Help with Rates of Change Homework

[lufbrajames]

Homework Statement

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 which 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

Homework Equations

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)

4)

theta = A(theta - theta 0) / k


thanks for any help you can give =)

 

[Kurdt]

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

\frac{d\theta}{dt}=kA(\theta-\theta_a)

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.

 

[lufbrajames]

Thanks, I am still a bit confused about 4

from this equation:
\frac{d\theta}{dt}=kA(\theta-\theta_a)

i get

\theta=\frac{b}{k}+Ce^kt

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

is this at all right?

 

[marlon]

Thanks, I am still a bit confused about 4

from this equation:
\frac{d\theta}{dt}=kA(\theta-\theta_a)

i get

\theta=\frac{b}{k}+Ce^kt

Both integration variables need to be on one side each !
k and A are constants


\frac{d\theta}{dt}=kA(\theta-\theta_a)
\frac{d\theta}{\theta-\theta_a}=kAdt

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

ln ( \theta-\theta_a ) =kAt

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

you should be able to take it from here

marlon

 

[lufbrajames]

Thanks for the help so far, i think i mite be getting in now i have:

\frac{d\theta}{dt}=kA(\theta-\theta_a) \frac{d\theta}{\theta-\theta_a}=kAdt ln ( \theta-\theta_a ) =kAt ( \theta-\theta_a ) = e^{kAt}

and becuase it must satisfy \theta = \theta_0 and t = 0 then:

e^{kA} = \theta-\theta_aIs this correct?

 

[estel]

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.

 

[lufbrajames]

e^{kA} = \theta-\theta_a + C ??

also the first question says when the electricity supply is switched off, sory that i missed that bit out.

 

[estel]

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)

 

[lufbrajames]

ok i think i understand

Ce^{kAt} = \theta-\theta_a

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?

 

[lufbrajames]

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 \theta

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

3 sketch the graph of \theta against time

4 if \theta =\theta_0 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

\frac{d\theta}{dt} = kAW-(\theta - \theta_a)

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

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

this is what i have for part 4:

\frac{d\theta}{dt} = kAW-(\theta - \theta_a)

\frac{d\theta}{\theta - \theta_a} = kAW-dt

ln(\theta-\theta_a) = kAW-t

(\theta-\theta_a) = Ce^{kAW-t}

 

[lufbrajames]

anyone? I am rly stugling with this topic

 

[Kurdt]

( \theta-\theta_a ) = e^{kAt}

and becuase it must satisfy \theta = \theta_0 and t = 0 then:

e^{kA} = \theta-\theta_a


Is this correct?

For this bit you should get:

( \theta-\theta_a ) = Ce^{kAt}

then when \theta = \theta_0 at t=0 one would get,

C= \theta_0 - \theta_a

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.

\frac{d\theta}{dt} = W -kA(\theta - \theta_a)

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 \frac{d\theta}{dt}=0.

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.

 

[lufbrajames]

What do you mean solve the equation? there are no figures givend except for:

\theta = \theta_0 at t = 0

Assume \theta_0 = \theta_a

 

[Kurdt]

What do you mean solve the equation? there are no figures givend except for:

\theta = \theta_0 at t = 0

Assume \theta_0 = \theta_a

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 \theta(t) with the conditions given.

 

[lufbrajames]

oh rite, like i did before,

How do i rearange \theta - \theta_a = Ce^{W-kAt}

to get A, i have figures for W K and \theta_a and \theta

 

[Kurdt]

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

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

 

[lufbrajames]

\theta(t) = \frac{W+kA\theta_a}{kA} +C e^{kAt}

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

\frac{d\theta}{dt}=kA(\theta-\theta_a)

and solving this i got:

Ce^{kAt} = \theta-\theta_a

 

[Kurdt]

\theta(t) = \frac{W+kA\theta_a}{kA} +C e^{kAt}

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

\frac{d\theta}{dt}=kA(\theta-\theta_a)

and solving this i got:

Ce^{kAt} = \theta-\theta_a

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:

\frac{d\theta}{dt}+kA\theta=W+kA\theta_a

For the equation in your first post we see that a= kA and b=W+kA\theta_a

 

[Kurdt]

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

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

It was just missing a minus sign in the exponential.

 

[lufbrajames]

ok thanks for the help so far.

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

 

[Kurdt]

ok thanks for the help so far.

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

no. you can use \frac{W+kA\theta_a}{kA} = \frac{W}{kA} + \theta_a

if you wish

 

[lufbrajames]

How can i find the value for A? i have the follwing:

W = 50 watts (heat supplied to iron)
Operating temperature = 500 degrees C (\theta_0)
air temperature = 20 degrees C (\theta_a)
k = 1750 j/s/K/m^2

but i carnt seem to rearange it properly.

 

[Kurdt]

The value of A will be temperature dependent. The original question syas you may interpret it as being constant. I suggest if you really wish to find A that you consider the dimensions of k and solve it when the temperature is that of room temp. remember j/s = watts and you want m².