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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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150 is the area

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5 0
3 years ago
The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.
ehidna [41]

Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

So

we can say that acceleration= 0.6 x 9.81

acceleration,a=5.88\ \frac{m}{s^2}

We know that

a=\omega ^2A

So now by putting the values

a=\omega ^2A

5.88=\omega ^2 \0.15\times 10^{-3}

\omega =198.09\ \frac{rad}{s}

We know that

  ω= 2πN/60

198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

                                               

       

4 0
2 years ago
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

7 0
2 years ago
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