Question 14.13: Hydroturbine Design A retrofit Francis radial-flow hydroturb...

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, 𝜌 = 998.0 kg/m³.
Analysis   (a) We solve for the normal component of velocity at the inlet using Eq. 14–12,

\dot{V}=2 \pi r_1 b_1 V_{1,  n}=2 \pi r_2 b_2 V_{2,  n}                      (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)

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,

V_{2,  t}=\omega r_2  –  \frac{V_{2,  n}}{\tan \beta_2}         (14.45)

\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)

Equations 1 through 3 are repeated for the runner outlet, with the following results:

Runner outlet:    V_{1,  n} = 20.6  m/s,  V_{1,  t} = 3.63  m/s,   \beta_1 = 47.9°       (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,

\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)

\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 = 6.18 × 10^5  hp                                         (5)

Finally, we calculate the required net head using Eq. 14–44, assuming that \eta_{\text {turbine }}= 100 percent since we are ignoring irreversibilities,

\eta_{\text {turbine }}=\frac{\dot{W}_{\text {shaft }}}{\dot{W}_{\text {water horsepower }}}=\frac{\text { bhp }}{\rho g H \dot{V}}               (14.44)

H=\frac{ 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)

(b) When we repeat the calculations with no swirl at the runner outlet (\alpha_1 = 0°), the runner blade trailing edge angle reduces to 42.8°, and the output power increases to 509 MW (6.83 × 10^5  hp). The required net head increases to 86.8 m.
(c) When we repeat the calculations with reverse swirl at the runner outlet (\alpha_1 = −10°), the runner blade trailing edge angle reduces to 38.5°, and the output power increases to 557 MW (7.47 × 10^5  hp). 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_{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.