Let,

$D_1 :\textit{Diameter of impeller at inlet} = 2 \hspace {0.05cm} \times\hspace{0.05cm}R_1$

$D_2 :\textit{Diameter of impeller at outlet} = 2 \hspace {0.05cm} \times\hspace{0.05cm}R_2$

$N : \textit{Speed ofimpeller in rpm}$

$u_1 : \textit{Tangential blade velocity at inlet} = wR_1 = (\frac{2 \pi N}{60})R_1$

$u_2 : \textit{Tangential blade velocity at outlet} = wR_2 = (\frac{2 \pi N}{60})R_2$

$V : \textit{Absolute velocity}$

$V_r : \textit{Relative velocity}$

$V_f : \textit{Velocity of flow}$

$V_w ; \textit{Velocity of whirl}$

$\alpha_1 : \textit{Angle mode by absolute velocity V_1 at inlet}$

$\theta : \textit{Inlet angle of vane}$

$\phi : \textit{Outlet angle of vane}$

$\beta : \textit{Discharge angle of absolute velocity at outlet}$

$\textit{Angular momentum} = mass \times tangential velocity \times Radius$

$\textit{Angular momentum entering the impeller per sec} = m.V_{w1}.R_1$

$\textit{Angular momentum leaving the impeller per sec} = m.V_{w2}.R_2$

$\textit{Torque transmitted} = \textit{rate of change of angular momentum}\\ \hspace{3cm}= m.V_{w2}.R_2 - m.V_{w1}.R_1\\ \hspace{3cm}= \frac{w}{g}(V_{w2}.R_2 - V_{w1}.R_1)$

Since the work done in unit time is given by the product of torque and angular velocity

W.D per sec = Torque x W

$\hspace{1.8cm}=\frac{w}{g}(V_{w2}.R_2w - V_{w1}.R_1w)$

But $R_2w = u_2$ and $R_1w = u_1$

W.D per sec = $\frac{w}{g}(V_{w2}u_2.V_{w1}u_1)$

Work done by impeller per N weight of liquid per sec,

W.D = $\frac{1}{g}(V_{w2}u_2-V_{w1}u_1)$

But $V_{w1} = 0$ since entry is radial

W.D per N weight per sec = $\frac{V_{w2}.u_2}{g}$