Question 15.P.13: An axial flow pump running at 600 rpm deliver 1.4 m³/s again......
An axial flow pump running at 600 rpm deliver 1.4 m³/s against a head of 5 m. The speed ratio is 2.5. The flow ratio is 0.5. The overall efficiency is 0.83. Determine the power required and the blade angles at the root and tip and the diffuser blade inlet angle. Inlet whirl is zero.
β_{1} – Blade angle at inlet
β_{2} – Blade angle at outlet
α_{2} – Diffuser blade inlet angle
{\mathrm{Power}}={\frac{1.4\times10^{3}\times5\times9.81}{0.83\times10^{3}}}=82.73{\mathrm{~kW}}u_{t}=2.5\,{\sqrt{2g\ H}} \\ \quad =2.5{\sqrt{2\times9.81\times5}} \\ \quad =24.76\ \mathrm{m/s}, \\ V_{f}=0.5{\sqrt{2g~H}}\,=4.95~{\mathrm{m/s}} \\ D_{0}={\frac{24.76\times60}{\pi\times600}}=0.788~{\mathrm{m}} \\ Q={\frac{\pi\left(D_{0}{}^{2}-D_{1}{}^{2}\right)V_{f}}{4}}
∴ \qquad \qquad 1.4=\pi(0.788^{2}-D_{1}{}^{2})\,4.95Solving, D_{1} = 0.51\ {\rm m}
At Tip : From inlet triangle,
\beta_{1t}=\tan^{-1}\left({\frac{4.95}{24.76}}\right)=11.3^{\circ}
From outlet triangle, β_{2t} = V_{f} /(u – V_{u2})
u_{2}\ V_{u2} = gH, 24.76 × V_{u2} = 9.81 × 5, V_{u2} = 1.981\ {\rm m/s}
∴ \qquad \qquad \beta_{2\mathrm{t}}=\tan^{-1}\!\left({\frac{4.95}{24.76-1.981}}\right)={\mathrm{12.26^{\circ}}}\alpha_{2t}=\tan^{-1}\left(\frac{4.95}{1.981}\right)=68.2^{\circ}, can also be given as (180 – 68.2°)
At Root : \qquad u_{2}\ V_{u2} = gH, u_{2} = π × 0.51 × 600/60 = 16.02\ {\rm m/s} \\ ∴ \qquad \qquad V_{u2} = 5 × 9.81/16.02 = 3.06\ {\rm m/s}, V_{f} = constant
\beta_{1 \rm R}=\tan^{-1}\left({\frac{4.95}{16.02}}\right)=17.17^{\circ} \\ \beta_{\mathrm{2R}}=\tan^{-1}(4.95/(16.02-3.06)=20.9^{\circ} \\ \alpha_{2\mathrm{R}}=\tan^{-1}\left(\frac{4.95}{3.06}\right)=58.3^{\circ}
α values can also be given as (180 – 58.3)°.
