Question 16.SP.9: A given turbine runs at a maximum efficiency of 87% when dis......

(a)                 N=\frac{7200}{n}=\frac{7200}{150}=48 poles,

\mathrm{bhp}=\frac{\eta \gamma Q h}{550}=\frac{0.87(62.4)(200) 81}{550}=1600 \mathrm{hp}

(b) To solve for the speed, we make use of Eq. (16.16):

Eq. (16.16): n \propto h^{1 / 2} / D

So:             \frac{n_2}{n_1}=\left(\frac{h_2}{h_1}\right)^{1 / 2}\left(\frac{D_1}{D_2}\right)=\left(\frac{40.5}{81}\right)^{1 / 2}\left(\frac{D}{D}\right)

\begin{aligned}& n_{2}=n_{1}\left(\frac{1}{2}\right)^{1 / 2}=150(0.707)=106 \mathrm{rpm}\end{aligned}

Eq. (16.17): Q \propto h^{1 / 2} D^{2}

So:                \frac{Q_{2}}{Q_{1}}=\left(\frac{h_{2}}{h_{1}}\right)^{1 / 2}\left(\frac{D_{2}}{D_{1}}\right)^{2}=\left(\frac{40.5}{81}\right)^{1 / 2}\left(\frac{D}{D}\right)^{2}

\begin{aligned}& Q_{2}=Q_{1}\left(\frac{1}{2}\right)^{1 / 2}=200(0.707)=114 \mathrm{cfs}\end{aligned}

Eq. (16.18): \quad P \propto h^{3 / 2} D^{2}

So:                   \frac{P_{2}}{P_{1}}=\left(\frac{h_{2}}{h_{1}}\right)^{3 / 2}\left(\frac{D_{2}}{D_{1}}\right)^{2}=\left(\frac{40.5}{81}\right)^{3 / 2} \times 1

\begin{aligned}& P_{2}=1600\left(\frac{1}{2}\right)^{3 / 2}=566 \mathrm{hp} \end{aligned}

(c) N=7200 / n_{2}=7200 / 106=68 poles

Note: If the number of poles does not turn out to be an even number, the rotative speed must be changed somewhat. For example, suppose the head was 66 \mathrm{ft}, with all other data unchanged. Then n_{2}=150(66 / 81)^{1 / 2}=135.4 \mathrm{rpm}. For is case the number of poles N=7200 / 1354^{2}=53.2 which is impossible. So, we shall use 54 poles, in which case n=7200 / N=7200 / 54=1333 \mathrm{rpm}.

 n \propto h^{1 / 2} / D \quad \text { or } \quad n=K_n h^{1 / 2} / D      (16.16)

Q=K_Q h^{1 / 2} D^2      (16.17)

P=K_p h^{3 / 2} D^2      (16.18 )