Answer: 5.35m
Explanation:
By using energy equation:
[tex]\frac{P_1}{\gamma}+z_1+\frac{v_1^{2} }{2g} =\frac{P_2}{\gamma}+z_2+\frac{v_2^{2} }{2g}+h_{L}[/tex]
[tex]\gamma=specific weight[/tex]
[tex]v_{1} =\frac{Q_1}{A_1} =\frac{0.2}{\frac{\pi }{4} \times 0.24^2} =4.42 m/s\\v_{2} =\frac{Q_2}{A_2} =\frac{0.2}{\frac{\pi }{4} \times 0.2^2} =6.37 m/s[/tex]
[tex]h_{L}=\frac{P_1-P_2}{\gamma}+z_1-z_2+\frac{v_1^{2}-v_2^{2} }{2g}[/tex]
[tex]h_{L}=\frac{(8-7.3)\times 100 }{9.81} +0+5+\frac{4.42^2-6.37^2}{2\times 9.81}[/tex]
[tex]h_L=7.135+3.927\\h_L=11.062m[/tex]
exit velocity head = [tex]\frac{v_2^{2} }{2g}[/tex]=2.068m
head loss as a function of exit velocity head is=[tex]\frac{11.062}{2.068}[/tex]
[tex]h_L=K\times V_e[/tex]
head loss as a function of exit velocity head =5.35m
