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anonymous
 one year ago
integration problem
Heat Transfer
anonymous
 one year ago
integration problem Heat Transfer

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ill have to write out the problem. give me a tic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\frac{ V*\rho*C }{ hA })\frac{ dT _{s} }{ dt }+T _{s}=T _{f}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now volume, density and heat capacity are constant values but we need to integrate this to get \[T _{s}T _{f}=(T _{s}(0)T _{f})\exp(\frac{ t }{ \tau })\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0to yield the last equation i've provided

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where tau is the time constant and t is time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I figure doing this \[(\frac{ 1 }{ T _{f}T _{s} })dT _{s}=\frac{ hA }{ \rho*V*C _{s} }dt\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0What exactly is the problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me see if i can attach the file

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.0i need to integrate the last equation to obtain an expression for tau

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be reasonable to find the unit for the lumped bit and see if it has units of seconds?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3lets not even wry about the units because we're allowed to lump constants and w/o that it is looking messy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'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
 one year ago
Best ResponseYou've already chosen the best response.0yep, alright ganeshie8. Lets say we will lump them to say they are tau

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3if you prefer, call it with some other name like \(k\) or something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'll stick with tau for the time being! cheers

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah good, you have already separated, integrate and finish it off ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me write it out for my benifit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\frac{ 1 }{ T _{f} T _{s}})dT _{s}=\tau*dt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now how do we integrate the LHS, like what would our limits be? from Ts=Ts(0) to Ts=Ts?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it should be \[\int (\frac{ 1 }{ T _{f} T _{s}})dT _{s}=\int \frac{1}{\tau}dt\] right ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3evaluate the integrals both sides

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3maybe lets lump it with some other name, \(k\), because we don't know the expression for \(\tau\) yet

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[(\frac{ 1 }{ T _{f}T _{s} })dT _{s}=\color{red}{\frac{ hA }{ \rho*V*C _{s} }}dt\] lump them and call it \(\color{Red}{k}\) \[\int (\frac{ 1 }{ T _{f} T _{s}})dT _{s}=\int k ~dt\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3evaluate the integrals and solve \(T_s\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be: \[\ln(T _{f}T _{s})=kt\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3careful, \(T_s\gt T_f\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it should be : \[\ln T _{f}T _{s}=kt+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i should be a little more pedantic

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you will get wrong answer if you don't put the absolute bars

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[T _{f}T _{s}=e^{kt+C}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it should be : \[\ln T _{f}T _{s}=kt+C\] \[\ln T _{f}T _{s}=ktC\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we deal with two cases then

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3now rise both sides to power e

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3there are no two cases because you knw that \(T_s\gt T_f\), so \(T_fT_s = T_sT_f\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh right, yea thats fair. so we have \[T _{f}T _{s}=e ^{kt+C}=e ^{ktC}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But we can sub \[T _{s}T _{f}=T _{f}T _{s}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, because 34 = 43

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i get that far, what would be the likely thing to do next?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to make it look like equation 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[T _{s}=T _{f}e ^{kt+C}\] \[T _{s}=T _{f}+e ^{ktC}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im abit slow with the equation syntax sorry

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3so the general solution of given differential equation is \[T _{s}(t)=T _{f}+e ^{\color{red}{k}tC}\] where \(\large \color{Red}{k}=\color{red}{\frac{ hA }{ \rho*V*C _{s} }}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why is that the only general solution

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3but nobody leaves that arbitrary constant in the top, we can make it look better by substituting \(C_1=e^{C}\) \[T _{s}(t)=T _{f}+C_1e ^{\color{red}{k}t}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3plugin the initial condition and eliminate \(C_1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3say at \(t=0\), the temperature of sphere is \(T_s(0)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3plugin \(t=0\) in the general solution and solve \(C_1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, compare that with the given solution and you can get an expression for \(\tau\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so by plugging in the initial condition we have \[T _{s}T _{f}=(T _{s}(0)T _{f})e ^{kt}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perhaps we should lump the bits in equation 3 as tau so that when we rearrange it is infact 1/tau

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats the same idea hey

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.3Yes \(\large \tau = \frac{1}{\color{red}{k}} = \dfrac{1}{\color{red}{\frac{ hA }{ \rho*V*C _{s} }}}\) doesn't really matter, they all are same

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3we're done, don't need to integrate again

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yay, that was fun to watch :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just simply rearrange

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah how epic was that!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that helps me so much

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Haha too bad I could only give one medal, good job @chris00 :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one 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
 one year ago
Best ResponseYou've already chosen the best response.0it 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
 one year ago
Best ResponseYou've already chosen the best response.0thanks so much guys! moral support goes to astrophysics aha and well the medals obviously go to ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i think you know more about physics part of this than me, never got a chance to study the heat equation..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha, thanks but that integration saved me. You're the true hero today! cheers again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this 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
 one year ago
Best ResponseYou've already chosen the best response.0Could you not just solve for tau then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea it should be \[\frac{ t }{ \ln \left( \frac{ T _{s}T _{f} }{ T _{s(0)}T _{f} } \right) }=\tau\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yes, you shouldn't have to integrate again, I think that's good

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea, minor brain fade sorry haha

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\tau\) is supposed to be a constant right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the time constant, yes

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3doesn't make much sense to express it as a weird function of Ts, t etc..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\large \tau = \frac{1}{\color{red}{k}} = \dfrac{1}{\color{red}{\frac{ hA }{ \rho*V*C _{s} }}}\) simplify and this should be good enough

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\tau\) is not a function of time, it is a constant, doesn't vary over time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you think its silly to express it as a function of time then?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3they are not asking you to express it as a function of time

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Nice catch ganeshie

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but dont we simply rearrange equation 4 to obtain an expression for tau

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats what it says though

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3why 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
 one year ago
Best ResponseYou've already chosen the best response.3thats not just expression, thats how we're defining \(\tau\) after integrating.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh! I see what you mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was slightly confused since we already assumed that tau was that lump of vairables

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah 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
 one year ago
Best ResponseYou've already chosen the best response.0you would rather write equation 4 with the subject being time right? since we want to get a plot of ln (...) vs time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep i understand now what you mean by that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0time to write it up now! cheers everyone

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\large T _{s}(t) T _{f}=[T _{s}(0)T _{f}]e ^{t/\tau}\] where \(\tau =\color{red}{\frac{ \rho*V*C _{s} }{hA }} \)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0That was awesome nice work both of you xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm glad you find it interesting! that was epic the last 30mins haha
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