Question 16.SP.14: Two or more identical turbines are to be selected for an ins......

The total available power is

$\frac{\gamma Q h \eta}{550}=\frac{62.4(600) 350(0.90)}{550}=21,400 \mathrm{hp}$

Assume two turbines at an operating speed of $75 \mathrm{rpm}$ (96-pole generators for 60 -cycle electricity).

Eq. (16.21)        $N_{s}=\frac{75 \sqrt{21,400 / 2}}{350^{5 / 4}}=5.12$

Thus, if the operating speed is $75 \mathrm{rpm}$, use two impulse turbines with $N_{s}=5.12$, in which case from Fig. $16.14 \phi_{e} \approx 0.45$. The required wheel diameter of these

turbines is found from Eq. (16.20):

$D=\frac{153.3 \sqrt{350} \times 0.45}{75}=17.2 \mathrm{ft}$

A wheel diameter of $17.2 \mathrm{ft}$ is quite large; a smaller size is possible by increasing the rotative speed. If $n=100 \mathrm{rpm}, N_{s}=5.12(100 / 75)=6.8$ and $D=17.2 \times$ $(75 / 100)=12.9 \mathrm{ft}$. Other combinations of $N_{s}$ and $D$ could be used with other speeds; however, in accordance with Fig. 16.14, $N_{s}$ should be less than about $7.0$ if impulse wheels are selected. Another possible solution is four identical if impulse wheels are selected. Another possible solution is four identical turbines with $N_{s}=5.88\text {and}  D=10.7 \mathrm{ft}$ operating at 120 rpm .

Finally, let us explore the possibility of using Francis turbines. Assume wo Francis turbines operating at $600 \mathrm{rpm}$ (12-pole generators for 60-cycle electricity).

Eq. (16.21):             $N_{s}=\frac{600 \sqrt{21,400 / 2}}{350^{5 / 4}}=41$

According to Fig. 16.16, turbines with $N_{s}=41$ and $h=350 \mathrm{ft}$ will be safe against cavitation only if they are set one or more feet below tailwater. This may be impractical in some topographic situations.

To provide greater safeguard against cavitation, we might select a lowerspecific-speed machine, but then its efficiency may not be so good as indicated y Fig. 16.14. A good choice would be two Francis turbines with $N_{s}=30.8$ perating at $450 \mathrm{rpm}$. The required diameter of these turbines would be about Aft, assuming $\phi=0.75$.

Note: There are actually an infinite number of alternatives. The things to watch out for are: (a) freedom from cavitation (Fig. 16.16); (b) reasonably high efficiency (Fig. 16.14); and (c) size not too large (Eq. 16.19). Flexibility of choice achieved through variation in the number of units (and hence brake orsepower per unit) and in the operating speed. Variation in the draft-head setting also provides some flexibility.

$u_1=\phi \sqrt{2 g h}$      (16.19)

$D=\frac{153.3 \phi_e \sqrt{h}}{n}$              (16.20)

$N_s=\left(\frac{n_e \sqrt{\mathrm{bhp}}}{h^{5 / 4}}\right)_{\eta_{\max }}$       (16.21)