Question 12.P.9: Determine the maximum discharge through a circular pipe of 2......
Determine the maximum discharge through a circular pipe of 2 m diameter with a bed slope of 1/1000. Also determine the depth for maximum velocity and the corresponding discharge Chezy’s constant C = 60
Adopting Chezy equation (Refer problem 12.7)
A={\frac{R^{2}}{2}}\ (2\theta-2\sin2\theta),P=2R\theta,R=1,\,Q=2.69 radians
∴ \qquad\qquad A={\frac{1}{2}}~[2\times2.69-\sin\,(2\times2.69)]=3.083~\mathrm{m^{2}},P=2\times1\times2.69=5.38~\mathrm{m} \\ R_{h}=A/P=3.083/5.38=0.573\mathrm{~m} \\ Q=A C~\sqrt{R_{h}~S_{b}}~=3.083\times60~\sqrt{0.573\times\left({\frac{1}{1000}}\right)}=4.43~{\mathrm{m}}^{3}/\mathrm{s}
For maximum velocity: θ = 2.247 radians
A={\frac{R^{2}}{2}}\ (2\theta-2\ \sin\ 2\theta)={\frac{1}{2}}\ (\ 2\times2.247-\sin\ (2\times2.247)) \\ =2.7351\,{\mathrm{m}}^{2} \\ P=2 R\theta=2\times1\times2.247=4.494\ \mathrm{m} \\ R_{h}=A/P=2.7351/4.494=0.6086~\mathrm{m} \\ Q=2.7351\times60\times{\sqrt{0.6086\times{\frac{1}{1000}}}}\,=4.049{\mathrm{~m^{3}/s}}
In case of full area flow: A = π × 1² = πm²
P=\pi D\;\;\;\;∴\;\;\;\;R_{h}=\pi/\pi\;D=1/2=0.5
∴ \qquad\qquad {\bf Q}=\pi\times60~\sqrt{\frac{0.5}{1000}}\,=4.215~{\rm m}^{3}/{\mathrm{s}}(Note: Compare the flows for maximum discharge, maximum velocity and full flow)