A centrifugal pump lifts water against of 0.08 m, major losses in the suction side h,, = EN m and in the discharge side hin is 3 m. The pump impeller has a diameter of 0.15 m, exit width of0.015 m, exit blade angle of 22° and speed ofN= 20EN rpm The hydraulic efficiency is 0.85. If the flow at the entrance is radial, find the discharge, pump head, atmospheric pressure Pa. vapor pressure P, and maximum allowable height of the pump from the feeding tank water surface elevation if the pump is operating under T= 25 °C at altitude H= 20EN m. static !! o, = NPSH, H, =1.21x10*N !! $256 Ply =10.33(1-2.256×10H

Structural Analysis
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Chapter2: Loads On Structures
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Problem-3:
A centrifugal pump lifts water against a static head Az = VEN m. The pipe setup has a constant diameter
of 0.08 m, major losses in the suction side h,, = EN m and in the discharge side hip is 3 m. The pump
impeller has a diameter of 0.15 m, exit width of0.015 m, exit blade angle of 22° and speed ofN=20EN
rpm The hydraulic efficiency is 0.85. If the flow at the entrance is radial, find the discharge, pump head,
atmospheric pressure Pa. vapor pressure P, and maximum allowable height of the pump from the feeding
tank water surface elevation if the pump is operating under T= 25 °C at altitude H= 20EN m.
%3!
o, = NPSH, IH, =1.21x10 N
- P,)/y-hST-H,
!3!
!!
SM
(P
al-abs
$256
Par ly =10.33(1–2.256x10-H
(14 29-3985:(T+2334)
!!
Transcribed Image Text:Problem-3: A centrifugal pump lifts water against a static head Az = VEN m. The pipe setup has a constant diameter of 0.08 m, major losses in the suction side h,, = EN m and in the discharge side hip is 3 m. The pump impeller has a diameter of 0.15 m, exit width of0.015 m, exit blade angle of 22° and speed ofN=20EN rpm The hydraulic efficiency is 0.85. If the flow at the entrance is radial, find the discharge, pump head, atmospheric pressure Pa. vapor pressure P, and maximum allowable height of the pump from the feeding tank water surface elevation if the pump is operating under T= 25 °C at altitude H= 20EN m. %3! o, = NPSH, IH, =1.21x10 N - P,)/y-hST-H, !3! !! SM (P al-abs $256 Par ly =10.33(1–2.256x10-H (14 29-3985:(T+2334) !!
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