Find the rise in pressure in the impeller of a centrifugal pump through which water is flowing at the rate of 15 litres/s. The internal and external diameters of the impeller are 20 cm and 40 cm respectively. The widths of impeller at inlet and outlet are 1.6 cm and 0.8 cm. The pump is running at 1200 r.p.m. The water enters the impeller radially at inlet and impeller vane angle at outlet is 30°. Neglect losses through the impeller.

$$ u_1 = \frac{\pi D_1 N}{60} = \frac{\pi \times 0.2 \times 1200}{60} \approx 12.566 \, \text{m/s} $$ $$ u_2 = \frac{\pi D_2 N}{60} = \frac{\pi \times 0.4 \times 1200}{60} \approx 25.133 \, \text{m/s} $$

$$ V_{f1} = \frac{Q}{\pi D_1 b_1} = \frac{0.015}{\pi \times 0.2 \times 0.016} \approx 1.492 \, \text{m/s} $$ $$ V_{f2} = \frac{Q}{\pi D_2 b_2} = \frac{0.015}{\pi \times 0.4 \times 0.008} \approx 1.492 \, \text{m/s} $$

$$ \tan(\phi) = \frac{V_{f2}}{u_2 - V_{w2}} \implies u_2 - V_{w2} = \frac{V_{f2}}{\tan(\phi)} $$ $$ V_{w2} = u_2 - \frac{V_{f2}}{\tan(\phi)} = 25.133 - \frac{1.492}{\tan(30^\circ)} $$ $$ V_{w2} \approx 25.133 - 2.584 \approx 22.549 \, \text{m/s} $$

$$ \text{Pressure Head Rise} = \frac{P_2 - P_1}{\rho g} = \frac{1}{2g} \left[ (u_2^2 - u_1^2) - (V_{r2}^2 - V_{r1}^2) \right] $$ $$ \text{First, find relative velocities } V_{r1} \text{ and } V_{r2} $$ $$ V_{r1}^2 = V_{f1}^2 + u_1^2 = (1.492)^2 + (12.566)^2 = 2.226 + 157.904 = 160.13 $$ $$ V_{r2}^2 = V_{f2}^2 + (u_2 - V_{w2})^2 = (1.492)^2 + (2.584)^2 = 2.226 + 6.677 = 8.903 $$ $$ \text{Now substitute into the pressure head equation:} $$ $$ \frac{P_2 - P_1}{\rho g} = \frac{1}{2 \times 9.81} \left[ (25.133^2 - 12.566^2) - (8.903 - 160.13) \right] $$ $$ \frac{P_2 - P_1}{\rho g} = \frac{1}{19.62} \left[ (631.667 - 157.904) - (-151.227) \right] $$ $$ \frac{P_2 - P_1}{\rho g} = \frac{1}{19.62} [473.763 + 151.227] = \frac{624.99}{19.62} \approx 31.855 \, \text{m} $$