Seminar assignments - Take Home Challenge PDF

Preview

Summary

Missing mesh analysis at the last question...

Description

Part B By applying mesh and super mesh analysis:Label all mesh currents and identify super mesh:-

Deriving equations from Ohm’s Law, Kirchhoff’s Voltage Law and Kirchhoff’s Current Law:Equation 1:!!! = 7.8 A Equation 2:!!! − !! = 1.4 A Equation 3: −27 !! − !! − 0.1 !! − 0.11 !! − !! − 13 − 0.1 !! = 0 Equation 4: 13 − 0.11 !! − !! − 0.3 !! − 7.1 !! − 6.8 − 0.3 !! = 0 Solving the simultaneous equations for mesh currents with a calculator:!! = 7.8 A !! − !! = 1.4A 27.2 !! + 0.11 !! − 0.11 !! = 197.6 −0.11 !! + 7.81 !! = 6.2 Therefore:!! = 7.8 A !! = 7.245 A

!! = 5.845 A !! = 0.876 A Calculations on power delivery and consumptions: Power for the 7.8 A solar panel current source:! = !" ! ! ! ! !

= (7.8 − 7.245)!×27! = 14.985! = !"! = 14.985×7.8! = 116.883 W (Delivered)

Power for the 27Ω solar panel shunt resistance:! ! ! ! ! !

= !"! = 7.8 − 7.245 27 ! = 14.985! = !"! = 14.985× 7.8 − 7.245 ! = 8.317 W (Dissipated)

Power for the 0.1Ω resistor:! ! ! ! ! !

= !"! = 7.245×0.1! = 0.7245! = !"! = 0.7245×7.245! = 5.249W (Dissipated)

Power for the 1.4A current consumption by refrigerator:! ! ! ! ! !

= !"! = 13 + 0.11 5.845 − 0.876 ! = 13.547! = !"! = 13.547×1.4! = 18.966 W (Dissipated)

Power for the 0.11Ω battery internal resistance:! = !"! ! = 5.845 − 0.876 ×0.11! ! = 0.54659!

! = !"! ! = 0.54659× 5.845 − 0.876 ! ! = 2.716 W (Dissipated) Power for the 13V battery:! = !"! ! = 13× 5.845 − 0.876 ! ! = 64.597 W (Dissipated) Power for the 0.3Ω resistor:! ! ! ! ! !

= !"! = 0.876×0.3! = 0.2628! = !"! = 0.2628×0.867! = 0.2302 W (Dissipated)

Power for the 7.1Ω LED internal resistance:! ! ! ! ! !

= !"! = 0.876×7.1! = 6.2196! = !"! = 6.2196×0.876! = 5.448 W (Dissipated)

Power for the 6.8V LED voltage drop:! = !"! ! = 6.8×0.876! ! = 5.957 W (Dissipated) Power supplied must be equal to sum of all power dissipated for power balancing. Power supplied = 116.883 W Power consumed = 8.317 + 5.249 + 5.249 + 18.966 + 2.716 + 64.597 + 0.2302 + 0.2302 + 5.448 + 5.957 = 116.959 W The difference of 0.076 is treated as tiny and insignificant, due to rounding error.

Efficiency of power supply:The solar panel supplies 116.883 W The refrigerator uses 18.966 W. The battery uses 64.597 W The LEDs use 5.957 W Therefore, power transfer is considered to be effective from solar panel to refrigerator, battery and LEDs. Remaining power of 27.363 W is being consumed by resistors across the circuit.

Part C Using circuit simulation software to verify the calculations:Assumptions are made when using the circuit simulation software:1. Since the calculations are done with assuming all the electrical components are ideal, and the simulation software considers LED as non-ideal component, the LEDs are replaced with a voltage source to simulate that ideal environment. 2. The LEDs switch is closed. 3. Tiny discrepancies between calculated values and circuit simulation values due to rounding error.

Solar Panel Shunt Resistor Solar Panel Current Delivery

Refrigerator Current Consumption

Battery Internal Resistor

Battery

LED Internal Resistor

LED

Reference: http://www.falstad.com/circuit/

Part D From mesh analysis at Part B, the charging current of the battery is:!! − !! = 5.845 − 0.846! Current = 4.999!A If 4.999 A is flowing through the battery for 8 hours:4.999 A × 8 hours = 39.992 Amp-hours is stored in the battery Applying a new mesh and super mesh analysis by setting solar panel current source to 0 A:-

!! = 0 A! !! − !! = 1.4 A −27 !! − 0.1 !! − 0.11 !! − !! − 13 − 0.1 !! = 0! −27 !! − 0.11 !! + 0.11 !! = 13 ! 13 − 0.11 !! − !! − 0.3 !! − 7.1 !! − 6.8 − 0.3 !! = 0! −0.11 !! + 7.81 !! = 6.2 Solving the new simultaneous equations derived using a calculator:!! !! !! !!

= 0 A! = - 0.467 A! = - 1.867 A! = 0.768 A

The negative (-) sign shows that the current is flowing in the opposite direction (counterclockwise) to the assigned convention (clockwise). This would mean that the battery is powering the refrigerator and LEDs, instead of solar panel.

Finding the current drawn from battery at night:!! − !! = Current drawn from battery – 1.867 A – 0.768 A = – 2.635 A If 2.635 A of current is being drawn from the battery during night time for 16 hours:2.635 A × 16 hours = 42.16 Ah This clearly shows that the battery would run flat before the next day’s charging cycle begins. The difference of storage and drainage is:39.992 Ah – 42.16 Ah = 2.168 Ah The night-time hours without power:2.168 Ah ÷ 2.635 A = 0.822 hour This means that night-time with power is:16 hours – 0.822 hours = 15.178 hours

Using circuit simulation software to verify calculations:-

Battery Internal Resistance

Battery

LED Internal Resistor

LED

The assumptions made previously with circuit simulation software are applied. Reference: http://www.falstad.com/circuit/

Part E The desired current flow through the LEDs is 0.5 A. Based on the mesh analysis in Part B, !! is the current flowing through the LEDs. The new LEDs resistance, ! , using the related equations derived from Part B and setting !! to 0.5 A:!! = 5.845!!A! !! = 0.5 A 13 − 0.11 !! − !! − 0.3 !! − 7.1 !! − 6.8 − 0.3 !! = 0! ! 13 − 0.11 0.5 − 5.845 − 0.3 0.5 − ! 0.5 − 6.8 − 0.3 0.5 = 0! ! = 12.9759 Hence, the new resistor’s value is around 13Ω. Part F This circuit modelling is done with the original LED resistance value. A diode is integrated into the circuit as such:-!

The diode essentially acts as a one-way valve, which allows current to flow in one direction whilst blocking current flowing from the other direction. This arrangement allows certain amount of power to be delivered to the desired component, rather than being wasted on unused component. Using the circuit simulation software, the diode blocks current flowing through the 27Ω resistor as well as the two 0.1Ω resistor, thus power is not consumed by these components.

For the 0.1Ω resistor:-

For the 27Ω resistor:-

For the 0.1Ω resistor:-

Power consumed by the 27Ω resistor:! ! ! ! ! ! !

= !"! = !! − !! ! ! = [0 − (−0.467)]×27! = 12.609! = !"! = 12.609×0.467! = 5.888W

Power consumed by the 0.1Ω resistors:! ! ! ! ! !

= !"! = 0.467×0.1! = 0.0467! = !"! = 0.0467×0.467! = 0.0218 W

Total power consumed:5.888 + 0.0218 + 0.0218 = 5.932 W Therefore, with a diode placed in the circuit to avoid unwanted power wastage, an additional 5.932 W of power can be delivered across refrigerator and LEDs. Calculation based on the values given by the simulation software is carried out to find out the gain for refrigerator:-

Power without diode = 17.79 W

Power with diode = 17.864 W Comparing power efficiency of the refrigerator:(17.864 – 17.79) × 100% = 7.4% Hence, the refrigerator gains 7.4% of relative power with a diode integrated into the circuit. A suitable diode is found online. The diode is able to handle specified forward voltage and current in Olive’s system, as well as withstanding reverse voltage and current flow.

Reference: http://www.onsemi.com/PowerSolutions/product.do?id=NSR0115CQP6

Part G A new circuit is modelled accordingly on the simulator, with the original LED resistance value and without the diode specified in Part F. From the mesh analysis at Part B and circuit simulation, the current charging the battery with the LEDs switched on was:!! − !! = Charging current! 5.845 − 0.876 = 4.969A Simulation:-

With the new circuit and both the LEDs switched off, the amount of current charging the battery during daytime is:!!"##$%& = !! − !! ! !!"##$%& = 5.485 − 0! !!"##$%& = !5.485 A Simulation:-

This clearly shows that the new circuit is charging the battery at a higher rate, due to unused LEDs. So the new battery Amp-hours stored is:5.845 A × 8 hours = 46.76 Ah With the aid of the circuit simulator and both LEDs lights switched on, the current drawn from the battery during night-time is 3.27 A.

So, the night-time that Olive can run with both LEDs and refrigerator is:46.76 Ah ÷ 3.27 A = 14.299 hours ∴ Even though the battery is being charged at a higher rate in the new system, the current is being drawn is higher at night-time due to new parallel LEDs added. Olive can have only about 14.3 hours of power, leaving about 1.7 hours without power when the sun is not shining....