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3 years ago

6

As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de

signing a new irrigation canal that will be used by the English Pea Farmers Cooperative of Northern Minnesota. The canal will run from Basswood Lake to the pea fields located just south of town. The canal is to be 22 miles in length, unlined, excavated in stiff clay and must handle a flow rate of 13.15 m3/s over a slope of 0.2%. Your Boss wants you to calculate the necessary canal parameters and to also determine if the canal will fit within the 85-foot wide right-of-way established by the Lake County land-use office

1 answer:

katen-ka-za [31]3 years ago

5 0

yes it will

Explanation:

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How do you identify all sensors, functions, and where we can use them?

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Sensor/Detectors/Transducers are electrical, opto-electrical, or electronic devices composed of specialty electronics or otherwise sensitive materials, for determining if there is a presence of a particular entity or function. Many vehicles including cars, trains, buses etc. all use sensors to monitor oil temperature and pressure, throttle and steering systems and so many more aspects.

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2 years ago

Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe

Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe $\left ( D\right )=0.15m$

Velocity of water in pipe $\left ( V\right )=15cm/s$

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Pressure drop is given by hagen poiseuille equation

$\Delta P=\frac{128\mu \L Q}{\pi D^4}$

We have asked pressure Drop per unit length i.e.

$\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}$

Substituting Values

$\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}$

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3 years ago

What is the pressure at the bottom of a 25 ft volume of hydraulic fluid with a weight density of 55 lb/ft3 a. 114.6 psi b. 1375p

Assoli18 [71]

Answer:

d) 9.55 psi

Explanation:

pressure at the bottom is =ρgh

weight density is ρg=55 lb/ft³

h=25ft

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1 ft = 12 inch

pressure at bottom = $\frac{1375}{12^2}$

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2 years ago

Who plays a role in the financial activities of a company?

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Who plays a role in the financial activities of a company?

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3 years ago

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

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3 years ago