Question

Can you check my work? I don't think gauge pressure is correct.

Answer 1

Number 3 Is what we are looking at.

Answer 2

I've got a big problem with your jpg enclosure. For some reason it's enormous when I try to view it and read it. I was wondering if you've ever considered doing stuff on pptx. It's about as good as I've seen on any computer I've tried to use, although probably not geared towards sci/tech/eng. But, it does have some advantages ... like I may be able to read what you've written. Probably a good start.
Enclose is a blank one.
Careful about that tank full of explosive methane/marsh gas - many cows in your area ?

Answer 3

It might help if you emailed me directly on osprey2015@hotmail.com
May give some flexibility to the discussions/thinking.

Answer 4

@osprey hold ctrl+scroll forward and backward on the mouse wheel. That should zoom out.

Answer 5

My analysis for the gauge pressure was that the system comes to thermal equilibrium which means Tsur = Tmeth, right? So temp. is not changing, thus pressure is not changing either. I have an equation sheet I can post later with the equations that undergo different processes, like isobaric, isothermal, isometric, etc. I will BRB I am about to start a team meeting.

Answer 6

First just a quick check I got everything you wrote right. I took a peak and as always I started off with a dimensional analysis. I'm really confused with all the "K"'s you have in there hard to tell by which you mean "kilo" and by which "kelvin" so the next line is a guess:
$$\frac{m^3}{kg} = \frac{ \left[ \frac{kJ } { kgK} \right] [K] }{ [kPa] }$$
Just checking if this is what you meant. By the numbers that's what it looked like. From that point on everything you write is correct and so is the following task. I've googled tables and calculators online and the numbers check out.

According to wiki the gauge pressure is equal to absolute pressure minus atmospheric pressure, so 100kPa - 1 atm = -1325 pascals.

Answer 7

Just to post the sources too:
http://www.peacesoftware.de/einigewerte/methan_e.html -> used to calc densities.
http://www.engineeringtoolbox.com/methane-d_1420.html -> verify results of task a)
good thing about things "specific" is the fact they're usually not influenced a lot by things like temps, pressures etc... for small changes. Your task is 30°C at 100kPa and standard condition is 20°C at 101kPa.

I have no idea what gauge pressure is and for the purposes of full disclosure I am really bad at TDM the definition was taken from first link about gauge pressure and in it is the link to wiki pages.
http://physics.stackexchange.com/questions/20460/gauge-pressure-vs-absolute-pressure

Answer 8

But in the context of the problem I thought the pressure of the surroundings is equal to that of 1 atm which made me think that 1atm is about 100kPA.

Answer 9

me too, I totally tilted once I figured out that's not the case, but the task clearly states:

"the temperature and the ***absolute pressure*** of the surroundings in which the tank exists are 30°C and 100kPa, respectively"

I tried redoing the task by adding atm pressure, then by doing vacuum etc. But none of them produced any results that could be compared to values you can google for.

Answer 10

Okay so I went to her office hours. I need to use PV = mRT. I need the mols of methane though. R = 8.314 kJ/kmol

How do I find mols of methane?

Answer 11

I have a chart that shows molecular weight for methane is M = 16.043 kg/kmol

Answer 12

Once we use PV = mRT to find P we subtract that from our Pabs.

Answer 13

From what I remember chemistry a mol is
$$n = \frac{m}{M} = \frac{m}{\sum Ar_i}$$

where Ar is the atomic mass

Answer 14

m = mass and M = molecular weight?

Answer 15

yeah
also that number seems wrong, it should be more like 16g/mol

Answer 16

I have 16.043 kg/kmol in my chart. so I need to look up m?

Answer 17

what I need is n.

Answer 18

Yes, you need n. n is the molarity or whatever it's actually called.
But you have m - it's 20kg

Answer 19

you have in your problem 20kg/0.016kg/mol moles.

Answer 20

Yep... I see now.

Answer 21

I get about 57kPa to answer part c.

Answer 22

I got to run to class now. I will be back on later. THanks.

Answer 23

Okay, now I know what gauge pressure is after today's lecture. She explained it well in class today.

Answer 24

If you want I can explain to you with a image of my notes.

Answer 25

post it, no harm in learning something.

Answer 26

Wait a min. So Pg = Pabs - Patm

When I use PV = mRT and solve for P that is Patm? It has to be right because T I am using is the T of the surroundings.

Answer 27

yeah, but in kelvins, so it's like you're doing it in absolute.
When pressure is concerned you're doing it in absolute, which is for most of "earth-bound" pressure-o-meters as if you were doing it in bad vacuum, as far as I understand the task. But I'm as lost as you, I think.

Answer 28

Actually after turning in my homework assignment I was thinking about the problem over again. I must have been solving for Patm. Anyways, sorry have not been active. I've been really busy since school started.