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3 years ago

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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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Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase

ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite ( $\% V_{Gr}$ ) is determined by the following expression:

$\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%$

Where:

$V_{Gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{Fe}$ - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

$V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}$

$V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}$

Where:

$m_{Gr}$ , $m_{Fe}$ - Masses of the graphite and ferrite phases, measured in grams.

$\rho_{Gr}$ , $\rho_{Fe}$ - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

$\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%$

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

$m_{Gr} = \frac{2.5}{100}\times (100\,g)$

$m_{Gr} = 2.5\,g$

$m_{Fe} = 100\,g - 2.5\,g$

$m_{Fe} = 97.5\,g$

If $m_{Gr} = 2.5\,g$ , $m_{Fe} = 97.5\,g$ , $\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}$ and $\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}$ , the volume percent of graphite is:

$\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%$

$\% V_{Gr} = 91.906\,\%$

The volume percent of graphite is 91.906 per cent.

6 0

3 years ago

What flight patterns do groups of birds utilize and why?

lara31 [8.8K]

Birds mostly fly in a V because the lead bird cuts al of the wind and has the hardest work and then after a while the lead bird goes to the back and another bird takes his place cutting all the wind for the other birds

7 0

3 years ago

A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a

nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$

$T_{1}$ = 20+460 = 480°R

$T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}$ = 533.25°R

Pressure at the inlet of compressor at isentropic condition

$P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}$

$P_{a}$ = $(10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}$ = 14.45 psia

$P_{2}= 8P_{a} = 8(14.45) = 115.6 psia$

4 0

3 years ago

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Explain the conductivity results observed for ionic compounds in the solid state and in aqueous solution.i.Solid ionic compounds

Svetach [21]

Answer:

Aqueous solution of ionic compounds conduct electricity while solid ionic compounds don't.

Explanation:

Ionic compound conduct electricity when liquid or in aqueous solution that is resolved in water because the ionic bonds of the compound become weak and the ions are free to move from place to place.

Ionic compounds don't conduct electricity while in solid state because the ionic bonds are to strong and ions cannot move around with lack of space for movement which makes the electric conductivity zero.

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Document the XSS stored exploit script: Use the View Source feature of the web page and create a screenshot of the few lines cod

Natali [406]

Answer:

Hold on let me ask my brother

Explanation:

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