Answer:
a) [tex]h_L=1.17m[/tex]
b)[tex]h_L=1.52m[/tex]
c)[tex]P_1=1.721 kN[/tex]
[tex]P_2=2.236 kN[/tex]
Explanation:
velocities of the pipe;
velocity of small dia pipe
[tex]v_{small}=\frac{Q}{A_{small}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.15^2 }=8.52m/s[/tex]
velocity of larger dia pipe
[tex]v_{large}=\frac{Q}{A_{large}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.25^2 }=3.05m/s[/tex]
a) pressure head loss when water is flowing from large to smaller pipe
[tex]h_L=(\frac{1}{C_c} -1)^2 \times \frac{v^2}{2g}[/tex]
[tex]h_L=(\frac{1}{0.64} -1)^2 \times \frac{8.52^2}{2\times 9.81} = 1.17m[/tex]
b) pressure head loss when water flow from small pipe to large pipe
[tex]h_L= \frac{v^2}{2g}(1-\frac{A_{small}}{A_{large}} )^2[/tex]
[tex]h_L=\frac{8.52^2}{2\times9.81}(1-\frac{\frac{\pi}{4}\times 0.15^2}{\frac{\pi}{4}\times 0.25^2}} )^2= 1.52 m[/tex]
c) power loss in both cases are
[tex]P_1= \rho g Q h_{L1}= 1000 \times 9.81\times 0.15\times 1.17= 1.721 kN[/tex]
[tex]P_2= \rho g Q h_{L2}= 1000 \times 9.81\times 0.15\times 1.52= 2.236 kN[/tex]
