Team EE4 pm11. SSV Cases. VentuSolaris

Transcription

1 Team EE4 pm11 SSV Cases VentuSolaris Steven Dendooven, Jeroen Renders, Martijn Schaeken, Simon Putseys, Frederik Vercammen

2 I. INHOUD I. Inhoud... 1 II. List of figures... 2 III. List of tables Characteristics of the solar panel Analytical calculation Simulation List of assumptions and simplifications Bisection method Simulating solar panel behavior Simulation without solar panel Combined simulation Why a simulation? Impact test Introduction Force cell Amplifier Test results Conclusion Maximal velocity Half of the velocity Assumptions Stresses during impact Section B Section A Maximum allowed force I. Section B II. Section A Collision exercise

3 II. LIST OF FIGURES Figure 1: Graph of our measurements (Voltage and Current)... 4 Figure 2: Power graph (Power in function of Voltage)... 4 Figure 3: The characteristic of the current in function of the voltage Figure 4: The power delivered by the solar panel in function of the voltage Figure 5: The simplified characteristic of the current in function of the voltage for the solar panel Figure 6: Simplified characteristic of the power in function of the voltage Figure 7: Surface plot of the speed(mass and ratio) Figure 8: surface plot of the speed(mass and ratio) Figure 9: Contour of the surface plot Figure 10: The optimal mass of the SSV Figure 11: Position of the SSV in function of the time Figure 12: Speed of the SSV in function of the time Figure 13: Solar Panel Setup Figure 14: Current in function of Voltage Figure 15: Power in function of Voltage Figure 16: Power in function of Resistance Figure 17: Speed of the SSV Figure 18: Position of the SSV Figure 19: Setup of the total SSV Model Figure 20: 3D - plot of Gear ratio, Mass and Speed ball Figure 21: 3D - plot of Gear ratio, Mass and Height of the ball Figure 22: Optimal Gear ratio Figure 23: Speed in function of Position of the SSV Figure 24: Position of the SSV in function of Time Figure 25: Torque delivered by the motor during the race Figure 26: Measurement setup Figure 27: Electronic circuit impact sensor Figure 28: Result test Figure 29: Result test Figure 30: result test Figure 31: result test Figure 32: plot of tested distances Figure 33: Sankey Diagram full speed Figure 34: Sankey Diagram half speed Figure 35: Stresses on SSV during impact Figure 36: Section B Figure 37: Section A Figure 38: Situation before collision III. LIST OF TABLES Table 1: Measured voltage and current and calculated power and diode factor

4 SSV CASE CHARACTERISTICS OF THE SOLAR PANEL In the experiment we measured the voltage and current delivered by the solar panel. First we measured the short circuit current and open circuit voltage without a resistor connected to the solar panel. Our measurements aren t accurate because our solar panel was getting too hot. spanning stroom vermogen m-waardes 8,55 0,07 0,5985 1, ,51 0,08 0,6808 1, ,46 0,09 0,7614 1, ,4 0,09 0,756 1, ,35 0,09 0,7515 1, ,3 0,1 0,83 1, ,89 0,11 0,8679 1, ,7 0,11 0,737 1, ,61 0,11 0,6171 0, ,49 0,11 0,4939 0, ,75 0,12 0,45 0, ,45 0,12 0,294 0, ,57 0,12 0,1884 0, ,42 0,12 0,0504 0, ,3 0,12 0,036 0, Table 1: Measured voltage and current and calculated power and diode factor. We plotted these measurements, which resulted in the following graphs. Because of the error in our measurements the graphs are not very accurate. The current is too low and so is the power. 3

5 Power (W) Current (A) Measurements 0,14 0,12 0,1 0,08 0,06 0,04 0, Voltage (V) Figure 1: Graph of our measurements (Voltage and Current) Power 1 0,8 0,6 0,4 0, Voltage (V) Figure 2: Power graph (Power in function of Voltage) With these measurements we calculated the diode factor of our solar panel and the power delivered. We did this by filling in all the factors of the formula: All the variables we had to fill in this formula were given except for the voltage, the current and the diode factor. The short circuit current and the number of solar cells were measured by our team. Is was 10*10^(-8) Ampère, this is the reverse saturation current. The diode factor m and N, the number of solar cells, were to be decided by us. The thermal voltage was given to us and was 25,7 mv. Isc, the short circuit current was 0.45 A. The open circuit voltage we measured was 8.55V. When we filled in our different measurements of the current and the voltage we found different diode factors m. We calculated the average diode factor which turned out to be We plotted the graph of the 4

6 current in function of the voltage and we got the result. We optimized the graph with a better diode factor of 1.29 and the correct short circuit current of 0.9 A, given to us by the coaches. Figure 3: The characteristic of the current in function of the voltage. In this graph we can clearly see the short circuit current of 0.9 A and the open circuit voltage of 8.66 V. This is the graph for a diode factor of 1.29 which is the value we will use in our further calculations. This is the optimized diode factor. We calculated m in the following manner: We also plotted the power delivered by the solar panel in function of the voltage across the solar panel. Figure 4: The power delivered by the solar panel in function of the voltage. In this graph we see that the power reaches a maximum at 7.08 V. We found this value by deriving the equation of the Power curve and by deciding for which voltage it equals 0. To calculate the error on our calculations we need to search for the error on the digital multimeter. With this error we can calculate the final error on our calculation. We calculate the error on one measurement in the following manner: ( ) 5

7 ( ) ( ) When we calculate this for our working point U=7.09 V, Iout=1.01 A and Isc=1.03 A(the value of the open circuit current changed after we calculated this error to 0.9) we get: 1.2 ANALYTICAL CALCULATION When we simplify the characteristics of the solar panel we get the following results. Figure 5: The simplified characteristic of the current in function of the voltage for the solar panel. Figure 6: Simplified characteristic of the power in function of the voltage. When we assume elastic collision with the formula: We fill in the mass of the ball, 735±10 gram and we set the initial speed of the ball equal to zero so we get an equation for Vend,ball. Now we search an equation for V car in function of the mass of the car and the gear ratio. In order to plot an accurate graph of the speed of the car, we need to calculate the value for as much points as possible. The easiest way to do this is with an iterative calculation. This means it is one big calculation that consists of a lot of smaller parts that use the solutions of the previous parts. For these 6

8 calculations we use a lot of parameters. Some of these are measurable, but for some of them we have to make assumptions. The following is a list of all the parameters used for the calculation in Maple. The initial current supplied by the solar panel I0, the saturation current of the solar panel, the thermal voltage of the solar panel, the number of individual solar cells on the panel, the diode factor of the panel and the initial voltage supplied by the solar panel. All of these are parameters given by the manufacturer of the solar panel, except for the diode factor. The diode factor we measured during the first seminar. The next parameters are the air resistance, the frontal area of the SSV, the density of air, the calculation for the air drag force, the gravity, the rolling resistance constant and the calculation for the rolling resistance. The constants Cw, Crr, Rho and g are found in tables. The frontal area is measured. The following parameters are the torque constant, the gear ratio, the radius of the wheel on the driven axle, the mass of the SSV, the first time interval we want to simulate, the speed constant of the motor and the resistance of the armature of the motor. The torque constant, the speed constant and the armature resistance are given by the manufacturer of the motor. The gear ratio and mass are parameters we chose. After running the iteration and interpreting the result, we can adjust these parameters to improve the result. After a few times, we will come to an ideal value of both mass and gear ratio. The sequence of calculations for the first interval is the following: > > > > > > > > > > > Then we can calculate all the values for the second step of the iteration by replacing I0 by the calculated I1. This of course results in a2, V2 and I2. This value of I2 will then be used in the next step and so on. 7

9 We can run these calculations an infinite number of times, but this would be useless. Either at one point we will hit the ball, or the SSV will reach an equilibrium speed. The SSV will definitely not keep accelerating. We chose to keep iterating this calculation until the motor reaches its maximum rotational speed. This is 1120 rad/s, the inverse of the speed constant. This speed is reached in the sixth iteration, so after between 0.8 seconds and 1 second. The top speed of the SSV is then 5,23 m/s. This is of course with the chosen gear ratio of 6.5 and a mass of 1.2 kg. To understand the influence of the mass and the gear ratio on the overall behavior of the SSV during the race, we have to change the values of these parameters and study the difference the changes make. When we use a lower gear ratio we notice that it takes a longer time to reach the maximum speed of the motor and therefore the maximum speed of the SSV, so we have a lower acceleration but we increase the top speed. If we use a higher gear ratio we will reach this maximum earlier, but this lowers the top speed of the SSV. For example, when we change the gear ratio to 4, we reach our maximum speed after between 4 and 5 seconds, but the speed of the SSV is then 8,69 m/s. When we increase the gear ratio to 8, the maximum rotational speed of the motor is reached after between 0,8 and 1 seconds. The speed of the SSV at that point in time is 4,10 m/s. We also wrote a Matlab program that makes the same calculations. Because of the analytic calculations we can make important decisions about the car, like mass, height, material, shape, gear ratio etc. This is important so you don t waste time rebuilding multiple times. A disadvantage is us making assumptions and simplifications and it is an ideal case. 1.3 SIMULATION We simulated our SSV in Matlab for different masses and gear ratios going from 0.8 kg to 1.8 kg with steps of 0.1kg and the ratios going from 3.5 to 10 with steps of 0.5. We plotted a few graphs. These are shown below. Figure 7: Surface plot of the speed(mass and ratio). 8

10 Figure 8: surface plot of the speed(mass and ratio). Figure 9: Contour of the surface plot. Figure 10: The optimal mass of the SSV. 9

11 We can see clearly in the surface plot the speed of the ball will be the highest between a ratio of 6 to 9 and a mass of 0.9 kg to kg. This is even more visible in the contour plot. Where the red section is the area where the speed is at its highest. If we choose a gear ratio of around 6.5 the corresponding mass would be between 0.9 kg and 1.2kg. We choose for the largest mass of 1.2 kg. corresponding to these choses we get m/s as speed of the ball. We wanted to plot the position of the SSV and the speed of the SSV as well for our chosen ratio and mass. These are the results. Figure 11: Position of the SSV in function of the time. Figure 12: Speed of the SSV in function of the time. 10

12 In the next section we answer to the questions between the Matlab lines. a. Draw a flow chart of the relation between these files. Energy solver Func The Energy solver runs true the entire program and uses the smaller sub programs Func. This small program fills in all the variables of the solar panel, the motor and other important variables in to Energy solver. Func uses the program Energy-Func. This part is where the user fills in the variables that are being filled in by Func in Energy Solver. b. Explain the following line: [t,s] = ode15s(@energy_func,(t0:tf/100:tf),x0,options); What are t and s? ode15s is used to solve a differential equation, this is being done by using Laplace transformation to s with steps of 0.01s. x0 refers to the initial values to help solve the differential equation. These are defined in line 16 of the code. The reference to Energy_Func is used because the equation is found in that part. c. What is done here and why? The differential equation is solved for different values, in steps of 0.1 from 1 to 10. d. What is the function of this file? The variables are defined and the functions are worked out. e. What are dx, t and x? Why are they in this line? Does there exact name matter for the program? Their exact name doesn t matter for the program. Dx is defined at the bottom of the page, and is the energy function. For the other two variables t is the time, and x is the distance travelled by the SSV. f. Does the exact name of these parameters matter? Why (not)? 11

13 The exact name of the parameters doesn t really matter, as long as they are the same in every tab or every formula that is defined. g. Describe these parameters and give their units. Solar Panel: Isc=0.9 A; short circuit current Is=1e-8 A; Saturation current Ur= V; Terminal voltage m= 1.29 no dimension; diode constant N= 16 no dimension; number of cells DC Motor: w=420; R= 3.36 Ohm resistance of the armature Ce= V/rpm Air resistance: Cw= no dimension A= 0.20*0.15 m² area rho= kg/m³ density Rolling resistance: g=9.81 m/s²; gravity Crr= no dimension; drag resistance SSV: r = 0.03 m wheel radius h. What is x(2)? X(2) is the speed after two seconds, after two seconds the air resistance needs to be taken into account. i. What is TolFun? What is fzero and why do we call it here? What are sol and f? With fzero, the zero point of the function f is defined. j. Explain the energy equations. What is the difference? 12

14 In the first equation the air resistance isn't taken into account, this assumption is made because the speed of the SSV isn't that height until two seconds after the start. After two seconds the speed is high enough to take the air resistance into account, so a different equation is used after two seconds. k. What is the function of this file. How is it used? In this file de functions for the behavior of the solar panel and the DC Motor are melted together into one equation. Because the used variables are global, they can be used in different programs at one moment. To make the overall program a bit more comprehensible it is divided into three smaller programs each with their own function. l. What is f? f is constructed out of three parts, U is the open circuit voltage. The second term is the voltage of the solar panel and the third term is the voltage used by the DC Motor. If we take a decision based on only these graphs of the ideal mass and the ideal gear ratio we made the following decision as we mentioned before. We choose our gear ratio and mass being the following. When we use the speed of the ball at the end of the track right after it got hit by the car we can calculate the maximal height of the ball. 6.5 met This gives us a maximal height for the ball of m. When we compare these results with our results from the analytic calculations we notice the velocity and the height being higher in the Matlab simulation. When we simulate the behavior of the car we get an ideal mass of 1.2 kg and a gear ratio of 6.5, which is a lower mass than the one we calculated analytically. But it is a higher gear ratio than the analytically calculated one. A simulation is more accurate than an analytic calculation. The simulation uses less assumptions than we did during our analytic calculation. 1.4 LIST OF ASSUMPTIONS AND SIMPLIFICATIONS We assumed constant acceleration for the calculations of the gear ratio Gr and the optimal mass of the car. We also simplified the method to find the rotational speed of the motor. We estimated the working point of the motor characteristic and the solar panel characteristic. We didn t take the air resistance and the rolling resistance into account yet for these calculations. Because of this our car will drive slower during the race and the ball will get less high on the ramp. We also assumed elastic 13

15 collision between the ball and the car, so in a real life situation there will be energy loss in the collision. This assumption makes it possible to calculate the maximal height of the ball. We got a value for the short circuit current from our coach which we used in the collected. We do a sensitivity analysis on the value of the diode factor m. If we make m larger the working point of our solar panel and our motor changes, the ideal voltage will get larger but the current stays the same. The rotational speed of our motor would also increase. The torque delivered by the motor will stay the same because it is dependent of the current which stays the same. These two things result in an increased power delivered by the motor. The gear ratio will also increase a lot which will result in an increased speed when the car hits the ball. This will result eventually in a higher maximal height of the ball on the ramp. 1.5 BISECTION METHOD We search the intersection with the x-axis with the bisection method. As an example we do this with the following function. ( ) ( ( )) First we check if 0 lies in between the values of the interval. We see that the function hits 0 somewhere between those values so we take the average of the interval and calculate f(5). f(5) >0 and f(10)<0, so the function is zero in between [5,10]. So we take the average 7.5 for the next iteration We see that the function is zero in between [5,7.5] and iterate more. Eventually we find a value for x which is the following: Now that we are familiar with the bisection method we can use it for our personal values. We use our optimal mass and gear ratio to look at the first second of our speed and position characteristic. When we plot this we get figure 14 and figure 15. The equations of the position and the velocity are the following. They are found by using Matlab figures. 14

16 When we apply the bisection method for the first second with an interval of 0.1s to these equations we get the following results. First we calculate zero for the position function, so we check if zero lies in between 0 and 1. We can see that zero does indeed lie in the interval so now we can start iterating. First we take the average of the interval which is 0.5. x(0.5)>0 so zero lies on the left hand side of the average, so now we take the average of the interval [0,.5], which is x(0.25)<0 so now we see that zero lies on the right hand side of t=0.25. So again we take the average of the interval [0.25,0.5]. After this we iterate a few more times according to the same method until we find a value for t. Eventually we find a value for t where the position function is almost 0: Now we do the same for the speed function, we check if zero lies in between 0 and 1. Again we can see that zero does indeed lie in the interval so now we can start iterating. First we take the average of the interval which is 0.5. V(0.5)>0 so zero lies on the left hand side of the average, so now we take the average of the interval [0,.5], which is v(0.25)>0 so zero still lies on the left hand side of t=0.25. So again we take the average of the interval [0,0.25]. After this we iterate a few more times according to the same method until we find a value for t. Eventually we find a value for t where the position function is almost 0: 15

17 16

18 CASE SIMULINK 2.1 SIMULATING SOLAR PANEL BEHAVIOR. In order to find the optimal working point of our solar panel we made a simulation using Simulink, you can see the setup we used in the figure below. We used one solar cell and multiplied the diode factor (1.29) by 16. We added some sensors like a current and voltage sensor to measure the output, we also multiplied these to get the power. Figure 13: Solar Panel Setup We simulated this model for different resistance values varying between 0 and 100 ohm (4 values between 0 and 10 to make better plots). We plotted some nice graphs using matlab, like a power in function of voltage graph. This graph looked a lot like the graph we draw in maple in the first part of case SSV part 1. Our matlab code is pasted underneath: clear all; close all; %%% Solar Power Ir = 800 ; % solar irradiance [W/m^2] Is = 1e-8 ; % saturation current [A] Isc = 0.88 ; % short circuit current [A] Voc = 8.55 ; % open circuit voltage for one cell [V] Ir0 = 700 ; % irradiance used for measurements [W/m^2] m = 16*1.29 ; % diode quality factor % Because we only use one cell, we need to multiply our diode factor by 16 % Initialising vectors 17

19 V=[]; I=[]; P=[]; R_list=[ ]; % Simulating R values between 0 and 100 ohm. for i=1:length(r_list) R=R_list(i); sim('resistor_and_solar_panel',1); % Simulating Simulink model "Resistor_and_Solar_Panel.mdl" for 1 s end V = [V yout(end,1)]; I = [I yout(end,2)]; P = [P yout(end,3)]; figure(1) % Current in function of voltage plot plot(v,i,'*b'); ylabel('current [A]'); xlabel('voltage [V]'); axis([ ]); set(gcf,'color',[1 1 1]) figure(2) % Power in function of voltage plot plot(v,p,'*r'); ylabel('power [W]'); xlabel('voltage [V]'); axis([ ]); set(gcf,'color',[1 1 1]) figure(3) % Power in function of resistance plot plot(r_list,p); ylabel('power [W]'); xlabel('resistance [R]'); axis([ ]); set(gcf,'color',[1 1 1]) Below you can find our graphs we got using the simulation of the model, the first one is the Current in function of the voltage plot. The second one is the Power in function of the voltage plot. And the third one is the Power in function of the voltage, we plotted this one so we could clearly see which resistance value is ideal for delivering maximum power output. The first and second plot look a lot like the ones we plotted using our measured results, the main difference are the values. Because our measurements aren t correct our calculated values in the first part (case SSV part 1) are much lower than the values the Simulink simulation gave us. 18

20 Figure 14: Current in function of Voltage Figure 15: Power in function of Voltage 19

21 Figure 16: Power in function of Resistance We can clearly see that the maximum power output can be achieved with a resistor of 10 ohm in the series of resistors ranging from 0 to 100 ohm. We can t compare this value with our measured ones because our measurements aren t correct. 2.2 SIMULATION WITHOUT SOLAR PANEL. In order to simulate the backwards race we need to simulate the entire solar car model which you can find in the next section. If we would simulate this model we would simulate the race, in order to simulate the backwards race we need to change a few things. First and probably most importantly we need to set our irradiance to zero, because the irradiance is zero the solar panel doesn t deliver any power. Secondly we need to set our starting conditions for the SSV. We set an initial speed for our mass, in this way we simulate the mass hitting the SSV at the speed we want. What we want to simulate is when the ball drops from 1 meter high. We can calculate the starting speed of the SSV, by using the formula: or We get a starting value for the speed of the SSV of m/s. We change the initial speed of the SSV at the mass block, then we run the simulation for 15 seconds. When we run the simulation we get the following scope. 20

22 Figure 17: Speed of the SSV You can see that the speed of the SSV is negative, this is because the SSV is running backwards. The starting speed is m/s. At around 10 seconds the SSV crosses the 0 m/s point, in the simulation the SSV starts driving forward again, which isn t possible in real life. So after about 10 seconds the SSV comes to a stop. When we take a closer look on the scoop we can see that the zero point is after seconds. In the next scope you can see the distance travelled by the SSV. Figure 18: Position of the SSV Were the curve of the graph becomes horizontal the SSV reaches 0 m/s and stops moving, if we look on the position graph we can see that the SSV has traveled roughly 17 meters by that time. If we take a closer look we get a value of meter. 21

23 In order to simulate the backwards race we also had to change the signs of the air and rolling resistance, if we didn t change these they would contribute to the speed of the SSV.So if the SSV collides with the ball (which is coming from a height of 1 meter) it gets a starting speed of m/s and travels approximately meters. 2.3 COMBINED SIMULATION. In the combined simulation the previous two models are combined into one model, which represents our SSV. In the figure below you can see the setup we used to simulate the SSV. On the left side the solar model is connected to the DC Motor. On the right side the mechanical part of the SSV is simulated as well as some forces like air resistance and friction (rolling resistance). Figure 19: Setup of the total SSV Model We can now use this model to simulate the race with different gear ratio s and different masses. We can visualize these results using some nice plots in matlab. In order to find the most effective gear ratio and the best mass, which result in the highest velocity of the ball (and thus in the highest point the ball can reach) we will simulate gear ratio s between 3.5 and 10 using steps of 0.5. We will simulate masses between 0.7 kg and 1.5 kg using steps of 0.1 kg. In the graphs below you can see the results of the simulations of the model with the different gear ratio s and masses. The first graph is a 3 dimensional plot with the gear ratio on the x axes, the mass on the z axes and the speed of the ball on the y axes. The point where the speed of the ball is highest is selected and in this point we can find our optimal gear ratio and mass which are 6.5 for the gear ratio and 1.2 for the mass of the SSV. The highest possible speed of the ball is m/s which equals a maximum height of meter. This result is found using the formula: 22

24 In figure 2 you can find the height of the ball in a 3 dimensional plot like the speed of the ball. Here you can also see that the maximum height is meter, with a gear ratio of 6.5 and a mass of 1.2 kg. Figure 20: 3D - plot of Gear ratio, Mass and Speed ball Figure 21: 3D - plot of Gear ratio, Mass and Height of the ball 23

25 In this figure you can see a side view of the 3 dimensional plot, in this graph the optimal gear ratio is easier to see. Figure 22: Optimal Gear ratio Below you can see some more graphs which illustrate the speed of the SSV, the position of the SSV and the torque the motor delivers during the race. Figure 23: Speed in function of Position of the SSV 24

26 Figure 24: Position of the SSV in function of Time Figure 25: Torque delivered by the motor during the race We can see that the speed of the SSV reaches a maximum, and becomes constant, after a certain amount of time (roughly 15 seconds), this constant speed which we will call the terminal velocity is m/s. You can also see in the position in function of time graph that our SSV reaches the 10 meter point (where the ball will be hit) after 5 seconds. So we don t need to worry about the 20 second time limit for the race. 25

27 At the time the speed becomes constant, the torque also becomes constant. This is due to a number of factors, like air resistance, rolling resistance and others. These resistances are forces working against the SSV, at the terminal velocity (so when the speed is constant) there is an equilibrium being reached between the forces working against the SSV and the force the SSV provides. The power the SSV delivers is the same as the forces it needs to overcome in order to accelerate even more, so it can t accelerate anymore and the speed becomes constant. 2.4 WHY A SIMULATION? In a simulation you can easily change parameters and instantly get the results of your simulation with the new parameters. You can easily simulate different situations and use them to get the ideal values for your SSV. If we had to do these simulations in real life they would take much more time and would cost a lot more than the software we used. On the downside a simulation will never be as accurate as real life tests because you can t include all factors that are in play in real life in a simulation. So a simulation is in fact an ideal representation of real life, we will get better results with a simulation then we can get in real life. A simulation can provide a great guideline to help us build our car. 26

28 SSV CASE II 3.1 IMPACT TEST Introduction To get an idea of the size of the forces the SSV has to resist, we perform an impact test on the car. With the results of these measurements, we can calculate the force the metal frame of the SSV has to withstand and how much energy the frame absorbs. For this impact test, we use a force cell, an amplifier and a data-acquisition card to show the results of the measurement in Labview. The setup is shown in figure 1. Figure 26: Measurement setup Force cell For this impact test we use a force cell of type PCB200C20 (datasheet in appendix). The datasheet is available on the website of Piezotronics, the manufacturer. While doing the measurement, we have to keep in mind the cost of the infrastructure we use. The force sensor by itself costs about two thousand euros. In order to use the force sensor in a safe manner, we have to fully understand the values shown in the datasheet. The most important number in the datasheet is the maximum axial force the sensor can resist. This is kn. If the force on the sensor exceeds this value, we risk permanent damage and a hefty fee. The force sensor converts a physical impact in a voltage output. The voltage-force-ratio is 56,2 mv/kn. The maximum impact for accurate results is 88.6 kn. We attach a mass to the force sensor and we attach the whole to a cable, so we create a pendulum with the force sensor. The length of the cable is cm and the total weight of the mass is 780 g. We let go of the mass from a measured distance and analyze the impact by the graph, created by Labview. 27

29 Figure 2 shows the electronic circuit of the impact sensor. Figure 27: Electronic circuit impact sensor Amplifier We use an amplifier with a possible gain of 1x, 10x or 100x. This means that the signal, picked up by the sensor, will be sent to the data-acquisition card directly (1x), amplified by 10 or amplified by 100. By amplifying the signal, we can see more details in the graph. For example, when the sensor doesn t measure an impact, it still measures noise. This results in a very messy graph with short, small peaks. Because the peaks are very close to each other, we can conclude that the amplifier has a good resolution. We know that the voltage-force-ratio of the sensor is 56.2 mv/kn, and the range for accurate measurements is 88.6 kn. The maximum output voltage with a gain of 1 is then: 56.2 * 88.6 / 1000 = V. With a gain of 100x, this is 5 V. For accurate measurements, the output voltage as to be below 5 V. Test results For the test, we sling the pendulum from an accurately measured distance towards the thrust piece of the SSV. Our thrust piece is a rectangular piece of about 50 cm² and is about 1 cm thick. For the first collision, we swing the pendulum from a distance of 21.5 cm from the thrust piece. We set the gain of the amplifier to 100x. Labview now plots a graph. The signal has one clear peak with an amplitude of V. When we zoom in on the peak, we can derive a few things from the form of the peak. The maximum height of the peak indicates the highest force. The surface under the peak indicates the amount of energy that is absorbed by the thrust piece. The wider the base of the peak, the more energy is absorbed by the thrust piece. The maximum force is then (0.165*1000)/(56.2*100) = kn. Figure 3 shows the results of test 1. 28

30 Figure 28: Result test 1 For the second collision we sling the pendulum from a distance of 35 cm from the SSV. This results in a graph with a maximum of 0.58 V, again with a gain of 100x. With a voltage-force-ratio of 56.2 mv/kn, this results in a force of (0.58*1000)/ (56.2*100) = kn. Figure 4 shows the result in a graph. Figure 29: Result test 2 For the third collision we sling the pendulum from a distance of 60.5 cm from the SSV. This results in a graph with a maximum of 1.35 V, again with a gain of 100x. With a voltage-force-ratio of 56.2 mv/kn, this results in a force of (1.35*1000)/(56.2*100) = kn. 29

31 Figure 30: result test 3 For the last collision we sling the pendulum from a distance of 76.5 cm from the SSV. This results in a graph with a maximum of 1.9 V, again with a gain of 100x. With a voltage-force-ratio of 56.2 mv/kn, this results in a force of (1.9*1000)/(56.2*100) = kn. Figure 31: result test 4 The double peak in this graph is probably from a rebounce of the force sensor or an oblique impact. We see that the maximum values of our measurements are far below the accuracy limit of 5 V. We can conclude that our measurements are trustworthy. In comparison with the test results of other groups, our thrust piece absorbs a lot of energy. 30

32 Conclusion Our SSV has a rigid metal frame of hollow bars, welded together in a triangular shape. Our frame can absorb much bigger impacts than the ones we measured, but that won t be necessary for the race. When we switch the mass with the force sensor for a ball with the same weight, we can start experimenting for the real race. We can still make adjustments to the thrust piece, so we have maximum transfer of energy. The force on the frame during the test was much lower than the force on the frame during the race will be. In SSV case I, we calculated that the ball will reach a height of 1,16 m. For safety reasons and the limitations of the equipment, we didn t sling the mass from that height. In order to find the maximum force on the frame, we have to extrapolate to 1,16 m height, from the distances we did test. When we plot a graph of the forces we tested in function of the distance from where we dropped the mass, the result is shown in figure y = 0,0001x 2 + 0,0432x + 1,2338 Reeks1 Poly. (Reeks1) Figure 32: plot of tested distances From the equation of the trend line calculated by Excel, we can calculate the force on the frame if the height would be 1,16m. This results in a force of 876,85 N on the frame. We have to build our frame in a manner that it can resist this force without deformation or damage. 3.2 SANKEY-DIAGRAMS MAXIMAL VELOCITY We are using Sankey diagrams to display the power losses of the SSV graphically. In the first case, the calculations are made when the car is at maximal velocity. This is the constant velocity on an endless, flat road. In our simulations, this velocity corresponds to

33 To start the calculations, we assume the solar irradiance is 800 W/m^2. This is approximately the standard irradiance value of the Air mass 1,5 (AM 1,5) spectrum defined by the commission of the European community. Total incoming power First, the total power of the incoming sunlight is calculated. with: the irradiance and the solar panel surface area. Losses due to inefficiencies solar panel and reflection The power when the motor is at maximum speed, doesn t necessarily corresponds to the maximum power. In fact it will be less. The power loss in the solar panel is calculated in a backwards way starting from the maximum velocity which is The voltage and current will be calculated for this velocity. The power equals to the multiplication of voltage and current. Afterwards we calculate E, the back electromotive force. with: with with: = 7.67 Volt 32

34 Then U is substituted by in the solar panel equation: This equation is solved iteratively for I. Maple returned this result: Power equals Voltage times current: 1.70 W remains from the 38,94 W. This is only 4,37 %. Losses due to air drag The air drag is calculated by with this formula: = N With: coefficient of drag of SSV vehicle frontal area atmospheric density maximum velocity SSV Now, the power lost due to air resistance is: Losses due to rolling resistance The rolling resistance is calculated by this formula: = 0.41 N With: Crr = m = 1.2 kg coefficient of rolling resistance gravitational acceleration mass of SSV The power lost due to rolling resistance is: N Copper losses Losses due to the internal resistance is simply calculated with this formula: 33

35 With: Friction and iron losses Because the car isn t accelerating anymore, the remaining power is also lost due to friction and iron losses in the motor. Figure 33: Sankey Diagram full speed HALF OF THE VELOCITY When the SSV is driving at half the velocity, the power losses of the solar panel will be different because the power output of the solar panel will change. We calculate the power in the same way we did in the previous part. But this time the velocity is half of the maximum velocity. With: (calculated the same way as in 2.1) 34

36 Losses due to air drag This is calculated the same way as we did before, we only replace Vmax by Vhalf. Losses due to rolling resistance This is calculated the same way as we did before, we only replace Vmax by Vhalf. Copper losses With: Friction and iron losses We assumed that these do not depend on the velocity and thus will be the same as in 2.1. Power due to acceleration Since the car is still accelerating, there is power needed. The remaining power is 2.61 W. 35

37 Figure 34: Sankey Diagram half speed. Figure 35: Stresses on SSV during impact. 3.3 STRESSES ON SSV DURING IMPACT ASSUMPTIONS To calculate this problem we made some simplifications and assumptions. We assumed that both sections are circular and are made from steel (F50 C35) because we know the maximum allowable 36

38 stress from that material. These assumptions were made because that is the case with our SSV. For the measurements we also took the values of our own SSV STRESSES DURING IMPACT Section B The forces on section B are shown on figure 10. The normal stress is N A F 2 Pi* r N 1.91 mm² With F =338N the force of the impact and r=7.5mm the radius. Bending Moment M * y I With M the moment, I the moment of Inertia and y=10mm radius of the section. First we need to calculate the moment: M F * d M m * g * d M Nmm With m=0.145kg mass, g=9.81n/kg gravitational constant and d=40mm distance between force and section. Next, we need to calculate the moment of inertia I * Pi * r With r=10mm radius of cylinder. Therefor the stress due to bending is: MPa Shear stress VQ It 4 With V,shear force, I 7854mm and t=20mm thickness moment of inertia (which we already calculated earlier) 37

39 First we need to calculate Q: Q y. A With y the distance from the neutral line to the center of gravity. A is half of surface of the section, where the thickness is maximal. 4* r Pi * r² Q * 3* Pi 2 With r=10mm the radius of the section. Q mm³ Next we need to calculate the shear force: V m* g With m=0.145kg mass and g=9.81n/kg gravitational constant. V 1. 42N Therefor the shear stress is: 6.037*10 3 MPa The total stresses are then: MPa MPa MPa 6.037*10 MPa Section A The forces on section A are shown on figure 11. Figure 36: Section B Bending moment M * y With M moment due to F,I moment of inertia, y=20mm radius of section B. I First we need to calculate the moment: M F * x With F=338N and x=43mm the height of the car divided by two. M 14534Nmm 38

40 The moment of Inertia is: 1 4 I * Pi * r With r=20mm the radius of section B. 4 Therefor the stress due to bending is: N MPa mm² Normal stress And N A N mg With m=0.377kg mass of solar panel and g=9.81n/kg gravitational constant. N 3. 7N The surface A is: A Pi *r² With r=20mm the radius of the section B. Therefor the normal stress is: MPa Shear stress VQ It With V=338N shear force, and t=40 thickness. We calculate Q using the formula: Q y * A 4 I mm moment of inertia (already calculated earlier) With y the distance from the neutral line to the center of gravity. A is half of surface of the section, where the thickness is maximal. 4* r Pi * r² Q * 3* Pi 2 With r=20mm the radius of the section. Q mm³ Therefor the shear stress is: 0.359MPa The total stresses are then: 2.31MPa MPa MPa 0.359MPa 39 Figure 37: Section A

41 3.3.2 MAXIMUM ALLOWED FORCE The values of the maximum allowable stresses are found on the table 1. We use the values on the row of the two triangles and the column of FE50-C35. normal, allowed bending, allowed shear, allowed 120MPa 140MPa 100MPa I. Section B The maximum force is only determined by the normal stress. The stress due to bending and the shear stress are independent from the force on impact. Normal stress normal, allowed * A F max With A =176.7mm² F 21206N 21. 2kN max II. Section A The maximum force is determined by the shear stress as well as the stress due to bending. Stress due to bending bending, allowed F max * x * y I With x,y and I the same values as in part one. F N kN max Shear stress Fmax Q shear, allowed It With Q,I and t the same values as in part one. F 94248N kN max Therefor F is limited by the stress due to bending and F kN max max 40

42 Overall, we can conclude that the maximum force is determined by section A and is 20.46kN. Table 2: Allowable stresses of steel (Fe50C35) 41

43 3.4 COLLISION EXERCISE Figure 38: Situation before collision. (a) What is the speed of object B and C right after the collision. We assume B and C will collide inelasticly. After the collision B and C will stick together and move at the same speed. The Spring has no effect right after the collision. V bc is calculated by this formula: With We find: mass of B mass of C speed of B before collision speed of C before collision V bc = velocity afther collisions (b) What is the maximum speed of object A, when the potential energy is at its maximum? The potential energy is maximal when the spring is most compressed. This happens when the speed of A and BC is equal. With: 42

44 We find (c) What is the maximum potential energy in the spring? This question is solved with the principle of conservation of energy. We define two situations: Situation 1: Right after the collision. The spring is not yet compressed. The total amount of kinetic energy: With:, Situation 2: The spring is fully compressed at its maximum The total amount of kinetic energy: With: The difference between these kinetic energies is equal to the maximum potential energy stored in the spring due the law of conservation of energy. (d) Is it possible to change the direction of object A during the oscillation? This calculation is done by setting the speed of block A to to the other direction. We will test if this is possible.. After this point in time, block A moves The system has a momentum of 24. This is calculated by the next equation. 43

45 Due to conservation of momentum, the next equation is used when the speed of block A is. This concludes that is 4 when the speed of bock A is 0. Now we need to check if this is possible. Is is only possible when the kinetic energy is smaller or equal to 48 Joule. The kinetic energy is 48J so it is possible. But when A starts moving in the other direction, then block B and C would have exceed the speed of 4. This will result in a kinetic energy larger then 48 Joule, which is not possible. So object A can be stationary for a brief moment, but will never move in the other direction. 44

46 4. BIJLAGEN Product Specifications ENGLISH SI Performance Sensitivity (±15 %) 0.25 mv/lb 56.2 mv/kn Measurement Range (Compression) lb kn Maximum Static Force (Compression) lb kn Broadband Resolution (1) 0.30 lb-rms 1.3 N-rms [1] Low Frequency Response (-5 %) Hz Hz [3] Upper Frequency Limit Hz Hz [4] Non-Linearity 1 % FS 1 % FS [2] Environmental Temperature Range -65 to +250 F -54 to +121 C Temperature Coefficient of Sensitivity 0.08 %/ F %/ C Electrical Discharge Time Constant (at room temp) 2000 sec 2000 sec Excitation Voltage 20 to 30 VDC 20 to 30 VDC [5] Constant Current Excitation 2 to 20 ma 2 to 20 ma Output Impedance 100 Ohm 100 Ohm Output Bias Voltage 8 to 14 VDC 8 to 14 VDC Spectral Noise (1 Hz) lb/ Hz 0.17 N/ Hz [1] Spectral Noise (10 Hz) lb/ Hz N Hz [1] Spectral Noise (100 Hz) lb/ Hz N/ Hz [1] Spectral Noise (1000 Hz) lb/ Hz N/ Hz [1] Output Polarity (Compression) Positive Positive Physical Stiffness 63 lb/µin 11 kn/µm [1] Size - Diameter 1.5 in 38.1 mm Size - Height 0.5 in 12.7 mm Weight 3.1 oz 88 gm Housing Material Stainless Steel Stainless Steel Sealing Hermetic Hermetic Electrical Connector Coaxial Coaxial Jack Jack Electrical Connection Position Side Side Mounting Thread 1/4-28 Female Not Applicable 45