134a Refrigerant

Watch him squeal! We all read it... you gave an example of 'in a pinch' as 'such as on a rooftop in -30 degree weather'... notice you didn't say that the challenge rules defined -30, you were giving an example of 'in a pinch'. In fact, I believe you're in a pinch now! You in no way defined a new challenge.

No... you gave an example of 'in a pinch'. This in no way defined the exact terms of the challenge, and I can think of several scenarios where you'd need to replace the valve but the ambient was high. I also refer you to the original challenge, where you had no such condition... why do you attemp to put such a condition onto me, when you didn't make the same condition to Jeff? Says a lot about you, doesn't it? Says 'welch' to me...

Logical fallacies... means your brain no workie too good. Could be causing that eye rolling! See a doctor!

Keep squealing... makes me feel like I've accomplished something here.

you:

me: You don't have a clue about your limitations... imagine you, an HVAC tech, making pronouncements about atmospheric science! That shows *extreme* hubris... you deserve what you get.

Limitations! Own them! Or lie about it... welch. __ Steve .

You can't give up... even though you are digging the hole deeper and deeper. We all can read... and remember. Just think... this will live in the Google archives for thousands of years! Own your limitations!

Now you're very confused again... I can understand, since your head's been spinning a mile a minute since this started. Please try to keep this straight... I'll remind you as necessary. The 'challenge' involved the head pressure control... the 'argument' involved CFC's antagonism to the global ozone layer. When I refer to 'challenge', you should address the proper issue. All clear?

See what I mean... daubing the walls.

you:

me: You don't have a clue about your limitations... imagine you, an HVAC tech, making pronouncements about atmospheric science! That shows *extreme* hubris... you deserve what you get.

Limitations! Own them! Or lie about it... __ Steve .

You continue to top yourself! Are you not aware of the formula for a regular cylinder? Do you not agree that an area times a length gives you your cubic dimension? What about good old Mr. What'sHisName? Didn't he teach you about this? You claimed to have 'gone through Calculus'! Like corn through a goose, I replied... apropos.

Here's another hint... the liquid level forms a chord across the end, parallel to a diameter of the circular face. Draw a line from the center of the circular face to the edge of the tank where the liquid level is... a radius line. It should all be clear now... if you can figure out the area of a piece of 2D pie! __ Steve .

No... I think it's a victory when I can point out your silly mistakes... in your own avowed field of 'expertise'. Not only a welch, but provincial, too!

'buwahaha'? Sounds like baby talk.

And you seem to have forgotten that I was chief site tech at a radiotelescope for four years. You see, I *did* cryo... lots of it. Dry helium, two-stage piston refrigerators, thermocouple vacuum sensors, recirculating oil compressors.

15 Kelvins, *that's* refrigeration. __ Steve .

You made $1000 gross on Sunday? Well, I certainly must admit that I didn't make a dime on Sunday. Must put me into pretty hard company, for sure. __ Steve .

OK here's the equation for calculating the sticking volume of a storage tank. xx x x x------------x Liquid level which is variable, x x can go up or down x x x x x x xx

Imagine the above is a circle. I know the diameter of the circle, and I can measure from the top of the circle down to the liquid level. How can I use this information to derive the amount of liquid in the circle? I hope to make a formula I can use in a spreadsheet.

Thanks so much.

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Date: 3 Feb 1995 19:51:37 GMT From: Dr. Math Subject: Re: Volume of partially filled cylinder on its side?

Hello there!

Thanks for writing to Dr. Math! You asked a good question.

Before we start, let's go over the "law of sines" (we'll be using it shortly). Given a triangle ABC where A, B, and C are vertices, the following relation holds:

sin A sin B sin C -------- = ------- = ------- A B C

where sin A means sin of the angle at vertex A, and A (when it appears in the denominator) is the length of the side of the triangle opposite A.

Okay, now on to the problem. You did a nice job with your drawing on the computer, so I'm not going to draw another picture. Instead, you draw a picture on paper and label the following:

Call the distance from the top of the circle to the liquid level h. Call the radius of the circle r. (these are given quantities)

Now, draw in the center of the circle. Draw a line connecting the center of the circle to the top of the circle. Then draw two more lines from the center of the circle to the points where the liquid level lines intersect the circle. Note that these lines are just radii, so they have length r.

Now you have 3 triangles drawn, 2 of which are identical and make up the third. Look at one of the small triangles. Call the angle at the center angle z. One of the other angles is a 90 degree angle. Call the third angle y. Now, note that the distance between the center of the circle and the top of the circle is r (that's the definition of radius). We already said that the distance between the top of the circle and the liquid level was h, so that means the distance between the center of the circle and the liquid level must be r-h, right? So, our small triangle has a side of length r-h opposite the angle y, a side of length r opposite the 90 degree angle, and a side of length x opposite angle z. x is an unknown quantity we will need later on in the problem.

By the law of sines, we have the following relations:

r r-h x ----- = -------- = -----

• sin 90 sin y sin z

sin 90 = 1, so we have:

r-h ----- = r sin y

Solve this equation for y to get:

y = arcsin [(r-h)/r]

Note also that z = 90 - y.

And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] = (1/r)Sqrt(2rh - h^2)

(this is by trig identities and stuff...if you have questions, write back).

So, then equation * becomes: rx r = -------------- Sqrt (2rh - h^2)

So, x = Sqrt (2rh - h^2)

Now the problem is a lot easier. To find the area of the whole circle, let's first consider the area contributed by the segment we've been talking about that is of angle 2z from the center.

The area of this segment occupied by liquid is just the area of the 2 small triangles. We know the base of each triangle is x (see above) and the height is r-h. So, the total area contributed by this segment is x(r-h).

Now consider the area outside of the segment. That is, we want the area of a portion of the circle. If we measure z in degrees we want the portion of the circle that takes up 360 - 2z degrees of the circle. So, the fraction we are dealing with here is:

360 - 2z -------- 360

The area of the portion we want then is:

360 - 2z -------- (Pi)r^2 360

Now add these two areas together and you are done. Thus, the total area is going to be:

360 - 2(90 - arcs>

you stupid stool. were it not for low ambient the HMC wouldnt be needed AT ALL. its funny to watch you squirm and spin, hoping desperately that you can convince use that you actually know something about refrigeration. LOL! :-)

lol bullshit, liar. i CLEARLY stated -30 degrees. now squirm boy! :-)

being stump-stupid, you require further clarification. :-)

you pathetic little liar! i outlined it for you BEFORE you even knew what an HMC was.

lol.....youre actually trying to suggest that a URL including "proudliberal" is credible? BUWHAHAHA! :-)

lol, not a chance! :-)

your lies are there for one and all.

TRANSLATION --> "its lying time"

lol you find a real one (you know, not something youre LYING about, liar) and then you might have something. :-)

lol now THAT is pathetic. now youre claiming to be a refrigeration guy? BUWAHAHA! :-) you couldnt even give me an alternative to replacing an HMC but youre going to claim to be a cryo/refrigeration guy? LOL! :-)

i said i _grossed_ it. yeah, it was sorta slow. :-)

Exactly as I said... he can't give up. I've proved him wrong, beat him at his own challenge... even *proved* him a liar and a welch. Watch yourself in dealing with this person.

It's all he's got, folks... put money on it. __ Steve "i recognize my limitations and immediately own it (sic). " -- Nate .

That's why it's so important for you to weasel and welch your way into a modified spec for the challenge. As I stated in my first post after you revealed what the magic letters HMC stood for (a brand name, btw), it allows the cooling system more bandwidth, i.e. operation in larger extremes. Remember, this was before I googled HMC and found the proper name, as well as examples... I was *guessing*, and beat your silly challenge. If the head pressure control was leaking, and low ambient was not in the forecast, you could bypass it with two valves and a straight piece of pipe, to replace it after the part came in on order.

I believe the psychological phenomenon you're exhibiting is called 'transferrence'... read up on it.

Doesn't matter.. you didn't specify it in the original challenge, the only document that matters. You even used the term 'such as' when elaborating on a possible example of 'in a pinch'... we can flog this to death (if it's not already there!) and you'll never own up to your welchage. You are a complete spoilsport and loser.

I believe we've already proven who's 'stump-stupid' around here... do you do cryo?

You didn't 'outline' anything... you gave an example of 'in a pinch', even used the common example modifier 'such as'. All there for anyone to read. Watch Nate spin! Spin, Nate.... spin!

"proudliberal"

And why not? Your political proclivities have blinded you, we can all see that. You are not addressing the NOAA stuff, Nate... big science, just waiting for some HVAC tech to debunk! C'mon, you know your limitations... and your lack of a degree in atmospheric science, or any science at all, must not be one of them! Tell us again how the rest of the world is wrong, but you, Nate, with your knowlege of your limitations, must be right! __ Steve "i recognize my limitations and immediately own it (sic). " -- Nate .

Do you regard yourself as qualified in cryo? I mean, are you stupid, or a liar?

But I *did* give you an alternative to replacing a head pressure control, Nate... we all read it. I even specified that lower ambient was not allowable under the circumstances.

Go ahead, google back, see if you can find my early NG posts... they go back to '92. You'll find NRAO in there... ever hear of them? The VLBA? Chief site tech, Nate... four years, my own $6M site, 82 foot dish to maintain. Two complete, separate cryo systems, with

5 or 6 refrigerators on each. 15 Kelvins... that's some cold stuff, Nate... colder than you'll *ever* work on. I call that *refrigeration*. Why do you think I've been kicking your ass so bad? Who do you think you're messing around with? You're just an HVAC chump. __ Steve "i recognize my limitations and immediately own it (sic). " -- Nate .

You're way behind, Bill... catch up with the rest of the class.

We just covered two hint/facts:

1) We know the half-full condition, since it's easy to construct.

2) We considered that the liquid always forms a level line... considering the diameter of the circular face parallel to this line, we can construct a radius line from the center of the circular face to the edge of the liquid level. This final picture, with the diameter, the liquid level, and the radius line, should have been enough for an astute person to take it the rest of the way.

Let's elaborate on that last picture... add a diameter perpendicular to the half-full diameter. Now you can see the triangle formed by the liquid level, the radius line, and the vertical diameter. We know two sides and an angle of this triangle, so we know all we need to figure all sides and angles. We can use he Pythagorean Theorem to find the unknown side... A^2 + B^2 = C^2. We can find the area using

1/2 (base * height).

Now we need the area of the pie-shaped piece on top (or bottom, for the more-than-half-full case) of the triangle. Since we know the angle of the wedge (arcsin (base/hyp)) we can take the ratio of this to 360 degrees, times the area of the circular face, which we know since we know the radius of the tank. All the pieces are now in place.

For the less-than-half full case: Half he area of the circular face minus the quantity twice the area of the triangle plus twice the area of the pie wedge

For the more-than-half-full case: Half the area of the circular face, plus twice the area of the triangle, plus twice the area of the pie wedge.

Then times the length. Easy, wasn't it? __ Steve "i recognize my limitations and immediately own it (sic). " -- Nate .

But you're just pasting some Google stuff, like you always do... I thought you went through Calculus! You do remember the specifics of the challenge... no Google! Even a chimpanzee can wave a book over his head... doesn't give him the understanding in the book.

You lose! __ Steve "i recognize my limitations and immediately own it (sic)."... Nate .

not what they taught you in your seminars, eh? :-)

lol, you are a liar. you gave me links with "proudliberal" in them and you consider that a credible source of information? of course not, you just hope that if you tell the same LIE over and over that someone will believe you.

ha! as if ANYBODY in this newsgroup needs YOU of all people to advise anyone else.

yup. i presented fact, and you told lies.

i never claimed to be qualified in cryo. cryo does not fall under my refrigeration certification.

oh, but id bet im more qualified to work on cryo than you are to talk about it. :-)

and it was TOTAL BULLSHIT! :-) beating your car with a hammer is an "alternative" to changing the tire but that doesnt mean its correct.

and it doesnt mean you know SHIT about refrigeration. :-)

show me your certification or youre simply LYING. hell even NASA has janitors who can claim they work at NASA, doesnt mean that they have anything to do with space operations.

lol you think this because you actually believe your own lies.

a LIAR. :-)

lol, even with the mild temps ill gross at least $1500 today. :-)