Question 6.SP.6: A radial-flow turbine runner has 18 blades each 0.20.2 in th......
A radial-flow turbine runner has 18 blades each 0.2 in thick, with r_1=10 \mathrm{in}, \beta_1=65^{\circ}, r_2=6 in, and \beta_2=122^{\circ}. The depth B of the flow passage between the two sides of the turbine is 4 in. When rotating at 180 \mathrm{rpm}, the water flow rate is 7.5 \mathrm{cfs}. (a) For both entrance and exit conditions, tabulate values of m, A_c, and V_r, showing the method of calculation. (b) For the same. conditions, tabulate values of v_u, v, u, V_u, \alpha, and V, and draw and label velocity triangles. (c) Find the torque exerted by the water, and the horsepower delivered to the shaft. Assume that water enters and leaves the blades without
Eq. (6.30): m=1-\frac{n t}{2 \pi r}=1-\frac{18(0.2)}{2 \pi r}
Eq. (6.29): \begin{gathered}A_c=m 2 \pi r B=m 2 \pi r\left(\frac{4}{12}\right) \end{gathered}
From Eq. (6.28): V_r=v_r=\frac{Q}{A}=\frac{7.5}{m A_c}
Substituting for r-values and evaluating,
\begin{aligned} \begin{array}{llcccc} \text { Point } & r, \mathrm{ft} & m & A_c, \mathrm{ft}^2 & V_r=v_r \text {, fps } & \\ \hline 1 & 0.833 & 0.943 & 1.645 & 4.56 & \\ 2 & 0.5 & 0.905 & 0.947 & 7.92 \end{array} \end{aligned}
m=1-\frac{n t}{2 \pi r} (6.30)
A_c=m 2 \pi r B (6.29)
Q=A_{c 1} V_{r 1}=A_{c 2} V_{r 2} (6.28)
(b) Velocity triangles
\begin{gathered}\omega=\frac{2 \pi n}{60}=180\left(\frac{2 \pi}{60}\right)=18: 85 \mathrm{rad} / \mathrm{sec} \\v_u=\frac{V_r}{\tan \beta} ; \quad v=\frac{v_u}{\cos \beta} ; \quad u=r \omega ; \quad V_u=u+v_u ; \\\alpha=\tan ^{-1}\left(\frac{V_r}{V_u}\right) ; \quad V=\frac{V_u}{\cos \alpha}\end{gathered}
\begin{array}{clrrrr} \text { Point } & r, \mathrm{ft} & \beta & v_u, \mathrm{fps} & v, \mathrm{fps} & u, \mathrm{fps} \\ \hline 1 & 0.833 & 65^{\circ} & 2.13 & 5.03 & 15.71 \\ 2 & 0.5 & 122^{\circ} & -4.95 & 9.34 & 9.43 \end{array}
\begin{array}{clccr} \text { Point } & r, \mathrm{ft} & V_u, \mathrm{fps} & \alpha & V, \mathrm{fps} \\ \hline 1 & 0.833 & 17.83 & 14.3^{\circ} & 18.41 \\ 2 & 0.5 & 4.48 & 60.5^{\circ} & 9.10 \end{array}
(c) Torque
From Eq. (6.31):
\begin{aligned} T & =\rho Q\left|\Delta\left(r V_u\right)\right|=\rho Q\left(r_1 V_{u 1}-r_2 V_{u 2}\right) \\& =(62.4 / 32.2) 7.5(0.83 \times 17.83-0.5 \times 4.48) \\& =183.4 \mathrm{ft} \cdot \mathrm{lb}\end{aligned}
Eqs. (5.38) and (5.40):
P=T \omega / 550=183.4(18.85) / 550=6.28 \mathrm{hp}
T=\rho Q\left(r_1 V_1 \cos \alpha_1-r_2 V_2 \cos \alpha_2\right) (6.31)
\text { Rate of energy transfer }=\text { Power }=P=F V=T \omega (5.38)
\text { Horsepower }=P=\frac{\gamma Q h}{550}=\frac{Q \Delta p}{550} (5.40)