Question

As we learned, one must be careful, when placing a pump, to avoid cavitation, the likelihood of which is characterized by the Net Positive Suction Head. Using a pump to 'pull' fluid up (rather than push) can lead to cavitation if the available NPSH falls below the required NPSH (required by the pump). Well, using a pump to pull' fluid horizontally can also cause cavitation. Consider the sketch below. The pump is 2 m below the free surface of the reservoir. Fluid is moving from left to right. The atmospheric pressure is 101,000 Pa (absolute). The vapor pressure of water is 5000 Pa (absolute). The NPSH required by the pump is 3 m. The pipe is 10 cm in diameter, the flow velocity is 1 m/s, and the friction factor can be assumed to be 0.025. Neglect all local losses. L 1a (5 pts) Apply the energy equation, and the definition that: Ps Pvapor NPSHA + 7 29 7 to determine NPSHA in terms of the conditions at location 1 and the losses in the system. The's subscript refers to conditions at the suction' side of the pump, i.e. location 2 1b. (5 pts) By considering the critical case, where the NPSHA equals the required NPSH, or NPSHR, determine the maximum allowable pipe length L.

Answer 1

a) @ 2 Elevation at 1, 2,=2m at a, Z2 = 0 Pressure at 1 P = 101000 Pa ; at 2 P2 = Ps Velocity at 1, V=0 ; at 2. V = Vs = 1 m/s unit weight of water V=9810 N/m3 g = 9.81 m/s² Energo Head loss between (1 and 2 hi fLv² ago 0.025 XLX (1) ? 2x9.81X(0.1) h = 0.012TL

Energy equation between 0 and ① P + V ² Vi -th 29 Per + + Za the 29 Ps (1)? r toto.012TL 2x9.81 10 IOU O 9810 to+2= + 12.296 Ps Υ +0.0509 +0.01274 Ps 3 - 12.245 -0.0124L ... NPSHA = P + 1 - P vapour 5000 (12.245 -0.012 TL 272) + CO2 249.81 9810 NOSHA 11.786 -0.01271 NPS HA = NPSHR = 3m b) At coitical case 3m = 11.786 -0.012TL 11.786-3 L = 691,811 m 0.0127
Maximum length = 691.811 m