Thermodynamics

188 | ThermodynamicsP, kPa35023AIRFIGURE 4–32Schematic and P-V diagram forExample 4–10.V 1 = 400 LP 1 = 150 kPaT 1 = 27°CQ15010.4A0.8V, m 3Use actual data from the experimentshown here to obtain the specific heatof aluminum. See end-of-chapterproblem 4–179.© Ronald MullisenUse actual data from the experimentshown here to obtain the specific heatof aluminum. See end-of-chapterproblem 4–180.© Ronald MullisenEXPERIMENTEXPERIMENT(b) The work done could be determined by integration, but for this caseit is much easier to find it from the area under the process curve on a P-Vdiagram, shown in Fig. 4–32:Therefore,The work is done by the system (to raise the piston and to push the atmosphericair out of the way), and thus it is work output.(c) Under the stated assumptions and observations, the energy balance onthe system between the initial and final states (process 1–3) can beexpressed asNet energy transferby heat, work, and massChange in internal, kinetic,potential, etc., energiesThe mass of the system can be determined from the ideal-gas relation:The internal energies are determined from the air table (Table A–17) to beThus,DiscussionA 1V 2 V 1 2P 2 10.4 m 3 21350 kPa2 140 m 3 # kPam P 1V 1RT 1E in E out Q in W b,out ¢U m 1u 3 u 1 21150 kPa210.4 m 3 210.287 kPa # m 3 >kg # K21300 K2 0.697 kgu 1 u @ 300 K 214.07 kJ>kgu 3 u @ 1400 K 1113.52 kJ>kgQ in 140 kJ 10.697 kg2311113.52 214.072 kJ>kg4Q in 767 kJW 13 140 kJ¢E system⎫ ⎪⎬⎪⎭⎫⎪⎬⎪⎭The positive sign verifies that heat is transferred to the system.