Question 14.12: Hydroturbine Design A retrofit Francis radial-flow hydroturb...
Hydroturbine Design
A retrofit Francis radial-flow hydroturbine is being designed to replace an old turbine in a hydroelectric dam. The new turbine must meet the following design restrictions in order to properly couple with the existing setup: The runner inlet radius is r_{2}=8.20 ft (2.50 m ) and its outlet radius is r_{1} =5.80 ft (1.77 m ) . The runner blade widths are b_{2}=3.00 ft (0.914 m) and b_{1}=8.60 ft (2.62 m) at the inlet and outlet, respectively. The runner must rotate at \dot{n}=120 rpm (\omega=12.57 rad / s ) to turn the 60-Hz electric generator. The wicket gates turn the flow by angle \alpha_{2}=33^{\circ} from radial at the runner inlet, and the flow at the runner outlet is to have angle \alpha_{1} between -10° and 10° from radial (Fig. 14–99) for proper flow through the draft tube. The volume flow rate at design conditions is 9.50 \times 10^{6} gpm (599 m³/s), and the gross head provided by the dam is H_{\text {gross }}=303 ft (92.4 m). (a) Calculate the inlet and outlet runner blade angles \beta_{2} \text { and } \beta_{1} , respectively, and predict the power output and required net head if irreversible losses are neglected for the case with \alpha_{1}=10^{\circ} from radial (withrotation swirl). (b) Repeat the calculations for the case with \alpha_{1}=0^{\circ} from radial (no swirl). (c) Repeat the calculations for the case with \alpha_{1}=-10^{\circ} from radial (reverse swirl).
For a given set of hydroturbine design criteria we are to calculate runner blade angles, required net head, and power output for three cases—two with swirl and one without swirl at the runner outlet.
Assumptions 1 The flow is steady. 2 The fluid is water at 20°C. 3 The blades are infinitesimally thin. 4 The flow is everywhere tangent to the runner blades. 5 We neglect irreversible losses through the turbine.
Properties For water at 20°C, \rho=998.0 kg / m ^{3} .
Analysis (a) We solve for the normal component of velocity at the inlet using Eq. 14-12,
V_{2, n}=\frac{\dot{V}}{2 \pi r_{2} b_{2}}=\frac{599 m ^{3} / s }{2 \pi(2.50 m )(0.914 m )}=41.7 m / s (1)
Volume flow rate: \dot{V}=2 \pi r_{1} b_{1} V_{1, n}=2 \pi r_{2} b_{2} V_{2, n} (14–12)
Using Fig. 14–99 as a guide, the tangential velocity component at the inlet is
V_{2, t}=V_{2, n} \tan \alpha_{2}=(41.7 m / s ) \tan 33^{\circ}=27.1 m / s (2)
We now solve Eq. 14-45 for the runner leading edge angle \beta_{2} ,
\beta_{2}=\arctan \left(\frac{V_{2, n}}{\omega r_{2}-V_{2, t}}\right)=\arctan \left(\frac{41.7 m / s }{(12.57 rad / s )(2.50 m )-27.1 m / s }\right)= 8 4 . 1 { }^{\circ} (3)
Runner leading edge: V_{2, t}=\omega r_{2}-\frac{V_{2, n}}{\tan \beta_{2}} (14–45)
Equations 1 through 3 are repeated for the runner outlet, with the following results:
Runner outlet: V_{1, n}=20.6 m / s , \quad V_{1, t}=3.63 m / s , \quad \beta_{1}=47.9^{\circ} (4)
The top view of this runner blade is sketched (to scale) in Fig. 14–100.
Using Eqs. 2 and 4, the shaft output power is estimated from the Euler turbomachine equation, Eq. 14–39,
\begin{aligned}\dot{W}_{\text {shaft }}=& \rho \omega \dot{V}\left(r_{2} V_{2, t}-r_{1} V_{1, t}\right)=\left(998.0 kg / m ^{3}\right)(12.57 rads / s )\left(599 m ^{3} / s \right) \\& \times[(2.50 m )(27.2 m / s )-(1.77 m )(3.63 m / s )]\left(\frac{ MW \cdot s }{10^{6} kg \cdot m ^{2} / s ^{2}}\right) \\=& 461 MW = 6 . 1 8 \times 10^{5} hp\end{aligned} (5)
Euler turbomachine equation for a turbine: \dot{W}_{\text {shaft }}=\omega T _{\text {shaft }}=\rho \omega \dot{V}\left(r_{2} V_{2, t}-r_{1} V_{1, t}\right) (14–39)
Finally, we calculate the required net head using Eq. 14–44, assuming that \eta_{\text {turbine }}=100 percent since we are ignoring irreversibilities,
H=\frac{\text { bhp }}{\rho g \dot{V}}=\frac{461 MW }{\left(998.0 kg / m ^{3}\right)\left(9.81 m / s ^{2}\right)\left(599 m ^{3} / s \right)}\left(\frac{10^{6} kg \cdot m ^{2} / s ^{2}}{ MW \cdot s }\right)=78.6 m (6)
Turbine efficiency: \eta_{\text {turbine }}=\frac{\dot{W}_{\text {shaft }}}{\dot{W}_{\text {water horsepower }}}=\frac{\text { bhp }}{\rho g H \dot{V}} (14–44)
(b) When we repeat the calculations with no swirl at the runner outlet \left(\alpha_{1}=\right. \left.0^{\circ}\right) , the runner blade trailing edge angle reduces to 42.8°, and the output power increases to 509 MW \left(6.83 \times 10^{5} hp \right) . The required net head increases to 86.8 m.
(c) When we repeat the calculations with reverse swirl at the runner outlet \left(\alpha_{1}=-10^{\circ}\right) , the runner blade trailing edge angle reduces to 38.5°, and the output power increases to 557 MW \left(7.47 \times 10^{5} hp \right) . The required net head increases to 95.0 m. A plot of power and net head as a function of runner outlet flow angle \alpha_{1} is shown in Fig. 14–101. You can see that both bhp and H increase with decreasing \alpha_{1} .
Discussion The theoretical output power increases by about 10 percent by eliminating swirl from the runner outlet and by nearly another 10 percent when there is 10° of reverse swirl. However, the gross head available from the dam is only 92.4 m. Thus, the reverse swirl case of part (c) is clearly impossible, since the predicted net head is required to be greater than H_{\text {gross }} .
Keep in mind that this is a preliminary design in which we are neglecting irreversibilities. The actual output power will be lower and the actual required net head will be higher than the values predicted here.
