anonymous
  • anonymous
integration problem Heat Transfer
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Ill have to write out the problem. give me a tic
anonymous
  • anonymous
\[(\frac{ V*\rho*C }{ hA })\frac{ dT _{s} }{ dt }+T _{s}=T _{f}\]
anonymous
  • anonymous
Now volume, density and heat capacity are constant values but we need to integrate this to get \[T _{s}-T _{f}=(T _{s}(0)-T _{f})\exp(\frac{ -t }{ \tau })\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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
  • anonymous
to yield the last equation i've provided
anonymous
  • anonymous
where tau is the time constant and t is time
anonymous
  • anonymous
I figure doing this \[(\frac{ 1 }{ T _{f}-T _{s} })dT _{s}=\frac{ hA }{ \rho*V*C _{s} }dt\]
anonymous
  • anonymous
any suggestions?
Astrophysics
  • Astrophysics
What exactly is the problem?
anonymous
  • anonymous
let me see if i can attach the file
ganeshie8
  • 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
  • anonymous
i need to integrate the last equation to obtain an expression for tau
anonymous
  • anonymous
would it be reasonable to find the unit for the lumped bit and see if it has units of seconds?
ganeshie8
  • ganeshie8
lets not even wry about the units because we're allowed to lump constants and w/o that it is looking messy
anonymous
  • 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
  • anonymous
yep, alright ganeshie8. Lets say we will lump them to say they are tau
ganeshie8
  • ganeshie8
if you prefer, call it with some other name like \(k\) or something
anonymous
  • anonymous
i'll stick with tau for the time being! cheers
ganeshie8
  • ganeshie8
yeah good, you have already separated, integrate and finish it off ?
anonymous
  • anonymous
let me write it out for my benifit
anonymous
  • anonymous
\[(\frac{ 1 }{ T _{f} -T _{s}})dT _{s}=\tau*dt\]
anonymous
  • anonymous
now how do we integrate the LHS, like what would our limits be? from Ts=Ts(0) to Ts=Ts?
ganeshie8
  • ganeshie8
it should be \[\int (\frac{ 1 }{ T _{f} -T _{s}})dT _{s}=\int \frac{1}{\tau}dt\] right ?
anonymous
  • anonymous
yep!
ganeshie8
  • ganeshie8
evaluate the integrals both sides
anonymous
  • anonymous
why is it 1/tau?
ganeshie8
  • ganeshie8
maybe lets lump it with some other name, \(k\), because we don't know the expression for \(\tau\) yet
anonymous
  • anonymous
ok
ganeshie8
  • 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
  • ganeshie8
evaluate the integrals and solve \(T_s\)
anonymous
  • anonymous
would it be: \[-\ln(T _{f}-T _{s})=kt\]
ganeshie8
  • ganeshie8
careful, \(T_s\gt T_f\)
ganeshie8
  • ganeshie8
it should be : \[-\ln |T _{f}-T _{s}|=kt+C\]
anonymous
  • anonymous
sorry, i should be a little more pedantic
ganeshie8
  • ganeshie8
you will get wrong answer if you don't put the absolute bars
anonymous
  • anonymous
\[-|T _{f}-T _{s}|=e^{kt+C}\]
ganeshie8
  • ganeshie8
it should be : \[-\ln |T _{f}-T _{s}|=kt+C\] \[\ln |T _{f}-T _{s}|=-kt-C\]
anonymous
  • anonymous
so we deal with two cases then
ganeshie8
  • ganeshie8
now rise both sides to power e
anonymous
  • anonymous
yep
ganeshie8
  • 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
  • anonymous
oh right, yea thats fair. so we have \[|T _{f}-T _{s}|=-e ^{kt+C}=e ^{-kt-C}\]
anonymous
  • anonymous
But we can sub \[T _{s}-T _{f}=|T _{f}-T _{s}|\]
anonymous
  • anonymous
right?
ganeshie8
  • ganeshie8
Yes, because |3-4| = 4-3
anonymous
  • anonymous
yep
anonymous
  • anonymous
so i get that far, what would be the likely thing to do next?
anonymous
  • anonymous
to make it look like equation 4
ganeshie8
  • ganeshie8
isolate \(T_s\)
anonymous
  • anonymous
\[T _{s}=T _{f}-e ^{kt+C}\] \[T _{s}=T _{f}+e ^{-kt-C}\]
anonymous
  • anonymous
im abit slow with the equation syntax sorry
ganeshie8
  • 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
  • anonymous
why is that the only general solution
ganeshie8
  • 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
  • anonymous
yep thats right
ganeshie8
  • ganeshie8
plugin the initial condition and eliminate \(C_1\)
ganeshie8
  • ganeshie8
say at \(t=0\), the temperature of sphere is \(T_s(0)\)
ganeshie8
  • ganeshie8
plugin \(t=0\) in the general solution and solve \(C_1\)
anonymous
  • anonymous
ahhh!!!
anonymous
  • anonymous
so C1=Ts(0)-Tf
ganeshie8
  • ganeshie8
Yes, compare that with the given solution and you can get an expression for \(\tau\)
anonymous
  • anonymous
so by plugging in the initial condition we have \[T _{s}-T _{f}=(T _{s}(0)-T _{f})e ^{-kt}\]
anonymous
  • anonymous
perhaps we should lump the bits in equation 3 as tau so that when we re-arrange it is infact 1/tau
anonymous
  • anonymous
thats the same idea hey
anonymous
  • 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
  • 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
  • ganeshie8
we're done, don't need to integrate again
anonymous
  • anonymous
oh whoops
Astrophysics
  • Astrophysics
Yay, that was fun to watch :D
anonymous
  • anonymous
just simply re-arrange
anonymous
  • anonymous
yeah how epic was that!!!!
anonymous
  • anonymous
that helps me so much
Astrophysics
  • Astrophysics
Haha too bad I could only give one medal, good job @chris00 :)
anonymous
  • 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
  • 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
  • anonymous
thanks so much guys! moral support goes to astrophysics aha and well the medals obviously go to ganeshie8
ganeshie8
  • ganeshie8
i think you know more about physics part of this than me, never got a chance to study the heat equation..
anonymous
  • anonymous
haha, thanks but that integration saved me. You're the true hero today! cheers again
anonymous
  • 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
  • Astrophysics
Could you not just solve for tau then?
anonymous
  • anonymous
yea it should be \[\frac{ -t }{ \ln \left( \frac{ T _{s}-T _{f} }{ T _{s(0)}-T _{f} } \right) }=\tau\]
Astrophysics
  • Astrophysics
Yes, you shouldn't have to integrate again, I think that's good
anonymous
  • anonymous
yea, minor brain fade sorry haha
ganeshie8
  • ganeshie8
\(\tau\) is supposed to be a constant right
anonymous
  • anonymous
the time constant, yes
anonymous
  • anonymous
in seconds
ganeshie8
  • ganeshie8
doesn't make much sense to express it as a weird function of Ts, t etc..
ganeshie8
  • 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
  • anonymous
what do you mean?
ganeshie8
  • ganeshie8
\(\tau\) is not a function of time, it is a constant, doesn't vary over time
anonymous
  • anonymous
so you think its silly to express it as a function of time then?
ganeshie8
  • ganeshie8
they are not asking you to express it as a function of time
Astrophysics
  • Astrophysics
Nice catch ganeshie
anonymous
  • anonymous
but dont we simply re-arrange equation 4 to obtain an expression for tau
anonymous
  • anonymous
thats what it says though
ganeshie8
  • 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
  • ganeshie8
thats not just expression, thats how we're defining \(\tau\) after integrating.
anonymous
  • anonymous
oh! I see what you mean
anonymous
  • anonymous
i was slightly confused since we already assumed that tau was that lump of vairables
anonymous
  • anonymous
right
ganeshie8
  • 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
  • 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
  • anonymous
yep i understand now what you mean by that
anonymous
  • anonymous
thanks again!
anonymous
  • anonymous
time to write it up now! cheers everyone
ganeshie8
  • 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
  • anonymous
thanks again mate
Astrophysics
  • Astrophysics
That was awesome nice work both of you xD
anonymous
  • anonymous
i'm glad you find it interesting! that was epic the last 30mins haha

Looking for something else?

Not the answer you are looking for? Search for more explanations.