Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 102: 2-65

Answer

$V=0.0291\frac{m^3}{s}$

Work Step by Step

Here we are working with potential energy. First we calculate the power we are going to have: $W=(7hp*\frac{745.7W}{1hp})*0.82=4280.32W$ Knowing that: $W=mgh=\rho Vgh$: $V=\frac{W}{\rho gh}=\frac{4280.32W}{1000\frac{kg}{m^3}*9.81\frac{m}{s^2}*15m}=0.0291\frac{m^3}{s}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.