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Physics

Discussion in 'Going Pro' started by agwatts, Jan 16, 2008.

  1. Jim Lapenta

    Jim Lapenta Dive Shop

    # of Dives: 500 - 999
    Location: Canonsburg, Pa
    16,869
    8,560
    113
    Been there done that. Try making up some little anecdotes about them. PM Tammy( TsStormdiver) here on the board and ask her if she still has the ones I sent her. I don;t remember them exactly cause I was trying to come up with new and better ones.

    Things like:
    Boyle's Law- Sir Roberts gas volume goes down when the pressure on his stomach increases by the tightening of his belt. Stupid one yeah. I had a better one but don't recall it

    Charles' Law- Chucks pressure goes up when he gets hot under the collar.

    Dalton's Law- They call it the Dalton gang because all the members( gases) have a role in the deal and it stays the same no matter how deep they get themselves in. Percentages of each gas do not change just the number of molecules due to the density of the gas which is determined by the pressure exerted.

    Henry's Law- When Henry opens his coke all those little bubbles coming out tickle his nose. Solubility. And along with that- The more pressure Henry uses when he carbonates his coke the more bubbles he can get in
     
  2. Indian Valley Scuba

    Indian Valley Scuba Dive Shop

    # of Dives:
    Location: Harleysville, Pennsylvania, United States
    1,347
    88
    48
    The key is to work with someone who truly understands the concepts they are trying to teach, and can explain them using different methods, analogies, and approaches. Sadly, in the scuba industry, there are many instructors who managed to pass an exam because they memorized laws or rules, but truly don't have a handle on the subject matter, thus making it difficult to pass knowledge and understanding on to the next person. Make sure you are working with someone who truly understands what they are attempting to teach.
     
  3. scubasully

    scubasully Marine Scientist

    # of Dives: 200 - 499
    Location: Colorado
    150
    0
    16
    I'm a little confused on 3 separate things:

    1) why does the workbook show to multiply depth by .97 for determining gauge P in fw, but the Encyclo tells us to divide depth by 10.3 for freshwater???? Confused. :confused:

    2) Flexible and Non-flex containers are bugging me out. Ok, if it's a steel tank, the pressure can increase or decrease, but vol stays the same? And with a balloon, does vol correspond with pressure? P+=V- and P-=V+? B/c the balloon questions are Boyle's law and the tank temperature questions are in the Charle's law section.

    3) The concept of supersaturation is beyond me. I just keep thinking of that chart in the DM Manual about how once tissue compartments are full, they can't be filled anymore. Maybe it means that that would be true for a given depth, but then to go down deeper (increasing pressure), more gas could sneak in there? Or do I have it backwards - and it's when the tissues are saturated with the gas, and then diver ascends, and there's less pressure but gas is still taking a while to get out (why we do safety stops)???

    HELP.

    Sorry for the dumb questions, but my brain is addled with fever & flu, and I don't want to contaminate everyone at dive shop.

    I think I've got the rest straightened out. Seems that there are only about 8 types of problems that can show up:

    1) Salvage problems that mention water displacement, where you need to convert L to KG, Balance, then convert back

    2) Finding ATA problems ( /10 +1 for sw; /10.3 + 1 for fw)

    3) Pressure problems - figuring out Ambient/Absolute/Gauge

    4)Almost anything mentioning a balloon is about finding volume (P1 X V1=P2 X V2)

    5) Determining ata & air consumption @ depth

    6) Determining bar for temp change.
    I've got a great formula for this:
    T1-T2=T3
    T3 X .6=B2
    B1-B2= B3 (new bar for temp change)

    7)Solving for Partial Pressure (% x ata = PP)

    8) Solving for Surface Equivalency
    % @ depth X ata

    Am I missing any types of problems? I'm feeling more and more ready every hour. Just some fuzzy spots with the above-mentioned.

    Thanks in advance for your help.

    Jim, thanks for the great memory cues. I have a good one for Dalton's Law already, about a guy name Dalton in high school who mixed his alcohols, but no matter how much of what he drank, he got equally drunk! :rofl3:

    I also tell myself that Prince Charles is NOT hot, and that's how I remember Charles' law is the one with Temp.

    Henry and Boyle's I don't have cues for, but I like working out Boyle's law problems, so it's easy to remember the formulas, and poor Henry just has big ideas, no problems to work out.

    Do I need to know which law I'm applying for the exam? Or will it be enough to see the problem and know which formula to plug in? I've got the 2nd step down, but not the attaching of the formula to the law name, yet.

    Thanks!
     
  4. scubasully

    scubasully Marine Scientist

    # of Dives: 200 - 499
    Location: Colorado
    150
    0
    16
    Still haven't taken physics exam yet... been recovering from an evil flu.

    Was looking over answer sheet - though....made me nervous. What's up with all of those FILL IN THE BLANK questions? Guess I'm really going to have to think, huh?

    Any responses to my last post would be mucho appreciated! My nerves need calming.
     
  5. Xaryo

    Xaryo Barracuda

    # of Dives: 500 - 999
    Location: montreal Québec
    363
    20
    18
    When you dive, you live by the Master Rule PV=nRT (or if you want to be precise or have a degree in engeneering PV=ZnRT)

    For the physics exam there is one formula to know, an arranged version of the Master Rule

    (P1 X V1) / T1= (P2 X V2)/T2



    forget charles, boyles, henry or wathever
     
  6. ClayJar

    ClayJar ScubaBoard Supporter ScubaBoard Supporter

    # of Dives: 200 - 499
    Location: Baton Rouge, LA
    3,510
    86
    0
    You can forget Boyle, Charles, Gay-Lussac, and the "combined gas law" as long as you're fluent in the ideal gas law (with or without the compressibility factor, hehe), but you can't *just* go with the ideal gas law. You'll still need Henry's help with dissolved gases and Dalton's help with partial pressures.

    Of course, if you forget how to work Dalton's, you may need a bit more remedial assistance. Although it was not always obvious to everyone, "add up the pieces to get the total" is pretty easy to remember now. :biggrin:

    The two keys to remember when you've branded all of the formulas into your head, by the way, are these:
    1. ALWAYS work in *ABSOLUTES*. Absolute pressure and absolute temperature are mandatory, although you can certainly use whatever units you'd like. (Rankine for the win!)
    2. If you can't do gas laws while simultaneously reciting The Star-Spangled Banner backwards in your sleep, take everything up to the surface and back down. There's no reason you need to do this, but doing the two-step is an easy way to avoid stupid mental mistakes.
     
  7. Xaryo

    Xaryo Barracuda

    # of Dives: 500 - 999
    Location: montreal Québec
    363
    20
    18
    Rankine Rulezz!
     
  8. gulfdvr68

    gulfdvr68 Public Safety Diver

    # of Dives: 50 - 99
    Location: Panama City,FL
    31
    0
    0
    Ok folks I thought that I was fairly smart until chapter four the divemaster encyclopedia. Boyle is driving me crazy, I am on 5 days trying to understand it. I really do not get the formula.

    p1xv1=k=p2xv2???????

    I was able to answer the first question covering that chapter in the workbook however, every question after made me feel that I never read the section at all.
    So can someone please plug actual numbers in the formula so that I can see it work. The book does not make it clear enough for me.
     
  9. ClayJar

    ClayJar ScubaBoard Supporter ScubaBoard Supporter

    # of Dives: 200 - 499
    Location: Baton Rouge, LA
    3,510
    86
    0
    Okay, let's start from the top. (I've heard sung that the very beginning is a very good place to start. :wink:) This is a bit more background than you really *need*, but it might help you wrap your head around where it all comes from or how it works. (Otherwise, just skim it.)
    All the gas laws we use (Boyle's, Charles', the rarely-for-scuba Gay-Lussac's, and the combined gas law) are all special cases of a main law called the Ideal Gas Law. (All the college science students just smiled... or moaned.) The Ideal Gas Law is:
    PV = nRT
    P is absolute pressure, V is volume, n is the amount of stuff (e.g. number of moles), R is the Ideal Gas Constant (i.e. the *magic*! bit :biggrin:), and T is absolute temperature. When applied with a little algebra, this is the basis of it all.

    With Boyle's Law, you're keeping the temperature constant, so T doesn't change, and of course, since you're not losing gas, n stays the same as well. (R is a constant, so it's not changing.) A constant times a constant times another constant is nothing but... yep, a constant. (3 times 5 times 4 will *always* be 60, for example.) If you just lump all those bits together into one letter, call it k (just because -- you could call it "Bob" if you wanted), you get:
    PV = k
    The key here is that as long as you're keeping the temperature constant and not losing or gaining gas, this must be true for *ANY* pair of pressures and volumes.
    P[sub]1[/sub]V[sub]1[/sub] = k
    P[sub]2[/sub]V[sub]2[/sub] = k
    P[sub]3[/sub]V[sub]3[/sub] = k

    .
    .
    .
    Since k is the same for each of those (that's what Boyle's Law says, after all), and multiplying each pair equals k, obviously the pairs have to equal each other. (To show the algebra, if 2 * 12 = k, 3 * 8 = k, and 6 * 4 = k, then obviously 2 * 12 = 3 * 8 = 6 * 4. They're all equal to 24.) Anyway, so dropping the k, we have:
    P[sub]1[/sub]V[sub]1[/sub] = P[sub]2[/sub]V[sub]2[/sub] = P[sub]3[/sub]V[sub]3[/sub] = . . .
    There are infinitely many P/V pairs that you could write (1 * 24, 2 * 12, 1.5 * 16, 0.75 * 32, et cetera, ad infinitum, all equal 24), but when you're actually using Boyle's Law for diving problems, you're just going to have two pairs:
    P[sub]1[/sub]V[sub]1[/sub] = P[sub]2[/sub]V[sub]2[/sub]
    That's two pressures (1 and 2, before and after, shallow and deep, whatever you want to call them) and two volumes (ditto). In order to solve the problem, you will be given three out of four of these, although they could do things like giving you depths (which you'd have to convert to pressures) or making you use "bucket" as a unit of volume (1 bucket at the surface is half a bucket at 33fsw).
    So, now let's work a couple problems with Boyle's Law in it's useful form, P[sub]1[/sub]V[sub]1[/sub] = P[sub]2[/sub]V[sub]2[/sub]. Ignore the whirring sound as I make up the problems. :biggrin:

    1. You have a 1 cubic foot bubble of air in your BC as you step off the boat at sea level (1 atmosphere, absolute) and plummet to the sand 33 feet below to recover a student's weight belt. What is the volume of that bubble of air at 2 atmospheres absolute pressure as you struggle to pick up the belt?
    Okay, remembering our equation, P[sub]1[/sub]V[sub]1[/sub] = P[sub]2[/sub]V[sub]2[/sub], what are we looking for? We're looking for one of the volumes, and we'll just call that one V[sub]2[/sub] (you could call it number one, if you'd like, as long as you keep the pairs together). We want to get our unknown all by itself on one side, so drawing back on algebra (if anyone doesn't get this part, we can do an algebra review in the chat later):
    P[sub]1[/sub]V[sub]1[/sub] / P[sub]2[/sub] = V[sub]2[/sub]
    So, we should know the other three things, P[sub]1[/sub], V[sub]1[/sub], and P[sub]2[/sub]. Let's write them out, being sure we're in absolute pressure:
    P[sub]1[/sub] = 1 atmosphere, absolute (ata)
    V[sub]1[/sub] = 1 cubic foot (cf)
    P[sub]2[/sub] = 2 ata
    And plugging them in, we have:
    V[sub]2[/sub] = P[sub]1[/sub]V[sub]1[/sub] / P[sub]2[/sub] = (1 [-]ata[/-])(1 cf) / (2 [-]ata[/-])
    V[sub]2[/sub] = (1/2) cf

    V[sub]2[/sub] = 0.5 cubic feet
    Which is our answer. At twice the absolute pressure, our volume is half what it was at the surface.​

    2. A seagull managed to open your post-dive beer while you were away, and now it's completely flat. Luckily, you happen to have a big trash bag full of spare carbon dioxide and an old bicycle pump (amazing what you can find behind dive shops these days). You pour a pint of beer into the four pint pump chamber, top it off with CO[sub]2[/sub], and seal it up. Standing on the pump lever, you find that you can squeeze the piston down halfway (squeezing the pump chamber's volume down to two pints, total). What pressure of CO[sub]2[/sub] did you get up to?
    This one's a little trickier, but if you think about it a bit, it should make sense. As always, let's start off by asking what we're looking for. (In real problems, sometimes they even throw extra stuff in to confuse you. If you know what you're looking for, you can ignore any fluff.) So, what's the problem asking for? It wants a pressure, so we'll call it P[sub]2[/sub].

    Now that we know what we're looking for, we can figure out the three other pieces. For our V[sub]1[/sub], we know that the pump volume was four pints, but we poured a pint of beer in it first. So, the volume for the CO[sub]2[/sub] was *three* pints. Since it was open to the atmosphere (until we closed it up), our P[sub]1[/sub] is one atmosphere, absolute. To get our V[sub]2[/sub], we know that the pump chamber was down to two pints, and remembering that our pint of beer's still in there, the gas volume is one pint.
    V[sub]1[/sub] = 3 pints
    P[sub]1[/sub] = 1 ata
    V[sub]2[/sub]= 1 pint
    Doing our algebra again, we can plug in these three to find our answer:
    P[sub]2[/sub] = P[sub]1[/sub]V[sub]1[/sub] / V[sub]2[/sub] = (1 ata)(3 [-]pint[/-]) / (1 [-]pint[/-])
    P[sub]2[/sub] = (3/1) ata

    P[sub]2[/sub] = 3 ata

    Which is our answer. Of course, if they wanted it in gauge pressure, you'd just subtract the 1 atmosphere of ambient pressure and get 2 atm gauge pressure.​

    3. You want to have a birthday candle on your underwater cake, but obviously you have to keep it dry. You tape it to the inside bottom of a 5-gallon bucket, tip the bucket upside down, and start dragging it down to your birthday party, 50 feet deep. (You strapped plenty of lead to the bucket, of course.) At 50 feet down, what is the volume of air in the bucket, assuming you were *really* careful? :)
    So, I suppose you'd agree that we're looking for a volume, and of course, we'll call it V[sub]2[/sub]. That leaves two pressures and a volume that we should know.

    P[sub]1[/sub] is the surface, so that's 1 atmosphere, absolute. V[sub]1[/sub] is the five gallons of air in the bucket. What is P[sub]2[/sub]? Well, let's see... 33 feet equals one atmosphere, so 50 / 33 equals about 1.5 atmospheres. (I'm rounding slightly, but the party's really hopping, anyway.) So, P[sub]2[/sub] is 1.5 atmospheres.

    If you just said, "No, it's not!", you've just avoided the usual gotcha. Remember, we have to work in *absolute* pressure, so after we add the 1 atmosphere from our wonderful sky, P[sub]2[/sub] is 2.5 atmospheres, absolute. Now that we have that straight, let's plug the numbers in:
    V[sub]2[/sub] = P[sub]1[/sub]V[sub]1[/sub] / P[sub]2[/sub] = (1 [-]ata[/-])(5 gal) / (2.5 [-]ata[/-])
    V[sub]2[/sub] = (5/2.5) gal

    V[sub]2[/sub] = 2 gallons
    If the question also asked how negatively buoyant our bucket-o'-candle would be at 50 feet, assuming it was neutral at the surface, you should be able to figure that out. The bucket had 5 gallons of air in it, and now it only has 2 gallons of air, so it lost three gallons of buoyancy. That's about 0.4 cubic feet, which means the bucket is now over 25 pounds negative. :D

    4. You have 0.7 liters of air in an open two-liter bottle down at 89 feet. You want to let all the water out as you ascend, but you don't want to lose any air. At what depth will you have to screw the cap back on the bottle?
    This is a doozie, but if you pick at it, it's really not that hard. It's just written to be a bit intimidating to people who don't have a good handle on things. It shouldn't be, of course, as you attack it the same way as the others.

    So, first, out of our variables, which one are we looking for?
    If you read the problem, you see it ask "At what depth...?" Depths are just another way to think of pressures, so we're looking for a pressure. As usual, we'll call it P[sub]2[/sub]

    So, now that we know what we're looking for, let's write down what we already know.
    We start with 0.7 liters of air, so there's our V[sub]1[/sub]. We need to know our P[sub]1[/sub] to go with that, and we can find that from our depth of 89 feet. 89 feet divided by 33 feet for each atmosphere gives us 2.7 atmospheres, and then we remember that we have to add the sky. So, P[sub]1[/sub] is 3.7 ata. What's our V[sub]2[/sub]? Well, if we don't want any water in the bottle, it has to be filled completely with air, so we let the air expand to two liters: V[sub]2[/sub] = 2 liters.

    Now that we have all our numbers, we can plug them in:

    P[sub]2[/sub] = P[sub]1[/sub]V[sub]1[/sub] / V[sub]2[/sub] = (3.7 ata)(0.7 [-]liters[/-]) / (2 [-]liters[/-])
    P[sub]2[/sub] = (3.7 * 0.7 / 2) ata

    P[sub]2[/sub] = 1.3 ata
    So, we now now our P[sub]2[/sub], but the problem was asking for a depth. To find that, we subtract the one atmosphere of the sky to get gauge pressure in atmospheres, then we multiply that by the 33 feet of depth for each atmosphere of pressure.

    P[sub]2[/sub] = 1.3 ata = 1 atmosphere of sky + 0.3 atmospheres from the water
    0.3 atm * 33 feet per atmosphere = 9.9 feet

    We should cap the bottle at 9.9 feet on our way back to the surface.

    Well, that's about all the difficulty ramp there is. A four problem set isn't really enough to get a lot of practice in, but it might help illuminate things. If you get lost somewhere in there, we can fill in some extra problems to bridge the gaps, of course.

    Just remember the steps:
    1. Figure out what you're looking for.
    2. Find the three other pieces (ignoring any wild geese).
    3. Be sure you're in absolutes!
    (All gas laws work in absolutes, so convert to absolutes, use the gas laws, then convert back to gauge pressures, depths, or non-absolute temperatures once you're done with the gas law math.)
     
  10. AzAtty

    AzAtty Instructor, Scuba

    894
    6
    18
    What I want to see is the amount of time it takes to re-carbonate the beer at 3ATA. THAT would be useful math.
     

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