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Ivan
3 years ago
5

What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00

5?
Engineering
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

The velocity of flow is 10.0 m/s.

Explanation:

We shall use Manning's equation to calculate the velocity of flow

Velocity of flow by manning's equation is given by

V=\frac{1}{n}R^{2/3}S^{1/2}

where

n = manning's roughness coefficient

R = hydraulic radius

S = bed slope of the channel

We know that for an asphalt channel value of manning's roughness coefficient = 0.016

Applying values in the above equation we obtain velocity of flow as

V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s

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The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content
MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf  

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf  

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}

Therefore, V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}

V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since 1 yd^{3}= 27 ft^{3}

The embankment requires water of  6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

Unit_{weight}=\frac {17451720}{498594}=35.00186 lb

1 gallon is approximately 8.35 yd^{3} hence

\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}

That's approximately 4.2 gallons

7 0
3 years ago
while performing a running compression test how should running compression compare to static compression
algol [13]

Answer:

The idle speed of a running compression should be between 50-75 PSI and that is about half of the static compression.

Explanation:

The Running or Dynamic compression is used to determine how well the cylinder in an engine  is absorbing air, reserving it for the proper length of time, and releasing it to the exhaust. The static or cranking compression test is used to check the sealing of the cylinder. Before performing the running compression test, the static compression test is first performed to rule out other issues like bent valves.

The standard value for the static compression is given by;

Compression ratio * 14.7 = Manufacturers Specification

The running compression should always be half of the static compression.

5 0
3 years ago
Given a two-dimensional steady inviscid air flow field with no body forces described by the velocity field given below. Assuming
kolbaska11 [484]

Answer:

the pressure gradient in the x direction = -15.48Pa/m

Explanation:

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Brainliest need help
insens350 [35]

Answer:

answer c

Explanation:

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The metal control joints used to relieve stresses caused by expansion and contraction in large ceiling or wall expenses in inter
timurjin [86]

Answer:

It’s called Expansion Joints

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