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.
1 answer:
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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Answer:
w = 10.437 kips
deflection at 1/4 span 20.83\E ft
at mid span = 1.23\E ft
shear stress 7.3629 psi
Explanation:
area of cross section = 18*76
length of span = 32 ft
moment = 334 kips-ft
we know that
moment = load *eccentricity
334 = w * 32
w = 10.437 kips
deflection at 1/4 span
= 20.83\E ft
at mid span
shear stress
Answer:
x = 93.8 m.
Explanation:
During the entire the reaction time interval, the vehicle continues moving at the same speed that it was moving, i.e., 60 mi/hr.
In order to calculate the distance in meters, travelled at that speed, it is advisable first to convert the 60 mi/hr to m/seg, as follows:
Applying the definition of average velocity, we can solve for Δx, as follows:
Δx = 26.8 m/s* 3.5 s = 93.8 m