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Help! rates of change

  1. Dec 6, 2006 #1
    1. The problem statement, all variables and given/known data
    We need to know the temperature of the copper bit of a soldering iron varies with time after the power has been switched on. This is a first step to determine how long it takes for the temperature of the bit to reach the operating temperature at wich it can melt the solder. We assume all heat produced goes directly to the bit and none is lost to the air. i.e. the temperature of the bit, Theta = Theta (t), depends only on time.

    3 laws of physics:

    the rate of energy storage in the bit is the product of the mass m of copper, the specific heat c of copper and the rate of change in the bit.

    the rate of loss of heat from the bit to the air has the form kA(theta - Thate a) where theta a is the temprature of the air, A is the (constant) cross section of the bit, and k is a constant

    The heat traveling from the barrel to the bit is the sum of the heat loss from the bit and the heat stored in the bit (consesrvation of energy)

    1) To which value do you expect the temprature of the bit to settle?

    2) Sketch a graph of Theta with t (time)

    3) Write down the differential equation which describes the cooling process.

    4)Given that the solution tof the equation

    dtheta/dt + a theta = b

    where a and b are cpnstant, is

    theta = b/a + Ce^-at

    where c is a constant, write down the solution of your eqaution in part 3 which satisfies the initial condition theta = theta 0 at t = 0

    2. Relevant equations

    3. The attempt at a solution

    1) the temperaturewill settle at theta a the temperature of the air.

    2)i drew a graph that showed the temperature drop rapidly at first and then steadyout to nearly level at theta a

    3) (this is where i get really stuck!) i got the following

    dtheta/dt + k = A(theta - theta 0)


    theta = A(theta - theta 0) / k

    thanks for any help you can give =)
  2. jcsd
  3. Dec 7, 2006 #2


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    Question 1 and 2 seem ok to me. For question 3 you want something like this:


    For question 4 put the above equation in the same form as the one they give you and you can imply the solution. Or you could always try and solve it yourself but its a tricky one.
  4. Dec 9, 2006 #3
    Thanks, im still a bit confused about 4

    from this equation:

    i get


    when trying to put it into the form given in quation 4...

    is this at all right? :confused:
  5. Dec 9, 2006 #4
    Both integration variables need to be on one side each !!!
    k and A are constants


    Integrate left and right hand side to come to (make sure you take into account given initial conditions for theta and t !!!) :

    [tex]ln ( \theta-\theta_a ) =kAt[/tex]

    and if ln(a) = b <-> a = exp(b)

    you should be able to take it from here

  6. Dec 9, 2006 #5
    Thanks for the help so far, i think i mite be getting in now i have:



    [tex]ln ( \theta-\theta_a ) =kAt[/tex]

    [tex] ( \theta-\theta_a ) = e^{kAt}[/tex]

    and becuase it must satisfy [tex] \theta = \theta_0 [/tex] and t = 0 then:

    [tex] e^{kA} = \theta-\theta_a [/tex]

    Is this correct?
    Last edited: Dec 9, 2006
  7. Dec 9, 2006 #6
    You're missing a constant of integration, and theta has disappered.

    Also, jumping back to question 1 - if the powe has been turned ON, the bit won't settle to the ambient temperature, it will stop at the temperature such that the electrical energy being converted to thermal energy = energy lost to the air. Might want to check that.
  8. Dec 9, 2006 #7
    [tex] e^{kA} = \theta-\theta_a + C[/tex] ??

    also the first question says when the electricity supply is switched off, sory that i missed that bit out.
  9. Dec 9, 2006 #8
    put the constant on the side integrated with respect to t, so when you exponentiate, you get a constant multiplier on that side Qexp(kAt)
  10. Dec 10, 2006 #9
    ok i think i understand

    [tex] Ce^{kAt} = \theta-\theta_a [/tex]

    i also have to sketch a graph of the solution and compare it to my original sketch in part 2...

    would'nt they be exactly the same???

  11. Dec 10, 2006 #10
    after sketching the graph, the next part says

    Now we switch on the electricity supply.

    1. Let W be the (constan) heat per second supplied to the solderin iron, produce a differential equation for [tex]\theta[/tex]

    2. what is the steady-state temperature of the bit?

    3 sketch the graph of [tex]\theta[/tex] against time

    4 if [tex]\theta =\theta_0[/tex] at t = 0 find the solution for theta as a function of time, you may assume that theta_0 = theta_a in order to sketch the graph of temperature with time


    i think i have the answer to part 1

    [tex] \frac{d\theta}{dt} = kAW-(\theta - \theta_a)[/tex]

    does steady-state temperature mean average? or the temperature at which it stops rising?

    for the 3rd part i beleive the graph should climb slowly at first then shoot up rapidly???

    this is what i have for part 4:

    [tex] \frac{d\theta}{dt} = kAW-(\theta - \theta_a)[/tex]

    [tex] \frac{d\theta}{\theta - \theta_a} = kAW-dt[/tex]

    [tex] ln(\theta-\theta_a) = kAW-t[/tex]

    [tex] (\theta-\theta_a) = Ce^{kAW-t}[/tex]
    Last edited: Dec 10, 2006
  12. Dec 11, 2006 #11
    anyone? im rly stugling with this topic
  13. Dec 11, 2006 #12


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    For this bit you should get:

    [tex]( \theta-\theta_a ) = Ce^{kAt}[/tex]

    then when [tex] \theta = \theta_0 [/tex] at t=0 one would get,

    [tex]C= \theta_0 - \theta_a[/tex]

    For question 1 of the second set you're on the right lines so the rate of change of temperature wrt time will be the heat input minus the heat lost.

    [tex]\frac{d\theta}{dt} = W -kA(\theta - \theta_a) [/tex]

    For part two steady state means when an equilibrium is reached so in other words when the temperature does not change any more. Therefore you want to make [tex]\frac{d\theta}{dt}=0[/tex].

    Part 3 will show the temperature rising then reaching a plateau.

    For part 4 you will have to solve the equation and apply the conditions given.
    Last edited: Dec 11, 2006
  14. Dec 12, 2006 #13
    What do you mean solve the equation? there are no figures givend except for:

    [tex]\theta = \theta_0 [/tex] at [tex]t = 0[/tex]

    Assume [tex]\theta_0 = \theta_a[/tex]
  15. Dec 12, 2006 #14


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    The second set of questions asks you to come up with a differential equation in question 1. For part four it asks you to solve the equation (constructed in part 1) to find [tex] \theta(t) [/tex] with the conditions given.
  16. Dec 12, 2006 #15
    oh rite, like i did before,

    How do i rearange [tex]\theta - \theta_a = Ce^{W-kAt}[/tex]

    to get A, i have figures for W K and [tex]\theta_a[/tex] and [tex]\theta[/tex]
  17. Dec 12, 2006 #16


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    I'm not sure you've solved the equation properly. You should be coming out with:

    [tex] \theta(t) = \frac{W+kA\theta_a}{kA} +C e^{-kAt} [/tex]
    Last edited: Dec 12, 2006
  18. Dec 12, 2006 #17
    [tex] \theta(t) = \frac{W+kA\theta_a}{kA} +C e^{kAt} [/tex]

    how did you get this? does this mean that earlier in my thread, i have also solved an equation wrong?


    and solving this i got:

    [tex] Ce^{kAt} = \theta-\theta_a [/tex]
    Last edited: Dec 12, 2006
  19. Dec 12, 2006 #18


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    The equation you got previously was fine. What I did is rather than attempt to solve it normally which is tricky, is to compare it to the equation in question 4 of the first set and infer the solution. We can write the differential equation as:

    [tex] \frac{d\theta}{dt}+kA\theta=W+kA\theta_a [/tex]

    For the equation in your first post we see that [tex] a= kA [/tex] and [tex]b=W+kA\theta_a[/tex]
  20. Dec 12, 2006 #19


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    Oh I made a mistake in the post above. The equation should be

    [tex] \theta(t) = \frac{W+kA\theta_a}{kA} +C e^{-kAt} [/tex]

    It was just missing a minus sign in the exponential.
  21. Dec 12, 2006 #20
    ok thanx for the help so far.

    Does the kA on the bottom cancel with the kA on the top?
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