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Fluid mechanics report

Fluid mechanics report - can be determined α z 1 = α z 2...

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The selection of the ideal pump is based upon the following assumptions: 1. That the pressure at P 1 and P 2 is a gage pressure, that is at atmospheric pressure, therefore indicating the P 1 and P 2 are equal to zero. 2. That the shaft rotation rate, n = 20 rad/s. 3. That rate of flow, Q = 6.309 m/s 4. Using the continuity theorem, that implies that the density is conserved. Using the energy equation, + α + z 1 = + α + z 2 – h p + h f + h turbine Using P 1 = 0 and P 2 = 0 (due to the assumption that gage pressure equals zero), v 1 = 0, v 2 = 0, and h turbine = 0 (since there is no turbine present), and h f = . Therefore, find h f h f = equation 1 However, we have to first determine the average velocity of the pipe, V. Using Q = VA Q = VA V = = = 3.459 10 -3 Using the value of V as obtained above, and substitute into equation 1. Therefore, h f = = = 1.647 10 -5 By substituting the value obtained from h f , it is substituted into the energy equation and h p can be
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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 10-5 – 15.24 = 20.48 m To get C max , = 70 – 274500 Therefore, in order to find , (Q = 6.309 10-5 m 3 /s, shaft rotation, n = 20 rads-1 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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