Question 14.5: The impeller of a radial-flow water pump has an outer radius......
The impeller of a radial-flow water pump has an outer radius of 200 mm, average width of 50 mm, and a tail angle β_2 = 20°, Fig. 14–9. If the flow onto the blades is in the radial direction, and the blades are rotating at 100 rad/s, determine the ideal power. The discharge is 0.12 m³/s.
I
Fluid Description. We assume steady, incompressible flow and will use average velocities. Here \rho_w = 1000 kg/m³.
Kinematics. The power can be determined using Eq. 14–13, with \alpha_1 = 90°. First we must find U_2, \,V_{r2}, and \alpha_2. The speed of the impeller at the exit is
\dot{W}_{pump} = T ω = \rho Q (U_2V_{r2} \cot α_2 – U_1V_{r1} \cot α_1) (14-13)
U_{2}=\omega r_{2}=(100\,\mathrm{rad/s})(0.2\,\mathrm{m})=20\,\mathrm{m/s}Also, the radial component of velocity V_{r2} can be found from Q = V_{r2} A_2.
0.12\;{\mathrm{m}}^{3}/{\mathrm{s}}=V_{r2}[2\pi(0.2\,{\mathrm{m}})\,(0.05\,{\mathrm{m}})]\quad\quad\quad\quad\quad V_{r2}=1.910\,{\mathrm{m}}/{\mathrm{s}}As shown in Fig. 14–9,
V_{r2}=\,U_{2}\,-\,V_{r2}\cot20^{\circ}=20\,{\mathrm{m/s}}\,-\,(1.910\,{\mathrm{m/s}})\,\mathrm{cot}\,20^{\circ}=\,14.75\,{\mathrm{m/s}}Therefore,
\tan\alpha_{2}=\frac{V_{r2}}{V_{t2}}=\,\left(\frac{1.910\,\mathrm{m/s}}{14.75\,\mathrm{m/s}}\right);\quad\alpha_{2}=7.376^{\circ}Ideal Power
II
The ideal power can also be related to the ideal pump head by Eq. 14–15, \dot W_{pump} = Q\gamma h_{pump}. First we must find h_{pump} by using Eq. 14–19. This gives
Therefore,