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Tue Nov 14 16:02:43 2017, Kira, Summary, PEM, seismometer can testing

Tue Nov 21 16:00:05 2017, Kira, Update, PEM, seismometer can testing

Wed Nov 22 12:13:15 2017, Kira, Update, PEM, seismometer can testing

Wed Nov 22 14:47:03 2017, Kira, Update, PEM, seismometer can testing

Tue Nov 28 16:02:32 2017, rana, Update, PEM, seismometer can testing

Message ID: 13438 Entry time: Tue Nov 21 16:00:05 2017 In reply to: 13427 Reply to this: 13446

Author:	Kira
Type:	Update
Category:	PEM
Subject:	seismometer can testing

I performed a test with the can last week with one layer of insulation to see how well it worked. First, I soldered two heaters together in series so that the total resistance was 48.6 ohms. I placed the heaters on the sides of the can and secured them. Then I wrapped the sides and top of the can in insulation and sealed the edges with tape, only leavng the handles open. I didn't insulate the bottom. I connected the two ends of the heater directly into the DC source and drove the current as high as possible (around 0.6A). I let the can heat up to a final value of 37.5C, turned off the current and manually measured the temperature, recoding the time every half degree. I then plotted the results, along with a fit. The intersection of the red line with the data marks the time constant and the temperature at which we get the time constant. This came out to be about 1.6 hours, much longer than expected considering that onle one layer instead of four was used. With only one layer, we would expect the time constant to be about 13 min, while for 4 layers it should be 53 min (the area A is 0.74 m^2 and not 2 m^2).

Quote:

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.

Attachment 1:

cooling_fit.png 21 kB | Hide | Hide all

Attachment 2:

IMG_20171121_164835.jpg 3.320 MB | Hide | Hide all

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