BINGO! Thank you.
So in the spreadsheet it looks like =0.61121*2.71828^((18.678-(T/234.5))*(T/(257.14+T))) Where T is the cell in the spreadsheet that I enter the value for temperature.
As stated in the description of the equation, it's close but not 100% accurate. Good enough for what...
Thanks berkeman. Gave 'em a look. Nothing clicked. Perhaps it will become more clear to me as I learn.
So, from what I've read, and this part is starting to make sense... The statement "how much water air can hold" is more of an over simplified point of view than a statement of fact. It...
Hi guys and gals. Hobbyist, not a student. I've had great input here in the past. Thank you in advance for any direction and help figuring this out.
-Seth-
1. Homework Statement
I'd like to learn how to calculate relative humidity, saturation, and dew point under differing pressures and...
This isn't homework, I'm a hobbyist, not a student. Posting here because questions I've previously posted were moved to this section. My intent is to better understand humidity and it's role in compressed air for an automotive application. All help and direction is greatly appreciated. Thank...
Been trying to figure this out all day and the light bulb just came on. Just going to jot it down here for my future reference...
Figuring out pressure and pressure ratio with a known compression ratio/volume ratio.
Simple really. It's the ratio to the power of 1.4
10:1
10 to the...
Hey Cal,
Per the forum rules, I can not offer any help without proof that you've attempted to figure it out on your own. Perhaps you can share some specific details about your inquiry and particular steps you've taken to solve your question?
What is the total head?
Pipe ID?
Pipe length...
Woo Hoo! (doing a little victory dance) :biggrin:
In automotive engine vernacular, what you call volume ratio, is referred to as the compression ratio and is usually expressed as 3.15:1 (in this case). Until now, that's the only term I knew to use.
I was writing when you posted and missed your reply earlier. Thank you for the verification and input.
So, I can use the same equation to find the temperature upon decompression? Let's try it out!
k = 1.4 so the exponent is 0.286
P2 = 29.4 absolute (14.7 psig)
P1 = 73.5
T1 = 294...
If the above checks out, can I apply Charles's Law to find the volume once the heat of compression is removed?
If,
V1 = 31.75 cubic feet
T1 = 466
T2 = 294
Then, 31.75 / 466 x 294 = 20 cubic feet
So, the new compression ratio, assuming a constant pressure ratio of 5 is now 5:1...
Rap, I can't thank you enough.
Reading through Wikipedia and taking notes like a man on a mission, I came upon this equation which I believe converts the numbers I have so far into a volumetric compression ratio, or pressure ratio to compression ratio.
CR = T1 / T2 x PR
I checked it...
T1 = 70°F = 294.261kelvin
P1 = 14.7psi or 1 bar
P2 = 73.5 psi or 5 bar absolute pressure
I solved for the exponent first, 1.4 -1 = 0.4 / 1.4 = 0.286
Then solved the pressure ratio, 73.5 / 14.7 = 5
Pressure ratio of 5 to the power of 0.286 = 1.585
1.585 x 294.26 kelvin = 466 kelvin
Is...
Maybe it's appropriate for this to be in the homework area. I'm drawing on memory and can't seem to work the equation out in any way that makes sense. A little embarrassing to have to ask, but would one of you guys walk me through it? Please.
Thank you for your time. I went back to the reference I used to find the Heat Capacity Ratio, and you are completely correct. I mistakenly read Ar as Air. :rolleyes:
Is this the correct equation to use in finding temperature increase with compression?
T1 = 70°F = 294.261kelvin
P1 = 14.7psi or 1 bar
P2 = 73.5 psi or 5 bar absolute pressure
In reading about the value for k, I've found that most use 1.4, which is for dry air at 20°c. I've noted that...
Hello everyone. New guy, first post.
I typically frequent gasoline engine performance forums, but my inquisitive nature has led me beyond the scope of general knowledge available within that format. I appreciate any help in figuring out the following.
Thank you in advance!
If 100 cubic feet...