Question 16.10: The impeller of a centrifugal pump has outlet blade angle Φ ......
The impeller of a centrifugal pump has outlet blade angle Φ with the tangent at outer periphery. The pump has no diffuser. The velocity of flow is constant, and the absolute velocity at inlet is radial. Assuming flow through the impeller is frictionless, show that:
\frac{Rise of pressure head in m}{Work done/s/N} = \frac{1}{2}\left\lgroup 1+\frac{V_{f_{1}} cot Φ}{u_{1}}\right\rgroup , with usual notations
Therefore, we have to show that
{\frac{(p_{1}/\rho g)-(p/\rho g)}{(V_{\mathrm{w_{1}}}u_{1}/g)}}={\frac{1}{2}}\left\lgroup1+{\frac{V_{f_{1}}\cot\phi}{u_{1}}}\right\rgroupFrom the above velocity triangle at outlet (Figure 16.13), we get
{\frac{V_{w_{1}}u_{1}}{g}}={\frac{(u_{1}-V_{f_{1}}\cot\phi)u_{1}}{g}} (i)
When there is no loss in the impeller, we have
\frac{V_{w1}u_{1}}{g}=\frac{(u_{1}-V_{f_{1}}\cot\phi)u_{1}}{g}\left\lgroup\frac{p_{1}}{\rho g}+\frac{V_{1}^{2}}{2g}\right\rgroup -\left\lgroup\frac{p}{\rho g}+\frac{V^{2}}{2g}\right\rgroupor \frac{(u_{1}-V_{f_{1}}\cot\phi)u_{1}}{g}=\left\lgroup\frac{p_{1}}{\rho g}+\frac{V_{1}^{2}}{2g}\right\rgroup -\left\lgroup\frac{p}{\rho g}+\frac{V^{2}}{2g}\right\rgroup (ii)
From outlet diagram of Figure 16.13, we obtain
V_{1}^{2}=V_{w_{1}}^{2}+V_{f_{1}}^{2}=V_{f_{1}}^{2}+(u_{1}-V_{f_{1}}\cot\phi)^{2} (iii)
Applying Eq. (iii) in Eq. (ii), we get
\left\lgroup\frac{p_{1}}{\rho g}-\frac{p}{\rho g}\right\rgroup =\frac{(u_{1}-V_{f_{1}}\cot\phi)u_{1}}{g}+\frac{V_{f_{1}}^{2}}{2g}-\frac{V_{f_{1}}^{2}+(u_{1}-V_{f_{1}}\cot\phi)^{2}}{2g}or \left\lgroup\frac{p_{1}}{\rho g}-\frac{p}{\rho g}\right\rgroup =\frac{(u_{1}-V_{f_{1}}\cot\phi)}{g}[2u_{1}-(u_{1}-V_{f_{1}}\cot\phi)]
or \left\lgroup\frac{p_{1}}{\rho g}-\frac{p}{\rho g}\right\rgroup =\frac{(u_{1}-V_{f_{1}}\cot\phi)}{g}[u_{1}+V_{f_{1}}\cot\phi]
or \left\lgroup\frac{p_{1}}{\rho g}-\frac{p}{\rho g}\right\rgroup =\frac{(u²_{1}-V²_{f_{1}}\cot²\phi)}{g} (iv)
Again dividing Eq. (iv) by Eq. (i), we get
{\frac{(p_{1}/\rho g)-(p/\rho g)}{(V_{w_{1}}u_{1}/g)}}={\frac{(u_{1}^{2}-V_{f_{1}}^{2}\cot^{2}\phi)/g}{(u_{1}-V_{f_{1}}\cot\phi)u_{1}/g}}or {\frac{(p_{1}/\rho g)-(p/\rho g)}{(V_{w_{1}}u_{1}/g)}}={\frac{(u_{1}^{2}-V_{f_{1}}^{2}\cot^{2}\phi)/g}{(u_{1}-V_{f_{1}}\cot\phi)u_{1}/g}}=\frac{u_{1}+V_{f_{1}}\cot\phi}{2u_{1}}
or {\frac{(p_{1}/\rho g)-(p/\rho g)}{(V_{w_{1}}u_{1}/g)}}={\frac{1}{2}}\left\lgroup1+{\frac{V_{f_{1}}\cot\phi}{u_{1}}}\right\rgroup
