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iragen [17]
3 years ago
10

5.5 A scraper with a 275 hp diesel engine will be used to excavate and haul earth for a highway project. An evaluation of the jo

b-site conditions indicates the scraper will operate 40 min./hr. For this project it is anticipated that the total cycle time will be 20 min. for a round trip. Previous job records show the scraper operated at full power for the 1.5 min. required to fill the bowl of the scraper and at 80% of the rated hp for the balance of the cycle time. Calculate the gallons per hour for fuel consumption of the scraper.
Engineering
1 answer:
natulia [17]3 years ago
5 0

Answer: Fuel consumption per hour of the scrapper is 6 gal/hr

Explanation:

Given that;

Rated power = 275 hP

The scraper will work = 40min per hr

anticipated total cycle time = 20min

Previous Scrapper operated full power for 1.5min at 80% hP

First we find the Engine factor;

filling the bucket =  1.5/20 * 100% = 0.075

rest of cycle = (20-1.5)/20 * 80% =  0.74

so total engine factor = 0.075 + 0.74 = 0.815

now Time factor will be 40/60 = 0.6666 =

therefore operating factor = 0.6666 * 0.815 = 0.543

Now we know that for a diesel engine, a range fuel consumption = 0.04 gal/hP

so

Fuel consumption per hour of the scrapper is;

= 0.543 * 275 * 0.04

= 5.97 = 6 gal/hr

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ryzh [129]

Answer:

The Question is incomplete, the complete question is as follows:

<em>Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F? </em>

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

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Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

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Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

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Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

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Maximum density = 25 veh/mi/ln

Density = V​​​​​​p /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

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LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

V​​​​​​p = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * f​​​​​​HV * f​​​​​​P​​​) = 2668.5

f​​​​​​HV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (E​​​​​​T -1)) = 0.754

E​​​​​​T = 4.26 ~ 3.5

<em>For 4% upgrade and 10% trucks with E​​​​​​T = 3.5, length of the grade is Greater than 1.0 miles</em>

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