Question

So who wants to have fun with an integration problem?

Answer 1

-Question Part-
So I'm working on this heat transfer problem which applies to a rectangle 0 < x < 1 and 0 < y < 2 and I have it down all the way to:

\[\large T = \sum_{n=1}^{\infty} a_n sin(n\pi x)sinh(n\pi y) \]

Now I am to apply boundary a boundary condition to this that states: AT Y = 2

\[\large T = \begin{cases} T_o (1-2x) & 0 < x < 1/2 \\ 0 & 1/2 < x < 1 \end{cases} \]

And this WHOLE process is supposed to come out to:

\[\large T = T_o \sum_{0}^{\infty} c_n sin(n\pi x) \frac{cosh(n\pi x)}{cosh(2n \pi)}\]

Where \(\large c_n = 2\int_{0}^{1/2} sin(n\pi x)dx = \frac{2}{n\pi}(1 - \frac{2}{n\pi} sin(\frac{n\pi}{2}))\)

-Answer Part-

So I rewrote the sum:
\[\large T = \sum_{n=1}^{\infty} a_n sin(n\pi x)sinh(2n\pi ) \]

So next I multiplied through by \(\large sin(m\pi x)\) where m is some positive integer

\[\large Tsin(m\pi x) = \sum_{n=1}^{\infty} a_n sin(n\pi x)sin(m\pi x)sinh(2n\pi ) \]

Now I'm going to integrate from x = 0 to x = 1

\[\large \int_{0}^{1} Tsin(m\pi x)dx = \sum_{n=1}^{\infty} [a_n sinh(2n \pi) \int_{0}^{1} sin(n\pi x)sin(m\pi x)dx] \]

1 good thing to know is that:

\[\large \int_{0}^{1} sin(n\pi x)sin(m\pi x) = \begin{cases} 0 & m\cancel{=}n \\ 1/2 & m=n\end{cases} \]

So I can now write this whole thing as

\[\large \frac{1}{2}sinh(2m\pi)a_m = \int_{0}^{1} Tsin(m\pi x)dx\]

-----------------------NOW-------------------

Let's apply that boundary condition where \(\large T = T_o(1-2x)\) between 0 and 1/2
\[\large \frac{1}{2}sinh(2m\pi)a_m = \int_{0}^{1/2} T_o(1-2x)sin(m\pi x)dx \]

I'm supposed to solve this for \(\large a_m\) *Which would be the final \(\large c_n\)* which I did and got:
\[\large \frac{1}{m\pi}(1 - \frac{2}{m\pi} sin(\frac{m\pi}{2}))\]
*So close*

So I'm posting all this to check if I missed anything/did anything wrong

and ALSO how the heck do I get that final summation up there? Where the sinh is replaced by cosh/cosh

Thanks all! :D

Answer 2

jk

Answer 3

^_^ Look at all the nice symbols though XD

Answer 4

yeah thats all o^_^o

Answer 5

steady state heat conduction?

Answer 6

Yup, I used that fact to get down to this step :)

Answer 7

what are the complete boundary conditions?

Answer 8

as in, 4 sides, and i only see 1 BC

Answer 9

The other 3 sides are perfectly insulation, hence the one boundary condition :)

Answer 10

no, perfectly insulated is a boundary condition
\[dT/dx =0\]

Answer 11

and dT/dy=0 at the other 3 locations

Answer 12

Well I guess I should have been a bit more clear, there are 4 boundary conditions

T = 0 at x=0,1
T=0 at y=0
T = To(1-2x) at y = 2

I'm just at the step in the problem where I have already utilized the other b.c's

Answer 13

ok, thats not insulated, those are dirichlet BC much nicer and makes more sense

Answer 14

sorry for being picky, i need to know this and thus good practice since im currently doing a project on convective heat transfer and i havent done one of these problem in like 2 years

Answer 15

ACTUALLY no, you just made me find a mistake!

The B.C's are actually

T = 0 at x=0,1
dT/dy = 0 at y = 0 <-- mislooked that thinking T = 0
and T = To(1-2x), 0<x<1/2 at y = 2

And no, never apologize, that "pickiness" just helped out more than likely :D

Answer 16

back to basics
\[\alpha \left( \frac{d ^2T }{dx^2} +\frac{d^2 T}{dy^2}\right)=0\]

\[ \frac{d ^2T }{dx^2} =-\frac{d^2 T}{dy^2}=-\lambda^2\]

Answer 17

Yup, I guess I'll walk through the whole thing again, just for practice

Focus on the 'x' first

\[\large X'' + X\lambda^2 = 0\]

\[\large X = C_1 sin(\lambda x) + C_2 cos(\lambda x)\]

After b.c's (T=0 at x=0,1)

\[\large X = C_1 sin(n\pi x)\]

Answer 18

Now focus on the 'y'

\[\large Y'' - Y\lambda ^2 = 0\]

\[\large Y = C_3sinh(n\pi y) + C_4 cosh(n\pi y)\]

Now B.C (dT/dy = 0 at y = 0)

\[\large \frac{dT}{dy} = 0 = C_3 n\pi cosh(n\pi y) + C_4 n\pi sinh(n\pi y)\]

When y = 0 we find C3 is 0

So \(\large Y = C_4 cosh(n\pi y)\)

Answer 19

Everything looking good so far?

Answer 20

yup

Answer 21

got stuck trying to remember breaking T into a function of X and Y :P

Answer 22

Yeah, nifty little thing XD

Alright so lets see...from here I guess its that long process I originally went through

\[\large T = a_n sin(n\pi x) cosh(n\pi y)\]

where \(\large a_n = C_1 C_4\)

I can now write it as

\[\large T(x,y) = \sum_{n=1}^{\infty} a_n sin(n\pi x) cosh(n\pi y)\]

Answer 23

So now applying that boundary condition

\(\large T = T_o(1-2x), 0<x<1/2\) at y = 2

Answer 24

\[\large T_o(1-2x) = \sum_{n=1}^{\infty} a_n sin(n\pi x) cosh(2n\pi)\]

Answer 25

So i guess I'll do that same thing multiplying through by sin(m*pi*x) to simplify this a bit

Unless there's a way you would take here?

Answer 26

Ehh, we'll see what happens

So

\[\large T_o(1-2x)sin(m\pi x) = \sum_{n=1}^{\infty}[a_n cosh(2n\pi) sin(n\pi x)sin(m\pi x)]\]
\[\large T_o\int_{0}^{1}(1-2x)sin(m\pi x) = \sum_{n=1}^{\infty}[a_n cosh(2n\pi) \int_{0}^{1}(sin(n\pi x)sin(m\pi x))dx]\]

Answer 27

Got cut off...but that is just

\[\large \int_{0}^{1} sin(n\pi x) sin(m\pi x)dx\]

Which = 0 when m =/= n and equals 1/2 when m = n

So lets make all n = m

Answer 28

\[\large T_o\int_{0}^{1}(1-2x)sin(m\pi x) = \frac{1}{2}a_m cosh(2m\pi)\]

Answer 29

It's looking worse and worse, but is it looking correct so far?

Answer 30

Because let's say I keep going, now throwing in the fact x is between 0 and 1/2

\[\large T_o \int_{0}^{1/2}(1-2x)sin(m\pi x) = \frac{1}{2}a_m cosh(2m \pi)\]

Answer 31

This is just solving for \(\large a_m\) at this point

So we would have

\[\large 2T_o \int (1-2x)sin(m\pi x) = a_m cosh(2m\pi )\]

Answer 32

\[\large 2T_o(\int_{0}^{1/2} sin(m\pi x) dx - 2\int_{0}^{1/2} x sin(m\pi x)dx)= a_m cosh(2m\pi) \]

Answer 33

do yourself a favor and make a_mcosh(2mpi) = C_m

Answer 34

*Already found another mistake coming up which accounts for my factor of 2 i was off* and let's NOT write out all of that integration by parts stuff...

\[\large \frac{2T_o}{m\pi}(-cos(m\pi x)|_{0}^{1/2}) - \frac{2T_o}{m\pi}(xcos(m\pi x) - \frac{sin(m\pi x)}{m\pi})|_{0}^{1/2}\]

Answer 35

Uh oh...maybe I didn't fix my factor of 2 lol, I'll wait to see what you're typing :)

Answer 36

the book im using atm, has a much nicer looking form for finding Cn which would be your a_n*cosh

\[C_n= \frac{\int\limits_{a}^{b}w(x)f(x)\phi_n(x)}{N(\lambda)}\]
where N(lambda) is the norm which happens to be 2/L for the boundary conditions given

w(x) is a weighting function which i believe is just 1 in this case but necessary for different geometries

phi_n is just sin (n pi x)

REALLLLY WISH I COULD PUT THESE THINGS INLINE

f(x) would be the inhomogenous BC

Currently just doing the integration on paper

Answer 37

oh god this integral sucks

Answer 38

Lol if this doesn't work out I'm gonna be using that XD


so lets see

\[\large \frac{2T_o}{m\pi}(-cos(m\pi x)|_{0}^{1/2}) - \frac{2T_o}{m\pi}(xcos(m\pi x) - \frac{sin(m\pi x)}{m\pi})|_{0}^{1/2}\]

is now

\[\large -\frac{2T_o}{m\pi}[cos(m\pi x)|_{0}^{1/2} + (xcos(m\pi x) - \frac{sin(m\pi x)}{m\pi})|_{0}^{1/2}]\]

\[\large cos(m\pi x)|_{0}^{1/2} = cos(\frac{m\pi}{2}) - 1 \]
\[\large xcos(m\pi x)|_{0}^{1/2} = \frac{1}{2}cos(\frac{m\pi}{2})\]
\[\large \frac{sin(m\pi x)}{m\pi})|_{0}^{1/2} = sin(\frac{\frac{m\pi}{2}}{m\pi})\]

So the full thing *And yeah i know, this sucks XD i don't see how this will simplify but...

\[\large -\frac{2T_o}{m\pi}(cos(\frac{m\pi}{2}) - 1 + \frac{1}{2}cos(\frac{m\pi}{2}) - sin(\frac{\frac{m\pi}{2}}{m\pi})) = C_m\]

Answer 39

when i did it, it doesnt simplify out

Answer 40

Yeah I'm not seeing this as working out XD There's no way

Answer 41

and i just realized i made an integration error, ugg

Answer 42

Not as bad as i think i did!

Answer 43

For the love of god check my integration by parts -_- i think i used my OLD result not what the updated one was..1 sec >.<

Answer 44

this certainly beats having to make a 30 presentation on cold chain packaging

actually the presentation might be easier
but procrastinationnnnn

Answer 45

\[\large \int xsin(m\pi x)dx = -\frac{x}{m\pi}cos(m\pi x) - \int \frac{1}{m\pi}sin(m\pi x)dx\]

\[\large = -\frac{1}{m\pi}xcos(m\pi x) + \frac{1}{m\pi}cos(m\pi x)\]

Answer 46

Nothing means you're in your senior year more than messing up integration XD

Answer 47

if it makes you feel any better, im in grad school

Answer 48

crap i still messed up...that second (m*pi) in the denominator should be squared right?

I don't trust myself anymore XD

Answer 49

not due tomorrow?

Answer 50

Of course due tomorrow lol

Answer 51

pity, wouldve suggested taking a nap

Answer 52

Oh the magic words lol, nah that'll come later in the semester when I think about how much of a hit I can take by not doing an assignment haha

Answer 53

That is STILL wrong...my god... XD

Okay okay...this is the last time!

Answer 54

\[T_o \int\limits_{0}^{.5}(1-2x)\sin(n\pi x) dx\]
gonna leave off the T, just cause

u(x)= 1-2x v'(x)= sin(n pi x)
u'(x)= -2 v(x)= -1/n*pi cos(n*pi*x)

\[= (1-2x)\frac{1}{n\pi} \cos(n \pi x) |_{0}^{.5}- \frac{2}{n\pi}\int\limits\limits_{0}^{.5} \cos(n\pi x) dx\]

Answer 55

its a bit different than the way you did it, you distributed and then integrated

Answer 56

And you just distributed the negative sign through everything so that is correct i believe

Answer 57

Also, much neater to do that way^!! thank you lol

Answer 58

too many negative signs to know what you're talking about

Answer 59

XD just take my word lol...So we have

\[\large (1-2x)\frac{1}{n\pi} \cos(n \pi x) |_{0}^{.5}- \frac{2}{(n\pi)^2}\sin(n\pi x)|_{0}^{0.5})\]

Answer 60

\[= (1-2x)\frac{1}{n\pi} \cos(n \pi x)|_{0}^{.5} - \frac{2}{(n\pi)^2} \sin(n\pi x) |_{0}^{.5}\]

Answer 61

first half cancels nicely

Answer 62

Woo finally on the same page!

\[\large -\frac{1}{n\pi} - \frac{2}{(n\pi )^2}(sin(\frac{n\pi}{2}))\]

Answer 63

multiply that by To/2,

close to whatever you're suppose to get?

Answer 64

Okay, let's see if this works out

\[\large -2T_o[\frac{1}{n\pi} + \frac{2}{(n\pi)^2}sin(\frac{n\pi}{2}) = a_n cosh(2n\pi )]\]

Answer 65

Wait, To/2 ?

Answer 66

Too many damn formulas on this page XD

Answer 67

no no sorry, 2To

Answer 68

you're good

Answer 69

this is the exact reason why i never answered Cn for my exams, always wrote up to the step but the integration takes a while

Answer 70

So let's factor out \(\large \frac{1}{n\pi}\)

\[\large \frac{-2T_o}{n\pi}[1 + \frac{2}{n\pi}sin(\frac{n\pi}{2})]\]

Answer 71

Hmm, off by a sign somewhere

Should be:
\[\large \frac{2}{n\pi}(1 - \frac{2}{n\pi}sin(\frac{n\pi}{2}))\]

Answer 72

at least we were close

Answer 73

but HELL of a lot closer!!! Probably just lost a sign along the way!

You are awesome!!

Answer 74

thanks for the review, it was enlightening

Answer 75

Granted more of a review in integration than heat transfer lol, still appreciated!

Answer 76

college senior?

Answer 77

Indeed, Mechatronics engineering

Answer 78

senior project?

Answer 79

well, not the problem but do you have one?

Answer 80

Not this specifically, but i'm doing something regarding data collection in a turbine engine...had to find something related to my field (aerospace and defense)

Answer 81

cool
well, best of luck with the rest of the future integrations and project

Answer 82

Haha i appreciate it! And best of luck getting that Masters!

Answer 83

wanna find the mistake?
@ganeshie8

Answer 84

Ha found it XD remember when you said too many negatives to know what you're talking about? yeah...there was a hidden one we missed...gonna see if it fixes the result

Answer 85

And it does! woo! Thanks again!