Model testing and specific speed of turbine

Given (prototype): $N=140$ rpm, $H=30$ m, $P=15000$ kW, $\eta_o=0.9$. Model: $D_m/D=0.2$, $H_m=6$ m.

(a) Specific speed of prototype

$$N_s=\frac{N\sqrt P}{H^{5/4}}=\frac{140\times\sqrt{15000}}{30^{1.25}}=\frac{140\times122.47}{70.21}=244.2$$

(This is in the fast Francis turbine range, 180 to 300.)

(b) Discharge of prototype

$$Q=\frac{P}{\rho gH\eta_o}=\frac{15\times10^{6}}{1000\times9.81\times30\times0.9}=56.63\ \text{m}^3/\text{s}$$

(c) Model

For geometrically similar turbines (same $\psi=gH/N^2D^2$, $\phi=Q/ND^3$, same efficiency):

$$\frac{N_m}{N}=\sqrt{\frac{H_m}{H}}\cdot\frac{D}{D_m}\ \Rightarrow\ N_m=140\times\sqrt{\frac{6}{30}}\times\frac{1}{0.2}=140\times0.4472\times5=313\ \text{rpm}$$ $$\frac{P_m}{P}=\left(\frac{H_m}{H}\right)^{3/2}\left(\frac{D_m}{D}\right)^2\ \Rightarrow\ P_m=15000\times0.00358=53.7\ \text{kW}$$ $$\frac{Q_m}{Q}=\left(\frac{D_m}{D}\right)^2\sqrt{\frac{H_m}{H}}\ \Rightarrow\ Q_m=56.63\times0.01789=1.013\ \text{m}^3/\text{s}$$

Check: $N_{s,\text{model}}=\dfrac{313\sqrt{53.7}}{6^{1.25}}=244.2$, the same as the prototype ($244.2$), as required for similarity.

Answer: (a) $N_s=244.2$; (b) $Q=56.63$ m³/s; (c) model speed $313$ rpm, power $53.7$ kW, discharge $1.013$ m³/s.