The h Bar

0celo7

2:04 AM

THESE TABLES ARE IN POUNDS MASS

WHO THE HELL USES THAT

Semiclassical

oof

source and publication date?

0celo7

@Semiclassical Intro to Thermal Sciences by Schmidt and others, 1993

Semiclassical

:S

0celo7

@Semiclassical thankfully the pounds mass simply cancel

but I am still triggered

Semiclassical

yeah

is it maybe more an engineering source?

0celo7

2:07 AM

definitely

Semiclassical

that might explain it, then

0celo7

no one actually likes pounds mass though

they serve only to trigger

Semiclassical

well, tbh, if you asked me how much I weighed, I'd answer in lbs

...oh, but mass

bleh

imperial units are screwy

0celo7

@Semiclassical I got the right answer. Yolo

@Semiclassical you'd answer in pounds FORCE

because that's how much you WEIGH

Semiclassical

right.

0celo7

2:10 AM

if someone asked your MASS you wouldn't give POUNDS MASS because they are EVIL

Semiclassical

lol

0celo7

cap are necessary

Semiclassical

tbf you'd give the same number (sans units) for both mass and weight

0celo7

I almost got a B in my statics course because the final was entirely in imperial units and I forgot the conversion

Semiclassical

which at one level is convenient, and at another level is just silly

0celo7

2:11 AM

then I remembered my ultra smart calculator can do it for me B)

Semiclassical

it's sorta reminiscent of the various temperature systems

scientifically, no one likes Fahrenheit

0celo7

I do

I tend to use horse blood in my experiments a lot so F is convenient

Semiclassical

...

you've got issues man

0celo7

@Semiclassical tell me about it

Semiclassical

but in terms of describing human temperatures, I like Fahrenheit because, while 32 F may be the freezing point of water, 0 F is my freezing point

0celo7

2:13 AM

@Semiclassical Ok, so here's what I might do re research

I have to write an honors thesis anyway

Maybe my literature review should roll into that

So I can make my summaries and then use it for credit later

Semiclassical

kill two birds with one stone, yeah

hmm. i'm forgetting. has the fortran stuff you've had to do been because of the fluids course?

0celo7

@Semiclassical Knowing my advisor he has something completely different in mind for that, however :/

@Semiclassical No, because of my fortran course

Semiclassical

they make you do a fortran course?

0celo7

almost all nuclear-relevant code is in fortran

Semiclassical

hmm, true

0celo7

2:16 AM

and it's more like a general numerics class with a month of awful fortran

and then you do everything in fortran

and it's bad

Semiclassical

mostly I was curious if you'd had to do any Lagrangian hydrodynamics stuff

0celo7

fortran isn't that bad but now we have to use this automatic code-checking software and the formatting is very finicky

and formatting in fortran is a meme

78 format ("Step Size:", 1X, F10.6, 1X, "Numerical Result:", 1X, F10.5, 1X, "An. Res.:", 1X, F9.4, 1X, "Err[%]:", F11.5, /)

, for example

just to print some numbers

Semiclassical

...

0celo7

@Semiclassical So I evaluated the pressure drop in a pipe as 1.5 times the pressure on the bottom of the Pacific

Semiclassical

...lol

0celo7

2:19 AM

But I can't find where the error is

It's a massive fucking heat flux over the pipe lol

Semiclassical

can you do some dimensional analysis to figure out what the relevant scales would be?

0celo7

@Semiclassical I get 3psi when I just account for the viscous flow

then I add in the heat term and get 18,000 psi

Semiclassical

well, if you're only looking at the heat term, that does limit the number of dimensionful parameters involved

so dimensional analysis seems like it might be a good sanity check

0celo7

the heat flow term in Bernoulli's eqn is $$-\frac{\dot Q}{\dot m}\rho$$

Semiclassical

just for comparison, what's the viscous term?

0celo7

2:22 AM

$\dot m =\rho vA_\times$, so that's $$-\frac{\dot Q}{v A_\times}$$

@Semiclassical $$\frac{1}{2}\rho v^2 f,$$where $f$ is some friction term

Semiclassical

hmm

0celo7

usually like $$f \frac{L}{D}+C_f$$

Semiclassical

kk

0celo7

$f$ depends on the material, $L$ is the length, $D$ the diameter, and $C_f$ depends on the valves, etc.

Semiclassical

so the heat term should amount to $\frac{1}{A_x}\frac{\Delta Q}{\Delta x}$

and energy density is pressure

0celo7

2:24 AM

yeah something like that

Semiclassical

what I had in mind was to write the ratio of the heat term to the viscous term as on the order of $\frac{\dot{Q}\rho/\dot{m}}{\rho v^2 f}$

and somehow use that to get an estimate of how much bigger the heat term is than the viscous term

0celo7

@Semiclassical You're thinking too theoretically

Semiclassical

probably

0celo7

I have a MASSIVE heat exchange of 13kW in a tiny-ass pipe

Semiclassical

oof

0celo7

2:28 AM

not tiny ass-pipe mind you

Semiclassical

that'd be even more oof

0celo7

lol

Semiclassical

what's the dimensions of the pipe?

0celo7

diam is 10cm and length is 84cm

so the cross-sectional area is small

it just blows up the 13kW

I think the 13kW is right because we had a similar problem in class with a larger pipe that got 32kW

so I'm not sure what's wrong

Semiclassical

something's wrong. the heat term should be 1/A dQ/dx

but Q is energy, not power.

0celo7

2:31 AM

$$\frac{dQ}{dx}=\frac{dQ}{dt}\frac{dt}{dx}=\frac{\dot Q}{v}$$

Semiclassical

seems like you'd need a velocity in there as well

right.

0celo7

yeah I said 2m/s flow

or maybe I only thought I did

but the 2 doesn't really help

Semiclassical

hm. that seems pretty fast.

0celo7

@Semiclassical hence the MASSIVE 13kW

Semiclassical

lol true

0celo7

2:33 AM

@Semiclassical the heat flux in the pipe is 500,000 W/m^2 lol

Semiclassical

i'm getting more like 830,000

(13 kW) * 1/(pi*25 cm^2) * 1/(2m/s)

but meh. order of magnitude is what's important here

0celo7

You want $\dot Q=q'' \pi DL$

Semiclassical

hrm

fair

0celo7

$\pi DL$ is the area of the pipe

Semiclassical

heat flux is what here? the units of pressure should be energy / meters^3

0celo7

2:46 AM

no that's energy density

Semiclassical

pressure is energy density

0celo7

heat flux is heat (energy/time) through a surface (length^2)

Semiclassical

right.

so you'd need to multiply heat flux times velocity to get heat density

0celo7

ok the velocity is 2m/s

it doesn't change the ridiculous pressure I get

Semiclassical

so 1 million J/m^3

or 145 psi

0celo7

2:50 AM

wtf

I must be doing something wrong

Semiclassical

i'm not sure you are. 145 psi isn't that huge

0celo7

@Semiclassical how did you get that

Semiclassical

I'll be honest, I used WA

0celo7

@Semiclassical Did you account for the area

WA?

Semiclassical

WolframAlpha

0celo7

2:52 AM

o

Semiclassical

but, if the heat flux is 1/2 MW/m^2

0celo7

what was your calculation exactly?

Semiclassical

and the flow velocity is 2 m/s

0celo7

you have to account for the mass flow somehow

so the cross-sectional area comes in

Semiclassical

hmm

well, dimensionally at least, (1/2 MW/m^2)/(2m/s) = 250 kJ/m^3

(i should've had that not 1000 kJ)

plus, the heat flux is already power per unit area

so I think the cross-sectional area would've had to be used in computing that?

0celo7

2:55 AM

heat flux was a given

the problem was to find the length of the pipe needed to raise the temp by 20C

Semiclassical

ah

0celo7

then some other crap

Semiclassical

ugh.

for my own reference, 1psi ~ 7000 pascals = 7 kJ/m^3

0celo7

@Semiclassical ok Alex agrees with my 13.1kW

Semiclassical

hmm

0celo7

2:58 AM

And Luke

Ok my first parts are correct

so how the heck am I getting a black hole at the end of the pipe :/

maybe it's not wrong, just insane

I don't knoooooooow

Evan

Anyone feel like talking about steradians?

0celo7

no

what even are those

Evan

Google m8

Semiclassical

so then 1/A * \dot(Q)/v = 1/(pi*25cm^2) * (13 kJ/s)/(2m/s) =...

units of solid angle

0celo7

@Evan no thanks

Semiclassical

3:00 AM

so 2pi steradians = solid angle of the upper hemisphere etc

which, btw, does not mean i'm interested

0celo7

@Semiclassical so Hausdorff measure on the sphere?

Semiclassical

surface area divided by R^2

0celo7

well, unit sphere

Semiclassical

yeah

it's not very interesting

0celo7

that's what I thought

Evan

3:01 AM

It's interesting in the context of rigid body dynamics and holonomy of parallel transport.

So forget I even asked.

Semiclassical

mmkay. good night.

@0celo7 If I compute 1/A* dot(Q)/(flow velocity) using the numbers I gave above

0celo7

yeah

are you on your phone?

Semiclassical

nah

my main problem right now is that my internet connection is being an arse

then that comes out as 830000 J/m^3 = 120 psi

0celo7

ok you must be a god

how is that even possible

Semiclassical

for reference: WolframAlpha

that's better

0celo7

3:05 AM

lmao

Semiclassical

i'm taking the cross section as pi*(d/2)^2 = pi*(5 cm)^2 to be clear

0celo7

yeah

Semiclassical

i could well be missing some factors, of course

0celo7

you must be a god, one sec

finishing another problem

Semiclassical

dimensional analysis, man

0celo7

3:06 AM

@Semiclassical we're using the same equations...

Semiclassical

huh

0celo7

ok 120 psi makes me feel better about life

lemme try again...

Semiclassical

for comparison, the pressure at the bottom of the ocean is about 16,000 psi (per google)

so it's about a factor of 100 different

by comparison, that's about 1000 times bigger than atmospheric pressure

0celo7

ok I still get the wrong answer

just shoot me now

Semiclassical

what's the actual computation you do?

0celo7

3:09 AM

wait

@Semiclassical what did I say the diameter was?

Semiclassical

10 cm

0celo7

I meant 1 (hides)

Semiclassical

42 mins ago, by 0celo7

diam is 10cm and length is 84cm

lol

0celo7

typo!

Semiclassical

mmkay

0celo7

3:12 AM

10mm :P

Semiclassical

yeah, that pushes it up by a factor of 100 to 12000 psi

and then, yeah, bottom of the ocean :P

0celo7

@Semiclassical So...do I write that down with a note that it's probably wrong?

Semiclassical

good question. lemme check something

0celo7

or do home water heaters have bottom of the ocean pressures

Semiclassical

if the cross sectional area is pi/4 cm^2 and the flow velocity is 2m/s

then the flow rate should be (2m/s)(pi/4 cm^2) = 150 cm^3/s = 150 mL/s of water

which doesn't seem that huge

scratch head

I dunno man

that velocity seems reasonable

but that final pressure is just wtf

0celo7

3:17 AM

so what I get is 8.3e7 Pa

the end of the pipe is that much higher than the inlet

that's yuge

Semiclassical

yuuuge

0celo7

trump yuge

Semiclassical

for comparison, this source gives the operating pressure in a hot water system at home as about 125 psig (cleaverbrooks.com/reference-center/boiler-basics/…)

(I don't know what the g in psig means)

0celo7

gage

aka gauge

not related to Yang-Mills, although similar concept

Semiclassical

right, relative to atmospheric

0celo7

3:19 AM

mhm

Semiclassical

i mean, we've got two numbers

dot(Q) = 13 kW

and Av = flow rate = 150 cm^3/s

the latter doesn't seem unreasonable to me.

0celo7

the 13kW is probably correct

Semiclassical

hrm

0celo7

the chance of three people getting the same wrong answer independently is low

Semiclassical

well, unless the probability of making the same error is high

hmmmmm

something is goofy but I don't know what.

13 kW doesn't seem unreasonably huge, and neither does 150 cm^3/s

but their ratio does.

about the only thing I can think of is that we're interpreting energy density as pressure when we shouldn't

0celo7

3:25 AM

that has to be it somehow...

because I was given info about the pipe friction but that number is waaaaaaay smaller than this boy

Semiclassical

right.

waait

0celo7

so maybe I should give two answers: one without the Qdot term and one with, and appeal to the mercy of the TA?

Semiclassical

isn't the usual way one uses Bernoulli's law in such situations to compare the pressure at both sides of the tube?

0celo7

yes

Bernoulli has an enthalpy term in its full form

Semiclassical

then what's dot(Q) on both ends of the tube?

0celo7

3:28 AM

you use the dot(Q) accumulated along the length

Semiclassical

hrm.

presumably you've got a model problem for this

0celo7

nope! He forgot to do this in class

wrote it on the board and never got to it

Semiclassical

hrm

then i'm dubious that's the right way to use it

something rings false there, as imprecise a statement as that is

0celo7

Semiclassical

to be clear, I'm not disputing that there's an enthalpy term

i'm questioning whether it's being used correctly in this context.

i mean, you've got fluid entering at one end and leaving from the other

0celo7

3:33 AM

it says "heat added to control volume"

that seems pretty clear

Semiclassical

yeah, it does.

pZombie

Jim is on planet X, while McCoy is inside his rocket ship, just passing by Jim locally at v = 0.6c.

0celo7

@Semiclassical so write two answers and pray to the TA gods?

Semiclassical

idk man

write them both and explain why neither seem satisfactory

0celo7

the '19 nuclear group chat was flooded by someone asking for class notes and killed it

Bernardo Meurer

3:38 AM

@0celo7 Is your birthday today or tomorrow?

0celo7

today

@BernardoMeurer can't

I'm in the lab doing hw

pZombie

Jim is on planet X, while McCoy is inside his rocket ship, just passing by Jim locally at v = 0.6c. There is a planet Y with Jim's mom on it, 5 lightseconds away as measured by Jim in the same rest frame, with synced stopwatches. Jim told his mom to kill the cat when the watch counter will display 2 seconds. Dr McCoy passes Jim just about when Jim's stopwatch counter is at 0 seconds (2 seconds before his mom is supposed to kill the cat).

Spock also being around, asks if the cat is alive still. McCoy says "it's dead Jim" because for him, planet Y is at a distance of 4 lightseconds with the stopwatch on his simultaneity axis as observed by him being at 3 seconds, displaying 1 second post Jim's mom murdering the cat. Jim insists the cat is alive still. Who is right?

some spacetime diagrams depicting the above imgur.com/mn0X2Pq

John Rennie

4:40 AM

@pZombie it's a meaningless question.

pZombie

@JohnRennie thought so. It seems meaningless to ask in the context of SR, but then one has to ask if SR does describe the "world out there" or if it is only good for local predictions. Philosophically speaking, it works for local predictions but it didn't get the methaphysics right

John Rennie

I'm afraid I think that's meaningless too :-)

SR allows you to choose a coordinate system in which to describe the universe.

But what you're doing to use two different coordinates

And unsurprisingly the descriptions don't match

But there are invariants that are the same in all coordinate systems. For example proper time is an invariant so the lifetime of the cat, as measured by the cat, will be the same for all observers.

pZombie

@JohnRennie SR is a method to predict local experiences.

John Rennie

That's not a description of SR that I recognise.

pZombie

@JohnRennie if or not the world out there is as SR describes it, does not bother a physicist, as long as the local predictions based on the theory are correct

@JohnRennie as you said, it is meaningless for Spock to ask if the cat or his mom is dead or alive. Not so meaningless for a metaphysicist

John Rennie

4:48 AM

SR isn't restricted to local predictions. To take the example I mentioned above: the lifetime of the cat, as measured by the cat, will be the same for all observers.

pZombie

the time interval the cat will be alive however has nothing to do with spock asking if the cat is alive or dead while they meet locally at the same spacetime location

if neither Jim or McCoy can answer this question meaningfully, then what does it say about relativity?

where and when is the cat while Jim and McCoy are at the same spacetime location?

John Rennie

I suspect you have rediscovered the Andromeda paradox.

Which isn't a paradox of course.

pZombie

@JohnRennie Can you make a statement about if someone you know, who is living at a distance to you, is alive or dead right now?

because according to you, you cannot

if neither Jim nor McCoy can make a statement about the cat being alive or dead, then you cannot make any statement about others who are not local to you either

John Rennie

You can make statements about other non-local observers, such as their lifespan, but you can't make statements that rely upon simultaneity because that isn't a meaningful concept in SR.

It's fair to say you cannot make some statements about others who are not local to you

It is not correct to say you cannot make any statement about others who are not local to you

pZombie

@JohnRennie I am asking about a specific statement. So you agree that you cannot tell if anyone you know as a person, that isn't local to you currently, is alive or dead.

Which as i said, is perfectly fine with physicists, because all a theory has to do is to deliver the proper results locally. That's all we can do as individuals anyway, not being omniscient. The results of an experiment can be checked only locally.

John Rennie

5:07 AM

I can tell if some other person is dead or alive because in my frame that question has an answer. It's just that the answer is different in different frames.

Even in Newtonian physics there are frame dependent properties - kinetic energy for example.

But we don't claim Newtonian physics is local because different observers cannot agree about kinetic energy.

pZombie

Newtonian physics is very different. All observers in absolute spacetime agree on the exact where and when some other observer is while they are here and now

John Rennie

You are presumably saying that being dead or alive is fundamentally different to having a certain kinetic energy, but i disagree.

You are attaching an unwarranted significance to the time axis.

pZombie

What i consider significant is two observers communicating locally being able to agree on if some third observer far away is alive or dead.

they might disagree on his inertial mass, or coordinates, but they shouldn't disagree on which instance of him is "active"

John Rennie

You are of course free to consider anything you want as significant. But I wouldn't do so when you're sitting your physics exam.

2

pZombie

of course not. I will just pretend to be solipsistic when passing a physics exam, not concerned about if others even exist. Just some simulation doing some maths and throwing results my way locally

0celo7

5:20 AM

Lmao

John Rennie

:-)

Semiclassical

I'm just going to leave this here...

John Rennie

@pZombie there is often conflict between what we think makes sense and theories of modern physics. Quantum mechanics is an obvious example. But the universe doesn't care what we think.

pZombie

@JohnRennie If i told you that all of physics started with metaphysics, you wouldn't believe me

John Rennie

I would tell you I don't consider that a useful statement

Physicists - those of us paid to do physics that is - re the ultimate pragmatists

If the predictions of our theories match experiment then we are are happy bunnies

Semiclassical

5:23 AM

ehhhh

I'm actually someone who doesn't consider metaphysics to be a dirty word in physics, so long as it is used correctly

John Rennie

Many, maybe most, of us have argued about the philosophical implications of physics, but we tend to do it in the bar after many drinks and not when trying to do our day job.

Captain Bohemian

what does Spin(d−1, 1) mean?

Semiclassical

I mean, it's not as though we just randomly do experiments and take data with the belief that said data will magically follow certain patterns.

we do have some regulative principles and concepts which guide us in formulating theories and drawing conclusions from data

even there, though, i'm using the word 'metaphysics' in a fairly broad way.

i don't, for instance, think the difference between living/dead matter is relevant in the context of theoretical physics

...meh, i'm rambling

0celo7

I agree with @Semiclassical more than @JohnRennie. Blindly following equations doesn't seem to be right

Semiclassical

ye olde empiricism vs. rationalism dilemma

pZombie

5:32 AM

My view of the world is that of Descartes. "I think therefore i am". I = the experiencing entity. I experience the world as the entity. This is the metaphysical part. Physicists translated this part into the "here and now" based on some theory of how space and time is shaped

the theory of space and time can change at any point, but the experience, the metaphysical part remains

John Rennie

@0celo7 should you ever decide to work in nuclear engineering let me know so I can move away as far as possible from any power stations :-)

Semiclassical

I'm not sufficiently educated in him to really say this, but my mindset tends far more towards Kant than Descartes.

@JohnRennie engineering != physics :P

pZombie

So following the above, the question now would be, if there are others out there, also experiencing, like i do. Or if i am alone. For physicists this might not be an interesting question, but if it was, then the question would be, if theories that work for solipsism would also always work for those who believe in others existing as well.

Bernardo Meurer

5:48 AM

@JohnRennie What if I decide to work in nuclear engineering?

John Rennie

OMG

Colonising other planets just became a priority :-)

Anonymous

You'd make free open source bombs :D

John Rennie

Do atomic bombs count as proprietary malware? :-)

Bernardo Meurer

@JohnRennie Hmmm

Good question

I guess so, but I'd make them open source

OS malware :)

pZombie

I'd rather you open source self-replicating, self-modifying, AI controlled, invisible to the human eye, swarms of C4 carrying nano drones able to multi connect and explode on their target

I imagined that would happen years ago, but according to this article life.spectator.co.uk/2017/09/… it is actually a thing now

Bernardo Meurer

6:03 AM

To be honest the worst thing I could open source are my thoughts

So y'all gotta thank me I don't keep an active blog

It's the only reason the world still somewhat works

John Rennie

I've occasionally wondered about writing a blog, but my thoughts revolve around food and smutty humour so I suspect it would have limited entertainment value.

Anonymous

6:20 AM

It's a pain to make MathJax work on blogger and wordpress. SE is the only one that does it right (with a text entry box and an instant preview box). I wish they allowed us to host personal blogs on SE.

Captain Bohemian

it turns out your mind interwines with food so much.

parvin

6:36 AM

hello guys!

can some one tell me what the notation for "half" is in physics?

like, how do you show it when you wanna point to "half of a system"?
system is not a number so I can not write 1/2

David Z

6:56 AM

@parvin I'm really not sure what to tell you without having more detail about what this system is

parvin

I'm trying to sketch a box with two sides, filled with two different liquids, in a stable condition

if the left side has liquid A and right side B

and if A is half of B

how can I show this?

1/2 B | B ?

because B is a name I can not do that, like writing 1/2 NaCl !

Anonymous

Why do you even want to write something like that? What's the problem in writing it out in words?

parvin

I want to print the sketch on a cup and give it to my friend, as Bday present ! :)
I want it to be totally formulas

John Rennie

@parvin What do you mean by A is half of B? The volume of the liquid A is half the volume of the liquid B?

Or the volume of A is half of the total volume?

parvin

do you know about "er" and "epr"?
I don't know much either but I know that "er =epr"
it's not exactly an equation tho.
here, right side would be er and left would be epr
but because they are equal I have to some how show half of this and half of that

Anonymous

7:07 AM

What...

Anonymous

ER=EPR

ER=EPR is a conjecture in physics stating that entangled particles are connected by a wormhole (or Einstein–Rosen bridge) and may be a basis for unifying general relativity and quantum mechanics into a theory of everything. == Overview == The conjecture was proposed by Leonard Susskind and Juan Maldacena in 2013. They proposed that a non-traversable wormhole (Einstein–Rosen bridge or ER bridge) is equivalent to a pair of maximally entangled black holes. EPR refers to quantum entanglement (EPR paradox). The symbol is derived from the first letters of the surnames of authors who wrote the first paper...

parvin

look, I'm not a physician I just want to sketch some thing relevant to physics, from the things my friend has told me about, so don't be mad at me I'm not familiar to these stuff

all i want to do is to some how show "half" of a "name"

that's all

guess the best way is to write $1/2$

David Z

The problem is just that we can't understand what you're asking

not well enough to answer, anyway

parvin

can't you even say how you show "half of a room" in math?

David Z

Sure we can, we'd write "half of a room". But you said that's not what you want, so I'm not sure how to proceed further.

John Rennie

7:11 AM

@parvin $$V = \tfrac{1}{2} V_\text{room} $$

parvin

@JohnRennie yes ! why not! that's better
thank you!

David Z

We would have gotten there several minutes ago if you'd answered John's question ;-)

parvin

yeas the problem is that I myself am not so sure about what i want to sketch, so..i confuse you too !

David Z

Ah, yeah that happens sometimes... but it's still very useful to consider the feedback people are giving (even in the form of questions)

parvin

I do ! i have to explain with details. i'll do that next time

Fawad

7:20 AM

👎😾

parvin

:/

Abcd

8:24 AM

@JohnRennie Why is reputation not awarded for upvoted comments?

John Rennie

@Abcd comments are meant to be used only for clarification of answers or questions. They aren't regarded as being important and people shouldn't answer in comments (though we often do :-)

David Z

8:44 AM

In a sense, we don't want to reward people for posting comments.

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