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The level of math required for diving is fairly basic so you shouldent have much of a problem.

Is there a particular question your strugling with?

ATB

Mark

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Q1 What volume of air in a lifting bag will hold a mooring with a volume of 20 litres and a mass of 100 kg neutrally bouyant in freshwater ?

The answer is 83 litres how is this calculated?

Q2 Your club RIB has a design burst pressure of 2 bar . You pump it to the recommended 1.75 bar pressure on a cool morning with a temperature of 7 c . The day turns out to be hot and sunny .How hot must the boat get to be at risk of bursting should the relief valves fail ?

the answer to this is 47 c also how is this calculated ? Help much appreciated. Thanks .

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Possibly the rise in temp 47c - 7c = 40 cHere is two of the questions from the 2008 exam

Q2 Your club RIB has a design burst pressure of 2 bar . You pump it to the recommended 1.75 bar pressure on a cool morning with a temperature of 7 c . The day turns out to be hot and sunny .How hot must the boat get to be at risk of bursting should the relief valves fail ?

the answer to this is 47 c also how is this calculated ? Help much appreciated. Thanks .

pressure required to burst 2.0 bar

Filled pressure 1.75bar

2.0 - 1.75 = 0.25

0.25bar / 40c = 0.00625 per 1degree C

So air heated by 1 degree increases its pressure by 0.00625bar

just a thought

its also called Boyles Law

http://www.scribd.com/doc/2280505/Expansion-of-Gas

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Or to get that without knowing the 47 degree answer...Possibly the rise in temp 47c - 7c = 40 c

pressure required to burst 2.0 bar

Filled pressure 1.75bar

2.0 - 1.75 = 0.25

0.25bar / 40c = 0.00625 per 1degree C

So air heated by 1 degree increases its pressure by 0.00625bar

just a thought

its also called Boyles Law

Expansion of Gas

Pressure/Temp = a constant for a fixed volume of gas.

so, Pressure (morning)/Temp(Morning) = constant = Pressure (bursting)/Temp(Bursting temp)

Temp is in Kelvins (Normal temp + 273), so...

1.75/280 = 0.00625 so know we know the constant in this question, so to get bursting temp, we need to calculate

2/Burst Temp = 0.00625 (2 because that is the burst pressure). Re-arrange that, and it gives 2/0.00625=Burst temp

=320K or 47C (320-273)

I haven't got a scoobies about lift bags unfortunately.

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To be neutrally buoyant you need the weight (downwards force) to be equal to the buoyancy (upwards). To do it properly you'd have to work in Newtons because Kg are not a unit of force but I presume they don't go to all that detail because it's not something we work with on a daily basis.

So you need a buoyancy equal to 100 Kg. Part of that buoyancy is already there and it comes from the volume of the mooring. Archimedes principle states that an object totally or partially immersed in a fluid will be acted upon by an upwards force (buoyancy) equal to the weight of the volume of liquid displaced by the object, meaning, it's immersed volume.

In this case, the mooring has a volume of 20 l. Freshwater has a density of 1 Kg/l. This means that the mooring will have an intrinsic buoyancy of 20 Kg.

But since its weight is 100 Kg, you still need to apply an extra 80 Kg of uplift to keep it neutral. That means 80 l.

This is the reasoning, which I don't think it's wrong although I get a different answer.

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pV = constant, for a fixed temperature and amount of gas (no gas lost or added)

and the Gay-Lussac's law:

P / T = constant, for a fixed volume and amount of gas.

Instead of memorising these two laws you can work with the Ideal Gas Law:

pV = nRT

n is related to the amount of gas, which for these exercises is always a constant and R is another constant, so you have:

pV = constant x T

From this you can work both the above laws, fixing either the temperature or the volume.

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i agree.

To be neutrally buoyant you need the weight (downwards force) to be equal to the buoyancy (upwards). To do it properly you'd have to work in Newtons because Kg are not a unit of force but I presume they don't go to all that detail because it's not something we work with on a daily basis.

So you need a buoyancy equal to 100 Kg. Part of that buoyancy is already there and it comes from the volume of the mooring. Archimedes principle states that an object totally or partially immersed in a fluid will be acted upon by an upwards force (buoyancy) equal to the weight of the volume of liquid displaced by the object, meaning, it's immersed volume.

In this case, the mooring has a volume of 20 l. Freshwater has a density of 1 Kg/l. This means that the mooring will have an intrinsic buoyancy of 20 Kg.

But since its weight is 100 Kg, you still need to apply an extra 80 Kg of uplift to keep it neutral. That means 80 l.

This is the reasoning, which I don't think it's wrong although I get a different answer.

so that means the lift bag and shackle needs 3lts of gas to also be neutral.

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Eeek....Q1 What volume of air in a lifting bag will hold a mooring with a volume of 20 litres and a mass of 100 kg neutrally bouyant in freshwater ?

The answer is 83 litres how is this calculated?

Well I can see 20L of displacement so in fresh water 20Kg of up-thrust so we need a further 80Kg of pull from the bag for neutral.

I can see that 80L of bag displacement won't quite work as 80L of air weighs 96gms so we need a teeny bit more but 3Kg more? Even a hundred meters down there's only a kilogram of gas in the bag.

If we were told something about the geometry of the bag and were quoting an answer in surface litres I can see it being more as the gas pressure in the whole bag is the absolute pressure at the gas/water interface which must be under the water level to get lift but we aren't.

I must have failed.

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