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torisob [31]
2 years ago
5

Joe Bruin has a big lawn in front of his house that is 30 meters wide and 20 meters long. Josephine makes him go out and mow the

grass every weekend, so several years ago, he bought a fancy John Deere riding mower , which he loves to ride around his yard. The mower has a gasoline engine that is rated at 20 horsepower. To mow his lawn each week, Joe needs to use the mower for about 30 minutes, and the average power used by the mower during this period is 5 horsepower. Joe has a very healthy lawn that generates a lot of lawn clippings, and after doing the EPSS 101 biofuels lab, Joe gets the idea that 3 if he converted his mower to run on biofuelds instead of gasoline, then maybe he could use biofuels generated from his lawn clippings to power his mowera. Calculate the weekly fuel energy requirements for Joe’s riding mower. (Hint. Don’t forget to include the efficiency of the mower’s internal combustion engine)b. Calculate the chemical energy content of the clippings produced by Joe’s lawn each week, assume solar insolation is 5.62 kWH per square meter per day (Hint. Start with the amount of sunlight the lawn absorbs each week and then use the photosynthetic efficiency of corn to calculate the chemical energy content of the grass that grows each week)c. Use your answer from Part b to calculate the mass of grass clippings produced by Joe’s lawn each week. Given that the density of grass clippings is 667 pounds per cubic yard, how many 50-gallon trash cans would be required to hold the clippings? (Hint: Convert the energy content you calculated for Part b to dry biomass content, and then assume that the grass clippings are 80% water)d. Joe’s first idea is to use solar power to dry the grass clippings. Then, he would mod his riding mower to install a Stirling engine in place of the gasoline engine, and then power the engine by burning the grass clippings. Compare the fuel energy that could be provided by burning the grass clippings in this manner each week to the energy needs of Joe’s riding mower. Independent of the significant air pollution that would be generated by burning the clippings, would this scheme work? Could the clippings power Joe’s mower? Would there be any energy left over for other purposes?e. Josephine works in a biology lab at UCLA and they have recently come up with a magic strain of bacteria that converts cellulose (the dominant organic component of grass clippings) into glucose. To create ethanol fuel, Joe and Josephine put the wet grass clippings into barrels, and then add an equal mass of water, plus a cup of the magic bacteria, and then they let it sit for two weeks while the cellulose is converted to glucose. Then, Joe takes the resulting glucose solution and adds yeast to ferment the solution to produce a 15% ethanol solution, which takes an additional week. Then, every week, he burns additional grass clippings in a large still to distill the 15% ethanol solution to create a 95% ethanol solution. The resulting distillate is clean-burning fuel, which Joe uses to power his riding mower, which he has, of course, modified to run on ethanol. Ignoring the fact that the magic strain of bacteria doesn’t really exist, will this scheme work? Would the clippings provide enough ethanol to power Joe’s mower? Would there be any additional ethanol left over for other purposes, like powering Joe’s car, or to create alcoholic beverages to 4 drink when Joe’s friends come over to watch UCLA football games on Saturday afternoons?f. If Joe could find a way to use biofuels to power his mower, would this whole operation be sustainable? Discuss the needs of Joe’s lawn for water and nutrients. Discuss the demands on Joe’s time. Do you think Joe would have enough time to run his home biofuel operation and also have time to invite his friends over to watch football games on weekends?
Engineering
1 answer:
zysi [14]2 years ago
5 0

<u>Explanation:</u>

5 Horsepower for 30 mins,

(5)(745.7) = 3.7285 KW power delivered

General Efficiency of IC engine = 20%

Power required = \frac{3 \cdot 7285}{0 \cdot 2}=18 \cdot 6425 kw

Energy required per week,

=P × Time = 18.64 × 60 × 30 = 33.5565 MJ

Lawn area = (30) (20) = 600m^{2}

let sunlight hours be 8 hours

Hence, solar power input on lawn,

=5.62×3600 = 20232 kJ/m^{2}/day

energy input in lawn = (600) (20232) (7)

                                  = 84974.4 mJ/week

Chemical efficiency by photosynthesis = 4%

Chemical content in grass = (84974.4) (0.04)

                                            = 3398.97 mJ

Mass of the clippers  \(=(30)(20)(1 \cdot 096)^{2}(667)\)

                                  \(=478632 \cdot 33\) pounds

Removing water content,

dried grass clippings \(=95726.46\) pound

                                    = 11533.25 gallons

Trash cans repaired  

                                     =\frac{11533}{50} =230.66\\=231 cans

By burning the gas, total energy input = 3398.97 MJ × 0.2

                                                                = 679.794 MJ

Efficiency of steeling engine  =  20%

Energy output by engine = 679.794 ×0.2

                                          = 135.96 mJ

Energy required by mover = 33.5565 mJ

Hence, Energy (output) ⇒ energy required

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Answer:

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Explanation:

#include<stdio.h>

int main()

{

int i, time_worked, over_time, overtime_pay = 0;

for (i = 1; i <= 10; i++)

{

 printf("\nEnter the time employee worked in hr ");

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printf("\nTotal Overtime Pay Of 10 Employees Is %d", overtime_pay);

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Output :

Enter the time employee worked in hr 42

Enter the time employee worked in hr 45

Enter the time employee worked in hr 42

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Enter the time employee worked in hr 52

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6 0
2 years ago
simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The
Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

q(x)=\frac{x}{L} q_{0}

σ = 120 MPa = 120*10⁶ Pa

L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\

We can see the pic shown in order to understand the question.

We apply

∑MB = 0  (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6   (↑)

Then, we apply

v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x

Then, we can get the maximum bending moment as follows

M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\  x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m

then we get  

M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}

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We use the formula

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⇒ M = σ*I/y

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y=\frac{h}{2} =\frac{0.3m}{2}=0.15m

If M = Mmax, we have

(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4}   }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}

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Answer:

oof

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I don't know but please don't report me

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If you don't report me!

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