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1
“SolarBullet”
Report of SSV: Part 1
Mei Lang Xiong
Toran Chhantyal
Sudip Khadka
Vivek Varma
Anand Swaroop
Elnaz Ghavanlou
2nd Bac. Group 223EM
2
Table of Content
INTRODUCTION 3
DETERMINING SOLAR PANEL CHARACTERISTIC 3
DIODE FACTOR 3 PROCEDURE 4 U-‐I CURVE 5 U-‐P CURVE 5 ERROR ANALYSIS 7
OPTIMAL MASS 8
CALCULATION MASS 8 MAXIMAL HEIGHT 9
OPTIMAL GEAR RATIO 10
SIMULATION SSV FOR 10 DIFFERENT GEAR RATIOS & 10 DIFFERENT MASSES MATLAB 10 DISPLACEMENT CURVE & SPEED CURVE 10 MAXIMAL HEIGHT 12
THE GRAPH SHOWS THE DIFFERENT HEIGHT THAT CAN BE ACHIEVED BY THE BALL WITH RESPECT TO DIFFERENT MASS AND GEAR RATIO COMBINATION OF THE SSV. 13
INTERSECTION MOTOR CHARACTERISTIS AND NONLINEAR SOLAR PANEL 14
BISECTION METHOD TO FIND THE INTERSECTION OF MOTOR AND SOLAR CHARATERISTICS 15 TIME INTERVAL (T = 0,1S) -‐ OPTIMAL GEAR RATIO(GR): 11 15 FIRST SECOND DISPLACEMENT CURVE 16 FIRST SECOND VELOCITY CURVE 17 FINDING ZERO USING BISECTION METHOD 18
CASE SIMULINK 19
SIMULATION SOLAR PANEL CONNECTED TO A RESISTANCE BETWEEN 10 Ω AND 100 Ω 19 COMBINATION MODEL OF YOUR SOLAR PANEL WITH THE MODEL OF SSV AND SIMULATION RACE 21
3
Introduction
Solar panels consume light energy, which is delivered by the sun and is transformed into electrical energy. Solar panels are applied in many electronic devices, including powering individual gadgets and vehicles.
The solar panel consists of solar cells and is also defined as photovoltaic cell. Every individual photovoltaic cell participates in conversion of sunlight into electricity. Arrangement of every individual cells in a group is called a “panel” and collection of two or more panels is called an array. Another important component of the small solar vehicle is the DC Motor. The basic function of a DC motor is to convert electrical energy into mechanical work. The mechanical energy generated by the DC motor is transferred to wheels through gears. In order to build and design our vehicle we first need to determine designing factors and this will be discussed in part I.
Determining solar panel characteristic
Diode Factor
To determine the diode factor of the solar panel, we measured the voltage and the current with varying resistance at room temperature. Therefore we obtain the diode factor m for each resistance by this formula:
4
Resistance (Ohm)
Current (A)
Voltage (V)
Diode factor (m)
Power (watt)
Current from motor
Power (m)
133,71 0,07 9,36 1,2986 0,66 0,0779 0,73 116,63 0,08 9,33 1,2962 0,75 0,0998 0,93 103,44 0,09 9,31 1,2953 0,84 0,1138 1,06 92,80 0,10 9,28 1,2931 0,93 0,1338 1,24 84,09 0,11 9,25 1,2909 1,02 0,1527 1,41 70,92 0,13 9,22 1,2908 1,20 0,1705 1,57 65,71 0,14 9,20 1,2901 1,29 0,1819 1,67 53,88 0,17 9,16 1,2914 1,56 0,2034 1,86 45,60 0,20 9,12 1,2934 1,82 0,2233 2,04 41,23 0,22 9,07 1,2919 2,00 0,2462 2,23 36,08 0,25 9,02 1,2941 2,26 0,2671 2,41 29,70 0,30 8,91 1,2970 2,67 0,3067 2,73 23,54 0,37 8,71 1,3064 3,22 0,3608 3,14 20,76 0,41 8,51 1,3130 3,49 0,3980 3,39 18,45 0,42 7,75 1,2076 3,26 0,4602 3,57 17,90 0,42 7,52 1,1718 3,16 0,4671 3,51 15,79 0,42 6,63 1,0331 2,78 0,4776 3,17 1,78 0,45 0,80 0,1304 0,36 0,4800 0,38 AVG 1,2047
3,57
Procedure
We started adjusting the current and the resistance on the solar panel. Due to that we could measure the corresponding voltages. By applying then the formula above we let excel calculate the diode factor m. The adjusted current with its corresponding voltage gives us then the power consumed. On the other hand, if we take the current calculated from the diode factor equation and substitute the diode factor we have, we obtain the current from the motor. This resulting current mulitplied by the measured voltage give us a new power consumed by the motor.
Th highlighted data corresponds to the maximum power we can obtain from the solar panel. This power is the ideal power meaning that in reality we cannot easily obtain this.
5
U-‐I curve
Figure 1
The number of rotations decreases when we take the results from the calculated m factor.
U-‐P curve
Figure 2
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0 2 4 6 8 10
curren
t
Voltage
Current vs voltage
I vs U
I vs U from m
motor rpm
Motor rpm(frm m)
motor rpm(n=0)
n=0 n=7118.272
n=8006.665
0
0,5
1
1,5
2
2,5
3
3,5
4
0 1 2 3 4 5 6 7 8 9 10
Power
Voltage
Power vs Voltage
Power
Power(m)
6
Determination of n (rpm):
𝑈 = 𝑖 𝑅 +𝑛
1120
with Resistance R = 3.32 Voltage U = 8.51 Current i = 0.41 n = defines number of revolutions per second by the motor
7
Error analysis Mean Value of m (we left out the last measurement, because this causes a big change in the average)
𝑋1 + 𝑋2 +⋯ . .𝑋𝑛
𝑛= 1.2986 + 1.2962 + 1.2953 + 1.2931 +⋯+ 1.0331
17= 1.2047
Standard Deviation = [!!
(𝑥! − 𝑥!"#$)!]!!
D₁ = 1.2986 -‐ 1.2047 = 0.0939
D₂ = 1.2962 -‐ 1.2047 = 0.0915
D₃ = 1.2953 -‐ 1.2047 = 0.0906
D₄ = 1,2931 -‐ 1.2047 = 0.0884
D₅ = 1,2909 -‐ 1.2047 = 0.0862
D6 = 1,2908 -‐ 1.2047 = 0.0861
D7 = 1,2901 -‐ 1.2047 = 0.0854
D8 = 1,2914 -‐ 1.2047 = 0.0867
D9 = 1,2934 -‐ 1.2047 = 0.0887
D10 = 1,2919 -‐ 1.2047 = 0.0872
D11 = 1,2941 -‐ 1.2047 = 0.0894
D12 = 1,2970 -‐ 1.2047 = 0.0923
D13 = 1,3064 -‐ 1.2047 = 0.1017
D14 = 1,3130 -‐ 1.2047 = 0.1083
D15 = 1,2076 -‐ 1.2047 = 0.0029
D16 = 1,1718 -‐ 1.2047 = -‐0.0329
D17 = 1,0331 -‐ 1.2047 = -‐0.1716
!.!"#" ²! !.!"#$ !! !.!"!# !! !.!""# !! !.!"#$ ²! !.!"#$ ²! !.!"#$ ²! !.!"#$ ²
! !.!""# ²! !.!"#$ ²! !.!"#$ ²! !.!"#$ ²! !.!"!# ²! !.!"#$ ²! !.!!"# ²! !!.!"#$ ²! !!.!"!# ²!"
12
= 0.0931 m factor = 1,2047 ± 0.0931
8
Optimal Mass
Calculation mass
Power supplied to wheel is P = 𝜏w(𝜔w)
= F(rw)(𝜔w)
= F(Vw)
= M(a)(vw) with M = mass of the car
= M !"!"
(vw)
P = M !"!"
!"!"
(vw)
Considering Vw = V, we integrate the equation on both sides
𝑝!! 𝑑𝑠 = 𝑀!! 𝑣!dv
P(s) = M !!
!!
!
v = 3P(!)!" !!
= vin,c (*)
Based on the conservation of energy and conservation of momentum we obtain that
Vf,b = !"!(!!",!)Mc!!! (**) with Vf,b = final velocity of ball
We substitute * in this equation
= !"!
!!!Mb !"(!)!"
!!
= ! !!"!! !
!!
!"!𝑀𝑏
To optimize to get the ball as high as possible, we need a maximum velocity. That velocity can be determined by derivating the relation between the velocity of the car and the velocity of the ball.
dvf,! !"!
= 2 3P s !!!!! !
!!"!!"
(!!!!!)! = 0
9
=> !!Mc(Mc+Mb) = Mc
=> 2(Mc+Mb) = 3Mc
=> Mc = 2Mb
Maximal height We can determine the maximal height with potential and kenetic energy relation. Therefore: KEinitial + PEinitial = KEfinal + PEfinal Mboule ∗ Vi!
2 + MBoule ∗ g ∗ height = Mboule ∗ Vf !
2 + MBoule ∗ g ∗ height As the initial height and Vf(final velocity) are equal to zero, the above equation become : Vi!
2 = g ∗ height
ℎ𝑒𝑖𝑔ℎ𝑡 = Vi!
g ∗ 2
We assume that no energy is lost during the collision, this means that the ball will reach a lower height than expected. Also we don’t consider that we will have an issue with the friction force. That means that the car will be slower at the end, before it touches the ball. This also causes a less high position for the ball.
10
Optimal Gear Ratio
Simulation SSV for 10 different gear ratios & 10 different masses MATLAB
M (kg)\G ratio 2 3 4 5 6 7 8 9 10 11 12 13 0.735 1.717 2.345 2.837 3.256 3.626 3.952 4.190 4.227 4.076 3.831 3.567 3.317 1.100 1.361 2.092 2.626 3.069 3.456 3.802 4.111 4.352 4.438 4.351 4.154 3.915 1.300 -‐ 1.918 2.475 2.927 3.319 3.668 3.984 4.258 4.433 4.446 4.318 4.114 1.410 -‐ 1.817 2.387 2.845 3.239 3.590 3.907 4.189 4.397 4.458 4.372 4.193 1.420 -‐ 1.808 2.379 2.838 3.232 3.582 3.899 4.182 4.392 4.458 4.375 4.199 1.430 -‐ 1.798 2.371 2.830 3.224 3.575 3.892 4.175 4.387 4.458 4.379 4.205 1.440 -‐ 1.789 2.363 2.822 3.217 3.567 3.885 4.168 4.382 4.458 4.383 4.211 1.470 -‐ 1.761 2.339 2.800 3.195 3.546 3.864 4.148 4.368 4.456 4.393 4.227 1.480 -‐ 1.751 2.330 2.792 3.187 3.538 3.856 4.141 4.363 4.454 4.395 4.233 1.830 -‐ -‐ 2.041 2.522 2.926 3.280 3.599 3.889 4.147 4.337 4.401 4.337
Table 1
The best gear ratio is 11. This is because at that instant the speed is maximal namely 4.458. This corresponds then with a mass of 1.430 kg.
Displacement curve & Speed curve
Figure 3: Gear ratio 11 & 1.42kg
Figure 4: Gear ratio 11 & 1.43kg
11
Figure 5: Gear ratio 11 & 1.44kg
Figure 6: Gear ratio 11 & 1.47kg
Figure 7: Gear ratio 11 & 1.48kg
12
Maximal height
M (kg)\G ratio 2 3 4 5 6 7 8 9 10 11 12 13
0.735 0.150184
0.280269
0.410354
0.540405
0.670051
0.796115
0.894934
0.910852
0.846627
0.748141
0.648334
0.560683
1.1 0.094419
0.222987
0.351543
0.480068
0.608635
0.736669
0.861328
0.96518
1.003998
0.964777
0.879546
0.781176
1.3 -‐ 0.187502
0.312148
0.436804
0.561394
0.685759
0.808894
0.923894
1.001622
1.007288
0.950295
0.862559
1.41 -‐ 0.168288
0.290434
0.412554
0.534732
0.656716
0.777928
0.894166
0.985197
1.012801
0.974093
0.895885
1.42 -‐ 0.166564
0.288469
0.410375
0.532265
0.654001
0.775007
0.891452
0.983212
1.012815
0.975772
0.898616
1.43 -‐ 0.164841
0.286528
0.40822
0.52981
0.651315
0.772117
0.888588
0.981043
1.012823
0.977415
0.90131
1.44 -‐ 0.16313
0.284569 0.406
0.527382
0.64866
0.769257
0.885603
0.978872
1.012792
0.979024
0.903853
1.47 -‐ 0.158002
0.278736
0.399478
0.520173
0.640805
0.760857
0.87684
0.972476
1.011831
0.983527
0.91087
1.48 -‐ 0.156305
0.276791
0.397286
0.517727
0.6381
0.757915
0.873978
0.970369
1.011161
0.98458
0.91312
1.83 -‐ -‐ 0.212233
0.32426
0.43626
0.548281
0.660025
0.770858
0.876389
0.958574
0.98715
0.958914
Table 2
From the table above, obtained from the Matlab simulation, we get the maximum height the boule can reach is 1.012823 m. These values were calculated using the potential and kinetic energy equation describe in the above analytic part. But the value was calculated ignoring the energy loss in the collision and other losses, we expect, in real experiment to get a lower value than this.
13
The table above will also be illustrated with the graph below.
The graph shows the different height that can be achieved by the ball with respect to different mass and gear ratio combination of the SSV.
14
Intersection motor characteristis and nonlinear solar panel
𝑦 = 12 + sin(
!!)*𝑒sin
𝑥3
x y 1.0000 1.1650 2.0000 2.0617 3.0000 2.8140 4.0000 2.9033 5.0000 2.1194 6.0000 0.8503 Average 7.0000 -‐0.2229 6.5000 6.5000 0.2525 6.7500 6.7500 -‐0.0036 6.6250 6.6250 0.1201 6.6875 6.6875 0.0572 6.7188 6.7188 0.0265 6.7344 6.7344 0.0114 6.7422 6.7422 0.0039 6.7461 6.7461 0.0001 6.7480 6.7480 -‐0.0017 6.7471 6.7471 -‐0.0008 6.7466 6.7466 -‐0.0003 6.7463 6.7463 -‐0.0001 6.7462 6.7462 0.0000 Table 3
Figure 8
-‐0,5000 0,0000 0,5000 1,0000 1,5000 2,0000 2,5000 3,0000 3,5000
0,0000 1,0000 2,0000 3,0000 4,0000 5,0000 6,0000 7,0000 8,0000
y
x
y(x)
15
Bisection method to find the intersection of motor and solar charateristics
Solar Car parameters:
M = 1.43 kg
α = 0.1253 rad
g = 9.81 N/kg Crr = 0.012 Ce*Ф = 8.9285*10-‐4 V/rpm r = 0.04 m Cw = 0.5 A = 0.02 m2 𝜌(air) = 1.293 kg/m3 (density)
𝐼 𝑡 = 𝐼sc−𝑒! ! !! ! ∗ !!
!∗!∗!!!!
Isc = 0.48 A; Is = 10e-‐8 A/m2 m = 1.1 kg N = 16; Thermal Voltage Ur = 0.0257 V; Ra = 3.32 Ω For the motor: E(t) = Ke × ω = CE.Ф × 60 × v(t) x !"#$ !"#$%
!"!
TIME INTERVAL (t = 0,1s) -‐ Optimal Gear Ratio(GR): 11 Initial conditions, at t = 0 s:
s(0) = 0; v(0) = 0; I(0) = 0.48 A then,
E(0)= CE.Ф × 60 × v(0) × !"#$ !"#$%!!"
= 0
a(0) = g × Crr + 60 x Ce*𝐺𝑒𝑎𝑟 𝑟𝑎𝑡𝑖𝑜2𝜋𝑟
x !! = 0.069m/s2
16
Table 4
First second displacement curve
Figure 9: 1st second displacement curve
0
0,5
1
1,5
2
2,5
0 0,2 0,4 0,6 0,8 1 1,2
Displacemen
t[m]
fme(s)
Displacement Vs Time
Time(s) I(t)[A] s(t)[m] v(t)[m/s] E(t)[V] a(t)[m/s2] 0 0.48 0 0 0 6.69E-‐01 0.1 0.479995422 3.34E-‐03 6.69E-‐02 0.156792572 0.668959256 0.2 0.479993134 2.01E-‐02 1.34E-‐01 0.313578644 0.66889481 0.3 0.479990845 6.36E-‐02 2.01E-‐01 0.47034961 0.66878991 0.4 0.479987183 1.47E-‐01 2.68E-‐01 0.627095992 0.668642329 0.5 0.479981689 2.84E-‐01 3.34E-‐01 0.783807784 0.668451348 0.6 0.479972534 4.88E-‐01 4.01E-‐01 0.940474815 0.668214011 0.7 0.479961548 7.73E-‐01 4.68E-‐01 1.097086221 0.667933371 0.8 0.479945068 1.15E+00 5.35E-‐01 1.253631852 0.667603489 0.9 0.479923096 1.64E+00 6.02E-‐01 1.410100169 0.667224446 1 0.479891968 2.24E+00 6.68E-‐01 1.566479647 0.666790327
17
First second velocity curve
Figure 10: 1st second velocity curve
During each time interval of 0.1s during the whole process of 1 second, we have assumed that the acceleration remains constant for that interval. The influence of this can be clearly seen in our graph. Hence, the position of the SSV increases exponentially with time while velocity increases linearly with time.
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0 0,2 0,4 0,6 0,8 1 1,2
velocity[m
/s]
fme(s)
Velocity vs fme
18
Example Calculation for Table1: For the final part i.e at 1 second.
T E(t) I(t) f(I(t)) 1.0000000000 1.5664796470 0.0000000000 -‐0.4799969081 0.48 0.0001080799 0.24 -‐0.2399815177 0.36 -‐0.1199552644 0.42 -‐0.0599304558 0.45 -‐0.0299133008 0.465 -‐0.0149031984 0.4725 -‐0.0073977144 0.47625 -‐0.0036448571 0.478125 -‐0.0017683987 0.4790625 -‐0.0008301620 0.47953125 -‐0.0003610417 0.479765625 -‐0.0001264811 0.479882813 -‐0.0000092006 0.479941406 0.0000494396 0.479912109 0.0000201195 0.479897461 0.0000054594 0.479890137 -‐0.0000018706 0.479893799 0.0000017944 0.479891968 -‐0.0000000381 Table 5
Finding zero using bisection method
Finding the zero of the function F(I(t)) which is given by
We take upto 6 decimal places to increase our approximation. For the last value of I i.e I=0.479891968 A in table 2, we get corresponding value of f(I(t)) to be 0. Therefore,we take this value of I for table 1.
19
Case Simulink
Simulation solar panel connected to a resistance between 10 Ω and 100 Ω With simulation of behavior of solar panel connected to a resistance between 10 ohm and 100 ohm with an increment in 5 ohm in Simulink, we have drawn the following graph
Figure 11: voltage VS current
Figure 1
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0 2 4 6 8 10
curren
t
Voltage
Current vs voltage
I vs U
I vs U from m
motor rpm
Motor rpm(frm m)
motor rpm(n=0)
n=0 n=7118.272
n=8006.665
20
Comparing voltage versus current graph from Simulink to the one from experimental observation, we can conclude that our experimental values more or less coincide with the Simulink values we get. As we can see from the graph the open circuit voltage for both cases happens to be about 9.45V.
Figure 12: voltage VS power
Figure 2
0
0,5
1
1,5
2
2,5
3
3,5
4
0 1 2 3 4 5 6 7 8 9 10
Power
Voltage
Power vs Voltage
Power
Power(m)
21
Comparing two graphs of power versus voltage, we can see that the maximum power from the solar panel is not the same for Simulink and experimental observation. For experimental we have maximum power of about 3.5 watt whereas for Simulink the value is approximately 4 watt. So we have deviation of almost 0.5 watt between the two cases.
Combination model of your solar panel with the model of SSV and simulation race
We calculated the optimal mass and optimal gear ratio using Matlab file in Simulink. We obtained optimal mass to be 0.75 kg and gear ratio as 5, which contradict our analytical calculations where we obtained optimal mass as 1.47kg and gear ratio of 11. The graph below is from Simulink with mass of 0.75 kg and gear ratio of 5.
Figure 13 & 14
Importance of Simulink simulation
The Simulink allow us to create same environment where our SSV would run in reality. This helps a lot to tackle real time obstacles and we can take into account many influencing factors. The Simulink interface is very user friendly and one can easily observe the ongoing phenomena.