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Entry  Tue Nov 14 16:02:43 2017, Kira, Summary, PEM, seismometer can testing time_const.png
    Reply  Tue Nov 21 16:00:05 2017, Kira, Update, PEM, seismometer can testing cooling_fit.pngIMG_20171121_164835.jpg
       Reply  Wed Nov 22 12:13:15 2017, Kira, Update, PEM, seismometer can testing cooling_fit_1.png
          Reply  Wed Nov 22 14:47:03 2017, Kira, Update, PEM, seismometer can testing 
             Reply  Tue Nov 28 16:02:32 2017, rana, Update, PEM, seismometer can testing 
Message ID: 13427     Entry time: Tue Nov 14 16:02:43 2017     Reply to this: 13438
Author: Kira 
Type: Summary 
Category: PEM 
Subject: seismometer can testing 

I made a model for our seismometer can using actual data so that we know approximately what the time constant should be when we test it out. I used the appendix in Megan Kelley's report to make a relation for the temperature in terms of time.

\frac{dQ}{dt}=mc\frac{dT(t)}{dt} so T(t)=\frac{1}{mc}\int \frac{dQ}{dt}dt=\frac{1}{mc}\int P_{net}dt and P_{net}=P_{in}-P_{out}

In our case, we will heat the can to a certain temerature and wait for it to cool on its own so P_{in}=0

We know that P_{out}=\frac{kA\Delta T}{d} where k is the k-factor of the insulation we are using, A is the area of the surface through which heat is flowing, \Delta T is the change in temperature, d is the thickness of the insulation.

Therefore,

T(t)=\frac{1}{mc}\int_{0}^{t}\frac{kA}{d}[T_{lab}-T(t')]dt'=\frac{kA}{mcd}(T_{lab}t-\int_{0}^{t}T(t')dt')

We can take the derivative of this to get

T'(t)=\frac{kAT_{lab}}{mcd}-\frac{kA}{mcd}T(t), or T'(t)=B-CT(t) 

We can guess the solution to be

T(t)=C_{1}e^{-t/\tau}+C_{2} where tau is the time constant, which we would like to find.

The boundary conditions are T(0)=40 and T(\infty)=T_{lab}=24. I assumed we would heat up the can to 40 celcius while the room temp is about 24. Plugging this into our equations,

C_{1}+C_{2}=40, C_{2}=24, so C_{1}=16

We can plug everything back into the derivative T'(t)

T'(t)=-\frac{16}{\tau}e^{-t/\tau}=B-C[16e^{-t/\tau}+24]

Equating the exponential terms on both sides, we can solve for tau

\frac{16}{\tau}e^{-t/\tau}=16Be^{-t/\tau}, \frac{1}{\tau}=B, \tau=\frac{1}{B}=\frac{mcd}{kA}

Plugging in the values that we have, m = 12.2 kg, c = 500 J/kg*k (stainless steel), d = 0.1 m, k = 0.26 W/(m^2*K), A = 2 m^2, we get that the time constant is 0.326hr. I have attached the plot that I made using these values. I would expect to see something similar to this when I actually do the test.

To set up the experiment, I removed the can (with Steve's help) and will place a few heating pads on the outside and wrap the whole thing in a few layers of insulation to make the total thickness 0.1m. Then, we will attach the heaters to a DC source and heat the can up to 40 celcius. We will wait for it to cool on its own and monitor the temperature to create a plot and find the experimental time constant. Later, we can use the heatng circuit we used for the PSL lab and modify the parts as needed to drive a few amps through the circuit. I calculated that we'd need about 6A to get the can to 50 celcius using the setup we used previously, but we could drive a smaller current by using a higher heater resistance.

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time_const.png
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