A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

integration problem Heat Transfer

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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)

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    to yield the last equation i've provided

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    where tau is the time constant and t is time

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    any suggestions?

  9. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What exactly is the problem?

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let me see if i can attach the file

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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 \(\color{Red}{\tau}\)

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  17. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  19. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let me write it out for my benifit

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  23. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep!

  25. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    evaluate the integrals both sides

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why is it 1/tau?

  27. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  29. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  30. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    evaluate the integrals and solve \(T_s\)

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  32. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    careful, \(T_s\gt T_f\)

  33. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry, i should be a little more pedantic

  35. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  37. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so we deal with two cases then

  39. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    now rise both sides to power e

  40. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep

  41. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  42. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  43. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  44. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right?

  45. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  46. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep

  47. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  48. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    to make it look like equation 4

  49. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    isolate \(T_s\)

  50. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  51. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im abit slow with the equation syntax sorry

  52. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  53. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why is that the only general solution

  54. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  55. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep thats right

  56. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    plugin the initial condition and eliminate \(C_1\)

  57. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  58. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  59. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ahhh!!!

  60. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so C1=Ts(0)-Tf

  61. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  62. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  63. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  64. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thats the same idea hey

  65. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  66. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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

  67. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  68. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh whoops

  69. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yay, that was fun to watch :D

  70. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    just simply re-arrange

  71. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah how epic was that!!!!

  72. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that helps me so much

  73. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  74. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  75. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  76. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  77. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  78. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  79. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  80. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Could you not just solve for tau then?

  81. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  82. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  83. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea, minor brain fade sorry haha

  84. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  85. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the time constant, yes

  86. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    in seconds

  87. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  88. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  89. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what do you mean?

  90. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  91. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  92. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  93. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Nice catch ganeshie

  94. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  95. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thats what it says though

  96. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  97. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  98. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh! I see what you mean

  99. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  100. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  101. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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 \(k\) :)

  102. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  103. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep i understand now what you mean by that

  104. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks again!

  105. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    time to write it up now! cheers everyone

  106. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  107. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks again mate

  108. Astrophysics
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That was awesome nice work both of you xD

  109. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.