Question 14.1: The axial-flow pump in Fig. 14–4a has an impeller that is ro......
The axial-flow pump in Fig. 14–4a has an impeller that is rotating at 150 rad/s. The blades are 50 mm long and are fixed to the 50-mm-diameter shaft. If the pump produces a flow of 0.06 m³/s, and the lead blade angle of each blade is β_1 = 30° and the tail blade angle is β_2 = 60°, determine the velocity of the water when it is just on a blade and when it just leaves it. The average cross-sectional area of the open region within the impeller is 0.02 m².
Fluid Description. We assume steady ideal flow, using average velocities.
Kinematics. Using the mean radius to determine the velocity of the midpoint of the blades, we have
Also, since the flow is known, the axial velocity of the liquid through the blades is
\begin{array}{c c}Q\,=\,V A\,;&&0.06\,{\mathrm{m}}^{3}/{\mathrm{s}}\,=\,V_{a}\bigl(0.02\,{\mathrm{m}}^{2}\bigr)\\&&V_{a}=3\;\mathrm{m/s}\end{array}The kinematic diagram for the water as it just encounters a blade is shown in Fig. 14–4b. As usual, two sets of components for V_1 are established. They are V_1 = V_{t1} + V_a and V_1 = U + (V_{rel})_1. Using trigonometry, one way we can determine V_1 is as follows:
V_{t1}=7.50\,\mathrm{m/s}-(3\,\mathrm{m/s})\cot\,30^{\circ}=2.304\,\mathrm{m/s}\\\tan\alpha_{1}=\frac{3\,\mathrm{m/s}}{2.304\,\mathrm{m/s}},\quad\alpha_{1}=52.47^{\circ}\\\mathrm{3\,m/s}\,=\,V_{1}\,\mathrm{sin}\,52.47^{\circ},\,V_{1}\,=\,{3.78\,\mathrm{m/s}}The kinematic diagram for the water just leaving the blade is shown in Fig. 14–4c. We can determine V_2 the same way we determined V_1, but here is another way to do it.
\tan60^{\circ}={\frac{3\;{\mathrm{m/s}}}{7.50\;{\mathrm{m/s}}\,-\,V_{t2}}},\;\;V_{t2}=5.768\;{\mathrm{m/s}}Therefore,
V_{2}=\sqrt{(V_{a})^{2}+(V_{t2})^{2}}=\sqrt{(3\,{\mathrm{m}}/{\mathrm{s}})^{2}+(5.768\,{\mathrm{m}}/{\mathrm{s}})^{2}}=6.50\,{\mathrm{m}}/{\mathrm{s}}