Question 14.15: Application of Turbine Affinity Laws A Francis turbine is be...

We are to design a new hydroturbine by scaling up an existing hydroturbine. Specifically we are to calculate the new turbine diameter, volume flow rate, and brake horsepower.
Assumptions   1  The new turbine is geometrically similar to the existing turbine. 2  Reynolds number effects and roughness effects are negligible. 3  The new penstock is also geometrically similar to the existing penstock so that the flow entering the new turbine (velocity profile, turbulence intensity, etc.) is similar to that of the existing turbine.
Properties   The density of water at 20°C is 𝜌 = 998.0 kg/m³.
Analysis   Since the new turbine (B) is dynamically similar to the existing turbine (A), we are concerned with only one particular homologous operating point of both turbines, namely, the best efficiency point. We solve Eq. 14–38b for D_B,

\frac{H_{ B }}{H_{ A }}=\left(\frac{\omega_{ B }}{\omega_{ A }}\right)^2\left(\frac{D_{ B }}{D_{ A }}\right)^2             (14.38b)

D_{ B }=D_{ A } \sqrt{\frac{H_{ B }}{H_{ A }}} \frac{\dot{n}_{ A }}{\dot{n}_{ B }}=(2.05  m ) \sqrt{\frac{104  m }{75.0  m }} \frac{120  rpm }{120  rpm }=2.41  m

We then solve Eq. 14–38a for \dot{V}_{ B } ,

\frac{\dot{V}_{ B }}{\dot{V}_{ A }}=\frac{\omega_{ B }}{\omega_{ A }}\left(\frac{D_{ B }}{D_{ A }}\right)^3           (14.38a)

\dot{V}_{ B }=\dot{V}_{ A }\left(\frac{\dot{n}_{ B }}{\dot{n}_{ A }}\right)\left(\frac{D_{ B }}{D_{ A }}\right)^3=\left(350  m ^3 / s \right)\left(\frac{120  rpm }{120  rpm }\right)\left(\frac{2.41  m }{2.05  m }\right)^3=572  m ^3 / s

Finally, we solve Eq. 14–38c for bhp_B,

\frac{ bhp _{ B }}{ bhp _{ A }}=\frac{\rho_{ B }}{\rho_{ A }}\left(\frac{\omega_{ B }}{\omega_{ A }}\right)^3\left(\frac{D_{ B }}{D_{ A }}\right)^5             (14.38c)

bhp _{ B }=\operatorname{bhp}_{ A }\left(\frac{\rho_{ B }}{\rho_{ A }}\right)\left(\frac{\dot{n}_{ B }}{\dot{n}_{ A }}\right)^3\left(\frac{D_{ B }}{D_{ A }}\right)^5

=(242  MW )\left(\frac{998.0  kg / m ^3}{998.0  kg / m ^3}\right)\left(\frac{120  rpm }{120  rpm }\right)^3\left(\frac{2.41  m }{2.05  m }\right)^5= 5 4 8  MW

As a check, we calculate the dimensionless turbine parameters of Eq. 14–61 for both turbines to show that these two operating points are indeed homologous, and the turbine efficiency is calculated to be 0.942 for both turbines (Fig. 14–115).
As discussed previously, however, total dynamic similarity may not actually be achieved between the two turbines because of scale effects (larger turbines generally have higher efficiency). The diameter of the new turbine is about 18 percent greater than that of the existing turbine, so the increase in efficiency due to turbine size should not be significant. We verify this by using the Moody efficiency correction equation (Eq. 14–63), considering turbine A as the “model” and B as the “prototype,”

C_H=\text { Head  coefficient }=\frac{g H}{\omega^2 D^2}         C_Q=\text { Capacity  coefficient }=\frac{\dot{V}}{\omega D^3}
                                                                                                              (14.61)
C_p=\text { Power  coefficient }=\frac{\text { bhp }}{\rho \omega^3 D^5}       \eta_{\text {turbine }}=\text { Turbine  efficiency }=\frac{\text { bhp }}{\rho g H \dot{V}}

\eta_{\text {turbine,  prototype }} \cong 1-\left(1-\eta_{\text {turbine,  model }}\right)\left(\frac{D_{\text {model }}}{D_{\text {prototype }}}\right)^{1 / 5}                     (14.63)

Efficiency correction:

or 94.4 percent. Indeed, the first-order correction yields a predicted efficiency for the larger turbine that is only a fraction of a percent greater than that of the smaller turbine.
Discussion   If the flow entering the new turbine from the penstock were not similar to that of the existing turbine (e.g., velocity profile and turbulence intensity), we could not expect exact dynamic similarity.