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garri49 [273]
3 years ago
11

A seawall with an opening is used to dampen the tidal influence in a coastal area (and limit erosion). The seawall is 2.5 m long

(in the direction perpendicular to the page), the mean sea level on the sea side is 2.2 m above the centroid of the slot. The mean sea level on the land side of the wall is 1.4 m above the centroid of the slot. The sea side tide fluctuates +0.6 m, and the landward tide fluctuates +0.4 m.
The slot extends the entire length of the wall and is estimated to have a discharge coefficient of 0.80.


If the volume of seawater moving from the sea side to the land side of the wall is to be no more than 6 000 m' in 18 hours, what is the maximum allowable height of the slot?


Note: the discharge coefficient is the ratio of the actual flow rate over the predicted flow rate. It mostly accounts for the fact that the average velocity is different from the velocity.
Engineering
1 answer:
raketka [301]3 years ago
3 0

Answer:

The maximum allowable Height of the slot is 11.685mm

Explanation:

The explanation is attached. The approach used is Bernoulli's equation

Download pdf
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A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.
Nataliya [291]

Answer:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

Explanation:

We can assume that the general formula for the drag force is given by:

D= C_D \frac{\rho}{2}V^2 A

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D1} \frac{\rho}{2}V^2 (wh) V

We can assume the same drag coeeficient C_{D1}=C_{D2}=C_{D} and we got:

\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D} \frac{\rho}{2}V^2 (wh) V

\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

\frac{l}{h}=\frac{5}{1.7}2.941

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And we can calculate the power difference like this:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

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Answer:

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