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 43: 1-46

Answer

a) $ 1200 kN/m^2$ b) $1200000 \dfrac{kg}{m \times s^2}$ c) $1200000000 \dfrac{kg}{km \times s^2}$

Work Step by Step

a) $P = (1200 kPa)(\dfrac{1 \frac{kN}{m^2}}{1 kPa}) = 1200 kN/m^2$ b) $P = (1200 kPa)(\dfrac{1 \frac{kN}{m^2}}{1 kPa}) (\dfrac{1000kg\times m/s^2}{1kN})= 1200000 \dfrac{kg}{m \times s^2}$ c) $P = (1200 kPa)(\dfrac{1 \frac{kN}{m^2}}{1 kPa}) (\dfrac{1000kg\times m/s^2}{1kN})(\dfrac{1000 m}{1 km})= 1200000000 \dfrac{kg}{km \times s^2}$
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.