2010%20BRAE%20236%20HW10

2010%20BRAE%20236%20HW10 - To calculate this we must...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
BRAE 236 - HW10 Given: In the sketch below, static lift = 120 feet, drawdown = 15 feet, pump bowls set at 180 feet Discharge pressure = 45 psi Column and Discharge Head losses = 5 psi Find: a) TDH - Total Dynamic Head b) WHP at Q = 225 GPM Formulas: TDH of a Well Pump = Static Lift + Drawdown + Surface Discharge Pressure + Column and Discharge Head Losses WHP = Q * TDH / 3960 Solution: Q = 225 GPM Static Lift = 120 ft Drawdown = 15 ft Discharge Pressure = 45 psi Column and Discharge Head Losses = 5 psi a) TDH = 120 ft + 15 ft + 45 psi + 5 psi =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: To calculate this we must convert psi into ft --> 2.31 ft = 1 psi Discharge Pressure = 45 psi ==> 45 psi x 2.31ft/1psi = 103.95 ft Column and Discharge Head Losses = 5 psi ==> 5 psi x 2.31ft/1psi = 11.55 ft TDH = 120 ft + 15 ft + 103.95 ft + 11.55 = 250.5 ft b) WHP = 225 GPM x 250.5 ft / 3960 = 14.233 hp Answer: a) 251ft b) 14.2hp Discharge Pressure Static Lift Drawdown Ground Surface Well Casing Column Pipe Pump Bowls Discharge Head Motor Static Water Level...
View Full Document

This note was uploaded on 11/07/2010 for the course BRAE 236 taught by Professor Styles during the Fall '08 term at Cal Poly.

Ask a homework question - tutors are online