It is easiest to create a table of values and calculate the energy consumption piece by piece.

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1 Energy Efficiency in Smart Homes Solutions Question #21: A family s home office typically consists of the following items: a desktop computer, LCD monitor, modem, and a printer. The devices run for 5 hours, 5 hours, 24 hours, and 0.2 hours respectively. What percentage of the office s daily energy consumption can be attributed to phantom power? Solution: We ll use the following equations as a foundation to calculate energy consumption: E t = Daily Energy Consumption = 1 (W ont on + W off t off ) E on = 1 W ont on E off = 1 W offt off It is easiest to create a table of values and calculate the energy consumption piece by piece. Product W on W off t on t off E T E on E off Desktop Computer LCD Monitor Modem Printer Total The percentage we re looking for is the amount of energy that can be attributed to phantom power, Eoff. Therefore, we d like to know this value: E off % Contibution = E off E T 100% 100% = ( ) 100% = 18. 7% E T

2 Question #22: An LCD TV that is off 3 times as long as it is on is plugged into a power strip that reduces the standby power to zero. By what percentage has the energy consumption of the TV been cut? Solution: Let s start by rewriting our givens. We know the on and standby wattage of the LCD TV are 70 watts and 3 watts respectively. The time is not given, but we do know that the TV is off 3 times as long as it is on, so assuming that t is the amount of time on, then 3t is the amount of time off. Let s name two functions, E(t) will be the total energy consumption and Ep(t) will be the total energy consumption with the power strip using the following equation as our starting point. Daily Energy Consumption = 1 (W ont on + W off t off ) Our first equation, E(t) will be easily defined: E(t) = 1 79 (70t + 3(3t)) = t Since the power strip reduces the phantom power to zero, the second term will disappear in Ep(t). E p (t) = 1 70 (70t + 0) = t Now, in order to find the percentage that the energy consumption has been cut. Let s take the following quotient: 70 E p (t) E(t) 100% = t % = 100% = % 79 t 79 It follows that the power strip reduces the energy consumption by % or 11.4%.

3 Question #23: Calculate the maximum efficiency of the solar panel. Solution: In order to calculate the maximum efficiency, we need to apply the following equation. η max = P max E IRF A We have all of the information, but not in the correct units. First, we need the maximum power to be in watts, not kilowatts, so kw = 130 W. Also, we need to calculate the area of the solar panel itself: A = (58.5 in)(26.5 in) = in 2 We would like this to be in m 2. We know 1 m = in. Let s use dimensional analysis: in 2 ( 1 m in ) 2 = m 2 Now we can plug these values into the equation for maximum efficiency. (130 W) η max = ( W = 13% m 2) ( m2 ) Question #24: If the family decides to dedicate 225 ft 2 on their roof for an array of solar panels. What is the maximum efficiency of the system? Solution: We need to first determine how many solar panels will fit on the family s roof, but we need common units first: in 2 ( 1 ft 12 in ) 2 = ft 2 The family has 225 square feet to work with, so the most solar panels we can fit will be: 225 ft2 # of panels = ft2 = 20 where x is the floor function. Which means that the total area of the array is (20)( m 2 ) = Using the efficiency equation again: (130 W) η max = ( W = 0. 65% m 2) ( m2 )

4 Question #25: What are the minimum dimensions (2x by 3x by 1.5) needed to acquire the standard 20% rate for a power rating of 0.5 kw? Solution: To start, 0.5 kw = 500 W. We need to solve for x in order to resolve the unknown dimensions. First, we calculate the area as a function of x. A(x) = (2x)(3x) = 6x 2 Now, let s enter our information into the efficiency equation: (500 W) (0.2) = ( W m 2) (6x2 m 2 ) Let s simplify: 0.2 = 1 12x 2 2.4x 2 = 1 x = Now, we turn to the original expressions for the dimensions: 2 ( 1 1 ) by 3 ( ) by 1.5 = 1. 2 m by 1. 9 m by 1.5 m

5 Question #26: A homeowner decides to replace the lights in a few rooms with more energy efficient bulbs after testing CFLs with a few light. The homeowner had 12 incandescent bulbs, 3 CFLs, and LED track lighting with 3 bulbs. After replacing the incandescent bulbs with CFLs, how much energy was saved over the course of the year? Solution: We re given all that we need to determine how much energy is saved. We re assuming the light bulbs run about 5 hours per day. We need to calculate two quantities, yearly energy consumption before and after replacing the bulbs (Eb and Ea), and find their difference. E saved = E b E a Both Eb and Ea will be calculated similarly using the energy consumption equations as before, but we need to adjust this for yearly cost, so the whole expression will be multiplied by 365. [( ) + (3 14 5) + (3 10 5)] E b = 365 = kwh / yr. [( ) + (3 10 5)] E a = 365 = 438 kwh / yr. E s = E b E a = ( kwh / yr) (438 kwh / yr. ) = kwh / yr.

6 Question #27: Given the situation in the previous problem, how many years will it take to recoup the cost of replacing the incandescent bulbs based on the energy saved? Solution: In order to determine the time it will take to pay off the bulbs with the money saved, we d be interested in solving the following equation: Cost of Replacing = Money Saved from Replacing We replaced 12 bulbs at $3.95 per bulb. In doing so, we lowered our wattage from 60 to 14, a difference of 46 watts. Then, they run for 5 hours per day, and we are charged $ per kwh. The variable t is a factor of days. (60 14 watts)(5 hours) 12($3.95) = ( ) ( $ kwh ) t $47.4 = (0.23 kwh) ( $ kwh ) t $47.4 $(0.23)(0.1284) = t t = days We were asked to find the amount of time in years, so divide t by 365. The result will be approximately 4.4 years.

7 Question # 28: What luminous flux would we expect for an arbitrary bulb rated at some power, W, and luminous efficacy, n? Solution: This question is a matter of determining the mean luminous flux. From the table, we can pick out those values: Bulb Type Power Luminous Efficacy Luminous Flux Incandescent CFL LED To calculate the mean (denoted μ), we can use the following official equation amounting to adding all of the fluxes together and dividing by the amount of fluxes. μ = n i=1 (φ V) i n = lumens

8 Question #29: Choose the best option for the homeowner using a decision matrix if the criteria are weighed as 20%, 35%, and 45% respectively. What is the score of the winning option? Solution: We ll set up a table to easy calculate the scores of each alternative. Criteria Weight Incandescent CFL LED Yearly Cost 20% Lumens / Watt 35% 0 1 2/3 Yearly Energy Consumption 45% Calculating the rating of each piece is relatively simple, first we need to know each of the values we find important: Yearly cost: I: (1.25) = $8.75 CFL: (3.95)(1) = $3.95 LED: (35.95)(1) = $35.95 Lumens per watt: I: 15 lumens CFL: 60 lumens LED: 45 lumens Yearly Energy Consumption: I: 365 ( (60)(5) ) = kwh CFL: 365 ( (14)(5) ) = kwh LED: 365 ( (10)(5) ) = kwh Now we calculate the ratings: Yearly cost (min): I: = 0.85 CFL: = 1 LED: = 0 Lumens per watt (max): I: = 0 CFL: = 1 LED: = 2/3 Yearly Energy Consumption (min): I: = 0 CFL: = 0.92 LED: = 1 Finally, we multiply the ratings by the weights to find. Yearly cost I: (20)(0.85) = 17 CFL: (20)(1) = 20 LED: (20)(0) = 0 Lumens per watt I: (35)(0) = 0 CFL: (35)(1) = 35 LED: (35) ( 2 3 ) = 70/3 Yearly Energy Consumption I: (45)(0) = 0 CFL: (45)(0.92) = 41.4 LED: (45)(1) = 45 To find the total scores, we apply this equation: P i = m P ij W j j=1

9 Which essentially tells us to sum up the scores for each alternative. P I = P CFL = 3 P Ij j=1 3 P CFLj j=1 W j = = 70 W j = = 96.4 P LED = 3 P LEDj j=1 W j = = The winner is the CFL bulbs with 96.4 total points.

10 Question #30: Consider a light bulb that sells for $8.00 per bulb. If the homeowner decides to use this bulb in 5 sockets and has a budget of $100 per year, determine the necessary lifetime of the bulb to make the switch a good financial decision when considering only the cost of the bulbs. Solution: To determine the minimum lifetime, let s set up an inequality using the floor function: Up Front Cost + Replacements Budget 5($8.00) + 5($8.00) 8760 x $100 Since a floor function is involved, we need to be careful naming solutions. Let s instead solve: 5($8.00) + 5($8.00)n = $100 This will give us the bounds on the floor function and which one to use to determine a solution. $40.0n = $60 n = 1.5 We find that n is a rational number, not an integer. Since we re dealing with an inequality, we should solve for x knowing that n cannot equal 2. Let s find out which value of x forces n to < 2 x 4380 < x After simplifying, we see that x cannot be 4380 hours or the homeowner will exceed his/her budget, so 4381 will be the minimum lifetime we re looking for.