CEE575 - Homework 1. Resistive Sensing: Due Monday, January 29

Transcription

1 CEE575 - Homework 1 Resistive Sensing: Due Monday, January 29 Problem 1: Planes A metallic wire embedded in a strain gage is 4 cm long with a diameter of 0.1 mm. The gage is mounted on the upper surface of an airplane wing. Before strain is applied (airplane is parked), the initial resistance of the wire is 60 Ω. The plane then takes off, stretching the wire 0.15 mm, and increasing its electrical resistivity by Ωm. If Poisson s ratio for the wire is 0.3, find the change in resistance in the wire due to the strain to the nearest hundredth ohm. Figure 1: Plane Gage. 1

2 Problem 2: Detecting cracks You re a mechanical engineer working at a manufacturing center. There is a really important piece of machinery, which has developed a crack on a motor. It can still work well, but if the crack gets too big the machine will get damaged. You want to measure the crack as it is propagating and then shut the machine off if it gets too big. You recall that electricity takes the path of least resistance, so you pick up a wire, some tape, and build a crack sensor (see picture). You glue your sensor over the crack. As the crack grows, it will break strands of the sensors, which changes its overall resistance. a) Derive a symbolic expression for R, the total resistance of your sensor as a function of the crack length c L, wire resistivity ρ, and wire cross sectional area A. You can assume that the crack starts right before the first strand of your sensor. Plot R as a function of c L (this does not need to be exact, just a general behavior). You can ignore any strain effects. b) How would you increase the resolution of your sensor if you wanted to measure crack propagation more precisely? Figure 2: Crack Sensor. 2

3 Problem 3: Quarter bridge vs. half bridge In class we have introduced the single point Wheatstone Bridge (also known as a Quarter Bridge) and the Half Bridge. In the ambient (unloaded) case assume that all resistors and sensors have the same nominal resistance R 1 = R 2 = R 3 = R 4 = R. Show that the half bridge has twice the sensitivity (twice the output) of the quarter bridge. The sensors are indicated by the red arrows. Figure 3: Quarter bridge and half bridge setups. Problem 4: Cantilever load cell A load cell is constructed by mounting strain gages to the surface of a cantilever beam (Figure 4). Gages 1 and 2 are mounted to the top of the beam, gages 3 and 4 to the bottom. The beam cross section has a moment of inertia I and a Young s modulus E. In the unloaded state, all gages have the same resistance R 1 = R 2 = R 3 = R 4 = R, and the gage factor is G f. A load F is applied a distance L 1 from gage 2 and 4. For all Wheatstone bridge setups, it can be shown that the relation between bridge output voltage V 0 and the input Voltage can be linearized to take the following form: V 0 V = ( 1 4 ) [ R1 R 1 R 2 R 2 + R 4 R 4 R 3 R 3 This simplification applies to small changes in resistance, which is almost every problem we will care about in this course. Using the above information: ] (1) 3

4 a) What is the output voltage V 0 in terms of the input voltage V for the below cantilever setup? As we ve done in class, your expression should contain references to material properties E, I, c (distance to beam center, sometimes also written as half the beam thickness t/2), and G f. b) If the distance L 1 is increased, how does the bridge output change? c) What is the advantage of this design? Give an example application. d) (extra credit) show how equation 1 was derived. Figure 4: Force measurement setup. Problem 5: Photo-resistors and getting the most out of measurement setup A conductor s resistance can also be affected by radiation (in this case, more specifically, visible light). Photoresistors (also often called phototransistors or Cadmium sulphide photoconductive photocells) are simple resistors that alter resistance depending on the amount of light they are exposed to. Light is measured in lux, the SI illuminance unit ( Photoresistors are probably the most common sensor used in measuring light in every day applications, the most affordable (only a dollar a pice), and the easiest of all radiation sensors to implement. A variety of Photoresistors are shown in figure 5. The use of this sensor is highly dependent upon the application. Using the sensor usually involves anticipating which levels of illuminance the sensor will experience. This often involves exposing the sensor to the brightest environment 4

5 it will see and then measuring the resistance at this stage - call it R b. This is done for the darkest anticipate illuminance too, by measuring the dark resistance R d (you can measure this by just covering the sensor with your hand). The idea is that future readings of resistance will then fluctuate between R d and R b across the lifetime of the sensor deployment. a) You are using a simple voltage divider circuit (fig 5) to measure the resistance of the sensor. Given a known regular resistor R in series (see figure), and given that you know R d (dark resistance) and R b (bright resistance), derive an expression for V 0 (R), the maximum difference in potential that your voltage divider will output between dark and bright instances. b) Given that R d = 3300Ω, and R b = 780Ω, Plot V 0 (R) for a series of values of R, starting at R = 100Ω and going up to R = 5000Ω, in 100Ω increments. What do you notice? c) This question was asked in class, so let s build on it: building a voltage divider involves selecting the value of the series resistor R. You can match the resistors so that your sensor and regular resistor have the same resistance, but that s not always useful. Here, we want to select this resistor to maximize the potential difference V 0 (R). This will reduce the need to amplify our output voltage signal, and will make it easier to discern voltage readings. Derive an expression for the optimal value of R that will maximize this voltage difference. d) What is the optimal value of R (in Ohms) given the values of R d and R b above? Figure 5: Photoresistors and voltage divider circuit. 5

6 Problem 6: Robotic hands You re building a robot, which can pick things up using human-like hands. Robots can obviously be super strong, so you don t want it to squeeze too hard. You design some simple load cells, by using small aluminum cylinders and implanting them into the fingers of the robot (see figure). A cylinder has a stiffness k of 1,000,000 N/m and height of 1cm. In other words, a force of 1 million N has to be applied to compress the cylinder 1 m (this is effectively includes what you need to know about its stress/strain properties). A metal film strain gauge with a nominal resistance of 1 kω and a gage factor G f of 2.5 is attached to the unstrained cylinder, measuring in the direction the force is applied. Figure 6: Robots. a) Since the robot will be working with humans, you want to ensure that the robot does not push with a force greater than 120N on each finger. Find the resistance R of the gage that corresponds with this force. b) Assuming you choose to use a simple voltage divider to measure the resistance of the strain gage for forces between 0N and 120N, use the results you obtained in problem 5 to determine the optimal size of the required resistor that you would need to use in the circuit. Now go online 6

7 (digikey.com, sparkfun.com, or any other site) and find a suitable resistor for this application. Provide the model number and link. There will be many options, so don t get overwhelmed. c) Your robot has been hired to work in a steel mill, which can get really hot. We know that metals expand or contract under extreme temperatures. Assuming you have a lot of money, propose a design that could help offset the temperature effect so that your measurement is just giving you force. Problem 7: Trees Trees and other plants soak up water and nutrients (called the sap) from the ground and move it up their trunk to their leaves. The flow of water in a tree is difficult to measure, since there is relatively little fluid. It turns out, however, that we can measure temperature of the water as is moves through the tree and use this to estimate flows. In class we learned about thermistors, which change resistance based on a change in temperature. More specifically, the temperature T at the location of the thermistor can be obtained by T = T amb + G f R where T amb is the ambient temperature (starting temperature) at which the resistor has a nominal resistance R, and G f is the thermistor factor. We can combine two thermistor to measure flow. Figure 7 shows a sensor which is called a thermoanemometer (in the case of a tree, a sapflow sensor). It is composed of three small needles that are pushed into a tree (two thermistors and one heater). The two thermistors are R 0 and R s and in between these two sensors is a heating element is positioned. The sensor operates as follows: The first temperature sensors R 0 measures the temperature of the flowing water in the tree. A heater warms up the water in the middle of the tree and the elevated temperature is measured by the second thermistor R S. When the water flows in the tree trunk, heat dissipation increases due to forced convection. Because the heater is positioned closer to ther s sensor, that detector will register a higher temperature. The higher the rate of flow, the higher the heat dissipation and the lower the temperature that will be registered by the R s sensor. Heat loss can be measured and converted into the flow rate of the medium. A fundamental relationship for this thermometry setup is based on a modified version of King s law, which states v = K ρ ( dq dt 1 T s T 0 ) 1.87 (2) where K is a calibration constant, ρ is the density of the liquid flowing through the tree, T s is the temperature of thermistor R S, T 0 is the temperature of sensor R 0, and dq dt is the thermal loss to the flowing medium over time. The law of conservation of energy demands that the electric power in the heating 7

8 element be equal to thermal loss to flowing medium. across a heater/resistor is given by The electric power W W = V 2 = dq R h dt where R h is the resistance of the heating element, and V is the voltage across of the heating element. For this problem, refer to figure 7. R 0, R s, R 1 and R 2 all have the same nominal (ambient) resistance R (e.g. R 0 = R s = R 3 = R 4 = R). R 1 and R 2 are just part of your measurement circuit and are not sensitive to environmental conditions. a) Derive an expression for the temperature difference T s T 0 that only depends on the input voltage V the measured voltage V 0, the sensor factor G f, and the nominal resistance R. b) Now plug this relation into equation derived in part (a) to obtain a relation for the flow velocity v based on the measured voltage V 0 of your circuit. c) Plot flow velocity v as a function of the output voltage of your circuit V 0 (whatever scale looks best to you), for pure water ρ = 1000kg/m 3 and sap water (sugar water) ρ = 1150kg/m 3. Let K = 45000, V = 5V olts, R h = 10kΩ, G f = 0.02, and R = 1kΩ. Figure 7: Water flow in a tree. 8