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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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What are the three elementary parts of a vibrating system?
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the three part are mass, spring, damping

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At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

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Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

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\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

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The shear stress on the vertical plane is zero

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3 years ago
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