starting speed of pump if manometer efficiency is 70%

Given :

$D_{1}$=30cm=0.3 m

$D_{2}$=60cm =0.6 m

$n_{mano}$=0.7

tan$\phi=\frac{Vs_{2}}{u_{2}-Vw_{2}}$

i.e tan45$^{\circ}=\frac{2}{U_{2}-Vw_{2}}$

$VW_{2}=u_{2}-2$

$n_{mano}=\frac{g.Hm}{(u_{2}-22)u_{2}}$

0.7=$\frac{9.81\times H_{m}}{(u_{2}-2)u_{2}}$

Hm=0.0714(u2-2)u2

Let N be the minimum speed of the pump required for starting

Vane velocity at exit $V_{2}=\frac{\pi D_{2}N}{60}=\frac{\pi\times 0.6\times N}{60}$=0.03142 N

Vane velocity at inlet $u_{1}=\frac{\pi D_{1}N}{60}=\frac{\pi\times 0.3\times N}{60}$=0.01571 N

Condition for minimum starting speed is

$\frac{u_{2}^{2}-U_{1}^{2}}{2g}=H_{m}=0.0714(u_{2}-2)U_{2}$

$\frac{(0.03142N)^{2}-(0.01571N)^{2}}{2\times 9.81}$=0.0714(0.08142N-2)0.03142N

0.01682N=(0.03142N-2)

N=137.0rpm