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SAN JOSÉ STATE UNIVERSITY E10 Introduction to Engineering E10 Introduction to Engineering | Energy Power |
Page 1 of 5 HW #1: Energy/Power Homework Spring 2022 Due: Wed. Feb. 16, 2022 – upload to Canvas by 11:59 pm
Total: 30 points
Name #1:__ _____ Student ID:_____________________________ Name #2:_______
___ ____________________ Student ID:_________ ____________________ Circle LECTURE Section: Sect.01 (12:00) or Sect.02 (1:30) You should complete the homework with a partner
. Partners MUST work TOGETHER to complete it. This DOES NOT mean one person does one problem and another person does another problem. Upload only one file per group. Make sure both names are written on the first page (above). Both partners will get the same score. No late homework will be accepted
Grading policy: •
Put BOTH names on the homework and upload ONE solution. No more than two people may be partners on a homework set. If two sets of homework indicate that there has been collaboration across teams, both teams will earn zero credit. Feel free to come to office hours if you need help. •
No late homework will be accepted
. Make sure you allow enough time to upload (30 min. before the due date) •
Work must be shown. No credit will be given when no work is shown—even for a correct answer
! •
Write down the correct equation first, then plug in the values for the variables. •
Proper units must be included in all answers. •
Some problems do not have a unique numerical answer. For these types of problems, grading is based on the accuracy of your calculation (use of correct units, etc.), your approach, and whether your numerical answer is within a reasonable range. For example, you need to use the energy contained in a candy bar to solve a problem in this homework assignment. You will need to do some research to find a reasonable value for the energy contained in a candy bar. Problems: 1
. Mark ‘N’ if a wrong type of units is used. Mark ‘Y’ otherwise. (1 point each)
a)
N A typical size banana gives you about 2.5 watts of energy. b)
N It takes 10kW to keep a 10W light bulb on for one hour. c)
_N__ It takes 100 HP to lift a 1000 kg weight from ground to the 4
th
floor. 2.
A world class runner can run long distances at a pace of 10 km/hour. That runner expends 700 kilocalories of energy per hour. a)
Find the power (in Watts [W]) the runner is exerting while running. (3 points)
1 Kilocalories = 4184J = 700 Kilocalories / h =700*4184J / 3600s 815.56W
SAN JOSÉ STATE UNIVERSITY E10 Introduction to Engineering E10 Introduction to Engineering | Energy Power |
Page 2 of 5 7 bars 815.56 J/s
Answer: b)
Find the total energy (in Joules [J]) exerted by the runner in a 10 km run. (3 points)
v = 10 km/hr, W = 700 kilocalorie / hr v = d / t t = d / v = 10km / 10km / hr = 1hr 700 kilocalorie* 4184 J / kilocalorie = 2,928,800 J Answer:
c)
How many Milky Way (Original Single 52.2g) chocolate bars does the runner need to buy to supply the amount of energy to complete a half-marathon (13.1 miles)? (Cite your source for the number of calories in a Milky Way bar) (3 points)
1 bar = 236 kilocalories
(
https://www.calorieking.com/au/en/foods/f/calories-in-chocolate-bars-original-milky-
way-bar/ympalrS0TxCwImvuQP3vEw
) 1 mile = 1.60934 km, 13 km = 13 miles*1.60943km/mile = 20.92km Given: Runner uses 700 Kilocalories per hour, v = 10km/hr Distance = velocity*time, time=distance/velocity t=20.92km/10km/hr = 2.092 hr 2.092hr*700kilocalories/hr=1,464kilocalories 1464 kilocalorie / 236 kilocalorie / bar = 6.2 bars Answer: 3.
You are to determine the power rating of the motor in an elevator system. The elevator (with a full load) weights 2000 kg and is required to move upward 2 m/sec at constant speed. The lifting mechanism is 80% efficient (i.e., 20% of the power generated by the motor is wasted in the mechanism, e.g., for overcoming friction). What is the minimum output power rating in HP
of the motor that meets the requirement? (4 points) 2,928,800 J
SAN JOSÉ STATE UNIVERSITY E10 Introduction to Engineering E10 Introduction to Engineering | Energy Power |
Page 3 of 5 65.68HP 1 HP = 746J/s PE/s = mgh/s =2000kg*9.8m/s/s*2m/s =39,200J/s HP = 39,200J/s/746J/s = 52.55 HP Efficiency= out power/in power 80% = 52.55 HP / xHP xHP = 52.55/0.8 = 65.68 HP
Answer: 4.
The power requirement of an electronic portable device is 1.5Amp at 12 volts. What is the minimum number of AA size batteries required for powering this device? Assume one AA battery is capable of supplying up to 1.5 V and 1 Amp of current. Draw a picture of your answer. (Hint: How should these batteries be connected? Series? Parallel? Both?) (4 points) P = V*I 1.5Amp * 12V = 18W Series: P = (V1+V2…+Vn) * I Parallel: P = V * (I1 + I2 + I3…+ In) To meet voltage requirements, 12V/1.5V/AA = 8 AA batteries in series. To meet current requirements, connect a pair of 8 AA batteries in series in parallel 1.5V * 8 = 12V P = 12V * (1Amp+1Amp) = 24W To meet the power requirement we need a pair of 8 cells connected in series that are then connected in parallel. 16 batteries Motor I
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SAN JOSÉ STATE UNIVERSITY E10 Introduction to Engineering E10 Introduction to Engineering | Energy Power |
Page 4 of 5 Answer: 5.
(a) In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed lighting the lab during the 3 hour lab period. Calculate the amount of energy saved if all bulbs are replaced by CFL (spiral) bulbs (P=25W). (3 points) W = J/s Wtotal = 50*100W + 30*60W*(3600s*3) = 73,440,000 J Wtotal CFL = 25W*(50+30)*(3600s*3) = 21,600,000 J Saved = 73,440,000-21,600,000 = 51,840,000 J Answer: (b) Continuing from (a), calculate the amount of natural gas (ft
3
) needed to provide the energy to keep the lights (long + regular) in the E391 lab on overnight for a month (10 hours per day for 30 days)? In other words, how much extra energy (in terms of cubic feet of natural gas) does SJSU have to purchase when we don’t turn off the lights in a lab. (
Note
: you’ll have to look up the energy in one ft
3
of natural gas
and CITE YOUR SOURCE
) (3 points) 1 (ft^3) Natural gas provides 1000 BTU energy
(
https://www.tulsagastech.com/measure.html
) 1000 BTU = 1,055,000 J Need 21,600,000 J for 3 hours / 3 = 7,200,000 J / hr 7, 200,000 J/hr * 10hr * 30 = 2,160,000,000 J 1.)
73,440,000 J 2.)
51,840,000J
SAN JOSÉ STATE UNIVERSITY E10 Introduction to Engineering E10 Introduction to Engineering | Energy Power |
Page 5 of 5 2,160,000,000 J / 1,055,000 J/ft^3 = 2,047.39 ft^3
Answer:
6.
Use Excel and the following empirical relationship to plot the fuel consumption in km/liter for a particular automobile. Note V is the speed of the car in kph and the given relationship is valid for 25≤ V≤ 100 kph. Include your properly labeled graph in your result. Around what speed are you getting the best fuel efficiency? (4 points) Fuel Consumption
(km/l) = !"#" ∗ &
’!"(&
!.##
Fuel efficiency is best at around 40kph 2,047.39 ft^3