## anonymous one year ago integration problem Heat Transfer

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. 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)

5. anonymous

to yield the last equation i've provided

6. anonymous

where tau is the time constant and t is time

7. anonymous

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

8. anonymous

any suggestions?

9. Astrophysics

What exactly is the problem?

10. anonymous

let me see if i can attach the file

11. 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 $$\color{Red}{\tau}$$

12. anonymous

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

13. anonymous

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

14. ganeshie8

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

15. 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

16. anonymous

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

17. ganeshie8

if you prefer, call it with some other name like $$k$$ or something

18. anonymous

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

19. ganeshie8

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

20. anonymous

let me write it out for my benifit

21. anonymous

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

22. anonymous

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

23. ganeshie8

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

24. anonymous

yep!

25. ganeshie8

evaluate the integrals both sides

26. anonymous

why is it 1/tau?

27. ganeshie8

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

28. anonymous

ok

29. ganeshie8

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

30. ganeshie8

evaluate the integrals and solve $$T_s$$

31. anonymous

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

32. ganeshie8

careful, $$T_s\gt T_f$$

33. ganeshie8

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

34. anonymous

sorry, i should be a little more pedantic

35. ganeshie8

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

36. anonymous

$-|T _{f}-T _{s}|=e^{kt+C}$

37. ganeshie8

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

38. anonymous

so we deal with two cases then

39. ganeshie8

now rise both sides to power e

40. anonymous

yep

41. ganeshie8

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

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

43. anonymous

But we can sub $T _{s}-T _{f}=|T _{f}-T _{s}|$

44. anonymous

right?

45. ganeshie8

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

46. anonymous

yep

47. anonymous

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

48. anonymous

to make it look like equation 4

49. ganeshie8

isolate $$T_s$$

50. anonymous

$T _{s}=T _{f}-e ^{kt+C}$ $T _{s}=T _{f}+e ^{-kt-C}$

51. anonymous

im abit slow with the equation syntax sorry

52. ganeshie8

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

53. anonymous

why is that the only general solution

54. ganeshie8

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

55. anonymous

yep thats right

56. ganeshie8

plugin the initial condition and eliminate $$C_1$$

57. ganeshie8

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

58. ganeshie8

plugin $$t=0$$ in the general solution and solve $$C_1$$

59. anonymous

ahhh!!!

60. anonymous

so C1=Ts(0)-Tf

61. ganeshie8

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

62. anonymous

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

63. anonymous

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

64. anonymous

thats the same idea hey

65. anonymous

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

66. 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

67. ganeshie8

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

68. anonymous

oh whoops

69. Astrophysics

Yay, that was fun to watch :D

70. anonymous

just simply re-arrange

71. anonymous

yeah how epic was that!!!!

72. anonymous

that helps me so much

73. Astrophysics

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

74. 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

75. 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

76. anonymous

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

77. ganeshie8

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

78. anonymous

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

79. 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

80. Astrophysics

Could you not just solve for tau then?

81. anonymous

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

82. Astrophysics

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

83. anonymous

yea, minor brain fade sorry haha

84. ganeshie8

$$\tau$$ is supposed to be a constant right

85. anonymous

the time constant, yes

86. anonymous

in seconds

87. ganeshie8

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

88. 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

89. anonymous

what do you mean?

90. ganeshie8

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

91. anonymous

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

92. ganeshie8

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

93. Astrophysics

Nice catch ganeshie

94. anonymous

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

95. anonymous

thats what it says though

96. ganeshie8

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

97. ganeshie8

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

98. anonymous

oh! I see what you mean

99. anonymous

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

100. anonymous

right

101. 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 $$k$$ :)

102. anonymous

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

yep i understand now what you mean by that

104. anonymous

thanks again!

105. anonymous

time to write it up now! cheers everyone

106. ganeshie8

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

107. anonymous

thanks again mate

108. Astrophysics

That was awesome nice work both of you xD

109. anonymous

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