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3 years ago

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Since you became discouraged not being able to find a job in the San Diego area, you enlarged the area in which you looked for a

job with the result that the City of Porterville, in Tulare County, offered you a position as junior engineer. You are in the process of interviewing with the senior engineer for the city who has presented you with a situation to test your knowledge level. If water flows with a velocity of 9.85 ft/s in a rectangular channel that is 3 meters wide and has a depth of 9.84 feet.
Required:
What is the change in depth and water surface elevation that will result if a gradual contraction in the channel width to 8.5 feet occurs?

1 answer:

nordsb [41]3 years ago

5 0

Answer:

need points 48986

Explanation:

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Answer:

Explanation:

Given

scale i.e. $L_r=1:15$

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$(Re)_m=(Re)_p$

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Properties of air

$\nu _{air}=1.57\times 10^{-4} ft/s$

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$\nu _{sea}=1.26\times 10^{-5} ft/s$

$\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )$

$V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )$

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3 years ago

If a 110-volt appliance requires 20 amps, what is the total power consumed?

juin [17]

Answer:

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Explanation:

Use the given relation between current, voltage, and power to find the power requirement:

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2 years ago

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3 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

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3 years ago

The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsq

Citrus2011 [14]

Answer:

Given that

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LHS of above given equation have dimension $[M^oL^{1}T^{-1}]$ .

Now find the dimension of RHS

Dimension of P = $[ML^{-1}T^{-2}]$ .

Dimension of d= $[M^{0}L^{1}T^{0}]$ .

Dimension of μ = $[ML^{-1}T^{-1}]$ .

Dimension of L= $[M^{0}L^{1}T^{0}]$ .

So

$\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}$

$\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]$

It means that both sides have same dimensions.

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4 years ago