3541354 80 kpa 200 kpa 748 k this will give an

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: eady-flow energy equation. However, first we need to determine the exit temperature. For the isentropic process of an ideal gas we have: T2s T1 P2s P1 (k 1)/k or T2s T1 P2s P1 (k 1)/k (950 K) 0.354/1.354 80 kPa 200 kPa 748 K This will give an average temperature of 849 K, which is somewhat higher than the assumed average temperature (800 K). This result could be refined by reevaluating the k value at 749 K and repeating the calculations, but it is not warranted since the two average temperatures are sufficiently close (doing so would change the temperature by only 1.5 K, which is not significant). Now we can determine the isentropic exit velocity of the air from the energy balance for this isentropic steady-flow process: 2 1 → h1 ein 0 2 eout h2s 2 2s 2 cen54261_ch07.qxd 11/18/03 9:57 AM Page 319 319 CHAPTER 7 or h2s) 2(h1 2s 2Cp, av(T1 2(1.099 kJ/kg · K)[(950 T2s) 748) K] 1000 m2/s2 1 kJ/kg 666 m/s (b) The actual exit temperature of the air will be higher than the isentropic exit temperature evaluated above, and it is determined from h1 h1 h2a h2s Cp, av(T1 Cp, av(T1 950 950 T2a 748...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online