Does the pressure take into consideration the area of the cylinder?

AI Thread Summary
The discussion centers on the interpretation of pressure in a thermodynamic context, specifically regarding a cylinder containing gas and its effect on a spring attached to a piston. Participants debate whether the area of the cylinder should be included in calculations involving pressure and spring compression. The equation PA - P atm A = K spring change in X is analyzed, with confusion arising over how to properly apply it and whether to "cancel" terms. Clarifications are made that the area is inherently part of the pressure calculation, as pressure is defined as force per unit area, and the distinction between gauge and absolute pressure is also highlighted. Ultimately, the conversation emphasizes the importance of understanding the relationship between pressure, area, and force in these calculations.
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Not a question to be solved, just an understanding of interpretation required.

In thermodynamics when a question is asked regarding cylinder pressures, so say a cylinder contained a gas pressure of 600Kpa, when a question asks for you to conclude say How Much A Spring is Compressed, does the pressure take into consideration the area of the cylinder?

Now the above may not sound briliant, but if the compressed gas was pushing on a piston which had a spring attached, so the spring constant was say 4.8 kN/m, does the following solution seem to be worked out correctly.

PA - P atm A = K spring change in x

(600 000 - 100 000)pie x 0.05 squared = 4800 change in x = 0.818m

I know the maths is ok, I am concerned about my interpretation of the equation above, and whether PA should have the area of the cylinder included or is already included in the pressure?

Please advise if you can

David
 
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Hi David! :wink:

I'm not sure what you're asking about, but if you mean how do we convert the pressure in a cylinder to the force at the end of the cylinder (which can push a piston which can compress a spring, for which we need to know the force), then it's force = pressure x cross-section area. :smile:
 


tiny-tim said:
Hi David! :wink:

I'm not sure what you're asking about, but if you mean how do we convert the pressure in a cylinder to the force at the end of the cylinder (which can push a piston which can compress a spring, for which we need to know the force), then it's force = pressure x cross-section area. :smile:
Hi tiny-tim, very much appreciated for your response to my question, and I appreciate that I may not have worded it very well.

The question asks how much would a spring be compressed if a gas pressure in a cylinder 600Kpa was acting on a piston, the spring K constant being 4.8 KN/m. The equation given is; PA - P atm A = Spring constant x change in distance X

Using this equation P = Pressure and "A" I think = area subtract pressure of the atmosphere multiplied by the area, which then = spring constant "4.8 KN/m".

The part I cannot see being used is any area calculation, which is why I thought that in this subject "Thermodynamics", just maybe the "Pressure" only is used, as the cylinder only gives the "Diameter" and no other measurements?

Thanks

David
 
Hi David! :wink:
Auto Engineer said:
The question asks how much would a spring be compressed if a gas pressure in a cylinder 600Kpa was acting on a piston, the spring K constant being 4.8 KN/m. The equation given is; PA - P atm A = Spring constant x change in distance X

The part I cannot see being used is any area calculation … as the cylinder only gives the "Diameter" and no other measurements?

Yes it is … it's a cylinder, so you know its cross-section is circular, so the given diameter tells you what the area is. :smile:
 


tiny-tim said:
Hi David! :wink:


Yes it is … it's a cylinder, so you know its cross-section is circular, so the given diameter tells you what the area is. :smile:

Hi tiny-tim, thanks for your replies.

Using the equation as given; PA - P atm A = K spring x change in X. Can I justify just cancelling the A's on the left hand side to read; P - P atm = K spring x change in X.

The example given shows; (600 000 - 100 000)pie x 0.05squared = 4800 change in X

The answer is = 0.818m, but if I include the areas I get 0.2m which is 0.6m different?

Is it the correct procedure therefore to cancel like terms?

Many thanks

David
 
Auto Engineer said:
Using the equation as given; PA - P atm A = K spring x change in X. Can I justify just cancelling the A's on the left hand side to read; P - P atm = K spring x change in X.

Is it the correct procedure therefore to cancel like terms?

Hi David! :smile:

Don't use the word "cancel"!

Except in actual fractions (like 4A/3A = 4/3), say what you're actually doing, and add "on both sides" …

in this case, say "dividing by A on both sides", and then you'll immediately see that you get P - P atm = K spring x change in X ÷ A
 


I know the maths is ok, I am concerned about my interpretation of the equation above, and whether PA should have the area of the cylinder included or is already included in the pressure?
P is pressure...look at the units/dimensions..its obvious that u include A??
Also...
Whether u subtract Patm from 600kpa depends on whether the given pressure is gauge or absolute pressure...that should b mentioned in th problem itself...I don't think u can guess that unless sumthings provided on that...rest should b ok...
 
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tiny-tim said:
Hi David! :smile:

Don't use the word "cancel"!

Except in actual fractions (like 4A/3A = 4/3), say what you're actually doing, and add "on both sides" …

in this case, say "dividing by A on both sides", and then you'll immediately see that you get P - P atm = K spring x change in X ÷ A

Hi tiny-tim, thanks for that input, if what you say I divide both sides by A, then the textbook answer of 0.818m would be incorrect because using the method of division where the right hand side of the equation is divided by A, the result shows; 0.643m

Have I understood you correctly?

David:smile:
 


nanunath said:
P is pressure...look at the units/dimensions..its obvious that u include A??
Also...
Whether u subtract Patm from 600kpa depends on whether the given pressure is gauge or absolute pressure...that should b mentioned in th problem itself...I don't think u can guess that unless sumthings provided on that...rest should b ok...

Hi nanunath, my limited understanding of Thermodynamics is that unless the question specifically says the pressure is "gauge", then it will be absolute, which in my case is absolute.

Thanks

David:smile:
 
  • #10


Ok...

but the other thing..
The answer is .2m ..not .6..reason :The same as u said earlier..
but I think this question isn't a part of thermodynamics...
is that all given in question?
 
  • #11
Auto Engineer said:
Using the equation as given; PA - P atm A = K spring x change in X. Can I justify just cancelling the A's on the left hand side to read; P - P atm = K spring x change in X.

The example given shows; (600 000 - 100 000)pie x 0.05squared = 4800 change in X

The answer is = 0.818m, but if I include the areas I get 0.2m which is 0.6m different?
Auto Engineer said:
… if what you say I divide both sides by A, then the textbook answer of 0.818m would be incorrect because using the method of division where the right hand side of the equation is divided by A, the result shows; 0.643m

Have I understood you correctly?

Hi David! :smile:

I don't understand what you think is wrong …

(600 000 - 100 000)pie x 0.05squared = 4800 change in X does give 0.818m,

and the area (π 0.052) is in there. :confused:
 
  • #12


tiny-tim said:
Hi David! :smile:

I don't understand what you think is wrong …

(600 000 - 100 000)pie x 0.05squared = 4800 change in X does give 0.818m,

and the area (π 0.052) is in there. :confused:

Hi tiny-tim, and nanunath,

Because one person is saying I am right with an answer of 0.2 and the textbook answer says 0.818, then I must conclude that the equation;

PA - P atm A = K spring muliplied by change in length X I am somewhere not completing understanding the idea being used. So in my endevour please allow me to make sure we are all singing from the same hymm sheet?

The question asks;

A 10 cm diameter cylinder contains a gas pressurised to 600Kpa. A frictionless piston is held in position by a stationary spring with a spring constant of 4.8 KN/m.

How much is the spring compressed?

The equation used is;

PA - P atm A = K spring change in X.

So my understanding of the above equation is as follows;

P = Pressure in pascals
A = Area in metres squared
atm = atmospheric pressure
K = the spring constant
x = springs change in length.

so the equation says;

PA - P atm A = K spring change in X. Putting in the data

(600 000 - 100 000)pie x 0.05 squared = 4800 multipled by X
500 000 x 7.85 x 10 -3 (power three) = 4800 multipled by X

divide both sides by 4800 = 0.818m

Now, if I complete the solution this way;

PA - P atm A = K spring change in X

(600 000 x pie x 0.05 squared) - (100 000 x pie x 0.05 squared) = 4800 x change in length.

This way I get a completely different answer?

Now there is also;

PA - P atm A = K spring chnage in X. Divide both sides by A, this gives;

P - P atm A = K spring change in X / A

which also gives a completely different answer?

How do I know which method is the right one to use?

Thanks

David:smile:
 
  • #13
Hi David! :smile:

(have a pi: π and try using the X2 tag just above the Reply box :wink:)
Auto Engineer said:
PA - P atm A = K spring change in X. Putting in the data

(600 000 - 100 000)pie x 0.05 squared = 4800 multipled by X
500 000 x 7.85 x 10 -3 (power three) = 4800 multipled by X

divide both sides by 4800 = 0.818m

Now, if I complete the solution this way;

PA - P atm A = K spring change in X

(600 000 x pie x 0.05 squared) - (100 000 x pie x 0.05 squared) = 4800 x change in length.

This way I get a completely different answer?

But that gives the same answer …

show us how you get it to be different. :confused:
 
  • #14


tiny-tim said:
Hi David! :smile:

(have a pi: π and try using the X2 tag just above the Reply box :wink:)


But that gives the same answer …

show us how you get it to be different. :confused:


Hi tiny-tim

if you divide both sides by A, you will end up with 4800/pie x A on the right hand side, which then shows a different solution.

Thanks

David:smile:
 
  • #15


Hi tiny-tim, just having to nip out now for about an hour, will check your response when I return. Thank you very much for both your patience and your help, much appreciated.

David
 
  • #16
Auto Engineer said:
if you divide both sides by A, you will end up with 4800/pie x A on the right hand side …

Yes, but it should still give the same solution. :confused:

Show us how exactly you get a different solution.
 
  • #17


tiny-tim said:
Yes, but it should still give the same solution. :confused:

Show us how exactly you get a different solution.

Hi tiny - tim, this is how I get the different answer.

PA - P atm A = K spring change in X (divide both sides by A)

P - P atm A = K spring/A multiply by change in X (now put in data)

(600 000 - 100 000)pi x 0.05 squared = 4800 / pi x 0.05 squared. multiplied by change in X

500 000 x 7.85 x 10 -3 (minus three is a superscript) = 611154.98. change in X

3926.9908 = 911154.98 change in X (divide both sides by 911154.98)

X = 6.43 x 10 -3 (minus three superscript) answer in standard form.

Hope this helps

David:smile:
 
  • #18
Auto Engineer said:
P - P atm A = K spring/A multiply by change in X (now put in data)

(600 000 - 100 000)pi x 0.05 squared = 4800 / pi x 0.05 squared. multiplied by change in X

But you still have A on the LHS. :confused:
 
  • #19


tiny-tim said:
But you still have A on the LHS. :confused:

yes I agree, but my understanding is that when using transposition of formula, what you do to one side you do to the other, so as I invented A on the right hand side, then I must invent "A" on the left hand side when I divided both sides, but when cancelling like terms on the left hand side, one can only cancel one "A" with one "A" which leaves a remainder of one "A" on the left hand side?

I don't know of an alternative way?

Thanks

David:smile:
 
  • #20
Auto Engineer said:
yes I agree, but my understanding is that when using transposition of formula, what you do to one side you do to the other, so as I invented A on the right hand side, then I must invent "A" on the left hand side when I divided both sides, but when cancelling like terms on the left hand side, one can only cancel one "A" with one "A" which leaves a remainder of one "A" on the left hand side?

I don't understand :confused:

you had (a + b)A = X, or aA + bA = X

so you divide both sides by A, to give a + b = X/A

(and what do you mean by "invent"?)
 
  • #21


tiny-tim said:
I don't understand :confused:

you had (a + b)A = X, or aA + bA = X

so you divide both sides by A, to give a + b = X/A

(and what do you mean by "invent"?)

Hi tiny-tim.

Let's stay with the original equation. We have;

PA - P atm A = K spring, change in X.

Notice the equation above is like a see saw where the pivot is the equals sign "=". On the left hand side we have; PA - P atm A. And on the right hand side we have; K spring, change in X

Now if we wish to divide both sides of this equation by "A", then we must add "A" to both sides. By doing this we get;

PA - P atm A = K spring, change in X
A A

By the above equation if we now cancel the like terms on the left hand side, "A" numerator will cancel with one "A" the denominator, which will leave a remainder;

P - P atm A = K spring, change in X
A

So that if we put our data into this new equation, we get a new result which is not the same as the original = 0.818m. In the above example the answer given is; 0.643m

Hope this helps

PS. the forum platform shows the "A"s to be at the far left, these denominators should be on each side of the equals sign, ie. A = A, not AA on the left?

David:smile:
 
  • #22
Auto Engineer said:
Now if we wish to divide both sides of this equation by "A", then we must add "A" to both sides. By doing this we get;

PA - P atm A = K spring, change in X
A A

By the above equation if we now cancel the like terms on the left hand side, "A" numerator will cancel with one "A" the denominator, which will leave a remainder;

P - P atm A = K spring, change in X
A

Hi David! :smile:

(just got up :zzz:)

Let's just put that into CODE tags to preserve the spacing: :wink:

Code:
[U]PA - P atm A[/U] = [U]K spring, change in X[/U]
         A                   A

By the above equation if we now cancel the like terms on the left hand side, "A" numerator will cancel with one "A" the denominator, which will leave a remainder;

P - P atm A = [U]K spring[/U], change in X
                A

ok, now I see what you've done … you've only "cancelled" one of the As on the LHS …

you can't do that! … you must "cancel" both of them:

(PA - PatmA)/A = P - Patm

that's how brackets work! :smile:
 
  • #23


I hope this helps... :)
 

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  • #24


tiny-tim said:
Hi David! :smile:

(just got up :zzz:)

Let's just put that into CODE tags to preserve the spacing: :wink:

Code:
[U]PA - P atm A[/U] = [U]K spring, change in X[/U]
         A                   A

By the above equation if we now cancel the like terms on the left hand side, "A" numerator will cancel with one "A" the denominator, which will leave a remainder;

P - P atm A = [U]K spring[/U], change in X
                A

ok, now I see what you've done … you've only "cancelled" one of the As on the LHS …

you can't do that! … you must "cancel" both of them:

(PA - PatmA)/A = P - Patm

that's how brackets work! :smile:



Hi tiny-tim

Thank you for your advice and input to my "lack of understanding" the equation?

I always seem to rememeber "Transposition of formula" when I think something needs to be moved or changed, what I never remember is "Factorisation" which is what you have just explained:smile:

Thank you for ALL your help, very much appreciated:smile:

David, just got home from work:smile:
 
  • #25


Nick Bruno said:
I hope this helps... :)

Hi Nick,

Thank you for your advice, much appreciated. I think I have made most of it out, but on the far left before the = signs the writing is a little difficult to read.

Thanks

David:smile:
 
  • #26


Show us how exactly you get a different solution.

Sorry...for my earlier reply...but .2 is the answer if .05 in diameter...which I took wrong in earlier reply...the ans is .818=.0245*4...
So I guess he also was getting a different answer because he was takin it as diameter while calculation...just 1 of the silly mistakes...
 
  • #27


soory/...I think that above reply was really late to make some sense..actually I read the 1st page and replied...srry
 
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