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Message ID: 1952     Entry time: Mon Jun 29 11:13:21 2015
Author: Megan 
Type: Misc 
Category: Seismometer 
Subject: Time constant for heating the seismometer frame 

To determine the time constant of the system, a model of the heat flows through the system was made, starting with the basic ΔQ=mcΔT.

dQ = mc\cdot dT

\frac{dQ}{dt}=P=mc\frac{dT}{dt}

T(t)=\frac{1}{mc}\int P dt

Where the value P is the net power flowing through the system, or Pin-Pout. Pin is the power supplied by the heaters, and Pout is the power lost radiatively through the insulation. Pin is a known value, while Pout can be calculated via the K-factor, the thickness of the insulation, the area of a side, and the temperature difference between the two sides. Taking all this into account, the differential equation becomes:

T_{Al}(t)=\frac{1}{mc}\Big[P_{in}t-\int_{0}^{t}P_{out}dt\Big]

T_{Al}(t)=\frac{1}{mc}\Big[P_{in}t-\int_{0}^{t}KA_{side}d_{insul}(T_{Al}(t)-T_{lab})dt\Big]

This is just a differential equation where the temperatue of the frame (Tal(t)) is related to the integral of itself. The equation can be rearranged such that the temperature is related to the derivative of the temperature (as differential equations typically are). 

T'_{Al}(t)=A-BT_{Al}(t)

where A and B are defined as follows: A=(1/mc)(Pin-KAsidedinsulTlab), and B=(KAsidedinsul)/(mc). Solving the differential equation via Mathematica yields:

T_{Al}(t)=\frac{A}{B}+C_1e^{-Bt}

Therefore the time constant tau is just the reciprocal of B:

\tau = \frac{1}{B}=\frac{mc}{KA_{side}d_{insul}}

The units of tau do work out to be time, given that the units of K are taken to be [E]/[A]/[T]/[t]/[d]. Plugging in numbers to get an estimation of the order of magnitude gives: 

\tau=\frac{(10kg)(0.9\frac{J}{g\cdot K})(1000\frac{g}{kg})}{(58.121\frac{J}{K\cdot s\cdot m^2\cdot m})(0.75m^2)(0.0254m)}\approx\boxed{2.25\text{ hours}}

This value is an estimation so far; within the next couple days I will go measure the frame and get more accurate values. I will also recheck the calculation, because I think that the time constant should have some dependence on the power input to the system.

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