Answer:
Following are the proving to this question:
Explanation:
[tex]\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}[/tex]
using the energy equation for entry and exit value :
[tex]\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}[/tex]
where
[tex]\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}[/tex]
[tex]= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\[/tex]
[tex]= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\[/tex]
[tex]\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\[/tex]
[tex]\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})[/tex]
L.H.S = R.H.S
