Thermodynamics

184 | ThermodynamicsAIR at 300 Kc v = 0.718 kJ/kg · KR = 0.287 kJ/kg . Korc v = 20.80 kJ/kmol . KR u = 8.314 kJ/kmol . Kc p = 1.005 kJ/kg . K{c p = 29.114 kJ/kmol . KFIGURE 4–29The c p of an ideal gas can bedetermined from a knowledge ofc v and R.{where a 28.11, b 0.1967 10 2 , c 0.4802 10 5 , andd 1.966 10 9 . From Eq. 4–30,From Eq. 4–25,c v 1T2 c p R u 1a R u 2 bT cT 2 dT 32T 2¢u c v 1T2dT 1Performing the integration and substituting the values, we obtainThe change in the internal energy on a unit-mass basis is determined bydividing this value by the molar mass of air (Table A–1):¢u ¢uM¢u 6447 kJ>kmol6447 kJ>kmol 222.5 kJ>kg28.97 kg>kmolwhich differs from the tabulated value by 0.8 percent.T 1 31a R u2 bT cT 2 dT 3 4 dT(c) The average value of the constant-volume specific heat c v,avg is determinedfrom Table A–2b at the average temperature of (T 1 T 2 )/2 450 K to beThus,¢u c v,avg 1T 2 T 1 2 10.733 kJ>kg # K231600 3002K4 220 kJ>kgc v,avg c v @ 450 K 0.733 kJ>kg # KDiscussion This answer differs from the tabulated value (220.71 kJ/kg) byonly 0.4 percent. This close agreement is not surprising since the assumptionthat c v varies linearly with temperature is a reasonable one at temperatureintervals of only a few hundred degrees. If we had used the c v value atT 1 300 K instead of at T avg , the result would be 215.4 kJ/kg, which is inerror by about 2 percent. Errors of this magnitude are acceptable for mostengineering purposes.EXAMPLE 4–8Heating of a Gas in a Tank by StirringAn insulated rigid tank initially contains 1.5 lbm of helium at 80°F and50 psia. A paddle wheel with a power rating of 0.02 hp is operated withinthe tank for 30 min. Determine (a) the final temperature and (b) the finalpressure of the helium gas.Solution Helium gas in an insulated rigid tank is stirred by a paddle wheel.The final temperature and pressure of helium are to be determined.Assumptions 1 Helium is an ideal gas since it is at a very high temperaturerelative to its critical-point value of 451°F. 2 Constant specific heats can beused for helium. 3 The system is stationary and thus the kinetic and potentialenergy changes are zero, KE PE 0 and E U. 4 The volume ofthe tank is constant, and thus there is no boundary work. 5 The system is adiabaticand thus there is no heat transfer.