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azamat
3 years ago
7

A rectangular channel 6 m wide with a depth of flow of 3m has a mean velocity of 1.5 m/sec. The channel undergoes a smooth, grad

ual contraction to a width of 4.5 m. (a) Calculate the depth and velocity in the contracted section (Hint: y2 is between 2.6~2.9m). (b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction. In each case, identify any assumptions that you made.

Engineering
1 answer:
k0ka [10]3 years ago
3 0

Answer:

The answer for (1) 2.07 m/s (2) F= 63.8kN

Explanation:

Solution

By applying the energy equation from the approach  section 1 to the contacted section 2 with negligible head losses and assuming a horizontal bottom

we have y₁ + V₁²/ 2g = y₂ +  q₂²/2gy₂²

Where

q is =flow rate

y = the depth

g = the acceleration

Kindly find an attached copy of the final solution for this question.

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Explanation:

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we have δL = 12.5x10⁻⁶

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putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

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12.5X10⁻⁶ *115-15 * 20

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When you arrive at an intersection with a stop sign in your direction, if there is no marked stop
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Answer:

Explanation:

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Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

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Z = 549.21Ohms

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3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

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= 0.2557

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1/6.28×f×4.8×10^-6 = 4.082f

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1 = 0.000030144×f×4.082×f

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