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ME224 Lab 5 ME224 Lab 5 - Thermal Diffusion (This lab is adapted from “IBM-PC in the laboratory” by B G Thomson & A F Kuckes, Chapter 5) 1. Introduction The experiments which you will be called upon to do in this lab give you a chance to apply timing concepts and to review 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 variety of other phenomena. In this lab you will learn to instrument and analyze experimental data to compute the diffusion coefficient, thermal conductivity and heat capacity of copper. 2. 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 conductivity. The specific heat indicates how much heat is added to a mass of material for a specified temperature rise. The thermal conductivity indicates how fast the thermal energy is transported through the material. Figure 1. Copper rod schematic. Page 1 ME224 Lab 5 Consider the flow of heat in a rod as shown in Figure 1. The specific heat C of a material is the ratio of the amount of heat added dQ (Joules) to the resulting rise in temperature dT (degrees Kelvin) per unit mass dm (kg); thus ⎛ dQ ⎞ 1 C =⎜ ⎟ ⎝ dT ⎠ dm For a rod of cross-sectional area A, the volume 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 p dTdm = C p dTρdV (1) = C v dTρAdz where Cv = Cp is the volumetric 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. The 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 dT/dz, the thermal conductivity κ (W/m·K) and the cross-sectional area: ⎛ dT ⎞ P = −κA⎜ ⎟ ⎝ dz ⎠ (2) There 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, convection or radiation. The net heat gain per unit time dQ/dt in the piece of rod between z and z+dz is given 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 dt ⎝ ∂z ⎠ (3) Combining Equations (1), (2) and (3) gives the differential equation for heat flow in a rod ⎛ T⎞ ⎛ ∂T ⎞ ∂ ⎟ Cv ⎜ ⎟ = κ⎜ ⎜∂ 2⎟ ⎝ ∂t ⎠ ⎝ z ⎠ 2 (4) This 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 B2 are constants). T (t , z ) = B1 + ( ) ⎧ − z 2 Cv ⎫ B2 exp ⎨ ⎬ t 1/ 2 ⎩ 4κt ⎭ (5) Page 2 ME224 Lab 5 Figure 2. Solution for the temperature distribution along the rod. The 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 T vs. distance z of the solution at various times after the impulse. These are shown in Figure 2. In this figure, T=T(t, z)-TS ( where TS – 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 values 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). This half-space rod is the configuration that you will study experimentally. To obtain a theoretical expression convenient for analyzing a quantitative experiment, it is useful to relate the constant B2 in Equation (5) to the total heat Q added to the rod (from z=0 to z=∞) by integrating Equation (1). Consider T as the excess temperature above T(0), i.e., T=T(t, z)-TS = T′ -TS; integrating Equation (1) from temperature TS to T′ gives dQ = C v ρAdz (T '−Ts ) = C v ρAdzT (6) To integrate from z=0 to z=∞, use Equation (5) to describe the variation of temperature at any z and t. Then Page 3 ME224 Lab 5 C ⎞ ⎛ B2 exp⎜ − z 2 v ⎟ 4κt ⎠ ⎝ Q = ∫ Cv A dz 1/ 2 t z =0 ∞ = Cv A = Cv A t 1/ 2 t1/ 2 ∞ C ⎞ ⎛ B2 ∫ exp⎜ − z 2 v ⎟dz 4κt ⎠ ⎝ 0 B2 π 1 / 2 ⎛ 4κt ⎞ = B 2 (πκC v ) (7) 1/ 2 ⎟ ⎜ 2 ⎜⎝ C v ⎟⎠ 1/ 2 A Solving for B2 and inserting into Equation (5) yields T (t , z ) = 1 Q A (πκC v )1 / 2 C ⎞ ⎛ exp⎜ − z 2 v ⎟ 4κt ⎠ ⎝ + Ts 1/ 2 t (8) As written Equation (8) is not an optimum form for displaying some of the important features it contains. It is often very helpful, particularly for purposes of recognizing the domain of behavior in a given 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 R0 and T0. For displaying the change in temperature T as a function of time t at a fixed z, Equation (8) can be written in terms of a characteristic time t1 and a characteristic temperature T1 as 1/ 2 T ⎛ t1 ⎞ ⎛ t ⎞ = ⎜ ⎟ exp⎜ − 1 ⎟ T1 ⎝ t ⎠ ⎝ t ⎠ z2 t1 = C v 4κ 2Q T1 = AzC v π 1 / 2 (9) T = T (t , z ) − Ts where T is the excess temperature. Equations (9) immediately show several important points. First, the variation of temperature with time at a constant z can he related to just two parameters t1 and T1. Second, the characteristic time scale t1 is proportional to z2; this is a general property of diffusion phenomena. Page 4 ME224 Lab 5 Activity 1: Graphing the Heat Diffusion Equation 1. The first equation in (9) is the normalized diffusion equation. Generate a graph of T/T1 as a function of t/t1 from t/t1 = 0.1 to t/t1 = 10. Note that the independent variable t/t1 is inverted both times it appears. Include this figure in your writeup. 2. Your plot should peak at t/t1 = 2 and T/T1 = 0.43. Derive these values analytically from the first equation in (9) and include the proof in your writeup. 3. When you collect your data, it should have this shape, but the axes will not be scaled properly (i.e., the peak will not occur at 2, nor will it have a value of 0.43). You will need to normalize your data by finding scaling factors T1 and t1 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. The 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. The 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 conductivity, Equation (2), which can be combined to form the differential equation, Equation (4). However for purposes of numerical integration, it is best to leave them separate and write them in this form: ∆T = ∆Q C v A∆z ∆Q = −κA∆T (1’) ∆t ∆z (2’) where ∆z is assumed to approach zero. Now break up the length of the rod (Figure 1) into Nz pieces of length ∆z each and consider the ith piece; the heat flowing into this piece in the time ∆t will be: Qin = κA(Ti −1 − Ti ) ∆t ∆z (10) If the temperature in element i-1 is hotter than in the element i then Qin will be positive. The heat flowing out of the piece will be: Qout = κA(Ti − Ti +1 ) ∆t ∆z (11) The difference of the two is the heat gained or lost in the element: ∆Qi = Qin − Qout (12) This heat changes the temperature of the element in proportion to its heat capacity: Page 5 ME224 Lab 5 ∆Ti = ∆Qi C v A∆z (13) and so ∆Ti new = ∆Ti old + ∆Ti (14) 4. Experimental Setup and Program Development Enclosed in a glass jar to reduce convective heat losses Figure 3. Experimental apparatus. The apparatus for these experiments is illustrated in Figure 3. In the top of the copper rod (8.25 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 520 HEXFET in a manner similar to that used in the Thermistor Experiment. After generating a short pulse of heat by momentarily turning on the HEXFET, the computer will measure the increase in temperature at two positions down the rod using two thermistors. The thermistor positions are as shown on Figure 3 (z1 = 30 mm and z2 = 60 mm). A plot of the temperature vs. time at each of these thermistors will yield values 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 average them to get a more reliable result. Page 6 ME224 Lab 5 5. Voltage Amplifier The change in temperature of each thermistor from an initial temperature (T(t, z) – T0) is the significant quantity to measure in this experiment. However the temperature increments and thus the voltage changes are very small; if the ADC is connected directly to the thermistor, the changes are close to the step size of digitization. To overcome this problem an amplifier is used to boost the voltage 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 voltages VA (output) (pin 6), V1 (pin 2) and VT (pin 3) is given by V A = G (VT − V1 ) (15) For the circuit components used, the gain G is equal to 21. Equation (15) is the equation corresponding to a differential amplifier circuit as discussed in class (see handouts) The amplifier output (VA) 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 voltage of the amplifier, the potentiometer R1 should be set so that the output voltage of the circuit starts near the lowest voltage before a heat pulse is applied. This will allow the greatest voltage swing as the thermistor heats up without exceeding the 3V limit. Use the oscilloscope or the multimeter to monitor the output voltage of each amplifier. Set the potentiometers (one for each amplifier-thermistor combination) so that the amplifier outputs are about 0.20 V before you start each run. When this is done each potentiometer R1 has been adjusted to be essentially the same resistance as the thermistor resistance RT before a temperature pulse is applied. Since the amplifier gain is 21, the Page 7 ME224 Lab 5 change in the output voltage ∆VA will be 21 times greater than the change in the thermistor voltage ∆VA. Activity 2: Heat Impulse VI Write a VI that will turn on the heater for a specified time. The VI should output a 0 to the DAQ digital output until a button is pushed. Then 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. Activity 3: Amplifier Check Before developing 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 voltage output rise and then slowly fall. It should start above 0 V and should NOT exceed 3 V. Do the same to check thermistor 2, 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 voltages at once, you will need to make a new task in MAX. Choose Analog input and voltage, then choose both ai0 and ai1 as the analog inputs from the DAQ. Give it a name, then set both Voltage0 and Voltage1 to have 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. The output of the read block is now an array containing both the ai0 and ai1 values. They can be split easily by using the “Array to Cluster” block followed by the “Unbundle” block, and wiring the top two outputs of “Unbundle.” Activity 4: Heat Flow Real-Time Plot 1. The next task is to make thermistor ADC measurements at specified times. To do this, modify the previous two programs. Put in time delays as indicated by Figure 6. Note that a sample is taken before the heater is turned on. This records the baseline ADC reading. The heating of the rod then changes the ADC reading from this starting value. Page 8 ME224 Lab 5 Read initial value of each thermistor. Apply heat pulse For 500 msec For I = 1 to 40 Wait 0.5 sec Read value of each thermistor Save data to a file Figure 6. Thermistor data acquisition algorithm. 2. 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. Turn the heater on and quickly measure the resistance of the heater and the voltage across it. 5. Use these values to compute the power dissipated by the heater. 6. This value 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 voltage changes at the thermistor are small enough so that their behaviors are adequately described by differentials. Thus: (change in amplifier output voltage) = (gain) x (change in the input voltage) dV A = GdVT (16) The relationship between VT and RT is similar to the thermistor experiment, i.e., VT/V0 = R1/(R1 + RT) with V0 = 5 V. The relationship between dVT and thermistor resistance changes dRT can be obtained by differentiation; the result (which you should work out) is dVT dR =− T V0 R1 ⎞ ⎛ 1 ⎟⎟ ⎜⎜ ⎝ 1 + RT R1 ⎠ 2 (17) Page 9 ME224 Lab 5 Noting that R1 and RT are adjusted to be nearly equal at the outset gives dVT dR =− T V0 4 RT (18) The next task is to relate a change in the thermistor resistance to a change in temperature. The relation between thermistor resistance and temperature is RT = R0 exp(T0/Ta) as discussed in the Thermistor Experiment. Differentiation of RT with respect to temperature Ta gives T dT dRT =− 0 a RT Ta Ts (19) where Ta is the absolute temperature (K) (not the excess temperature, T(t, z) – TS) and dTa is a small temperature change due to the heat pulse. Thus if dTa is small, it can be approximated by the measured temperature change of the apparatus (i.e., the excess temperature) and Ta can be approximated by room temperature. Appropriately combining Equations (16), (18) and (19) gives the result dTa = Ta 4 Ts dVa G T0 V0 (20) Where: Ta ≅ Ts is the room temperature. As Equation (20) shows, the change in output voltage in volts is not important, only its ratio with V0. This ratio dVA/V0 is equal to the ratio of the change in ADC units to the ADC full-scale reading. Activity 5: Thermal Conductivity and Specific Heat of Copper 1. We are only interested in the change of temperature, so we need to subtract the initial value to get a change of the thermistor reading (dVa) versus time t. 2. Use equation (20) to get a change in temperature (dTa) versus time t (T0 = 3440 K for the GB32J2 thermistor, V0 = 5 V). 3. Plot the curve of dTa versus time for the two thermistors (you will notice that the curves you plotted look similar to the one you got in Exercise 1). 4. Normalize your curves to be the same as the one you got in Exercise 1. Find a value of T1 so that the curve peaks at T/T1=0.43 and a value of t1 so that the peak occurs at t/t1=2 for each curve. 5. Use these estimates (T1, t1) to draw curves on your graphs of the data and check the fits. Then you may want to change your estimates and try other fits. 6. When you are satisfied with your values of T1 and t1, use them to calculate via Equation (9) the thermal conductivity κ and volume heat capacity Cv. Calculate a set of values using the data from each thermistor and average the values. 7. Calculate the diffusion constant D = κ/CV and the heat capacity C = CV/ρ, where ρ is the density. Page 10 ME224 Lab 5 8. Look up the values of D, κ, and CV in a RELIABLE source. Compare them to your measured values. 9. Make an estimate of the error made in differential evaluation 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 deriving Equation (5) we assumed that the time during which the heater was on (t) was very small in relation to the time the heat takes to diffuse down the rod (T1), i.e., it was an impulse of heat. In doing your experiments this approximation is valid 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 t1. Page 11