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torisob [31]
3 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]3 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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weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

volume  =  A'\times H

where A' = 61% of A

              = 0.61\times 2667 = 1626.87 sq ft

volume  =  1626.87 \times (\frac{14}{12} ft)

               =1898.015 ft^3

in\ m^3 = \frac{ 1898.015}{35.315} =   53.7457 m^3

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On the position time curve, if the slope of a tangent at a point is positive, that means:
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Answer:

  C. the object is moving forward

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A positive slope means position is increasing when time is increasing. Generally, increasing position is "moving forward."

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Air supplied to a pneumatic system is supplied through the C. Actuator

Explanation

Pneumatic systems are like hydraulic systems, it is just that these systems uses compressed air rather than hydraulic fluid.  Pneumatic systems are used widely across the industries. these pneumatic systems needs a constant supply of compressed air to operate. This is provided by an air compressor. The compressor sucks in air at a very high rate from the environment and stores it in a pressurized tank. the Air is supplied thereafter with the help of a actuator valve that is a more sophisticated form of a valve.

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3 years ago
Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba
Tpy6a [65]

Answer:

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Explanation:

The optimum cutting speed for the minimum cost

V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)

Where,

C,n = Taylor equation parameters

T_h =Tool changing time in minutes

C_e=Cost per grinding per edge

C_m= Machine and operator cost per minute

On comparing with the Taylor equation VT^n=C,

Tool life,

T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)

Given that,  

Cost of operator and machine time=\$40/hr=\$0.667/min

Batch setting time = 2 hr

Part handling time: T_h=2.5 min

Part diameter: D=73 mm =73\times 10^{-3} m

Part length: l=250 mm=250\times 10^{-3} m

Feed: f=0.30 mm/rev= 0.3\times 10^{-3} m/rev

Depth of cut: d=3.5 mm

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, C_e= \$20/15+2=\$3.33/edge

Tool changing time, T_t=3 min.

C= 80 m/min

n=0.130

(a) From equation (i), cutting speed for the minimum cost:

V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}

\Rightarrow 47.7 m/min

(b) From equation (ii), the tool life,

T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}

\Rightarrow T=53.4 min

(c) Cycle time: T_c=T_h+T_m+\frac{T_t}{n_p}

where,

T_m= Machining time for one part

n_p= Number of pieces cut in one tool life

T_m= \frac{l}{fN} min, where N=\frac{V_{opt}}{\pi D} is the rpm of the spindle.

\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc

So, the number of parts produced in one tool life

n_p=\frac {T}{T_m}

\Rightarrow n_p=\frac {53.4}{4.01}=13.3

Round it to the lower integer

\Rightarrow n_p=13

So, the cycle time

T_c=2.5+4.01+\frac{3}{13}=6.74 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc

(e) Total time to complete the batch= Sum of setup time and production time for one batch

=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times4.01}{457}=0.4387=43.87\%

Now, for the cemented carbide tool:

Cost per edge,

C_e= \$8/6=\$1.33/edge

Tool changing time, T_t=1min

C= 650 m/min

n=0.30

(a) Cutting speed for the minimum cost:

V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min [from(i)]

(b) Tool life,

T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min [from(ii)]

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc

n_p=\frac {7}{0.53}=13.2

\Rightarrow n_p=13 [ nearest lower integer]

So, the cycle time

T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc

(e) Total time to complete the batch=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.53}{275.5}=0.0962=9.62\%

Similarly, for the ceramic tool:

C_e= \$10/6=\$1.67/edge

T_t-1min

C= 3500 m/min

n=0.6

(a) Cutting speed:

V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}

\Rightarrow V_{opt}=2105 m/min

(b) Tool life,

T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc

n_p=\frac {2.33}{0.091}=25.6

\Rightarrow n_p=25 pc/tool\; life

So,

T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc

(e) Total time to complete the batch

=2\times60+ {50\times 2.63}=251.5 min=4.19 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.091}{251.5}=0.0181=1.81\%

3 0
3 years ago
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Radda [10]

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d) 0.1105 m³/s

Explanation:

See attached pictures.

3 0
3 years ago
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