Answer:
Height=14.25 m
Discharge=[tex]0.8155 m^{3}/s[/tex]
Explanation:
Bernoulli’s equation for the two points, let’s say A and B is given by
[tex]\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}+z_A=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}+z_B+h_{L(A-B)}[/tex]
But since the elevations are the same then
[tex]\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}+h_{L(A-B)} [/tex]
Since [tex]h_{L(A-B)}=\frac {0.9v^{2}}{2g}[/tex] then
[tex]\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}=\frac {100}{\rho 9.81}+\frac {(v_B)^{2}}{2g}+\frac {0.9v^{2}}{2g}[/tex]
From continuity equation
[tex]A_AV_A=A_BV_B=0.25\pi D^{2}V_A=0.25\pi d^{2}V_B hence [tex]D^{2}V_A= d^{2}V_B[/tex] and substituting 0.4 for D and 0.25 for d
[tex]V_B=\frac { D^{2}V_A}{d^{2}}=\frac { 0.4^{2}V_A}{0.25^{2}}=2.56V_A[/tex]
[tex]\frac {P_A}{\rho g}=\frac {100}{\rho 9.81}+\frac {(v_B)^{2}}{2g}+\frac {0.9v^{2}}{2g}-\frac {(v_A)^{2}}{2g}[/tex] and substituting [tex]V_B[/tex] with [tex]2.56V_A[/tex] we have
[tex]\frac {P_A}{\rho g}=\frac {100}{\rho 9.81}+\frac {(2.56V_A)^{2}}{2g}+\frac {0.9v^{2}}{2g}-\frac {(v_A)^{2}}{2g}[/tex]
Since [tex]\rho[/tex] is taken as 1 then
[tex]\frac {P_A}{9.81}=\frac {100}{9.81}+\frac {(2.56V_A)^{2}}{2*9.81}+\frac {0.9v^{2}}{2*9.81}-\frac {(v_A)^{2}}{2*9.81}[/tex]
v=6.49 m/s
Since discharge=[tex]AV=0.25\pi 0.4^{2}*6.49=0.8155 m^{3}/s[/tex]
[tex]H=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}=0.24+14.01=14.25 m[/tex]
