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 44: 1-61

Answer

$P_g = 278 kN/m^2 = 278 kPa$

Work Step by Step

$P_g = \dfrac{W}{A}$ $W = mg =(2000kg)(9.81m/s^2)$ $A = \dfrac{\pi D^2}{4} = \dfrac{\pi (0.3)^2}{4}$ $P_g = \dfrac{(2000kg)(9.81m/s^2)}{\dfrac{\pi (0.3)^2}{4}}\left(\dfrac{1kN}{1000\frac{kg\times m}{s^2}}\right) = 278 kN/m^2$ $P_g = 278 kN/m^2 = 278 kPa$
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