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Question 14.3: Determine the hydraulic efficiency for the axial-flow pump i......

Determine the hydraulic efficiency for the axial-flow pump in Example 14.2 if frictional head losses produced by the pump are 0.8 m.

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Fluid Description. We assume steady, incompressible flow through the pump.
Pump Head. We can use Eq. 14–16 and the results of Example 14.2 to determine the ideal pump head.

h_{\mathrm{pump}}={\frac{U(V_{t2}-V_{t1})}{g}}={\frac{7.50\,{\mathrm{m/s}}(5.768\,{\mathrm{m/s}}-2.304\,{\mathrm{m/s}})}{9.81\,{\mathrm{m/s}}^{2}}}=2.648\,{\mathrm{m}}

The actual pump head is determined from Eq. 14–11.

T = \rho Q(r_2V_{t2} – r_1V_{t1})                   (14–11)

(h_{\mathrm{pump}})_{\mathrm{act}}=(h_{\mathrm{pump}})\,-\,h_{L}=2.648\,\mathrm{m-0.8}\,\mathrm{m}\,=\,1.848\,\mathrm{m}

Hydraulic Efficiency. Applying Eq. 14–18 yields

\eta_{\mathrm{pump}}=\frac{(h_{\mathrm{pump}})_{\mathrm{act}}}{h_{\mathrm{pump}}}(100\%{{}})=\frac{1.848\;\mathrm{m}}{2.648\;\mathrm{m}}(100\%)=69.8{\%}

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