Evaluate the first moment of area around the water surface for the previous problem in which the depth is 1.2 m in a pipe with a diameter of D = 1.8 m (R = 0.9 m).

Question

Accepted Answer

Substituting D = 1.8 and β = 1.910633 rad in Equation A.14 gives, A[latex]h_{c}[/latex] = 0.94794 m³. An alternative is to numerically integrate Equation A.15. Using this option, cosβ is first evaluated as −0.33333 (or −1/3) and the integral is numerically evaluated as 0.65017 so A = (0.9)²1.8(0.65017) =0.94794 m³.

[latex]Ah_{c}=\frac{D}{2}\left(\frac{D^{2}}{6}\sin ^{3}\beta -A\cos \beta \right)[/latex] (A.14)

[latex]Ah_{c}= \int\limits_{Y}^{0}{R\left(\cos \alpha -\cos \beta \right)D\sin \alpha dy } =R^{2}D\int\limits_{0}^{\beta }{\left(\cos \alpha -\cos \beta \right)\sin ^{2}\alpha d\alpha }[/latex] (A.15)

Question A.2: Evaluate the first moment of area around the water surface f...

Related Answered Questions

Find the area of flow if the depth is Y = 1.2 m in a pipe with a 1.8 m diameter. ...

Verified Answer: