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Airida [17]
3 years ago
10

A pump is used to deliver water from a lake to an elevated storage tank. The pipe network consists of 1,800 ft (equivalent lengt

h) of 8-in. pipe (Hazen-Williams roughness coefficient = 120). Ignore minor losses. The pump discharge rate is 600 gpm. The friction loss (ft) is most nearly Group of answer choicesA. 15
B. 33
C. 106
D. 135
Engineering
1 answer:
Nataly_w [17]3 years ago
7 0

Answer:

h_f = 15 ft, so option A is correct

Explanation:

The formula for head loss is given by;

h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))

Where;

h_f is head loss due to friction in ft

L is length of pipe in ft

Q is flow rate of water in gpm

C is hazen Williams constant

D is diameter of pipe in inches

We are given;

L = 1,800 ft

Q = 600 gpm

C = 120

D = 8 inches

So, plugging in these values into the equation, we have;

h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))

h_f = 14.896 ft.

So, h_f is approximately 15 ft

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Artyom0805 [142]

Answer:

Tempered glass

Explanation:

Tempered or toughened glass is a type of safety glass processed by controlled

thermal or chemical treatments to increase its strength compared with normal glass. Tempering puts the outer surfaces into compression and the interior into tension . Such stresses cause the glass, when broken, to crumble into small granular chunks instead of splintering into jagged shards as plate glass (a.k.a. annealed glass) does. The granular chunks are less likely to cause injury.

4 0
3 years ago
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A Michelson interferometer operating at a 500 nm wavelength has a 3.73-cm-long glass cell in one arm. To begin, the air is pumpe
LekaFEV [45]

Answer:

The number of bright-dark fringe is 42

Solution:

As per the question:

Wavelength of light, \lambda = 500\ nm = 500\times 10^{- 9}\ m

Length of the glass cell, x = 3.73 cm = 0.0373 m

Refractive index, \mu = 1.00028

Now,

To calculate the bright-dark fringe shifts, we use the formula given below:

d_{m} = \frac{2x}{\lambda }\times (\mu - 1)

Now, substituting the appropriate values in the above formula:

d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)

d_{m} = 41.77 ≈ 42

6 0
3 years ago
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

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Answer:

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Explanation:

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