PHYS 352 Assignment #1 Solutions

Transcription

1 PHYS 352 Assignment #1 Solutions 1. Steinhart-Hart Equation for a thermistor In[1]:= a : In[2]:= b : In[3]:= In[4]:= d : ^ 8 t r_ : a b Log r d Log r ^3 ^ 1 Note: Mathematica Log function is natural logarithm. In[5]:= t 5000 Out[5]= The temperature would be K if the thermister measured 5 k. I see that some interpreted ln(r)^3 in the class notes as taking the cube of R and then taking the natural logarithm. This isn t correct though I admit it s a little ambiguous. If you look at the Steinhart-Hart equation, the coefficients are a polynomial expression as a function of ln(r). Also, if you note that ln(a^x) is x ln(a), then you d see the Steinhart-Hart equation would reduce to a very simple equation if it really were ln(r^3) and one can recognize that ln(r^3) isn t correct. I ve marked generously if you used ln(r)^3 though. In[6]:= In[7]:= twrong r_ : a b Log r d Log r^3 ^ 1 twrong 5000 Out[7]= Now, find dr/dt. Derivative of 1/T is -1/T^2 dt Derivative of a + b ln(r) + d ( ln(r) )^3 is [ b/r + 3 d ( ln(r) )^2 /R ] dr So, at this T = K, R = 5000 Ohms: In[8]:= Out[8]= ^2 b d Log 5000 ^ Thus, sensitivity at this temperature is dr/dt is -219 /K. Calibrate at T = K and T = K: In[9]:= Out[9]= NSolve t x , x x In[10]:= Out[10]= NSolve t x , x x

2 2 SolA1.nb If you assume it is linear between K with R = 6413 and K with R = 4134, then an equation could be T[R] = 10*(R-6413)/( ) + 20 in degrees Celsius. Below, the Steinhart-Hart equation is in blue and the linear approximation is in red. Below that, the graph is the difference between the real behaviour and the linear calibration assumption. In[11]:= Plot t r, 10 r , r, , Out[11]= In[12]:= Plot t r 10 r , r, , Out[12]= The maximum non-linearity error in this temperature range is 0.62 C and the linear calibration assumption gives a reading that is higher than the true temperature by this amount. In[13]:= Out[13]= Minimize t x 10 x , x 4134, x 6413, x , x Suppose you calibrate the thermistor by holding it at K and you find R = 5 k exactly. But, Steinhart-Hart tells us that the temperature should be K if R = 5 k. So, the equation is off from reality by 0.6 C at this temperature. You could continue to use the thermistor and the Steinhart-Hart equation knowing that there is a systematic error of plus/minus 0.6 C due to inaccuracy of this equation versus calibration. Or, you could correct any value you get from the equation by this amount though that leads to a question: do I correct by just subtracting 0.6 K from all Steinhart-Hart calculated values? Or is there a more complicated function I need to apply the correction? In order to deduce this function, you are going to need more calibra-

3 SolA1.nb 3 tion points. Basically you would be calibrating to deduce new Steinhart-Hart coefficients. I m looking for discussion points like the ones I ve mentioned here. 2. Design thermocouple to work at liquid nitrogen temperatures Things you should have in your design: - sketch of a thermocouple showing one junction to be immersed in cold liquid nitrogen temperatures and this is the unknown T - sketch should show the other junction and should call it the reference junction or room temperature or ambient - note: calling the reference the cold junction is a poor label because the probe junction is explicitly colder in this application you are designing and similarly don t call the unknown T junction hot junction - the question asked for you to use junction compensation so you will need something in your design that measures the ambient temperature of the reference junction, like a thermistor or an RTD - your junction compensation could then be just a measurement and then you should say that in software there will be a calculation that uses this measured reference T to determine the unknown T better - or your junction compensation could be a circuit that uses an IC temperature sensor, for example, to produce an output voltage in series with the thermocouple to simulate holding the reference junction at a known temperature (usually 0 C) - Wikipedia entry on Thermocouple (you must cite) says Type E chromel-constantan is well-suited to cryogenic use. Your design could pick Type T copper-constantan (suitable down to -200 C which is just below liquid nitrogen temperature...that s why they say -200 C; there is nothing magical that happens below that and it is just a common temperature to operate down to). You could pick Type K chromel-alumel which is also suitable down to -200 C. - I picked Type E, then looking at I read the mv for 22 C and -196 C and get that the signal would be minus = mv 3. Ratio pyrometer In[14]:= Quit In[1]:= b ^4 Out[1]= In[2]:= lambda Out[2]= 1.05 In[3]:= lambda Out[3]= 1.55 In[4]:= In[5]:= r t_ : lambda1 lambda2 ^5 Exp b lambda1 t 1 Exp b lambda2 t 1 r 1000 Out[5]=

4 4 SolA1.nb In[6]:= Plot r t, t, 500, Out[6]= What s the usable upper temperature range? There s no precise fixed answer here. What I see from the graph is that the ratio change per degree Kelvin starts to decrease (the slope starts becoming flat). If it s flat then you can t use ratio pyrometry any more since the ratio won t change much when the temperature changes. It s never perfectly flat, but the precision of the measurement will be worse the flatter it is. You can just eyeball a place on the graph where it starts to turn flat, around 1800 K say, and call this the upper range (that s what a manufacturer would do...no elaborate calculation needed). If you do want to get a little more elaborate: 0 In[7]:= Out[7]= In[8]:= D r t, t t t t t t t 2 t 2 Plot, t, 300, Out[8]= The derivative tells us how much the ratio changes per degree K. Around 1300 K the derivative is so the effect per degree is of order percent per degree K at 1300 K, larger at lower temperatures, and smaller at higher temperatures. You might not need plus/minus 1 degree K precision when you are measuring temperatures around 1000 K; but, undoubtedly a percent measurement is hard. That might be a place to declare the upper end of the range. What about the lower temperature range? The ratio method works at lower temperatures but the practical consideration is that the intensity of infrared radiation decreases the colder the object is. If the IR emitted by the object is faint, the detectors inside

5 SolA1.nb 5 the ratio pyrometer would need to be able to pick up this weak intensity over their noise floor which becomes harder at lower temperatures. The filters used to select 1.05 Μm and 1.55 Μm are narrow bandpass and how much they attenuate the incident desired signals starts to matter more too. 4. Design MEMS angular rate sensor The 2010 solutions describes the key points I m looking for here. - the Coriolis force causes a force perpendicular to the axis of rotation and the velocity of motion - a test mass is driven with vibrations (velocity) in one axis (say back and forth along the x-axis) and preferably driven at resonance - sense the force along the y-axis becomes sensitive to the angular velocity (rate of rotation) around the z-axis - the force can be sensed using a strain gauge (piezoresistive) or by measuring the position of a test mass attached to a spring - F = kx, then measure the position using capacitive techniques for example - you must find sensitivity specs online for your MEMS accelerometer (which is measuring the effect of the force due to acceleration on a test mass; then apply some simple calculation to deduce your angular rotation rate sensitivity