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#1
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question for ivymike: Ideal Gas Law
I was having a conversation (aka debate) with a friend a few days ago, and ran into a stumbing block concerning the nature of air density as a combined function of pressure and density. So since I'm so lazy and prone to not dealing with anything other than the simplest mathematical functions, could you describe what say a doubling in pressure with no change in temperature would do to air density? And conversely, what about doubling absolute temperature (I've got that you have to use Kelvin in the equation) without changing pressure. Will the first equal a doubling of density and the latter equal a halving of it, or is the result more complex than that?
And to further apply this to an example problem, we'll take my friend's temp probe readings and boost information to find density. At the time of the readouts the ambient air was 70 degrees F, 14.7 psi absolute. Manifold pressure however was 20.7 absolute and temp was 170 degrees F. What is the percent increase in air density inside the manifold relative to ambient? Thanks in advance, peace.
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'03 Corvette Z06 '99 Prelude SH |
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#2
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doubling pressure, with no change in temperature, such as by very slowly compressing the air, will result in doubled density. If you have a fixed volume, and you slowly add air until the pressure is doubled, you will accomplish the same effect on density.
doubling temperature, while maintaining pressure, will result in increased volume (2x) and reduced density (0.5x). We'll assume that you got the temp and pressure readings right, although the former can be very difficult to measure accurately with certain equipment. Outside air: 70degF = 529.67R 14.7psia rho (density) = 7.492E-2 lb/ft3 (1.2 kg/m3) Inside air: 170degF = 629.67R 20.7psia rho = 8.875E-2 lb/ft3 (1.422 kg/m3) Pressure ratio = 20.7/14.7 = 1.408 Temp ratio = 629.67/529.67 = 1.189 Density ratio = 1.422/1.2 = (20.7/14.7)*(529.67/629.67) = 1.185 remember, the ideal gas law says that for air, rho = P/(R*T), where: P is absolute pressure of air R is gas constant for air (~ 287 J/(kg*K)) T is absolute temperature of air Ooops, forgot to mention the percent increase: (1.422-1.2)/1.2 = 18.5%
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Come on fhqwhgads. I see you jockin' me. Tryin' to play like... you know me... |
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#3
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Thank you thank you. One thing though, I thought I was supposed to be using kelvin for the degrees, so I just converted F to K with an online calculator. My results were 294.26K for 70degF, and 355.37K for 170degF. How'd you generate your numbers? This way I don't have to keep asking when I need some math homework done
.
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'03 Corvette Z06 '99 Prelude SH |
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#4
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Kelvin and Rankine are both absolute temperature scales, so either one can be used in a temperature ratio. The spacing between K degrees is the same as the spacing between C degrees, and the spacing between R degrees is the same as that for F degrees.
To convert from C to K, add 273.16 To convert from F to R, add 459.67 To convert from R to K, divide by 1.8 Comparing figures: 294.26K * 1.8R/K = 529.67R 355.37K * 1.8R/K = 639.67R I converted units "behind the scenes" in my calcs above, which is why I have Rair in J/kg.K, and temp in Rankine, and still get a correct answer (unless I messed up somewhere)
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Come on fhqwhgads. I see you jockin' me. Tryin' to play like... you know me... |
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#5
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an example calc:
70degF = 529.67R = 294.26K R_air = 287 J/kg.K P = 14.7psia (must always use absolute pressure) = 1.014*bar = 1.014*10^5 Pa = 1.014*10^5 N/m^2
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Come on fhqwhgads. I see you jockin' me. Tryin' to play like... you know me... Last edited by ivymike1031; 07-26-2002 at 10:48 PM. |
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#6
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Thanks Professor.
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'03 Corvette Z06 '99 Prelude SH |
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#7
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are you gonna reference the site you took that off ivymike?
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3rd Gear Burnouts... Oh Yeah! |
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#8
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Did you mean the image? I made that myself as follows:
1) write out the equation in mathcad 2) copy the equation and paste into photoshop 3) select the text, delete the background 4) sample the background color of an AF post 5) apply the same color to the background of my image 6) flatten the image, change to indexed color mode, set pallete to match colors used 7) save as gif & upload. It would have been quite a coincidence to find a site that had Texan's example problem already solved...
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Come on fhqwhgads. I see you jockin' me. Tryin' to play like... you know me... |
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#9
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Quote:
also I've done similar things for my classes (i sell studyguides ), but why not simply make the background transparent if you're using a gif and avoid the hassle of AF's alternating colors?
__________________
![]() mmmmmm.... :jump: |
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#10
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I learned to use Mathcad while in college, because several professors required that all my homework be done in it. Once I had the hang of it, it sure made my homework for other classes easier too.
Mathcad is made by a company called Mathsoft, and is available at many software retailers. See http://www.mathsoft.com and http://www.mathcad.com for more info. I've heard a rumor that the software can also be found (by unscrupulous folk) on Kazaa. I would have liked to have saved my gif with a transparent background, but I couldn't remember how to do it. I don't really use photoshop all that often anymore (v4.0), and every time I converted to indexed color, PS added a white background. I just said "screw it" and grabbed the bg color from AF.
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Come on fhqwhgads. I see you jockin' me. Tryin' to play like... you know me... |
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#11
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In photoshop.. all u need to do to get transparent background is go to File-->New...-->Select 'Transparent' (Under Contents)
Assuming u did screenshot to take the formula, delete all the garbage around the formula first, then u can then use the Magic Wand tool to select the white background behind the formula, delete it(press DEL)... and then save... LoL, need further instructions, go to the Photoshop Forum... =D There will always be talented photochoppers there to help ya! heh... i think i'm going to get MathCAD now, i'm sick of doing math equations in Microsoft Word, not only does it keep crashing, but it sometimes doesn't let me save after i spend an hour putting in a big equation.. |
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#12
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Quote:
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