Thermodynamics

Chapter 7 | 361Alternative Solution The final temperature could also be determined fromEq. 7–42 by assuming constant specific heats for air:a T 2b a v k11bT 1 sconst. v 2The specific heat ratio k also varies with temperature, and we need to usethe value of k corresponding to the average temperature. However, the finaltemperature is not given, and so we cannot determine the average temperaturein advance. For such cases, calculations can be started with a k valueat the initial or the anticipated average temperature. This value could berefined later, if necessary, and the calculations can be repeated. We knowthat the temperature of the air will rise considerably during this adiabaticcompression process, so we guess the average temperature to be about 450 K.The k value at this anticipated average temperature is determined from TableA–2b to be 1.391. Then the final temperature of air becomesT 2 1295 K2182 1.3911 665.2 KThis gives an average temperature value of 480.1 K, which is sufficientlyclose to the assumed value of 450 K. Therefore, it is not necessary to repeatthe calculations by using the k value at this average temperature.The result obtained by assuming constant specific heats for this case is inerror by about 0.4 percent, which is rather small. This is not surprising sincethe temperature change of air is relatively small (only a few hundreddegrees) and the specific heats of air vary almost linearly with temperaturein this temperature range.Process: isentropicGiven: v1, T 1 , and v2Find: T 2T vr. .. .Tread v2v = 2r2 vv r11. .Tread 1 v r1. .FIGURE 7–39The use of v r data for calculating thefinal temperature during an isentropicprocess (Example 7–10).EXAMPLE 7–11Isentropic Compression of an Ideal GasHelium gas is compressed by an adiabatic compressor from an initial stateof 14 psia and 50°F to a final temperature of 320°F in a reversible manner.Determine the exit pressure of helium.Solution Helium is compressed from a given state to a specified pressureisentropically. The exit pressure of helium is to be determined.Assumptions At specified conditions, helium can be treated as an ideal gas.Therefore, the isentropic relations developed earlier for ideal gases areapplicable.Analysis A sketch of the system and the T-s diagram for the process aregiven in Fig. 7–40.The specific heat ratio k of helium is 1.667 and is independent of temperaturein the region where it behaves as an ideal gas. Thus the final pressureof helium can be determined from Eq. 7–43:P 2 P 1 a T k>1k122b 114 psia2a 780 R 1.667>0.667T 1 510 R b 40.5 psia