Question 12.15: A propeller turbine having an outer diameter of 5 m and a hu......

Given: Refer Figure 12.26. D_o=5  m;D_h=2.5  m ; S P=24 \times 10^6   W ; H=20  m ; N=150  rpm;V_{w 2}=0 ; \eta_o=0.87 ; \eta_b=0.94 .

From the equation for overall efficiency,

\eta_o=\frac{S P}{W Q H}=\frac{24 \times 10^6}{9810 \times Q \times 20}

∴  Discharge  Q=\frac{24 \times 10^6}{9810 \times 20 \times 0.88}=139  m ^3 / s

From the equation for discharge,

Q=\frac{\pi}{4}\left(D_o^2-D_h^2\right) \times V_{f 1}=\frac{\pi}{4}\left(5^2-2.5^2\right) \times V_{f 1}

∴    Flow velocity  V_A=9.44  m / s

Near the hub section,

u_1=\frac{\pi D_h N}{60}=\frac{\pi \times 2.5 \times 150}{60}=19.635  m / s

From the equation for hydraulic efficiency,

\begin{aligned}\eta_h &=\frac{V_{w 1}u_1}{g H}=\frac{V_{w 1}\times 19.635}{9.81 \times 20}\\∴                 V_{w 1}&=9.49   m / s\end{aligned}

\tan (180-\theta)=\frac{V_{f 1}}{u_1-V_{w 1}}=\frac{9.44}{(19.635-9.49)}=0.9305

∴         Inlet vane angle  \theta=180-42.9=137.1^{\circ}

From the outlet velocity triangle,

\begin{aligned}& V_{f 2}=V_{f 1}=9.44  m / s \\& u_2=u_1=19.635  m / s \\& \tan \phi=\frac{V_{f 2}}{u_2}=\frac{9.44}{19.635}=0.4808 \\&\end{aligned}

∴    Outlet vane angle  \phi=25.68^{\circ}

Near the outer edge of the runner,

\begin{gathered}u_1=\frac{\pi D_o N}{60}=\frac{\pi \times 5 \times 150}{60}=39.27  m / s \\\eta_h=\frac{V_{w 1}u_1}{g H}=\frac{V_{w 1}\times 39.27}{9.81 \times 20}=0.95 \\V_{w 1}=\frac{0.95 \times 9.81 \times 20}{39.27}=4.75   m / s \\\tan (180-\theta)=V_{f 1}\left(u_1-V_{w 1}\right)=\frac{9.44}{(39.27-4.75)}=0.2635\end{gathered}

∴            Inlet vane angle          \theta=(180-15.3)=164.7^{\circ}

From the outlet velocity triangle,

\tan \phi=\frac{V_{f 2}}{u_2}=\frac{9.44}{39.27}=0.24

∴            Outlet vane angle          \phi=13.52^{\circ}.