Question 16.8: Starting from the first principles, show that the theoretica......

Let V_{f} = Velocity of flow at inlet
u_{1}= Peripheral velocity of impeller at outlet
V_{f_{1}}= Velocity of flow at outlet
Φ = Impeller angle at outlet
The inlet and the outlet of the pump and the above notations are shown with other connected notations in Figures 16.11(a) and 16.11(b).

Now, applying Bernoulli’s equation at outlet and inlet of the impeller,
Energy at outlet = Energy at inlet + Work done by the impeller

or              {\frac{p_{1}}{\rho g}}+{\frac{V_{1}^{2}}{2g}}+Z_{1}={\frac{p}{\rho g}}+{\frac{V^{2}}{2g}}+Z+\mathrm{Energy~for~work~to~be~done}     (i)

As shown in the figure, inlet and outlet of the impeller are in the same vertical level, and hence Z = Z_{1} and energy needed to do work is

The expression given by Eq. (i) may now be written as

or             \frac{p_{1}}{\rho g}-\frac{p}{\rho g}\,=\,\frac{V^{2}}{2g}-\frac{V_{1}^{2}}{2g}+\frac{V_{w_{1}}u_{1}}{g}

but       \frac{p_{1}}{\rho g}-\frac{p}{\rho g} = Pressure rise

Hence,  Pressure rise = \quad\frac{V^{2}}{2g}-\frac{V_{1}^{2}}{2g}+\frac{V_{w_{1}}u_{1}}{g}

or            \mathrm{Pressure\ rise}={\frac{1}{2g}}(V^{2}-V_{1}^{2}+2V_{w_{1}}u_{1})          (ii)

From the outlet diagram,

and from the inlet diagram       V_{1}  =  V_{f}
Now, Eq. (ii) may be written as

or           \mathrm{Pressure~rise}=\frac{1}{2g}[V_{f}^{2}-V_{f_{1}}^{2}-(u_{1}-V_{f_{1}}\cot\phi)^{2}+2u_{1}(u_{1}-V_{f_{1}}\cot\phi)]

Since,      V_{w_{1}}=(u_{1}-V_{f_{1}}\cot\phi)      (from outlet velocity diagram)

or       \mathrm{Pressure\ rise}={\frac{1}{2g}}[V_{f}^{2}-V_{f_{1}}^{2}-(u_{1}^{2}+V_{f_{1}}^{2}\cot^{2}\phi-2u_{1}V_{f_{1}}\cot\phi)+(2u_{1}^{2}-2u_{1}V_{f_{1}}\cot\phi)]

or         \mathrm{Pressure\ rise}=\frac{1}{2g}[V_{f}^{2}+u_{1}^{2}-V_{f_{1}}^{2}(1+\mathrm{cot}^{2}\,\phi)]

Since,       (1+\cot^{2}\phi)=cosec^{2}\phi

Therefore,     \mathrm{Pressure\ rise}={\frac{1}{2g}}[V_{f}^{2}+u_{1}^{2}-V_{f_{1}}^{2}\mathrm{cosec}^{2}\phi]