ME224 Lab 5 - Thermal Diffusion

Transcription

1 ME4 Lab 5 ME4 Lab 5 - hermal Diffusion (his lab is adapted from IBM-PC in the laboratory by B G homson & A F Kuckes, Chapter 5) 1. Introduction he experiments which you will be called upon to do in this lab gie you a chance to apply timing concepts and to reiew the use of the ADC while learning about the phenomenon of diffusion. Specifically, you will be studying thermal diffusion but many of the concepts encompass a ariety of other phenomena. In this lab you will learn to instrument and analyze experimental data to compute the diffusion coefficient, thermal conductiity and heat capacity of copper.. Heat Flow Equation In this section you will explore some of the physical and mathematical considerations of one-dimensional heat diffusion. When heat is added to a material there are two parameters that affect the distribution of temperatures: the specific heat (or heat capacity) and the thermal conductiity. he specific heat indicates how much heat is added to a mass of material for a specified temperature rise. he thermal conductiity indicates how fast the thermal energy is transported through the material. Figure 1. Copper rod schematic. Page 1

2 ME4 Lab 5 Consider the flow of heat in a rod as shown in Figure 1. he specific heat C of a material is the ratio of the amount of heat added dq (Joules) to the resulting rise in temperature d (degrees Kelin) per unit mass dm (kg); thus dq 1 C = d dm For a rod of cross-sectional area A, the olume dv=a dz and dm=ρ dv where ρ is the density. So, the amount of heat added to the length dz of the rod is dq = C ddm = C dρdv p = C dρadz where C = C p is the olumetric heat capacity. When one end of the rod is hotter than the other, there will be a net flow of energy from the hot end to the cool end. he power P (watts) of this heat flow down the rod is the heat energy per unit time flowing past a point on the rod P=dQ/dt. For one-dimensional heat flow, P is proportional to the temperature gradient d/dz, the thermal conductiity κ (W/m K) and the cross-sectional area: p (1) d P = κ A () dz here is a minus sign because heat flows from higher to lower temperatures. In writing this equation, it is assumed that the rod is insulated; no heat escapes from the rod by conduction, conection or radiation. he net heat gain per unit time dq/dt in the piece of rod between z and z+dz is gien by the difference in the power flowing in at z and the power flowing out at z+dz, so dq P = P( z) P( z + dz) = dz (3) dt z Combining Equations (1), () and (3) gies the differential equation for heat flow in a rod = C κ (4) t z his equation has many solutions, depending on the initial and boundary conditions. For a quantity of heat added to the rod quickly (a heat pulse) at z = 0, the solution can be written as follows (B1 and B are constants). ( z C ) B ( t, z) = B1 + exp (5) 1/ t 4κt Page

3 ME4 Lab 5 Figure. Solution for the temperature distribution along the rod. he solution (5) describes the temperature at any point in the rod as a function of time after an impulse of heat has been added at z=0. Before proceeding further it is useful to examine the graphs of temperature s. distance z of the solution at arious times after the impulse. hese are shown in Figure. In this figure, =(t, z)- S ( where S initial temperature). At times near zero, the heat, and thus the excess temperature, is concentrated near z=0. As time progresses the heat diffuses away from z = 0 to larger and larger alues of z with the peak temperature decreasing in time. An important point is that since the solution is symmetric with respect to z, just as much heat diffuses up as down the rod. Since there is no heat flow across the cross section at z=0, cutting the rod at z=0 will not modify the form of the solution although now all the heat added flows in one direction (of course: from the hot to the cold). his half-space rod is the configuration that you will study experimentally. o obtain a theoretical expression conenient for analyzing a quantitatie experiment, it is useful to relate the constant B in Equation (5) to the total heat Q added to the rod (from z=0 to z= ) by integrating Equation (1). Consider as the excess temperature aboe (0), i.e., =(t, z)- S = - S ; integrating Equation (1) from temperature S to gies ( ) C ρadz dq = C ρ Adz ' = (6) o integrate from z=0 to z=, use Equation (5) to describe the ariation of temperature at any z and t. hen s Page 3

4 ME4 Lab 5 Q = C A = B 1/ t 1/ C A π = B 1/ t = B z= 0 B C A 0 exp z ( πκc ) 1/ A exp z 1/ t 4κt C Soling for B and inserting into Equation (5) yields ( πκc ) C 4κt dz C dz 4κt 1/ C exp z Q 1 t t z 4κ (, ) = + 1/ 1/ A t As written Equation (8) is not an optimum form for displaying some of the important features it contains. It is often ery helpful, particularly for purposes of recognizing the domain of behaior in a gien physical situation, to relate the quantities in an equation to physically significant parameters rather than simply measuring time in seconds, temperature in degrees centigrade, etc. You saw this before in the equation for the thermistor resistance as a function of temperature, the natural parameters there being R 0 and 0. For displaying the change in temperature as a function of time t at a fixed z, Equation (8) can be written in terms of a characteristic time t 1 and a characteristic temperature 1 as where is the excess temperature. 1 = t1 = t 1/ ( t, z) s t 1 exp t s (7) (8) z t1 = C 4κ (9) Q 1 = 1/ AzC π Equations (9) immediately show seeral important points. First, the ariation of temperature with time at a constant z can he related to just two parameters t 1 and 1. Second, the characteristic time scale t 1 is proportional to z ; this is a general property of diffusion phenomena. Page 4

5 ME4 Lab 5 Actiity 1: Graphing the Heat Diffusion Equation 1. he first equation in (9) is the normalized diffusion equation. Generate a graph of / 1 as a function of t/t 1 from t/t 1 = 0.1 to t/t 1 = 10. Note that the independent ariable t/t 1 is inerted both times it appears. Include this figure in your writeup.. Your plot should peak at t/t 1 = and / 1 = Derie these alues analytically from the first equation in (9) and include the proof in your writeup. 3. When you collect your data, it should hae this shape, but the axes will not be scaled properly (i.e., the peak will not occur at, nor will it hae a alue of 0.43). You will need to normalize your data by finding scaling factors 1 and t 1 to put the peak in this location. 3. Numerical Integration of the Heat Flow Equation General numerical integration of partial differential equations is a broad and difficult subject. he following will be a simple procedure that works in this case but must be used with care. It is really only meant to illustrate a general approach. he basic equations for the flow in a rod are the static equation for the heat capacity, Equation (1), and the dynamic equation with the thermal conductiity, Equation (), which can be combined to form the differential equation, Equation (4). Howeer for purposes of numerical integration, it is best to leae them separate and write them in this form: Q = (1 ) C A z t Q = κ A ( ) z where z is assumed to approach zero. Now break up the length of the rod (Figure 1) into N z pieces of length z each and consider the i th piece; the heat flowing into this piece in the time t will be: t Qin = κ A( i 1 i ) (10) z If the temperature in element i-1 is hotter than in the element i then Q in will be positie. he heat flowing out of the piece will be: t Qout = κ A( i i+1 ) (11) z he difference of the two is the heat gained or lost in the element: i Q = Q Q (1) his heat changes the temperature of the element in proportion to its heat capacity: in out Page 5

6 ME4 Lab 5 and so Qi i = (13) C A z new i = + (14) old i i 4. Experimental Setup and Program Deelopment Enclosed in a glass jar to reduce conectie heat losses Figure 3. Experimental apparatus. he apparatus for these experiments is illustrated in Figure 3. In the top of the copper rod (8.5 mm diameter) is set a resistor that is used as a heater. Current can be switched into the heater under program control using the IRF 50 HEXFE in a manner similar to that used in the hermistor Experiment. After generating a short pulse of heat by momentarily turning on the HEXFE, the computer will measure the increase in temperature at two positions down the rod using two thermistors. he thermistor positions are as shown on Figure 3 (z 1 = 30 mm and z = 60 mm). A plot of the temperature s. time at each of these thermistors will yield alues for the heat capacity and thermal conduction constants of copper and also demonstrate the functional dependence of heat diffusion on time and distance. Ideally, one thermistor would be sufficient to calculate these properties, but experiments are rarely ideal. Here you will want to calculate the material properties independently for each thermistor, then aerage them to get a more reliable result. Page 6

7 ME4 Lab 5 5. Voltage Amplifier he change in temperature of each thermistor from an initial temperature ((t, z) 0 ) is the significant quantity to measure in this experiment. Howeer the temperature increments and thus the oltage changes are ery small; if the ADC is connected directly to the thermistor, the changes are close to the step size of digitization. o oercome this problem an amplifier is used to boost the oltage change. On the protoboard attached to the experimental apparatus is an amplifier using a CA3140 operational amplifier as shown in Figure 4. Figure 4. Schematic of a CA3140 operational amplifier. It is not necessary to understand the details of this amplifier circuit except to note that the relationship between the three oltages V A (output) (pin 6), V 1 (pin ) and V (pin 3) is gien by V A ( ) = G V (15) For the circuit components used, the gain G is equal to 1. Equation (15) is the equation corresponding to a differential amplifier circuit as discussed in class (see handouts) he amplifier output (V A ) is constrained by the characteristics of the CA3140 to be between 0 V and +3 V. Since a rise in thermistor temperature will lead to a rise in the output oltage of the amplifier, the potentiometer R 1 should be set so that the output oltage of the circuit starts near the lowest oltage before a heat pulse is applied. his will allow the greatest oltage swing as the thermistor heats up without exceeding the 3V limit. Use the oscilloscope or the multimeter to monitor the output oltage of each amplifier. Set the potentiometers (one for each amplifier-thermistor combination) so that the amplifier outputs are about 0.0 V before you start each run. When this is done each potentiometer R 1 has been adjusted to be essentially the same resistance as the thermistor resistance R before a temperature pulse is applied. Since the amplifier gain is 1, the V 1 Page 7

8 ME4 Lab 5 change in the output oltage V A will be 1 times greater than the change in the thermistor oltage V A. Actiity : Heat Impulse VI Write a VI that will turn on the heater for a specified time. he VI should output a 0 to the DAQ digital output until a button is pushed. hen it should output a 1 for the specified time, then go back to 0 before the program ends so the DAQ output will not float at 1. Actiity 3: Amplifier Check Before deeloping a detailed program, write a simple program to see that the apparatus is functioning following the outline shown in Figure 5. Figure 5. Amplifier check algorithm When you run this program you should see on the oscilloscope the oltage output rise and then slowly fall. It should start aboe 0 V and should NO exceed 3 V. Do the same to check thermistor, the lower thermistor. You will need to let the apparatus cool down and reset the potentiometer between heat pulses. In order to measure both of the thermistor oltages at once, you will need to make a new task in MAX. Choose Analog input and oltage, then choose both ai0 and ai1 as the analog inputs from the DAQ. Gie it a name, then set both Voltage0 and Voltage1 to hae the range from -10 to 10 and use 1 sample on-demand. Hook this task constant to the start, read, and stop blocks as usual. Use the finger tool to change the drop box below the read block. Use multiple channels, single sample, 1D DBL. he output of the read block is now an array containing both the ai0 and ai1 alues. hey can be split easily by using the Array to Cluster block followed by the Unbundle block, and wiring the top two outputs of Unbundle. Actiity 4: Heat Flow Real-ime Plot 1. he next task is to make thermistor ADC measurements at specified times. o do this, modify the preious two programs. Put in time delays as indicated by Figure 6. Note that a sample is taken before the heater is turned on. his records the baseline ADC reading. he heating of the rod then changes the ADC reading from this starting alue. Page 8

9 ME4 Lab 5 Read initial alue of each thermistor. Apply heat pulse For 500 msec For I = 1 to 40 Wait 0.5 sec Read alue of each thermistor Sae data to a file Figure 6. hermistor data acquisition algorithm.. Combine the data gathering with plotting so that these unprocessed data are plotted as they are gathered, i.e., in real-time. 3. Send the collected data to a file for later analysis. 4. Once the data has been collected, unscrew the glass jar. urn the heater on and quickly measure the resistance of the heater and the oltage across it. 5. Use these alues to compute the power dissipated by the heater. 6. his alue will allow you to calculate the heat input. 6. Data Analysis Before proceeding to more data plots and analysis, here are some additional mathematical considerations. We will assume that the temperature and oltage changes at the thermistor are small enough so that their behaiors are adequately described by differentials. hus: (change in amplifier output oltage) = (gain) x (change in the input oltage) dv A = GdV (16) he relationship between V and R is similar to the thermistor experiment, i.e., V /V 0 = R 1 /(R 1 + R ) with V 0 = 5 V. he relationship between dv and thermistor resistance changes dr can be obtained by differentiation; the result (which you should work out) is dv 1 dr = (17) V R + R R1 Page 9

10 ME4 Lab 5 Noting that R 1 and R are adjusted to be nearly equal at the outset gies dv V 0 dr = (18) 4R he next task is to relate a change in the thermistor resistance to a change in temperature. he relation between thermistor resistance and temperature is R = R 0 exp( 0 / a ) as discussed in the hermistor Experiment. Differentiation of R with respect to temperature a gies dr 0 R d a = (19) a s where a is the absolute temperature (K) (not the excess temperature, (t, z) S ) and d a is a small temperature change due to the heat pulse. hus if d a is small, it can be approximated by the measured temperature change of the apparatus (i.e., the excess temperature) and a can be approximated by room temperature. Appropriately combining Equations (16), (18) and (19) gies the result 4 s dva da = a (0) G V Where: a s is the room temperature. As Equation (0) shows, the change in output oltage in olts is not important, only its ratio with V 0. his ratio dv A /V 0 is equal to the ratio of the change in ADC units to the ADC full-scale reading. Actiity 5: hermal Conductiity and Specific Heat of Copper 1. We are only interested in the change of temperature, so we need to subtract the initial alue to get a change of the thermistor reading (dv a ) ersus time t.. Use equation (0) to get a change in temperature (d a ) ersus time t ( 0 = 3440 K for the GB3J thermistor, V 0 = 5 V). 3. Plot the cure of d a ersus time for the two thermistors (you will notice that the cures you plotted look similar to the one you got in Exercise 1). 4. Normalize your cures to be the same as the one you got in Exercise 1. Find a alue of 1 so that the cure peaks at / 1 =0.43 and a alue of t 1 so that the peak occurs at t/t 1 = for each cure. 5. Use these estimates ( 1, t 1 ) to draw cures on your graphs of the data and check the fits. hen you may want to change your estimates and try other fits. 6. When you are satisfied with your alues of 1 and t 1, use them to calculate ia Equation (9) the thermal conductiity κ and olume heat capacity C. Calculate a set of alues using the data from each thermistor and aerage the alues. 7. Calculate the diffusion constant D = κ/c V and the heat capacity C = C V /ρ, where ρ is the density. 0 0 Page 10

11 ME4 Lab 5 8. Look up the alues of D, κ, and C V in a RELIABLE source. Compare them to your measured alues. 9. Make an estimate of the error made in differential ealuation of the temperature change. For doing this estimate, use the maximum change that can be measured using the amplifier circuit employed. Another consideration can be applied to the data analysis. In deriing Equation (5) we assumed that the time during which the heater was on (t) was ery small in relation to the time the heat takes to diffuse down the rod ( 1 ), i.e., it was an impulse of heat. In doing your experiments this approximation is alid as long as you make t = 0 on your graph correspond to the midpoint of the heating time and if the heating time is less than any t 1. Page 11