Air Receiver

Do your own homework. CR4 is not a homework cheat site; however, if you have questions about understanding concepts or how a portion of a problem is derived, these types of questions will be accepted.

• Do your own homework. CR4 is not a homework cheat site. While some here might relish the opportunity to sharpen up old rusty skills by working the homework problem, consider the following and consider it well. If you cheat on your homework by using someone else's answers, you are only cheating yourself, because the purpose of any homework or other academic assignments is to help you learn - by practice, repetition, and self-discovery.

Hey, I do homework for 10 bucks or 4 bottles of beer. Seriously, like what Lyn said you have to do your own homework.

But luckily, I'm generous today, like I felt your frustrations. Below is something to look at and ponder. Be polite next time, soft words break king's bone.

Please mind your words too. If you can't be arsed to put together sufficient information to enable readers to answer a comprehensive question first time, then I'd prefer it if you would take your business elsewhere, especially if your only approach is SHOUTING!

How rude.

Wrong information to even start with.

There is nothing like working pressure in kg.

Back to school!

Is this a plastic air receiver?

The answer for 1] 20 for 2] 30

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NOT!

You obviously don't understand English or English punctuation well enough to post here.

Questions:

1. What is kl.? It is an abbreviation for a pressure unit I have never heard of.
2. Is 8.5 kg the mass of the contents of the receiver?

Please find someone literate in English to help you compose another post. It will not be coherent otherwise.

Maybe your teacher can tell you how to post your question on an engineering forum in English.

Don't worry about speaking good English. Engineering is supposed to be universal. I can make sense of kg. You mean pressure I guess, but you have left out the 'per unit area' bit. Most important.

If it is homework, and you just want to answer a question without insight or knowledge of the subject, then click the link given here, and put in some numbers to suit, and you will get an instant thickness. You have to decide what max steel stress to allow. It needs to be a few thousand psi to get some sensible answers. If you don't know then get someone to tell you - it depends on so many variables to do with pressure vessel design.

http://www.engineersedge.com/calculators/shell_internal_pres_pop.htm

My best guess to the thickness based on the values in you OP can be linked to the answer to all questions in the Universe - which is 42. This will give the number of Lu's in question. Where Lu is a length unit invented and dedicated to you design.

I wonder where OP sticks the kg in the calculator.

He did mean mm2 or cm2? That'd be the missing reference.

1]16 kl. Working pr;8.5 kg .size:dia;2200 .Ht:4500 mm

What do you suppose kl. means?

I think he was using a cell phone or tablet and when he typed the period and a space, the auto-complete program capitalized the next word, making it seem like he was saying that the working pressure was 8.5 kg, and that doesn't make sense.

I doubt if we'll ever know because it would be too much trouble for him to come back and clarify.

I guess the kg means pressure but a sort of typo, or just lack of knowledge, or shorthand, or laziness -with the area missing. As you say mm2 or cm2, or could even be m2, or km2 (where do we stop). When you know the missing 'area', you convert to psi for the formula.

If the OP doesn't realise this, or know this, then guessing parameters will produce an answer where 42 might be correct.

42 as an answer for both questions seems like over kill for at least one case!

I think as well we'll never know!

The only reasonable interpretation for pressure is 8.5 kg/cm^2. The kl might be kiloliters, but I didn't do the arithmetic.

My thoughts exactly. Pressure 8.5 kg/cm^2 is most likely by far. 1 kiloliter would be what most people call 1 m³, and that agrees pretty well with the vessel dimensions given. In any case the volume does not come into thickness calculation explicitly.

But a vessel that size containing air at 8.5 kg/cm² is not something to be taken lightly. OP should get hold of a design standard and work to it. For one thing, the ends are more likely to be semi-ellipsoidal than hemispherical.