Question

The system below uses a centrifugal pump to circulate water at 19°C, having a vapor pressure of 2320 Pa, from a low reservoir to a higher one through a cast iron pipe with a diameter of 35 cm and surface roughness of 0.25 mm. The minor head losses upstream of the pump are represented by EK1 = 0,6 and downstream by & K2 = 7. After several months of usage, it is noticed that the installation is noisy and consumes excessive amounts of energy. The pump is suspected to be incompatible with the system and you are asked to replace it with another from the manufacturer Ingersoll-Rand whose pumps' characteristic curves are provided below.

Answer 1

Please refer the following images for the solution.

Solution. 9) Applying Bernouli's equation of the pope bed and and free surface of water in the tank. Plzit vir P2 + Zzt V2² Sg zg sg zo PL P2 30 Atmospheric pressure head 58 (Assuming datturm heed) viz, V22 - 0 2 Z2 = 10+1-Ilm hl= f Luz EKL uz 2,=0 zod E = 0.25 mm=025aron 1 d: 35cm=0.35m} Relative roughness fiction factor $ 1,14 -0.86 In (4) VF 1.14-0.86 In O. 25x103 V7 0.35 ::f=0.01841 kow pepe Hp = 1lt f Luz + Therefore, above equation reduces to Hp = 11 the SKL v2 ago zg FL + zg - Hp=117 117 v2 0.01841X (2+1+25) +(0.6127 2x9.81 Hp = vz 0.35

Q(gal/min) (m3/s) Hp(m) Hp(ft) 4000 0.252361 14.18151 46.52725 8000 0.504722 23.72602 77.84128 12000 0.757083 39.63355 130.0313 16000 1.009445| 61.90409 203.0974 20000 1.261806 90.53764 297.0395 24000 1.514167| 125.5342| 411.8576 28000 1.766528| 166.8938 547.5517|

11 = 1170 t/min 800 50 40 NPSH NPSH, 11 700 36 in dia 65% 72% 7890 82% 20 85% 600 87% 88% Total head, ft 870 32-in dia. 500 3500 bhp 3000 bhp 400 28-in dia 6 500 bhp 300 2000 bhp 500 bhp 200 0 4 8 12 16 20 24 28 U.S. gallons per minute 1000

BER n = 710 r/min 400 25 NPSH 20 NPSH, ft 350 41-in dia. 725 15 %08 84% %98 10 300 38-in dia. %88 Total head, 11 86% 250 35-in dia. 84% 200 1500 bhp 150 1000 bhp 1250 bhp 100 0 4 8 28 12 16 20 24 U.S. gallons per minute 1000

Power consumed by the pump in kW P = 8 9 Q He 1000x 9.818 1.230x86.868 :p=117772511LW le 1177:725 kW]

d) Cavitation cavitation is formation of vapors in the pump suction line when A suction pressure becomes equal to or less than vapor pressure of the liquid flowing through the pipe. These vapors when comes in discharge line where pressure is very higher vaper pressure of the liquid, the vapor bubble collapse near the pipe material. Itching material. Itching action takes place on the pipe material. After I long period small cavities are formed on the pipe material, which is called as cavitation. Cavitation should be avoided for the following than on the reasons ) To prevent the pump from damage. i) To supply constant flow rate from pump To provide required discharge or pump head. prevent pipe material from damage.