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11Alexandr11 [23.1K]
3 years ago
7

forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fue

l. How long will it takee to fill the tank in the truck?
Engineering
1 answer:
Zina [86]3 years ago
5 0

Answer: 131.75minutes

Explanation:

First if all, we've to find the density of liquid which will be:

= Specific gravity × Density to pure water

= 0.91 × 8.34lb/gallon

= 7.59lb/gallon

Then, the volume that's required to fill the tank will be:

= Load limit/Density of fluid

= 40000/7.59

= 5270.1gallon

Now, the time taken will be:

= V/F

= 5270.1/40

= 131.75min

It'll take 131.75 minutes to fill the tank in the truck.

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A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the
Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1  - cos (\frac{28.65 SSD}{Rv}) )        .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = \frac{65^2}{15(e+0.10)}

solve it we get

e = 2%

3 0
3 years ago
Vehicles begin arriving at an amusement park 1 hour before the park opens, at a rale of four vehicles per minute. The gale to th
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Answer:

a) ≈ 30 mins

b) 8 vpm

Explanation:

<u>a) Determine how long after the first vehicle arrival will the queue dissipate</u>

The time after the arrival of the first vehicle for the queue to dissipate

= 29.9 mins ≈ 30 mins

<u>b) Determine the average service rate at the parking lot gate </u>

U = A / t

where : A = 240 vehicles , t = 30

U = 240 / 30 = 8 Vpm

attached below is a detailed solution of the given problems above

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Solve the problem with conditions if a wall has inner and outer surface temperatures of 16 and 6 C, respectively. The interior a
Angelina_Jolie [31]

Answer:

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Q = 20 \frac{W}{m^2}

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Q = 20 \frac{W}{m^2 K} \, 1 K

Q = 20 \frac{W}{m^2}

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