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Unformatted text preview: can be determined. + α + z 1 = + α + z 2 – h p + h f + h turbine 0 + 0 + 15.24 = 0 + 0 + 45.72 – h p + 1. 647 + 0 h p = 45.72 + 1.647 105 – 15.24 = 20.48 m To get C max , = 70 – 274500 Therefore, in order to find , (Q = 6.309 105 m 3 /s, shaft rotation, n = 20 rads1 70 274500 = 0 274500 C 2 = 70 C = ± 0.01597 whereby, C 0 = 0.01597 Ƞ max = 70 (0.01597) – 91500 (0.01597) 3 = 0.7452 Therefore, to find the head pressure ≈ 6.04 – 161 (0.01597) ≈ 3.4688 = ϕ = = 0.1478 C = Q = 0.01597 (20) (0.1478) 3 = 3.24 Efficiency: = whereby the power to water = mw s = γ Q h p = Therefore, using Ƞ max = 0.7452, 0.7452 = 0.7452 = Power input = 94.12 W Based on the dimensionless performance of the pump, and using 0 < C < 0.027 C = When C = 0, = 0 Therefore it is found that D p = invalid. When C = 0.027, = 0.027 D p = 0.124 m Whereby, D p 0.124 m...
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This note was uploaded on 11/19/2011 for the course HES 2340 taught by Professor Tomedwards during the Three '09 term at Swinburne.
 Three '09
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