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Stolb23 [73]
4 years ago
13

The contractor was awarded a contract to construct a parking lot for a small shopping center. The contractor plans to use single

pusher scraper for site grading with a chain loading method.
It has been decided there will be total 11 scrapers, and for each scraper the estimated cycle time is 7 mins.

a) how many pushers are required to serve a fleet of 11 scrapers.b) the expected production of the scraper fleet if only 1 pusher is available. (production of a single scraper assuming adequate pusher support is 317 BCY/hr.
Engineering
1 answer:
musickatia [10]4 years ago
4 0

Additional information

Loading Method Single Pusher Tandem Pusher

Back-Track 1.5 1.4

Chain or Shuttle 1.0 0.9

Answer:

(a) 3 back-trucks and 2 chains

(b) 2491 BCY/hr

Explanation:

(a)

The number of scrapers per pusher

Back-track=\frac {7}{1.5}=4.667

Chain=\frac {7}{1}=7

Number of pushers needed

Back-truck=\frac {11}{4.667}=2.357\approx 3

Chain=\frac {11}{7}=1.57\approx 2

(b)

Production=(Number of pushers* Number of scrappers* Production per scrapper)/ Required number of pushers

Production=\frac {1}{1.4}\times 11 \times 317 BCY/hr= 2490.7\approx 2491 BCY/hr

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A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is igno
olga55 [171]

Answer:

critical clearing angle = 70.3°

Explanation:

Generator operating at = 50 Hz

power delivered = 1 pu

power transferable when there is a fault = 0.5 pu

power transferable before there is a fault = 2.0 pu

power transferable after fault clearance = 1.5 pu

using equal area criterion to determine the critical clearing angle

Attached is the power angle curve diagram and the remaining part of the solution.

The power angle curve is given as

= Pmax sinβ

therefore :  2sinβo = Pm

                   2sinβo = 1

                   sinβo = 0.5 pu

                   βo = sin^{-1} (0.5) = 30⁰

also ;   1.5sinβ1 = 1

               sinβ1 = 1/1.5

               β1 = sin^{-1} (\frac{1}{1.5} ) = 41.81⁰

∴ βmax = 180 - 41.81  = 138.19⁰

attached is the remaining solution

The critical clearing angle = cos^{-1} 0.3372  ≈ 70.3⁰

3 0
3 years ago
(a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as
Llana [10]

Answer:

Explanation:

a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2

We are to show how Two-2-to -1 multiplexers could be connected to form 3-to-1 MUX

If AB = 00 select I_o

If AB = 01 select I_1

If AB = 1_(B is don't care), select I_2

However, the truth table is attached and shown in the first file below.

Also, the free- body diagram for 2- to - 1 MUX is shown in the second diagram attached below.

b) We are show how  two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.

The perfect illustration showing how they are connected in displayed in the third free-body diagram attached below.

Where ; I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7 are the inputs of the multiplexer and Z is the output.

c)  Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.

For  four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs, we have a perfect illustration of the diagram in the last( which is the fourth) diagram attached below.

Where ; I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7 are the inputs of the multiplexer and Z is the output

5 0
4 years ago
7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 v
Naddik [55]

Answer:

23.34 seconds

Explanation:

Flow rate = 1500

Arrival = 800 vehicle per hour

Cycle c = 60 seconds

Dissipation time = 10 seconds

Arrival time = 800/3600 = 0.2222

Rate of departure = 1500/3600 = 0.4167

Traffic density p = 0.2222/0.4167 = 0.5332

Real time = r

r + to + 10 = c

to = c-r-10 ----1

t0 = p*r/1-p ----2

Equate both 1 and 2

C-r-10 = p*r/1-p

60-r-10 = 0.5332r/1-0.5332

50-r = 0.5332r/0.4668

50-r = 1.1422r

50 = 1.1422r + r

50 = 2.1422r

r = 50/2.1422

r = 23.34 seconds

7 0
3 years ago
A work cell is currently operated 2000 hr/yr by a human worker who is paid an hourly rate of $23.00, which includes applicable o
sashaice [31]

Answer:

a) 25000 pcs/yr

b) cost per part produced is $ 1.84 per pc

c) Cpc = $ 1.365 per pcs

Explanation:

a)

the relation to calculate the number of parts produced annually by manual process is;

Q = Hw / Tc

Hw is the hourly rate ( 2000hr/yr) and Tc is the cycle time ( 4.8 min)

so we substitute

Q = (2000 × 60) / 4.8

= 120000 / 4.8

= 25000 pcs/yr

b)

cost per part produced

the relation to calculate the cost per part produced is expressed as;

Cpc = $23(Hw) / Q

Cpc is the cost per part produced

so we substitute

Cpc = $23(2000) / 25000

= 46000 / 25000

= $ 1.8 per pc

therefore cost per part produced is $ 1.84 per pc

c)

for the robot cell, at a service life of 4 years and a 10% rate of return , the factor is expressed as;

f = [r(1 + r)^t] /  [((1 + r)^t ) - 1 ]

our rate r = 10% = 0.1 and our t = 4

so we substitute

f = [0.1 (1 + 0.1)^4] /  [((1 + 0.1)^4 ) - 1 ]

f = 0.14641 / 0.4641

f = 0.3155

now we find the total cost

TC = 120000(0.3155) + 2000(0.3) + 2500

TC =  $ 40,960

next we find the parts produced annually

Q = (2000 × 60) / 4

Q = 120,000 / 4

Q = 30000 pcs/yr  

finally we find the cost per part produced;

Cpc = TC / Q

we substitute

Cpc = 40,960 / 30000

Cpc = $ 1.365 per pcs

5 0
3 years ago
Question 2) A material that is malleable can be defined as:
Reil [10]

Answer:

A.

Explanation:

Able to return to it's original shape after distortion.

7 0
3 years ago
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