Answer
$T_2=576\ K$
Work Step by Step
For isentropic processes:
$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{k-1}{k}}$
Given $k=1.395\ @T_{avg}=450\ K,\ P_2=1000\ kPa,\ P_1=100\ kPa,\ T_1=27°C$
$T_2=576\ K$
Thermodynamics: An Engineering Approach 8th Edition
by Cengel, Yunus; Boles, Michael
Published by
McGraw-Hill Education
ISBN 10:
0-07339-817-9
ISBN 13:
978-0-07339-817-4
Chapter 7 - Entropy - Problems - Page 403: 7-85
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
