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Mechanics of Fluids

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618 Fluid machines<br />

velocities may be assumed equal at inlet and outlet. If the mechanical<br />

efficiency is constant, calculate the hydraulic and overall efficiencies<br />

<strong>of</strong> the turbine, the outlet angle <strong>of</strong> the guide vanes, and the rotor blade<br />

angles at inlet and outlet if there is no whirl at outlet.<br />

Solution<br />

Maximum hydraulic efficiency occurs for minimum pressure loss, that<br />

is, when<br />

dp1<br />

= 2.38Q − 1.43 = 0<br />

dQ<br />

∴ Qopt = 1.43/2.38 = 0.601 m 3 · s −1<br />

and minimum p1 ={1.19(0.601) 2 − (1.43 × 0.601) + 0.47} MPa<br />

= 40.4 kPa ≡<br />

40.4 × 103 Pa<br />

= 4.12 mH2O<br />

1000 × 9.81 N · m−3 Power specific speed refers to conditions <strong>of</strong> maximum overall efficiency,<br />

that is, to maximum hydraulic efficiency if mechanical<br />

efficiency is constant. Then<br />

(gH) 5/4 = ωP1/2<br />

ϱ1/2 =<br />

�P<br />

(69.1 rad · s−1 )(200 × 103W) 1/2<br />

(1000 kg · m−3 ) 1/20.565 rad<br />

whence H = 39.67 m.

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