so your saying the cooler that they store the dry ice in is a powered refrigerator, like a meat freezer? thus keeping it longer? So if i put in a deep freezer, with no other food, would that be ok then?
I am working on my new 3d printer and i came accross an interesting situation. According to the manufacture, my stepper motors need to have a voltage of roughly 0.95 volts.
I checked mine and they were 0.81 volts, hence the not so smooth prints. Ok, so i increased the voltage to the...
Those are interesting formulas. Now i just need to find all the other numbers and calculate density in my area haha.
I know pressure can effect volume, but where is that in the basic density equation? What substitutes in for volume that has P involved?
Thanks for all the jelp
Ok, we all know that density is mass/volume. So if air is 1.22kg/m3,If we increase pressure, volume will change, therfore density can change.
So how does one calculate density of air at higher elevations. Where i currently live, at 4700ft above sea lvl, im guessing air density is not 1.22kg/m3...
true, but if i take a room at 72F and put a wall in the center to divide it in half, the temp on both sides would be the same 72F. the volume of air would be diffeernet than the original value, but the density would be the same and so would the specific volume as well as the mass.
yeah, the volume of air changes(decreases), and not the mass since none escaped. so density would be different then.
yes, "no matter what" is a strong statement and that is why i am confused haha. because all the basic physics stuff i took is now being challenged in thermo haha
to answer your first question, i thought density never changes for a substance. that is why it is called an intensive property? its ALWAYS 1000kg/m3 or 1.22kg/m3, etc etc. because if it were to change, then all my problems i did for finding pressure in a multifluid manometer, would be wrong...
Ok, i am struggling to figure something out. I dont know why math is so much easier than physics haha.
ok, here is my struggle. I have two states, state 1 and 2, which i will call just 1 and 2.
Here is the problem statment:
R-134a at 400psia has a specific volume of
0.1384ft3/lbm. Determine the temp of the refridgerant based on the gerealized compresabilty chart, ideal gas law, and the steam tables.
The professor was doing this problem step by step:
Ok, so i answered all 3 parts...
I was assigned an example problem today and i can do the math, look at my tables and get the right answer no problem. What i want to understand, is why when specific volume increases for r134a, that it means its supper heated?
My given value for sv was .1384m3/kg which is above the .09 in the...
@Chestermiller Yup, there is no way i would have known to do it that way haha. i have seen the gc number before but never used it.
How did you go from 19.48lbf to 0.13psi?
@Wrichik Basu Yeah, im going to have to mezmorize those numbers. they were not in my book. To be honest, i like metric...
Awesome:) the 0.305 is what unit? Is that a foot per meter as in 1 foot = 0.305m? I just want to make sure i get the numbers right. and 454 g per pound, hmm. my book nor homework lecture notes do not have those numbers. Well, i have them now haha. onward and upward. ANything else i should know...
1. Homework Statement
The pressure drop in a duct is to be measured by a differential oil manometer. If the differential height between the two fluid columns is 5.7 inches and the density of oil is 41 lbm/ft^3, what is the pressure drop in the duct in mmHg? Report your answer to 1 decimal...
I see what you are saying, and the numbers come out exatly the same regardless if i use Pe or P/rho. Thats what is throwing me off. I have tried it both ways with exact same numbers. The solutions manual did it a different way and ended up same numbers haha.
Im not trying to contradict you or...
Ok, im stuck on understanding when to use a part of an equation and when not to.
e = (P/rho) + (Ke/m) + (Pe/m) where P is pressure, Ke and Pe are Kinetic and potential energy divided by mass.
Ok, if we look at P/rho and P is defined as rho(g)(h) then the rho's divide out and we are left with...
yes, Thank you:) that was interesting to see it done that way. After I continued to exploer a little bit, I did find this as well:
F=m(dv/dt) use chain rule end up with Fdx=mvdv then F = integration[mvdv] + c
so amazing how both of these equations take you to the same place. I love math!
Ok, I see the force in Pe, as in mass*gravity, and I see the distance h. So in that case work is being done because its a force*distance. that makes more sense. thank you.
Now, for the second half, where did the equation of 1/2mv2 come from again? I used to know it but I do not remember it anymore