.7/6= 11.7%
Team 2 = 0.7/14= 5%

258
Q

You have to pressurise a chamber to 250 fsw using 16% and 2% O2. PPO2 required is 0.4atm. How deep will you take the chamber on 16% O2?

a 7.8 ft
b 10.2 ft
c 9.6 ft
d 8.6 ft

A

c

3300 (0.4 - 0.21) - (250 x 2) / 16 - 2 = 9.07ft

259
Q

You vent a chamber from 60 msw to the surface. The O2% was 6%. What is the O2% and PPO2 at the surface?

a 6% and 6mb
b 6% and 210mb
c 21% and 210mb
d 6% and 60mb

A

d

% always the same if vented to surface.
PPO2 = 1x 6%=0.06bar
answer is in mbar so 0.06x1000= 60mb

260
Q

A diver at 340 fsw is breathing a 6% mix. What is the PO² in his mix ?

a 0.698 atm
b 0.678 atm
c 0.729 atm
d 0.619 atm

A

b

340+33/33x6%= 0.678atm

261
Q

What % of gas would be required to perform a dive to 140msw with a PPO2 of 600mb?

a 6%
b 4.3%
c 5.5%
d 4%

A

d

0.6/15= 0.04x100= 4%

262
Q

What is the PPO2 in a bell chamber at a depth of 396fsw, if the O2% is 3%

a 0.30 ats
b 0.36 ats
c 0.39 ats
d 0.45 ats

A

c

396+33/33=13x3%=0.39

263
Q

You have 100 bar of 5% O2. You require 200 bar of 7% O2. Mixes available 4% O2 and 10% O2. You are to use both 4% and 10% to make 5% up to 7%. How much 4% is required?

a. 10 - 14 bar
b. 14 - 15 bar
c. 16 - 17 bar
d. 20 - 25 bar

A

C

PS: Need to make richer mix first so find out how much 10% should be in the mix
Pressure Mix 1 = Final Pressure x (%Final Mix - %Mix 2) / (% Mix 1 - % Mix 2)
Pressure Mix 1 = 200 x (7 - 5) / (10 - 5) = 80bar of mix 1 (10%)So…..

100 bar of 5% = 5
80 bar of 10% = 8
deducting… 100 - 80 = 20 and 8 - 5 = 33/20 = 0.15 = 15% of 20 bar

Using triangle method

20 / 6 = 3.33 factor
5 x 3.33 = 16.66 bar of 4%

OR

Required 200 x 7% = 14
Have 100 x 5% = 5

14 - 5 = 9 and 200 - 100 = 100

9/100 = 0.09 = 9%
Need make 100 bar of 9% using 10% and 4%
Triangle = 16.66 bar of 4%

264
Q

You have to pressurise a DDC from surface to 55 msw using 2% O2 and 8% O2. The pO2 required is 400 mbar. How deep will you take the DDC on 8% O2?

a. 10 msw
b. 13.33 msw
c. 40 msw
d. 48.3 msw

A

b

Remember .21 on chamber

0.4-.21=.19/8%= 2.3-1x10= 13.3

265
Q

You vent a chamber from 75 msw to surface. If the O²% was 4% at 75 msw, what is the O² % and PO² at surface?

a. 4% - 340 mbar
b. 4% - 300 mbar
c. 4% - 40 mbar
d. 21% - 210 mbar

A

c

% dont change when vent a chamber to surface still 4% PPO2 - since chamber is on surface.
1x 4% = 0.04 bar = 40mb

266
Q

You have 120 bar of pure He and require 6% HeO2. Gas available is 100% O2. How much O2 is required?

a. 7 - 8 bar
b. 8 - 9 bar
c. 10 - 11 bar
d. 12 - 14 bar

A

a

120 / 94 = 1.27 factor
6 x 1.27 = 7.65 bar of 100% O2

267
Q

You require 190 bar of 10% O2. Mixes available 4% O2 and 14% O2. How much 4% is required?

a. 7.6 bar
b. 26.6 bar
c. 19 bar
d. 76 bar

A

d

190 / 10 = 19 factor
4 x 19 = 76 bar

268
Q

You are to use a 5% O² mix to establish a pO² of 400mb in a DDC. To what depth would you pressurise on 5%

a. 38 msw
b. 3.8 msw
c. 40 msw
d. 80 msw

A

a

Remember PPO2 chamber 0.21
0.4-0.21= 0.19/5%= 3.8x 10= 38m

269
Q

You pressurise a DDC to 95 msw using a 3/97 mix. What is the helium percentage at 95 msw?

a. 0.315%
b. 97%
c. 87.8%
d. 95%

A

c

9.5 x 97% / 10.5 = 87.7%

270
Q

A diver is working at 140 msw, using a gas reclaim system. What is his gas consumption?

A

75l/min

Gas consumption = Absolute pressure x 5 l/min Gas consumption

15 x 5 L/min= 75 l/min= 0.075 m3/min Gas reclaim

5l/m or 0.18ft/min”

271
Q
  1. A bibs gas contains 12% O2, at what depth would you change it over to prevent the partial pressure dropping below 0.5 bar during decompression ?

a 42 msw
b 55 msw
c 32 msw
d 15 msw

A

c

% x abs.pressure = ppo2

0.5/12%= 4.16 -1 x10= 31.16m

272
Q
  1. A chamber at a depth of 210 msw contained 1.5% O2. If vented to surface what would be the partial pressure of O2?

a 0.15 bar
b 0.015 bar
c 0.0015 bar
d 0.33 bar

A

b

1x1.5%= 0.015bar or 15 mbar

273
Q
  1. A chamber contains 6% O2. What would the pO2 be if it was vented to the surface?

a 6 bar
b 0.6 bar
c 0.06 bar
d 0.006 bar

A

c

1x6%= 0.06bar

274
Q
  1. What % O2 do we require to administer a treatment mix with a pO2 of 2.5 bar at 135 msw?

a. 13%
b. 15.5%
c. 17.2%
d. 18.5%

A

c

2.5/14.5= 0.1724x100= 17.2%

275
Q
  1. What does the reference ± 0.25% marked on a 300 msw gauge, refer to in depth?

a. 0.25 msw
b. 0.5 msw
c. 0.75 msw
d. 0.1 msw

A

c

0.25%x300= 0.75m

276
Q
  1. A DDC is at a depth of 106 msw the analysis at surface show the atmosphere to be 300 ppm CO2. What is the surface equivalent percent?

a. 0.03%
b. 0.318%
c. 0.348%
d. 0.3%

A

c

300/10.000 = 0.03x 11.6= 0.348%

277
Q
  1. What is the total metabolic oxygen consumption for 4 divers for 15 hours at a depth of 75 msw?

a. 255 litre
b. 1020 litre
c. 1800 litre
d. 15,300 litre

A

c

4x15x30=1800

Remember metabolic consumption regardless depth 0.5l/min or 30l/h or
720l/day or 25ft3.

278
Q
  1. What is the pO2 of a DDC at 75 msw, if the O2% is 4%?

a. 0.34 bar
b. 0.3 bar
c. 0.4 bar
d. 0.45 bar

A

a

8.5x4%= 0.34bar or 340mbar

279
Q
  1. Which gas would be required to perform a dive to 86 msw using a pO2 of 1.2 bar?

a. 12%
b. 12.5%
c. 13.4%
d. 14%

A

b

1.2/9.6= 0.125x100= 12.5%

280
Q
  1. Your company’s pO2 during decompression is 0.6 bar. At what depth would the O² reach 21%?

a. 18.6 msw
b. 28.6 msw
c. 2.9 msw
d. 30.0 msw

A

a

0.6/.21= 2.85-1x10= 18.6m

281
Q
  1. You require a treatment mix with a PPO2 of 2.4 ats. At a depth of 75 fsw. What % of gas would you require?

a 60%
b 21%
c 73%
d 100%

A

c

2.4 / 3.27 = 73%

282
Q
  1. Using a 500 msw gauge, what does the reference marked on the gauge ± 0.25% refer in depth?

a 1 msw
b 1.25 msw
c 1.5 msw
d 2.25 msw

A

b

283
Q
  1. A chamber at a depth of 75 msw shows a surface analysis of 600 PPMCO2. What is the CO2% and PPCO2 (in the chamber)?

a 0.5% and 0.5 bar
b 0.05% and 0.003 bar
c 0.06% and 0.005 bar
d 0.6% and 0.4 bar

A

c

600 / 10000 = 0.06%
600 x 8.5 = 5100 / 1.000.000 =
0.0051 bar

284
Q
  1. When increasing the PPO2 from 0.4 bar to 0.5 bar prior to decompression, how much depth would you increase by gauge using pure O2?

a 1.5 metres
b 10 metres
c 1 metre
d 0.5 metres

A

c

285
Q
  1. A gas contains 12% O2 , at what depth would you change it over to prevent the partial pressure dropping below 0.5 bar during decompression?

a 42 msw
b 40 msw
c 30 msw
d 32 msw

A

d

0.5/12%= 4.16 -1= 3.16x10= 31.6m

286
Q
  1. A chamber at a depth of 230 msw contains 1.5% O2. If vented to the surface what would be the partial pressure of O2?

a 150 mb
b 10 mb
c 210 mb
d 15 mb

A

d

1x1.5%= 0.015 x 1000= 15mbar

287
Q
  1. A chamber contains 12% O2. What would be the PPO2 if it was vented to surface?

a 14 mb
b 210 mb
c 120 mb
d 12 mb

A

c

1x12%= 0.12x1000= 120mbar

288
Q
  1. What % O2 do we require to administer a treatment mix with PPO2 of 2.4 bar at a depth of 70m?

a. 25%
b. 28%
c. 30%
d. 34%

A

c

2.4/8= 0.3 x 100= 30%

289
Q
  1. A DDC is at a depth of 106msw. The analysis at surface shows the atmosphere to be 300 PPM CO2. What is the surface equivalent percent?

a 0.03%
b 0.318%
c 0.348%
d 0.3%

A

c

300/10.000 x 11.6= 0.348%

290
Q
  1. You vent a chamber from 75 msw to the surface. If the % O2 was 4% at 75msw, what is the O2 % and PPO2 at surface?

a 4% 340mb
b 4% 300mb
c 4% 40mb
d 21% 210mb

A

c

% is the same when vented to surface
PPO2
1x4%= 0.04 x 1000= 40mbar

291
Q
  1. A body tissue can absorb 0.5 litres of helium when the PPHe is 2 bar. How much can it absorb if the PPHe is 5 bar?
A

1.25 litres

5 x 0.5 / 2 = 1.25 litres

292
Q
  1. A diver at 300 fsw is breathing a 5% mix. What is the PP02?
A
  1. 505 at

10. 09 x 5% = 0.505 at

293
Q
  1. A diver at 150 msw is breathing a 4% mix. What is his PP02?
A

640mb

16 x 4% = 0.64 bar

294
Q
  1. What is the diver’s PP02 if he is breathing a 7% mix at 85 msw?
A

665mb

9.5 x 7% = 0.665 bar

295
Q
  1. A diver is at 200 fsw. If he is breathing a 15% mix, what is his PP02?
A
  1. 06 at

7. 06 x 15% = 1.06 at

296
Q
  1. A diver is breathing an 18% mix at 60 msw. What is his PP02?
A

1.26 bar

7 x 18% = 1.26 bar

297
Q
  1. A chamber is at a pressure of 70 msw and the oxygen percentage 5.2%. What is the PP02 in the chamber?
A

416mb

8 x 5.2% = 0.416 bar

298
Q
  1. The PP02 in a chamber at 165 msw is 400mb. What is the oxygen percentage?
A
  1. 29%

0. 4 / 17.5 = 2.29%

299
Q
  1. The PP02 in a chamber at 400 fsw is 0.42 AT. What is the percentage of oxygen in the chamber?
A
  1. 20%

0. 42 / 13.12 = 3.2%

300
Q
  1. In a chamber at 85 msw, the oxygen percentage reading is 4.5%. What is the PP02 in the chamber?
A

428mb

9.5 x 4.5% = 0.427 bar

301
Q
  1. The C02 reading in a chamber at 90 msw is 250 ppm. What is the PPC02?
A

2.5 mb

250 x 10 = 2500 / 1000.000 = 0.0025 bar or 2.5 mb

302
Q
  1. The C02 reading in a chamber at 300 fsw is 150 ppm. What is the PPC02?
A

0.0015 ata

150 x 10.09 = 1513.5 / 1.000.000 = 0.0015a ata

303
Q
  1. During a saturation dive at 450 fsw, the divers require a PP02 between 0.5 and 0.7 AT. What is a suitable mix?
A

Anything between 3.4% and 4.7%

  1. 5 / 14.63 = 3.4%
  2. 7 / 14.63 = 4.7%
304
Q
  1. If the PP02 must lie between 1.2 and 1.4 bar, what is the greatest depth at which you could use a 12% mix?
A
  1. 6 msw

1. 4 / 12% -1 x 10 = 106.6 msw

305
Q
  1. The divers require a PP02 between 0.5 and 0.7 AT during a saturation dive at 320 fsw. What is a suitable mix?
A

Anything between 4.7% and 6.5%

306
Q
  1. During a saturation dive at 145 msw, the divers require a PP02 between 550mb and 750mb. What is a suitable mix?
A

Anything between 3.5% and 4.8%

307
Q
  1. A diver at 250 fsw is breathing a 15% mix. What is the pO2 in his mix?
A
  1. 287 ata

8. 57 x 15% = 1.28 ata

308
Q
  1. A diver at 125 msw is breathing a 4% mix. What is his pO2?
A

540mb

13.5 x 4% = 0.54 bar

309
Q
  1. A diver at 165 msw is breathing a 4% mix. What is the pO2 in his mix?
A

700mb

17.5 x 4% = 0.7 bar

310
Q
  1. A diver at 340 fsw is breathing a 6% mix. What is the pO2 in his mix?
A
  1. 678 ata

11. 3 x 6% = 0.678 ata

311
Q
  1. A diver at 60 msw is breathing an 18% mix. What is the pO2 in his mix?
A

1.26bar

7 x 18% = 1.26 bar

312
Q
  1. What is the pO2 in a chamber at 50 fsw if the oxygen percentage is 23%?
A
  1. 578 ata

2. 51 x 23% = 0.578 ata

313
Q
  1. The pO2 in a chamber at 108 msw is 400 mb. What is the oxygen percentage?
A
  1. 39%

0. 4 / 11.8 = 3.39%

314
Q
  1. The pO2 in a chamber at 327 fsw is 0.42 ata. What is the percentage of oxygen in the chamber?
A
  1. 85%

0. 42 / 10.90 = 3.85%

315
Q
  1. During a bounce dive to 80 msw, the divers require a pO2 between 1.2 and 1.6 bar. What is a suitable mix?
A

Anything between 13.3% and 17.7%

.1.2 / 9 = 13.3%
1.6 / 9 = 17.7%

316
Q

A 4.2m3 bell weights 3.85 tonnes, how much water will it displace ?

a) 4.16 m3
b) 4.08 m3
c) 3.96 m3
d) 3.74 m3

A

D

3.85 / 1.03 = 3.74 m3

317
Q

What weight is required to be added to a 5.04 tonne bell with a volume of 5.8 m3 to make it negatively buoyant by 500 kg?

a) 934 kg
b) 1234 kg
c) 1434 kg
d) 1634 kg

A

C

5.8 x 1.03 = 5.974 - 5.04 = 0.934 + 0,5kg = 1.434T or 1434Kg

318
Q

A pontoon is 20 m long x 15 m wide and 4m high it weighs 390 tonnes. What is the pontoon freeboard and how much extra weight can be added before the pontoon exceeds the 1m minimum freeboard requirement placed on it by the marine engineers?

FREEBOARD

a) 2.74 m
b) 1.63 m
c) 3.1 m
d) 1.1 m

WEIGHT

g) 537 tonne
f) 23 tonne
e) 750 tonne
h) 1000 tonne

A

A and G

390 / 20 x 15 x h x 1.03 = h = 1.26 m
4 - 1.26 = 2.74m

Weight 20 x 15 x 3 x 1.03 = 927 - 390 = 537 tonnes

319
Q

A diver on BIBS breathes 40 litre per minute. He has four cycles of 25 minutes at 40m. How much gas is used?

a) 20m3
b) 4m3
c) 16m3
d) 5m3

A

A

320
Q

A chamber has a floodable volume of 26 m3. How much gas is required to pressurise it to 136m?

a) 3,536 litres
b) 489.6m3
c) 379.6m3
d) 353.6m3

A

D

321
Q

On surface, the volume of an open bottom bell is 2,000 litres. If this bell is submerged to a depth of 69 msw, what has the gas volume been reduced to?

a) 289.9 litres
b) 253.16 litres
c) 28.99 litres
d) 25.316m3

A

B

322
Q

On decompression, the pO2 is maintained at 600 mbar. At what depth will the O2 level be at 21%?

a) 28.57 msw
b) 18.57 msw
c) 2.857 msw
d) on surface

A

B

323
Q

A chamber at 92m has O2 equal to 5.7% and PPO2 of 581 mb. If the chamber is vented to 74m what are the resulting O2 levels?

a) 6.9% and 581 mb
b) 5.7% and 581 mb
c) 5.7% and 479 mb
d) 6.5% and 479 mb

A

C

324
Q

What is the PPO2 of a DDC at 50 fsw if the O2 % is 21%?

a) 0.53 atm
b) 0.32 atm
c) 0.21 atm
d) 0.79 atm

A

A

325
Q

A diver is being treated for decompression sickness at 165 fsw using a gas with a PPO2 of 1.5 Bar. What % is the gas?

a) 30%
b) 20%
c) 25%
d) 8.6%

A

C

326
Q

A diver in the chamber at 50 fsw takes a Draeger tube reading of 0.5% (this is also know as surface equivalent pressure). What is your Draeger tube reading on the surface?

a) 0.125% (1250 ppm)
b) 0.1% (1000 ppm)
c) 0.25% (2500 ppm)
d) 0.2% (2000 ppm)

A

D

SEP = % / AP = 0.5 / 2.51 = 0.2% OR 2000PPM

327
Q

If you double the pressure (and therefore density) on a gas, the volume will be halved. Whose law is this?

a) Henry’s
b) Boyle’s
c) Dalton’s
d) Charles’

A

B

328
Q

A liquid has absorbed 1 litre of a gas on the surface. How much will it absorb if the pressure is increased to a gauge depth of 30 msw?

a) 1 litre
b) 3 litres
c) 4 litres
d) 30 litres

A

C

329
Q

A diver at a depth of 362ft is breathing a 5% mix. What is the PPO2?

a) 0.55 ATA
b) 0.47 ATA
c) 0.6 ATA
d) 0.72 ATA

A

C

330
Q

A chamber with an FV of 28m3 is pressurised to 113 msw. The temperature on arrival at depth is 37ºC. The temperature dropped to 27ºC. What is the new depth?

a) 92 msw
b) 109 msw
c) 100 msw
d) 126 msw

A

B

331
Q

A chamber at depth shows an oxygen level of 7.6% and a PPO2 readout of 0.42 Bar. At what depth is the chamber?

a) 18.1 msw
b) 45 msw
c) 42 msw
d) 55 msw

A

B

332
Q

If you pressurised a chamber with an FV of 1,225 ft3 to 237 ft, how much gas is required?

a) 2,903.25 ft3
b) 10,023 ft3
c) 29,032.5 ft3
d) 8,798 ft3

A

D

333
Q

What is the PPO2 of a DDC at 300 fsw if O2% is 5.2%?

a) 0.52 ATS
b) 0.4 ATS
c) 0.47 ATS
d) 1.56 ATS

A

A

334
Q

If you increase the depth of a chamber by 2m when adding pure oxygen, you will raise the PPO2 by how much?

a) 20 mb
b) 200 mb
c) 2 Bar
d) 400 mb

A

B

335
Q

How much air will be required to pressurise a chamber with an FV of 300 ft3 to a depth of 165 ft?

a) 1,800 ft3
b) 150 ft3
c) 18,000 ft3
d) 1,500 ft3

A

D

336
Q

A diver suffering from Arterial Gas Embolism is recompressed in order to reduce the size of the bubble causing the embolism. What law is this treatment based on?

a) Dalton’s
b) Henry’s
c) Charles’
d) Boyle’s

A

B

337
Q

How long will a bail out bottle of 4 litre pressurised to 276 bar last a diver at 95 msw breathing 35 l/min with the demand valve second stage set to 12 bar

a) 2.52 min
b) 2.76 min
c) 2.87 min
d) 3.01 min

A

B

1014 / 367.5 = 2.76 min

338
Q

Boyle’s Law states:-

a) at constant temperature, the volume of a given mass of gas is inversely proportional to the absolute pressure.
b) at constant temperature, the volume of a given mass of gas is inverselyproportional to its partial pressure.
c) at constant temperature, the volume of a gas will increase by a factor of two if the pressure doubles.

A

A

339
Q

A. What is the floodable volume of a 48 bottle quad? (Each bottle = 50 litres).

B. How much gas is required to pressurise a 22m3 chamber to 89 msw?

C. If the gas is supplied from a 48 bottle quad with a start pressure of 185 bar, what is the final pressure?

A

A. 48 x 50 = 2400 LTS = 2.4m3

B. 22 x 8.9 = 195.8m3

C. 2.4 x 185 = 444m3 - 195.8m3 = 248.2m3 / 2.4 = 103BAR

or

185 - (195.8/2.4) = 103 bar

340
Q

What is the gas consumption of a diver working at 108 msw for a duration of 1 hour 23 minutes?

a) 31,374 litres
b) 507.99 litres
c) 34.28 m3
d) 4,649.4 litres.

A

C

35 LTS x 83 MINS x 11.8 BAR (A) = 34279 LTS

341
Q

A diver has a 9 litre bail out at a pressure of 195 bar. If he is diving at a depth of 98 msw, how many minutes will his bail out last ? (Hat working pressure = 8 bar).

If the above diver breathes from his bail out for a period of 1.5 minutes what will the resulting pressure be?

A

195 - (10.8 + 8) + 176.2 x 9 = 1585.8LTS / 40 x 10.8 = 3.67 = 3 MINS 40 SECS

1.5 x 10.8 x 40 = 648 LTS / 9 = 72 BAR
195 -72 = 123 BAR

342
Q

A 10l bailout is charged to 300 bar. The temperature reaches 45°C. After lying on deck for half an hour the temperature drops to 18°C. What will be the pressure reading?

A

274.5 BAR

343
Q

After compressing six divers in a 20m3 chamber to 50msw the temperature reading is 90°F. After 5 minutes the temperature has dropped to 79°F. What will be the drop in depth of the chamber?

A

1.2 meters

344
Q

Convert 672 ppm to %. Is it:

a) 6.72%
b) 0.672%
c) 0.0672%
d) 0.00672%

A

C

345
Q

After pressurisation to a depth of 108 msw the temperature in a chamber drops from 37°C to 20°C. If the floodable volume of the chamber is 18m3 how much gas is required to return the chamber to the original storage depth?

A

New depth : 102 msw
Gas to blowdown 10.8m3
108 - 102 = 6msw

0.6 x 18M3 = 10.8M3

346
Q

What pressure is exerted at 2 atmospheres absolute (i.e.10 msw)

a) 29.4 psi(a)
b) 29.0 psi(g)
c) 0.89 psi(a)
d) 2 Bar(g)

A

A

347
Q

1 Bar is equivalent to

a) 760 mm Hg
b) 750 mm Hg
c) 770 mm Hg
d) 740 mm Hg

A

B

348
Q

A cylinder and piston arrangement contains gas at 10 psi. The temperature is held constant while the volume is reduced from 10 to 5 cubic feet by pushing down the piston. What is the resulting pressure of the gas cylinder? (Circle the correct answer)

a) 5 psi
b) 10 psi
c) 20 psi
d) 50 psi

A

C

349
Q

Air in an open bottom container at constant temperature occupies 24m3 as sea level. What is the volume of air at a depth of 20 msw?

a) 12 m3
b) 16 m3
c) 8 m3
d) 24 m3

A

C

350
Q

Given constant temperature, a flexible container of air with 50 litres volume at a depth of 3 msw. Now, the container has 20 litres at a certain depth in the sea. What was the pressure on the container at the that point?

a) 3.25 bar
b) 0.52 bar
c) 2.55 bar
d) 0.75 bar

A

A

P1V1 = P2V2 =1.3 x 50 = ? x 20 = 1.3 x 50/20 =
3.25 BAR(A)

351
Q

Given a constant temperature, what is the volume of air in a balloon at 20 msw if it contained 500 litres of air at sea level?

a) 16.7 l
b) 500 l
c) 250 l
d) 166.7 l

A

D

352
Q

How much air will be required to pressurise a chamber with a FV of 12m3 to a depth of 50 msw?

a) 72 m3
b) 12 m3
c) 60 m3
d) 600 m3

A

C

353
Q

A quad has 16 bottles each of 50 litre. How much air is in them at a pressure of 190 bar?

a) 15.2 m3
b) 0.8 m3
c) 152 m3
d) 15 m3

A

C

0.8m3 x 190 = 152m3

354
Q

We have 90 bar of 5/95 and we require 200bar of 8/92 heliox. You have 3% and 12% available, use both to make up the 5% to 8% How much 3% must be added to give the required mix?

a) 11 bar
b) 19 bar
c) 9 bar
d) 10 bar

A

B

200 x 8% = 16
90 x 5% = 4.5
11.5 / 110 = 10.45%

You need to make 110 bar of 10.45% using 3% and 12%

Triangle: 110 / 9 = 12.22 x 1.55 = 19 bar of 3% and 91 bar of 12%

Checking up your answer
5% x 90 = 4.5
12% x 91 = 10.9
23% x 19 = 0.57

4.5 + 10.93 + 0.57 = 16

355
Q

A diver at a depth of 25 msw breathes a 40/60 nitrox. What is the pO2?

a) 1.4 bar
b) 100 mbar
c) 0.14 bar
d) 1 bar

A

A

356
Q

If a chamber with a pO2 of 0.4 bar at 67 msw is vented to the surface, what is the resulting pO2?

A
  1. 4 / 7.7 = 5.2%

5. 2% x 1 BAR(A) = 0.052BAR or 52mb

357
Q

A Pontoon is 15m long by 4m wide and 2m deep. It draws a draft of 1.5m. What is the upthrust?

Upthrust = Volume x Density

a) 123.6 tonne
b) 120 tonne
c) 46.35 tonne
d) 92.7 tonne

What is the weight of the pontoon?
How much weight can be placed on the pontoon before it sinks?

a) 120 tonne
b) 30.9 tonne
c) 31 tonne
d) 59 tonne

A

D

1.5 x 15 x 4 = 90 m3 x 1.03te = 92.7t

Neutrally Bouyant weight of water displaced = weight of object
15 x 4 x 2 = 120 x 1.03te = 123.6t

B

0.5 x 15 x 4 = 30m3 x 1.03te =30.9t

358
Q

A diving bell displaces 5m3 of sea-water and weighs 4.8 tonnes. Is the bell positive or negatively buoyant?

A

U = V x D

5 x 1.03 = 5.15- 4.8= 0.35 + BOUYANCY

359
Q

A diving bell receives an up thrust of 14,250 lbs and weighs 6.5 ton. How much buoyancy is required to make it 300 lbs positive?

a) 300 lbs
b) 510 lbs
c) 610 lbs
d) 560 lbs

A

C

6.5 x 2240 = 14560 lbs - 14250 = 310 + 300 = 610 lbs

360
Q

What is the partial pressure of Nitrogen in Air at a depth of 132 fsw?

a) 3.95 ATA
b) 2.95 ATA
c) 4.95 ATA
d) 0.79 ATA

A

A

361
Q

How much 16/84 and 3/97 is required to make 200 bar of 10/90:

a) 146.2 bar 16% and 53.8 bar 3%
b) 92.3 bar 3% and 107.7 bar 16%
c) 53.8 bar 16% and 146.2 bar 3%
d) 107.7 bar 3% and 92.3 bar 16%

A

B

362
Q

You wish to make 180 bar of 15% using 20/80 and 5/95 oxyhelium. Whatpress of 5/95 do you require:

a) 80 bar 5/95
b) 30 bar 5/95
c) 120 bar 5/95
d) 60 bar 5/95

A

D

363
Q

We have 130 bar of 5/95 and we require 15/85 oxyhelium. How much O2 must be added to give the required mix:

a) 16.3 bar
b) 14.2 bar
c) 15.3 bar
d) 17.4 bar

A

C

364
Q
  1. A spare quad is filled with the following gases:

40 bar 15/85 O2 He
50 bar 2/98 O2 He
15 bar 3/97 O2 He
60 bar 10/90 O2 He

What will the % O2 be in the quad:

a) 8.15% O2
b) 9.1% O2
c) 10.3% O2
d) 12.6% O2

A

A

365
Q

You have a quad containing 70 bar of 4/96 O2 He. You wish to make 12/88 O2 He by adding some 23/77 O2 He. What will the final pressure be of the new mix:

a) 110.8 bar
b) 115.4 bar
c) 120.9 bar
d) 130.4 bar

A

C

366
Q

You have a bank of 20/80 O2 He and we require 180 bar of 12/88 O2 He. How can this be done by using 20/80 O2 He and pure He. How much of each gas must be added to create the new mix.

a) 72 bar He and 108 bar 20/80
b) 108 bar He and 72 bar 20/80
c) 90 bar He and 90 bar 20/80
d) 45 bar He and 135 bar 20/80

A

A

367
Q

You wish to increase the depth of a chamber from 105 msw to 153 msw using 2/98HeO2. If the pO2 is 0.4 at 105 msw, what will it be at the new depth, and what will be the % O2 be?

a) pO2 0.4 bar and 2.6%
b) pO2 0.4 bar and 2.45%
c) pO2 0.496 bar and 3.24%
d) pO2 0.496 bar and 3%

A

D

368
Q

Prior to decompression a 20 m3 chamber has a depth of 90 msw and chamber O2 % of 4.3%. How much O2 must be added to increase the pO2 to 600 mbar for final decompression?

a) 3.0 m3
b) 3.4 m3
c) 30 m3
d) 34 m3

A

B

369
Q

If a DDC has 8% O2 at 40 msw and has to be blown down to 130 msw, what will the new pO2 be if the pressurisation gas is 2% O2?

a) 1.58 bar pO2
b) 0.58 bar pO2
c) 1.530 bar pO2
d) 0.510 bar pO2

A

B

370
Q

After pressurisation of a 20 m3 chamber to 140 msw with 4 divers the chamber pO2 is 0.495 bar. How long will it take for the chamber to reach 0.415 bar if the metabolic the consumption rate of a diver is 0.5 l/min?

a) 53 h 20 min
b) 133 h 20 min
c) 12 h 20 min
d) 13 h 20 min

A

D

0.495 - 0.415 = 0.08 X 20 = 1.6m3 total consumption or

1600 litres

0.5 x 4 = 2l/min

1600 / 2 = 800 / 60 = 13.33 hours = 13h and 20 min

371
Q

During a split saturation a living chamber with a volume of 23 m3 is at a depth of 60 msw with a pO2 of 0.42 bar. The transfer lock is at depth of 118 msw with an O2 percentage of 3.2%. The transfer lock is then decompressed to 60 msw but is not equalised. How much O2 must be added to make a pO2 of 0.42 bar before the transfer lock can be equalised with the main living chamber?

a) 430 mbar
b) 228 mbar
c) 196 mbar
d) 100 mbar

A

C

372
Q

A chamber is at 120 msw and contains 410 mbar O2. Calculate the pO2 after thefollowing depth changes:

a) Vent to 90 msw.
b) Recompress (from 90 msw) to 120 msw using 2%

A

A. 3.15% X 10 = 315mb

B. 375 mb

373
Q

Six divers in sat for 10 days at 180 msw. What volume of O2 will they use(Metabolic consumption is taken as 0.5 l/min).

A

6 x 0.72 x 10 = 43.2 m3

374
Q

Pressurise a chamber from surface to 63 metres using 5/95 heliox.

  1. Give the pO2 and the % O2.
  2. Increase the storage depth to 100 metres using 2/98. Give the pO2 and % O2.
  3. Vent the chamber to 30 metres. Give the pO2 and % O2.
A
  1. ppo2 at 63m = 0.525 bar and O2 % is 7.2%
  2. pp02 at 100m = 0.599 bar and O2 % is 5.4%
  3. 4 x 5.4% = 0.216 bar and percentage still the same 5.4%
375
Q

There is 100 bar of pure Helium, how much pure Oxygen is required to make 8/92

a) 100 bar
b) 91.3 bar
c) 18.7 bar
d) 8.7 bar

A

D

376
Q

You wish to make 180 bar of 8% using 2/98 and 12/88 heliox. What pressure of 2/98 do you require?

a) 72 bar
b) 30 bar
c) 118 bar
d) 60 bar

A

A

377
Q

You wish to pressurise a 20 m3 chamber to a depth of 70 msw and establish a pO2 of 0.4 bar. Using 12/88 and 2/98 HeO2:

  1. To what depth would your initial pressurisation be using 12/88?
  2. What would be the total amount of 2% used to pressurise the chamber to 70 msw?

1a) 3 msw.
1b) 3.5 msw.
1c) 5 msw.
1d) 6 msw.

2a) 130 m3
2b) 135 m3
2c) 140 m3
2d) 145 m3

A

1 C

2 A

378
Q

One atmosphere is equal to the gauge pressure at a depth of: (circle the correct answer):

A. 10 feet
B. 14.7 feet
C. 33 feet
D. 66 feet

A

C

379
Q

If the gauge pressure is 29.4 psi, what is the absolute pressure:

A. 14.7 psi
B. 44.1 psi
C. 29.4 psi
D. 58.8 psi

A

B

380
Q

If a container is taken to a depth of 66 feet, what is the absolute pressure on thecontainer? (Circle the correct answer).

A. 1 atmosphere
B. 2 atmospheres
C. 3 atmospheres
D. 4 atmospheres

A

C

381
Q

If a balloon is taken to a depth of 66 feet, what will be the volume of the balloon?(Circle the correctanswer).

A. 100% - the original volume
B. 50% - 1/2 the original volume
C. 33% - 1/3 the original volume
D. 25% - 1/4 the original volume

A

C

382
Q

A cylinder is filled with a mixture of 10% oxygen and 90% nitrogen at sea level. What is the partial pressure of the oxygen? (Circle the correct answer).

A. 14.7 psi
B. 29.4 psi
C. 7.35 psi
D. 1.47 psi

A

D

383
Q

Air in an open bottom container at constant temperature occupies 48 cubic feet as sea level. What is the volume of air at a depth of 66 feet? (Circle the correct answer).

A. 12 cubic feet
B. 16 cubic feet
C. 18 cubic feet
D. 24 cubic feet

A

B

384
Q

I have 10% oxygen on line to BIBS. What depth will I change over to prevent thePPO2 falling below 0.16 Bar?

A

6 meters

385
Q

A DDC contains 12% oxygen at 93 fsw - what is the oxygen % in the DDC if ventedto surface, also what is the pO2 at surface?

A

Still 12%

PPO2 = 1 x 12% = 0.12 bar or 120 mb

386
Q

I need a pO2 of 0.4 in the chamber pressurising from surface using 5% O2, how deep would I go to achieve this?

A
  1. 4 - 0.21 = 0.19 bar
  2. 19 / 5% = 3.8 x10 = 38 meters

PS: remember took already out the surface pressure

387
Q

A DDC at a depth of 300 fsw contains 5% O2 if vented to 100 fsw, what is the pO2?

A

4.03 x 5% = 0.201bar or 201 mb

388
Q

A bell receives an upthrust of 16,000 lbs and weighs 7.3 tons, how much extrabuoyancy is required to make it positive by 500 lbs?

A

7.3 x 2240 = 16352 lbs havier

16352 + 500 = 16852 - 16000 = 852lbs extra

389
Q

In order to make 175 bar of 12% oxygen mix using 2% and 16% oxygen mixtures, how much 2% oxygen mix will be needed.

a. 44 bar
b. 46 bar
c. 48 bar
d. 50 bar

A

D

390
Q

A quad of heliox contains pure helium at 128 bar. We need 198 bar of 4.5% oxygenmix and have only pure helium and pure oxygen available.The pressure of oxygen needed to be added will be between:

a. 5-6 bar
b. 6-7 bar
c. 8-9 bar
d. 9-10 bar

A

C

4.5 x 1.98 = 8.91 bar

391
Q

Two heliox mixes 2% oxygen and 5% oxygen are used to pressurise a DDC from the surface to 65 msw. In order to arrive with a PPO2 of 400 mb the system must be pressurised with 5% oxygen to which one of the following depths:

a. 13 msw
b. 20 msw
c. 30 msw
d. 83 msw

A

B

392
Q

We have a He/O2 mixture containing 20% O2 and wish to add a mixture containing 2%O2 in order to obtain a new mixture containing 14% O2 at 200 bars, the start pressure of the 20% O2 mixture will have to be:

a. 80-90 bar
b. 110-120 bar
c. 130-140 bar
d. 145-150 bar

A

C

393
Q

The following group of questions (1-5) refer to the situation described below: A DDC is pressurised from surface to 140m on a variety of gas mixtures as follows:

From 0 m - 20 m on 6% O2
From 20 m - 57 m on 3% O2
From 57 m - 92 m on 2.5% O2
From 92 m - 140 m on 2% O2

Which one of the following is the correct PPO2 at the end of each stage of pressurisation.

  1. At 20m
    a. 0.12
    b. 0.33
    c. 0.41
    d. 0.532.

At 57 m

a. 0.23
b. 0.37
c. 0.44
d. 0.78

  1. At 92 m
    a. 0.32
    b. 0.35
    c. 0.53
    d. 0.92
  2. At 140 m
    a. 0.42
    b. 0.63
    c. 1.13
    d. 1.40
  3. Which one of the following is the correct % on arrival at 140 m
    a. 2.0% - 2.3%
    b. 2.8% - 2.9%
    c. 3.0% - 3.1%
    d. 4.1% - 4.2%
A
  1. B
  2. C
  3. C
  4. B
  5. D
394
Q

A DDC is pressurised from the surface to a depth of 88 msw with a mixture of 98%helium and 2% oxygen. The helium percentage at 88 msw is:

a. 88%
b. 92%
c. 86%
d. 98%

A

A

395
Q

A DDC has a floodable volume of 18.6 m3. The volume of gas needed to pressurise the DDC to 136 msw from surface is:

a. 234.36 m3
b. 252.96 m3
c. 266.56 m3
d. 271.56 m3

A

B

396
Q

Which of the following is the total volume of metabolic oxygen consumed by six divers at depth of 95 msw over a period of 18 hours:

a. 567 litres
b. 3,240 litres
c. 30,780 litres
d. 34,020 litres

A

B

397
Q

In feet gauge how much oxygen is required to raise the PPO2 of a chamber from 0.26ATS to 0.4 ATS when at an initial depth of 268 fsw:

a. 1.28 fsw
b. 4.66 fsw
c. 9.14 fsw
d. 42.14 fsw

A

B

0.14 / 0.03 = 4.66fsw

PS:
10cm of O2 will increase the PO2 by 10mb or 0.01 bar
1m of O2 will increase the PO2 by 100 or 0.1 bar
fsw of O2 will increase the PO2 by 0.03atm

398
Q

The atmosphere of a chamber at a saturation depth of 428 fsw is analysed at the surface for carbon dioxide. What is the percentage surface equivalent if the analyser records 400 ppm:

a. 0.558% surface equivalent
b. 0.040% surface equivalent
c. 0.518% surface equivalent
d. 0.400 surface equivalent

A

A

428 fsw = 13.96ata
13.96 x 400 / 10000 = 0.558%

PS: Check this question. because says analysed at surface.

399
Q

A DDC is at depth of 97 msw, the atmosphere when analysed at the surface show aCO2 concentration of 380 ppm, the percentage surface equivalent would be:

a. 0.004%
b. 0.046%
c. 0.406%
d. 4.060%

A

C

10.7 x 380 / 10000 = 0.406%

400
Q

A DDC is to be pressurised from the surface using a He/O2 mixture containing 3% O2,we wish to establish a PPO2 of 450 mb, the DDC would be pressurised to:

a. 1.5 msw
b. 70.0 msw
c. 80.0 msw
d. 150.0 msw

A

C

0.45 - 0.21 = 0.24 / 3% = 8 x 10 = 80m

401
Q

If a DDC containing 6% O2 is vented from a depth of 85 msw, to the surface the O2and the PPO2 on the surface will be:

a. 6% 60 mb
b. 6% 600 mb
c. 6% 720 mb
d. 21% 210 mb

A

A

PS: percentage still the same PPO2 is where you are at the moment

402
Q

A bounce dive will be carried out at a depth of 286 fsw, breathing a gas containingPPO2 of 1.2 ATS, which one of the following will be required:

a. 11.5%
b. 12.4%
c. 13.8%
d. 14.6%

A

B

1.2 / 9.66 = 12.4%

403
Q

If your pressurise a DDC from the surface to 297 fsw using a He/O2 mixture containing 3% oxygen, the helium percentage at 297 fsw will be:

a. 80.0%
b. 87.3%
c. 93.0%
d. 97.0%

A

B

9 x 97% = 8.73 / 10 = 87.3%

404
Q

A DDC with a floodable volume of 300 cu ft is to be pressurised from the surface to165 fsw. The volume of air required will be:

a. 300 cu. Ft
b. 1500 cu. Ft
c. 1800 cu. Ft
d. 2000 cu. Ft

A

B

5 x 300 = 1500cuft

405
Q

A DDC at 297 fsw is maintained at a PPO2 of 0.4 ATS. Which one of the followingwill be the PPO2 if the DDC is vented to 66 fsw.

a. 0.08 ATS
b. 0.12 ATS
c. 0.20 ATS
d. 0.40 ATS

A

B

297 / 33 + 1 = 100.4/10 = 4%66/33 + 1 = 3 x 4% = 0.12 ats

406
Q

We need 200bar of 10/90 using pure HE & 02

A

20 bar of O2
180 bar of He

Pressure Mix 1 = FP X ( %M2 - %MF ) / ( %M2 - %M1 )
Pressure Mix 1 = 200 X ( 100 - 10 ) / ( 100 - 0)
Pressure Mix 1 = 200 X 90 / 100
Pressure Mix 1 = 180bar
Pressure Mix 2 = 20bar

or

200 / 100 = 2 x 10 = 20 bar of O2
The rest is He

407
Q

We need to correct 150bar ( weak mix ) of 8/92 to 10/90. We have unlimited O2.

A

3.33 bar of 100%

Pressure Mix 1 = FP X ( %M2 - %MF ) / ( %M2 - %M1 )
150 = FP X ( 100 - 10 ) / 100 - 8
150 = FP X 90 / 92
50 X 92 = FP / 90
Therefore : 153.3 = FPSo 02 to add = 153.3 bars
150 bars = 3.3 bars 02

or

150/90 = 1.66 x 2 = 3.33 bar of 100%

408
Q

We need to correct 180 bar of 12/88 to 10/90, we have ample pure HE

A

36 bar of He

180/10 = 18 x 2 = 36 bar of He

409
Q

Mixing two mixes together 88bar of 10/90 & 112bar of 14/86

A

12.24%

Pressure Mix 1 = FP X ( %M2 - %MF ) / (%M2 - %M1)
88 = 200 x ( 14 - %MF ) / (14 - 10)
88 = 14 - % MF 200 / 4
88 X 4 = 14 - % MF / 200Â
352 = 14 - % MF / 200
1.76 = 14 - % MF
%MF = 14 - 1.76
%MF = 12.24% Ans Final Mix % = 12.24%

or

88 x 10% = 8.8
112 x 14% = 15.6
88.8 + 15.68 = 24.48/200 = 12.24%

410
Q

We have 150 Bar of 2/98 to correct to 3% using a 50/50 Heliox mix?

A

3.19bar of 50/50

Pressure M1 = FP X ( %M2 - %MF ) / ( %M2 - %M1 )
150 = FP X ( 50 - 3 ) / ( 50 - 2 )
150 = FP X 47 / 48
FP = 150 X 48 / 47
FP = 153.16 Bars
Ans 153.16 - 150 = 3.19 Bars of 50/50 to add

or

150 / 47 = 3.19 factor x 1 = 3.19bar of 50/50

411
Q

Mixing 3 gases ( or more )

80 bar 10/90,
40 bar 8/92,
60 bar 7/93

A

180 bar of 8.5%

80 x 10% = 8
40 x 8% = 3.2
60 x 7% = 4.2

Add all Equals 15.4 / 180 = 8.5% of 180 bar

412
Q

An onboard gas storage system has 30 kellys, each with an internal volume of 1.5 m3. If each kelly is pressurised to 180 bar, how much gas can be stored?

A

8100m3

30 x 1.5 x 180 = 8100m3

413
Q

A sat system consists of 3 chambers of 14m3 each, 1 tup 8m3 and 1 bell 6m3. How much gas is needed to blow the system down to 180 msw ?

A

1008m3

3 x 14 = 42 + 8 + 6 = 56m3
18 x 56 = 1008m3

414
Q

A bell contains 0.04% CO2 at 74msw. What is the PP of CO2 in the bell ?

A
  1. 0033%

8. 4 x 0.04% = 0.0033%

415
Q

After pressurisation to 145msw a chamber has a temperature of 32C. If the chamber cools to 27C, What will the depth be in the chamber?

A

142.6m

145 x 300 / 305 = 142.6m

416
Q

You want to make 200 bar of 23 % mix using 4% and 50 %. How much 4 % will you need?

A

117 bar of 4%

200/46 = 4.34 x 27 = 117bar of 4% and 83 bar of 50%

417
Q

A bounce dive is to be carried out using a PPO2 of 1200mb What % is needed at 75 msw?

A
  1. 1%

1. 2 / 8.5 = 14.1%

418
Q

Two divers are working out of the bell, each breathing 40l/m. They are at 95msw and are supplied by a quad containing 400m3 of available gas. How long can they work for?

A

7h and 56min

2 x 40 x 10.5 = 840l

400 000/ 840 = 476.19min

419
Q

What is the ppo2 in a 12% mix at 80msw?

A

1.08bar

420
Q

A chamber system has an internal volume of 42m3, how much gas is needed to blow it down to 185msw?

A

777m3

18.5 x 42 = 777m3

421
Q

The Co2 % read in the bell by the bellman is 4%. What is the PPCo2 in mb?

A

40mb

4/100 = 0.04 x 1000 = 40mb

422
Q

If the maximum PPo2 allowed is 1.6bar how deep can a diver work using 9% mix?

A
  1. 7m

1. 6/9% - 1 x 10 = 167.7m

423
Q

A chamber atmosphere contains 850 ppm of Co2. What is the Partial pressure of Co2 at 112msw?

A

0.0102bar

850 x 12.2 = 10370 / 1000000 = 0.0102bar

424
Q

A gas quad is pumped up to 200bar when the ambient temperature is 20C. During the day the temperature rises to 33C. What is the pressure in the quad?

A

208.8bar

425
Q

Which USN partial pressure table would you use for a dive to 200ft on a 20% mix?

A

?

426
Q

What volume of gas will a diver use for a dive lasting 6 hours at 140msw if he breathes 30 l/pm?

A

162m3

360 x 30 x 15 = 162m3

427
Q

A bounce dive is to be carried out using a PP02 of 1.4ats at 330ft. What mix is required?

A
  1. 7%

1. 4/11 = 12.7%

428
Q

A chamber is at a temperature of 34C after pressurisation to 300ft. What would its depth be if it cooled to 27C?

A

293ft

429
Q

What is the Ppo2 in a 6% mix at 290ft?

A

0.587ata

430
Q

A diver at 600ft breathes 1ft3 pm. How long can he work for if he has 6000ft3 of available gas.

A

5h and 12 min

6000/19.18 = 5.21h = 5h 12min

431
Q

You want to establish a ppo2 of 0.5 bar in a chamber at 100msw. You have 12% & 2% mixes available. What depth of 12% Should you put in?

A

9m

(500 - 210) - (100 x 2) / (12 -2) = 9m

432
Q

You want to establish a ppo2 of 0.45 in a chamber at a depth of 100msw. Gases available are 2/98 and 10/90, what depth would you use the two gases?

A

5m

(450 - 210) - (100 x 2) / (10 - 2) = 5m

433
Q

If 200 psi of oxygen is added to 2600 psi of 5%, what will be the final O2% of this mix?

(a) 10%
(b) 18%
(c) 11.8%
(d) 13.6%

A

C

Bell Diving Supervisors QUESTIONS (15 decks)

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