Question

integration problem
Heat Transfer

Answer 1

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

Answer 2

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

Answer 3

Now volume, density and heat capacity are constant values but we need to integrate this to get

\[T _{s}-T _{f}=(T _{s}(0)-T _{f})\exp(\frac{ -t }{ \tau })\]

Answer 4

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)

Answer 5

to yield the last equation i've provided

Answer 6

where tau is the time constant and t is time

Answer 7

I figure doing this

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

Answer 8

any suggestions?

Answer 9

What exactly is the problem?

Answer 10

let me see if i can attach the file

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

let me write it out for my benifit

Answer 21

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

Answer 22

now how do we integrate the LHS, like what would our limits be?

from Ts=Ts(0) to Ts=Ts?

Answer 23

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

right ?

Answer 24

yep!

Answer 25

evaluate the integrals both sides

Answer 26

why is it 1/tau?

Answer 27

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

Answer 28

ok

Answer 29

\[(\frac{ 1 }{ T _{f}-T _{s} })dT _{s}=\color{red}{\frac{ hA }{ \rho*V*C _{s} }}dt\]

lump them and call it \(\color{Red}{k}\)

\[\int (\frac{ 1 }{ T _{f} -T _{s}})dT _{s}=\int k ~dt\]

Answer 30

evaluate the integrals and solve \(T_s\)

Answer 31

would it be:

\[-\ln(T _{f}-T _{s})=kt\]

Answer 32

careful, \(T_s\gt T_f\)

Answer 33

it should be :

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

Answer 34

sorry, i should be a little more pedantic

Answer 35

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

Answer 36

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

Answer 37

it should be :

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

\[\ln |T _{f}-T _{s}|=-kt-C\]

Answer 38

so we deal with two cases then

Answer 39

now rise both sides to power e

Answer 40

yep

Answer 41

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

Answer 42

oh right, yea thats fair.
so we have

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

Answer 43

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

Answer 44

right?

Answer 45

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

Answer 46

yep

Answer 47

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

Answer 48

to make it look like equation 4

Answer 49

isolate \(T_s\)

Answer 50

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

Answer 51

im abit slow with the equation syntax sorry

Answer 52

so the general solution of given differential equation is
\[T _{s}(t)=T _{f}+e ^{-\color{red}{k}t-C}\]

where \(\large \color{Red}{k}=\color{red}{\frac{ hA }{ \rho*V*C _{s} }}\)

Answer 53

why is that the only general solution

Answer 54

but nobody leaves that arbitrary constant in the top, we can make it look better by substituting \(C_1=e^{-C}\)

\[T _{s}(t)=T _{f}+C_1e ^{-\color{red}{k}t}\]

Answer 55

yep thats right

Answer 56

plugin the initial condition and eliminate \(C_1\)

Answer 57

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

Answer 58

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

Answer 59

ahhh!!!

Answer 60

so C1=Ts(0)-Tf

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

thats the same idea hey

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

oh whoops

Answer 69

Yay, that was fun to watch :D

Answer 70

just simply re-arrange

Answer 71

yeah how epic was that!!!!

Answer 72

that helps me so much

Answer 73

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

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

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

Answer 78

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

Answer 79

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

Answer 80

Could you not just solve for tau then?

Answer 81

yea it should be

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

Answer 82

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

Answer 83

yea, minor brain fade sorry haha

Answer 84

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

Answer 85

the time constant, yes

Answer 86

in seconds

Answer 87

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

Answer 88

\(\large \tau = \frac{1}{\color{red}{k}} = \dfrac{1}{\color{red}{\frac{ hA }{ \rho*V*C _{s} }}}\)

simplify and this should be good enough

Answer 89

what do you mean?

Answer 90

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

Answer 91

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

Answer 92

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

Answer 93

Nice catch ganeshie

Answer 94

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

Answer 95

thats what it says though

Answer 96

why do you think below is not a good expression for \(\tau\) ?

\(\large \tau = \frac{1}{\color{red}{k}} = \dfrac{1}{\color{red}{\frac{ hA }{ \rho*V*C _{s} }}}\)

Answer 97

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

Answer 98

oh! I see what you mean