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 1 - Introduction and Basic Concepts - Problems - Page 46: 1-80

Answer

$P_{1}=99.48kPa$

Work Step by Step

$P_{atm}=(720mm Hg)*(\frac{0.1333kPa}{1mm Hg})=95.98kPa$ Solving the manometer: $P_{atm}+\gamma_{B}*h_{B}+\gamma_{A}*h_{A}=P_{1}$ $P_{1}=95.98kPa+20\frac{kN}{m^3}*0.15m+10\frac{kN}{m^3}*0.05m$ $P_{1}=99.48kPa$
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