Question 14.7: The guide vanes of a Francis turbine in Fig. 14–14a direct w......
The guide vanes of a Francis turbine in Fig. 14–14a direct water onto the 200-mm-wide runner blades at an angle of α_1 = 30°. The blades are rotating at 25 rev/min, and they discharge water at 4 m³/s in the radial direction, that is, toward the center of the turbine, α_2 = 90°, Fig. 14–14b. Determine the power developed by the turbine and the ideal head loss.
Fluid Description. We have steady flow onto the blades and assume water to be an ideal fluid for which \rho_w = 1000 kg/m³.
Kinematics. In order to determine the power, we must first find the tangential components of velocity onto and off the blades, Fig. 14–14b. Using the flow, the radial component of velocity of the water onto each blade, at r_1 = 1 m, is
Thus, from Fig. 14–14b, the tangential component is
V_{t1}=(3.183\,\mathrm{m/s})\cot30^{\circ}=5.513\,\mathrm{m/s}At the tail of the runner, V_{t 2} = 0, since only radial flow occurs.
Power. Applying Eqs. 14–25 and 14–26, we get \dot W_{turb} = \rho_wQ(r_2V_{t 2} – r_1V_{t1})\omega. Since we have a Francis turbine, the power output is therefore
T = \rho Q(r_2V_{r2} – r_1 V_{t1}) (14-25)
Francis turbine
\dot{W}_{turb} =T ω (14-26)
\dot{W}_{\mathrm{turb}}\,=\,(1000\,\mathrm{kg/m^{3}})(4\,\mathrm{m^{3}/s})\bigg[0\,-\,(1\,\mathrm{m})(5.513\,\mathrm{m/s})\bigg]\bigg({\frac{25\,\mathrm{rev}}{\rm min}}\bigg)\bigg({\frac{1\,\mathrm{min}}{60\,s}}\bigg)\bigg({\frac{2\pi\,\mathrm{rad}}{1\,\mathrm{rev}}}\bigg)\\\dot{W}_{\mathrm{turb}}=-57.74(10^{3})\,{\rm W}=-57.7\,{\rm k}{\rm W}The negative sign indicates the removal of energy from the water, which is expected.
Head Loss. The ideal head loss is determined from Eq. 14–27.
h_{\mathrm{turb}}=\frac{\dot{W}_{\mathrm{turb}}}{Q\gamma}=\frac{-57.74(10^{3})\,\mathrm{W}}{(4\,\mathrm{m}^{3}/{\bf s})(1000\,\mathrm{kg/m}^{3})(9.81\,\mathrm{m}/\mathrm{s}^{2})}\\=\,-1.47\,{\mathrm{m}}