Question

A water tower is fed from an underground water source (Figure 2). The difference in level is 67 m. The flow of water must be at least 55 m'/h. To provide this flow (55m/h), a pump must be driven at a speed of 690 RPM. It can turn faster, but without exceeding 900 RPM. Its efficiency is 81%. The pump is driven directly, therefore without gearbox, by an asynchronous motor, which itselif is supplied directly by the 400V/50 Hz three-phase network. The motor is rated as follows Efficiency: 92% Power factor (cosp): 0.84 . slip at nominal torque: 4.5% Simplifying hypotheses: The flow rate of the pump is proportional to its speed. It is assumed that the motor slip is equal to its nominal slip, even if the torque it furnishes is not exactly equal to its nominal torque p-1'000 kg/m3 g-9,81 m/s h- 67 m SUPPLY: v 400 V f- 50 Hz MOTOR: coso 0,84 92% PUMP: Q 255 mn Nmax . 900 RPM Q = 55 m'/h when N= 690 RPM 81% Figure 2, 04 Water tower detail i. Determine the number of pole required on the motor to satisfy the flowrate. 8 marks] ii. Following this choice, what is the exact flow rate of the pump? ii With the exact flowrate, find the pump power v. What should the rated power of the motor be? Possible values (standardized): 7.5 kW, 15 kW, 22 kW or 37 kW? v. Under these operating conditions, find the motor current? 2 marks] [6 marks] [5 marks] [4 marks]

Answer 1

Given .S = 'Sゾ、, Cos q = 0.gy び) poles ris N)= チ22-513 RPr1.

NJova AJvm f peley 20120 x 50 n, E 722-513 Given that the flow bote the pump is Paoportional t its sped. When No. poles p>8 Synchronous speed , Ns,ら0 RPM ? -Achal Speed N ienb No.s prepa tonol to speed Here -6905h

55 x 716-25 6go ris is t ero flos sote fell emac Canditioy in fisst step. GH) punp. Poses is givn 3600 e qlsec 57.09 3690 Oy)

)Rokedpouen Pozsen suppld to the pup Robed po 0.423 0.92 Rated posen Rated power 11-33 ko 15 kw (MOLer cassent P. (coe

Power Supplied to pumP 0.92 x 15 13.8 ko Pump poe pump 13.8 x lgar Actual flow ote O-0.81x755]-61.2198 [.nfu-0.8 N O Qi 6},2198 朽チ우 6S