Respuesta :
Answer:
A. 3.21ft
B. 3.51ft
C. 2.95ft
D. 1.5275ft
Explanation:
A) Q =212 cu.f/s
Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3
Where dc =critical depth, ft
Q= quantity of flow or discharge, ft3/s
B= width of channel, ft (m)
g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2
Now, from the question,
Q = 212 cu.f/s and b=6.5ft
Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)
To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft
B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)
From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2
Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft
C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)
Where A= (B + my)y and T=(B+2my)
Now from the question, B=6.5ft and m=5ft.
Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y
Now, let's plug the value of A and T into the initial equation to derive the critical depth ;
(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)
Which gives;
1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)
Multiply both sides by 6.5 + 10y to get;
1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]
Factorizing this, we get y = 2. 95ft
D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]
Where D is diameter of pipe and Ѳ is angle at critical depth in radians.
Angle not given, so we assume it's perpendicular angle is 90.
Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31
Therefore, dc = (6.5/2) (1 - cos (2578.31))
dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft
