ELEN280_MECH287 Fall 2023 Homework #2

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Santa Clara University *

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280

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Mechanical Engineering

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Jan 9, 2024

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Name : ELEN280 / MECH287 Homework Set #2 - FALL 2023 1. Sizing a Solar PV System The Rusk’s own a fairly large house near San Francisco and the following spreadsheet shows their month-to-month energy use for two years. Calculate the size (kW DC, ) of the solar PV system they will need in order to have a zero-dollar electricity bill annually (use the average given). Assume that you are using 14.9% efficient panels. Also assume that your inverter is 95% efficient. Use an average solar insolation of 5.5 kWh/m 2 /day for the location. Use a value of 0.8 as a de-rating factor for the solar panels. 43.9 kWh day ( AC ) ( 1 0.95 0.80 )( DC AC ) ( m 2 / day 5.5 kWh ) ( kWh m 2 suns / day )= 10.5 kW DC ( STC ) 1
Name : 2. Efficient Use of the Solar Resource – Comparing Biodiesel and Solar PV A gallon of diesel produces 139000 BTUs of energy when burned Conversion from kWh to BTUs: 3412 BTU/kWh Acre = 43560 sq. ft.; conversion from feet to meters 0.3048 m/ft Assume: 15% efficiency for solar PV panels, solar insolation average of 5.5 suns per day Compare the maximum amount of energy that can be recovered from the sun in the form of biodiesel versus electricity in one average year on one acre of land. Assume the values in the table are for biodiesel that can be produced in one year. The units for the “yield” column are gallons of biodiesel per acre. Maximum amount of energy that can be recovered from the sun in form of biodiesel = 15,000 gallons/acre/year 15,000 gallons acre / year ( 139,000 BTU gallon ) = 2.085 10 9 BTU acre / year 2.085 10 9 BTU acre / year ( kWh 3412 BTU ) = 6.11 10 5 kWh acre / year Maximum amount of energy that can be recovered from the sun via electricity = 5.5 suns/day 5.5 suns day ( kWh m 2 suns ) ( 0.15 ) 43560 ( ft 2 acre ) ( 0.3048 m ft ) 2 = 3,338.66 kWh day / acre 3,338.66 kWh day acre ( 365 days year ) = 1.22 10 6 kWh acre / year 2
Name : Therefore, electricity via solar PV panels generates about two times the energy than the maximum amount of energy that could be recovered in the form of biodiesel. 3
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Name : 3. Electricity Charges (a) You have net metering in your state and want to install solar PV system. Which of the tariffs above should you select maximize your ROI? Utilizing solar PV system would depend on solar peaks depending on the time of day, and thus I would select the Time of Use (TOU) residential rate to maximize ROI. (b) You have a commercial building using greater than 7500 kWh per month (i.e. Medium Service). Let’s say in one particular month that you use 11235 kWh and your bill comes to $947.56. What was your peak power that month? Medium Service cost = $0.026/kWh & $11.85/mo-kW, monthly use = 11,235 kWh $ 0.026 kWh ( 11,235 kWh ) + $ 11.85 ( x ) = $ 292.11 + $ 11.85 ( x kW ) = $ 947.56 $ 11.85 ( xkW ) = $ 947.56 $ 292.11 = $ 655.45 ( xkW ) = $ 655.45 $ 11.85 = ¿ 55.3 kW was the peak power that month. 4
Name : 4. Using the Solar Spectrum Below are the bandgaps of some materials used for solar cells. You now want to make a solar cell with three stacked PN-junctions. Pick three materials from the figure and explain very briefly the order in which you would stack them from top (facing the sun) to bottom (backside of the panel). Band gap energy and wavelength are inversely proportional to each other. To generate the maximum energy possible utilizing the materials possible, I would want the higher bandgap materials directly towards the sun, because those lead to a higher voltage generation and lower current, and the lower bandgaps on the bottom of the stack, as those generate lower voltages at higher currents. Therefore: First stack material, facing the sun: highest band gap energy: a-Si Second stack material, in the middle: medium band gap energy: CdTe Final stack material, on the underside: lowest band gap energy: CIS 5