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 17 - Compressible Flow - Problems - Page 890: 17-21

Answer

$ c= 173m/s$

Work Step by Step

We have the gas constant of R-134a is $ R = 0.08149 KJ/(KG.K)$ Specific heat ratio at room temparature is $K =1.108$ We know $c = \sqrt(KRT)$ substituting given values in the equation, we get $c = \sqrt(1.108 KJ/(KG.K) \times 0.08149 KJ/(KG.K) \times (60 +273)K \times ((1000m^2/s^2)/(1KJ/Kg)))$ $ c= 173m/s$
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