Take Home Exam 001

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Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge DC Circuit Design Challenge The following figure represents a system used to power certain elements in a caravan with a Solar Panel. Circuit Element Solar panel shunt resistance Solar panel current delivery Refrigerator current consumption Battery internal series resistance Battery internal voltage LED light internal series resistance LED light internal voltage drop Symbo l aΩ bA cA Value from student #9408363 27 Ω 7.8 A 1.2 A dΩ 0.18 Ω eV fΩ 12.8 V 7.6 Ω gV 6.8 V (a) Draw a circuit diagram of the system described above, with the LED light switch closed. Where possible, simplify the diagram by noting series and parallel resistances and sources. From the original figure above a circuit diagram can be formulated; By simplifying some of the circuit elements, the following circuit was produced to be used for analysis 1 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge (b) Using a circuit analysis technique of your choosing, calculate the power balance of the system. Show all working. Comment on the effectiveness of power transfer from the solar panels to the refrigerator, battery, and LED light. A mesh analysis can be performed over four sections, current in all of which assumed to be moving clockwise. Due to the current source, a supermesh was used (outlined with the dotted line) Analysis of the each of the meshes produced 3 equations to be solved simultaneously; 1.2=I 2−I 3 (Equation 1) Current Source: Supermesh: 0=0.1 I 2 +27 I 2 −(27 × I 1 )+ 0.1 I 2+ 0.18 I 3−0.18 I 4 +12.8 197.8=27.2 I 2+ 0.18 I 3−0.18 I 4 (Equation 2) Mesh 4: 0=−12.8+ 0.18 I 4 −0.18 I 3 +8.2 I 4 +6.8 6=−0.18 I 3 +8.38 I 4 (Equation 3) Solving on a Graphics Calculator for simultaneous equations, the following values were produced; I1 7.8 I2 7.2377 I3 6.0377 2 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge I4 0.845679 To complete a power balance over the system, power over all parts needs to be found, Power across the Solar Panel will equal the sum of all other parts. Resistors 0.1Ω Solar Panel P=VI 2 P=( I 2 ) × R P=VI P=((I 1−I 2) ×27)× I P=15.1821 ×7.8 P=118.42 P=( 7.2377 )2 ×0.1 P=5.23843 27Ω LED 0.18Ω Battery 8.2Ω Refrigerator P=VI 2 P=( I 1−I 2 ) × R 2 P=( 0.5623 ) ×27 P=8.53689 P=VI P=6.8 × I P=6.8 × 0.845679 P=5.7062 P=VI P=( I 3−I 4 ) 2 × R P=26.9571 ×0.18 P=4.85227 P=VI P=12.8 ×(I 3 −I 4) P=66.4579 P=VI 2 P=I 4 × R P=0.715173 × 8.2 P=5.86442 P=VI P=(12.8+0.18 × ( I 3−I 4 ) )× I P=13.7346 ×1.2 P=16.4815 ∑ Powe r ¿=∑ Powe r out PSolar Panel =PLED + PBattery + PRefrigerator +2× P0.1 Ω + P27 Ω + P0.18 Ω + P 8.2Ω PSolar Panel =5.7062+66.4579+16.4815+ ( 2× 5.23843 ) +8.53689+ 4.85227+5.86442 118.42calculated 118.376 The power transfer from the solar panel to the refrigerator is 13.9178% (( ) ) P Refrigerator ×100 PSolar Panel leaving 4.81861% to the LED and 56.1205% to the Battery. This means that the circuit and wire resistances are dissipating a quarter of the input energy through heat. Any reduction in these could improve the system (c) Use circuit simulation software to verify your calculations in (b). Show screen shots of the software that verify your results. The program EveryCircuit (http://everycircuit.com/app) was used to verify these values found in part b). It’s important to mention that in this simulation and the ones that follow, the LED is represented by a voltage difference rather than a diode. This is done in order to reflect theoretical conditions but in real life there would be some fluctuation in the voltage drop across the diode, and therefore the system. 3 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge Over the 0.1Ω resistors (I2) 7.24 in the simulation ~ the calculated 7.2377 Over the 8.2Ω resistors (I4) 0.846 in the simulation ~ the calculated 0.845679 For, I3 the simulation wouldn’t show I3 alone on the wire but can be verified by using the calculated values (I3-I4) to get current over the battery which gives 5.19202 ~ 5.19 on the simulation. Therefore I 3 is verified. (d) Olive mostly leaves her LED light turned on, day and night. During a sunny day, the solar panel delivers its rated current for 8 hours. How many Amphours are stored in the battery? (Initially assume the battery is discharged enough overnight to accept all of the next day’s charging). When the sun is not shining for the other 16 hours, the solar panel produces no current, but its shunt resistance remains connected. How many Amp-hours are drawn from the battery during this time? Comment on your result. Show all working. Use circuit simulation software to verify your calculations, supported by screen shots. Amp-hours for the Solar Panel; Amp-hours for the Battery; Amps × Hours 7.8 ×8=62.4 (I 3 −I 4) × 8 5.192021× 8=41.5362 During the night-time, the Solar Panel is inactive in the circuit and the current runs as shown below; This can be shown more clearly as; 4 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge This can be simplified as follows, with 3 meshes used (all assumed clockwise) From a mesh analysis, the following equations were determined; Current Source: 1.2=I 1−I 2 (Equation 1) Supermesh: Mesh 3: I (¿ ¿ 2−I 3)+12.8 0=27.2 I 1+ 0.18 ¿ −12.8=27.2 I 1 +0.18 I 2 −0.18 I 3 (Equation 2) 0=−12.8+ 0.18 ( I 3−I 2 ) +8.2 I 3 +6.8 6=−0.18 I 3 +8.38 I 4 (Equation 3) Using a Graphics Calculator, the following values were found; -0.455132 I1 -1.65513 I2 0.680439 I3 I1 and I2 are only negative because the current is flowing in the opposite direction to what was assumed whilst doing the mesh analysis. Current through the battery; I 3 −I 2 =0.680439−−1.65513=2.33557 These values can be verified below where current over the battery is 2.34; 5 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge When Solar Panel inactive for 16 hours, (from diagram) 2.33557 Amps run through the battery 2.33557 ×16=37.3691 Because this value is less than what is built up during the daytime (41.5362), the system works. 6 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge (e) During the day the LED light is very hot and you know this will significantly shorten its lifetime. You suggest changing resistor f in the LED light to limit the current in the LEDs to 0.5 A during the day. Calculate the new value of resistor f. Show all working. Using the simultaneous equations found from the initial mesh analysis Equation 1: 1.2=I 1−I 3 Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 I 4 Equation 3: 6=−0.18 I 3 +8.38 I 4 A new value for resistor f (7.6) by first separating the three resistors around the diode (0.3, 0.3 and f) in equation 3 to 0.6I 4 and fI4 making the new equations; Equation 1: 1.2=I 1−I 3 Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 I 4 Equation 3: 6=−0.18 I 3 +0.78 I 4+ f I 4 Letting I4 equal 0.5 Amps Equation 1: 1.2=I 1−I 3 Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 × 0.5 Equation 3: 6=−0.18 I 3 +0.78 ×0.5+ f ×0.5 Using a Graphics Calculator, the following value was found; f =13.3928 Therefore the new value of the resistor is 13.3928Ω This resistor, when simplified and summed with both 0.3Ω resistors in line on the diode, giving 13.9928Ω, the near value for f can be verified below from the circuit simulator; It can be seen that the new voltage over the diode (represented as a voltage difference) is 0.5A 7 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge (f) Show Olive how a diode can be used to improve the efficiency of the system during the night. Quantify the efficiency gains made from your design using either circuit simulation software (supported by screen shots) or manual circuit calculations. Look online for an appropriate diode to use in this application. Ensure that your diode model in the circuit is equivalent to the real diode that you have chosen. A diode may be used to stop current from flowing through the shunt resistor when the solar panel is not in use. Power is then not lost across a resistor that is serving no practical purpose when the sun is not shining. The diode would be placed between the shunt resistor and the refrigerator as shown below; The diode below would be effective in increasing efficiency as it allows 8A (greater than 7.2377A moving) to flow through it. With a voltage drop of 0.55 this diode will still allow the battery to charge up substantially during the day. http://www.futureelectronics.com/en/technologies/semiconductors/discretes/diod es/scottky-rectifiers/Pages/7024714-80SQ045NG.aspx?IM=0 8 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge The efficiency gains that would result can be calculated by Efficiency Gains= Power dissipated∈element Total power ∈the system To find the current that is flowing through the shunt resistor without the diode, a circuit simulator was used as shown above. 2 P=I R P=( 0.455 )2 ×(27+0.1+0.1) P=5.63108 W When the Solar Panel is inactive in the system, in the same way as d) the current is moving anti-clockwise, anti-clockwise and clockwise respectively for the meshes. The ‘Power in’ still equals ‘Power out’ at this state an ‘Power in’ is now coming from the battery. This is calculated below; P=VI P=12.8 ×2.34 P=5.63108 W Using the efficiency gains calculation Efficiency Gains= 5.63108 =0.188003=18.8 ℑ proved efficiency 29.952 The simulation below shows the use of this diode in the system during the daytime when the solar panel is active; 9 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge The diode would slightly reduce the current through the battery (this is considered negligible in the efficiency calculations). The next simulation, shows the system at night-time when the Solar Panel is inactive; It can be seen the current from the battery is now much less than without the diode, allowing the battery to last much longer. (g) Olive asks if she could have a second identical LED light (both resistor and diodes) in parallel with the first if she bothers to turn them off during the day. Assume that the meaning of “day” here are the 8 hours during which the solar panel delivers its rated current. Calculate how many night-time hours Olive can run the two lights (with the original resistor value) and still run her refrigerator 24 hours a day. The following circuit was created on the simulator. The 8.2Ω resistor was split up into two 0.3Ω resistors and one 7.6Ω resistor so that another could be added in parallel. Two switches were put in to turn off during the day. It can be seen that the battery current is increased which is good for charging should Olive remember to turn the LEDs off. To calculate how long the lights will work for, the new daytime battery amp-hours can be calculated. Amps × Hours 6.03 ×8=48.24 10 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge The following diagram shows the system at night without the Solar Panel It can now be seen that the battery current is 2.9A, the number of hours this can be sustained for is calculated as; Amps × Hours 2.9 ×hours=48.24 therefore; hours=16.63 Because this value is greater than the 16 hours of night-time, using the second light will be okay. However, it is important to consider that because the lights will only last a little bit longer than 16 hours if it was a cloudy day it would be recommended that only one light should be turned on at night. This is why having a second switch would be useful (not to mention if one of the lights break). If Olive decided to use the diode described in part f, this would further benefit the system for using two lights, and this is shown in the diagrams below in the daytime and night-time respectively. 11 Michael O’Brien n9408363 EGB120 DC Circuit Design Challenge During the night-time it the modified system would last for 19.6 hours ( 48.24 16.63−19.6 =hours ). This is a 17.8593% increase ( × 100 ) which is quite 2.46 16.63 substantial and Olive should consider this option. 12

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