- anonymous

integration problem
Heat Transfer

- katieb

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- anonymous

Ill have to write out the problem. give me a tic

- anonymous

\[(\frac{ V*\rho*C }{ hA })\frac{ dT _{s} }{ dt }+T _{s}=T _{f}\]

- anonymous

Now volume, density and heat capacity are constant values but we need to integrate this to get
\[T _{s}-T _{f}=(T _{s}(0)-T _{f})\exp(\frac{ -t }{ \tau })\]

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## More answers

- anonymous

This is the case for a sphere being heated to a temperature Ts(0) and suddenly subjected to an airflow of constant temperature Tf (at t=0)

- anonymous

to yield the last equation i've provided

- anonymous

where tau is the time constant and t is time

- anonymous

I figure doing this
\[(\frac{ 1 }{ T _{f}-T _{s} })dT _{s}=\frac{ hA }{ \rho*V*C _{s} }dt\]

- anonymous

any suggestions?

- Astrophysics

What exactly is the problem?

- anonymous

let me see if i can attach the file

- ganeshie8

I figure doing this
\[(\frac{ 1 }{ T _{f}-T _{s} })dT _{s}=\color{red}{\frac{ hA }{ \rho*V*C _{s} }}dt\]
lump them and call it \(\color{Red}{\tau}\)

- anonymous

i need to integrate the last equation to obtain an expression for tau

- anonymous

would it be reasonable to find the unit for the lumped bit and see if it has units of seconds?

- ganeshie8

lets not even wry about the units because we're allowed to lump constants and w/o that it is looking messy

- anonymous

i'm only concerned on the theortical notes section. I need to provide a detailed version of the theory behind this rather than just copy word for word in my report with what they have already provided us with

- anonymous

yep, alright ganeshie8. Lets say we will lump them to say they are tau

- ganeshie8

if you prefer, call it with some other name like \(k\) or something

- anonymous

i'll stick with tau for the time being! cheers

- ganeshie8

yeah good, you have already separated, integrate and finish it off ?

- anonymous

let me write it out for my benifit

- anonymous

\[(\frac{ 1 }{ T _{f} -T _{s}})dT _{s}=\tau*dt\]

- anonymous

now how do we integrate the LHS, like what would our limits be?
from Ts=Ts(0) to Ts=Ts?

- ganeshie8

it should be
\[\int (\frac{ 1 }{ T _{f} -T _{s}})dT _{s}=\int \frac{1}{\tau}dt\]
right ?

- anonymous

yep!

- ganeshie8

evaluate the integrals both sides

- anonymous

why is it 1/tau?

- ganeshie8

maybe lets lump it with some other name, \(k\), because we don't know the expression for \(\tau\) yet

- anonymous

ok

- ganeshie8

\[(\frac{ 1 }{ T _{f}-T _{s} })dT _{s}=\color{red}{\frac{ hA }{ \rho*V*C _{s} }}dt\]
lump them and call it \(\color{Red}{k}\)
\[\int (\frac{ 1 }{ T _{f} -T _{s}})dT _{s}=\int k ~dt\]

- ganeshie8

evaluate the integrals and solve \(T_s\)

- anonymous

would it be:
\[-\ln(T _{f}-T _{s})=kt\]

- ganeshie8

careful, \(T_s\gt T_f\)

- ganeshie8

it should be :
\[-\ln |T _{f}-T _{s}|=kt+C\]

- anonymous

sorry, i should be a little more pedantic

- ganeshie8

you will get wrong answer if you don't put the absolute bars

- anonymous

\[-|T _{f}-T _{s}|=e^{kt+C}\]

- ganeshie8

it should be :
\[-\ln |T _{f}-T _{s}|=kt+C\]
\[\ln |T _{f}-T _{s}|=-kt-C\]

- anonymous

so we deal with two cases then

- ganeshie8

now rise both sides to power e

- anonymous

yep

- ganeshie8

there are no two cases because you knw that \(T_s\gt T_f\),
so \(|T_f-T_s| = T_s-T_f\)

- anonymous

oh right, yea thats fair.
so we have
\[|T _{f}-T _{s}|=-e ^{kt+C}=e ^{-kt-C}\]

- anonymous

But we can sub
\[T _{s}-T _{f}=|T _{f}-T _{s}|\]

- anonymous

right?

- ganeshie8

Yes, because |3-4| = 4-3

- anonymous

yep

- anonymous

so i get that far, what would be the likely thing to do next?

- anonymous

to make it look like equation 4

- ganeshie8

isolate \(T_s\)

- anonymous

\[T _{s}=T _{f}-e ^{kt+C}\]
\[T _{s}=T _{f}+e ^{-kt-C}\]

- anonymous

im abit slow with the equation syntax sorry

- ganeshie8

so the general solution of given differential equation is
\[T _{s}(t)=T _{f}+e ^{-\color{red}{k}t-C}\]
where \(\large \color{Red}{k}=\color{red}{\frac{ hA }{ \rho*V*C _{s} }}\)

- anonymous

why is that the only general solution

- ganeshie8

but nobody leaves that arbitrary constant in the top, we can make it look better by substituting \(C_1=e^{-C}\)
\[T _{s}(t)=T _{f}+C_1e ^{-\color{red}{k}t}\]

- anonymous

yep thats right

- ganeshie8

plugin the initial condition and eliminate \(C_1\)

- ganeshie8

say at \(t=0\), the temperature of sphere is \(T_s(0)\)

- ganeshie8

plugin \(t=0\) in the general solution and solve \(C_1\)

- anonymous

ahhh!!!

- anonymous

so C1=Ts(0)-Tf

- ganeshie8

Yes, compare that with the given solution and you can get an expression for \(\tau\)

- anonymous

so by plugging in the initial condition we have
\[T _{s}-T _{f}=(T _{s}(0)-T _{f})e ^{-kt}\]

- anonymous

perhaps we should lump the bits in equation 3 as tau so that when we re-arrange it is infact 1/tau

- anonymous

thats the same idea hey

- anonymous

so how do we then get an expression for tau by integrating?
would it be
\[\int\limits_{?}^{?}(T _{s}-T _{f})dT _{s}=(T _{s(0)}-T _{f})\int\limits_{?}^{?}e ^{-kt}dt\]

- ganeshie8

Yes \(\large \tau = \frac{1}{\color{red}{k}} = \dfrac{1}{\color{red}{\frac{ hA }{ \rho*V*C _{s} }}}\)
doesn't really matter, they all are same

- ganeshie8

we're done, don't need to integrate again

- anonymous

oh whoops

- Astrophysics

Yay, that was fun to watch :D

- anonymous

just simply re-arrange

- anonymous

yeah how epic was that!!!!

- anonymous

that helps me so much

- Astrophysics

Haha too bad I could only give one medal, good job @chris00 :)

- anonymous

one thing, im not sure how proficient you are in energy balances, but would you think we need to construct the energy balance and simplify it to obtain equation 2 in the attached sheet

- anonymous

it simply says for us to carry out the integration and obtain an expression for tau. i wouldn't think we need to carry out an energy balance. perhaps i'll just state it but focus on the integration as part of my report for that section

- anonymous

thanks so much guys! moral support goes to astrophysics aha
and well the medals obviously go to ganeshie8

- ganeshie8

i think you know more about physics part of this than me, never got a chance to study the heat equation..

- anonymous

haha, thanks but that integration saved me. You're the true hero today! cheers again

- anonymous

this semester is quite jammed packed with maths i think. Im doing process control this semester so hopefully i'll be back here getting help with laplace transforms! whoop

- Astrophysics

Could you not just solve for tau then?

- anonymous

yea it should be
\[\frac{ -t }{ \ln \left( \frac{ T _{s}-T _{f} }{ T _{s(0)}-T _{f} } \right) }=\tau\]

- Astrophysics

Yes, you shouldn't have to integrate again, I think that's good

- anonymous

yea, minor brain fade sorry haha

- ganeshie8

\(\tau\) is supposed to be a constant right

- anonymous

the time constant, yes

- anonymous

in seconds

- ganeshie8

doesn't make much sense to express it as a weird function of Ts, t etc..

- ganeshie8

\(\large \tau = \frac{1}{\color{red}{k}} = \dfrac{1}{\color{red}{\frac{ hA }{ \rho*V*C _{s} }}}\)
simplify and this should be good enough

- anonymous

what do you mean?

- ganeshie8

\(\tau\) is not a function of time, it is a constant, doesn't vary over time

- anonymous

so you think its silly to express it as a function of time then?

- ganeshie8

they are not asking you to express it as a function of time

- Astrophysics

Nice catch ganeshie

- anonymous

but dont we simply re-arrange equation 4 to obtain an expression for tau

- anonymous

thats what it says though

- ganeshie8

why do you think below is not a good expression for \(\tau\) ?
\(\large \tau = \frac{1}{\color{red}{k}} = \dfrac{1}{\color{red}{\frac{ hA }{ \rho*V*C _{s} }}}\)

- ganeshie8

thats not just expression, thats how we're defining \(\tau\) after integrating.

- anonymous

oh! I see what you mean

- anonymous

i was slightly confused since we already assumed that tau was that lump of vairables

- anonymous

right

- ganeshie8

yeah ikr, it was a bad idea to let \(\tau\) as our letter for lumped constants, but sooner we realized that and changed it to \(k\) :)

- anonymous

you would rather write equation 4 with the subject being time right?
since we want to get a plot of ln (...) vs time

- anonymous

yep i understand now what you mean by that

- anonymous

thanks again!

- anonymous

time to write it up now! cheers everyone

- ganeshie8

\[\large T _{s}(t) -T _{f}=[T _{s}(0)-T _{f}]e ^{-t/\tau}\]
where \(\tau =\color{red}{\frac{ \rho*V*C _{s} }{hA }} \)

- anonymous

thanks again mate

- Astrophysics

That was awesome nice work both of you xD

- anonymous

i'm glad you find it interesting! that was epic the last 30mins haha

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