Sm2 28thru2 29

PROBLEM 2.28 KNOWN: N2 gas within a piston-cylinder assembly undergoes a compression where the p-V relation is pV^1.35 = constant. FIND: Determine the volume at the final state and the work. SCHEMATIC/GIVEN DATA: [Diagram of piston-cylinder] pV^1.35 = constant p1 = 0.2 MPa, V1 = 2.75 m^3 p2 = 2 MPa ENGR MODEL: 1. The N2 is the closed system. 2. The p-V relation during compression is specified. 3. Volume change is the only work mode. ANALYSIS: (a) p1 V1^1.35 = p2 V2^1.35 => V2 = (p1/p2)^(1/1.35) * V1 => n = 1.35. Thus, V2 = (0.2 MPa/2 MPa)^(1/1.35) * (2.75 m^3) = 0.5 m^3 (b) Since volume change is the work mode, Eq. 2.17 applies. Following the procedure of part (b) of Example 2.1 we have W = p1 V1 - p2 V2 / 1-n = (0.8 MPa)(0.5 m^3) - (0.2 MPa)(2.75 m^3) 40 (kN.m/1) = 1 kJ/103 N.m = -12.855 kJ PROBLEM 2.29 KNOWN: O2 gas within a piston-cylinder assembly undergoes an expansion where the p-V relation is p = AV^-1 + B. FIND: Determine the initial and final pressures and the work. SCHEMATIC/GIVEN DATA: [Diagram of piston-cylinder] p = A V^-1 + B A = 0.06 bar.m^3 B = 3.0 bar V1 = 0.01 m^3 V2 = 0.03 m^3 ENGR MODEL: 1. The O2 is the closed system. 2. The p-V relation during expansion is specified. 3. Volume change is the only work mode. ANALYSIS: (a) p1 = (0.06 bar.m^3)/(0.01 m^3) + 3.0 bar = 9.0 bar R2 = [(0.06 bar.m^3)/(0.03 m^3)] + 3.0 bar = 5.0 bar (b) Since volume change is the work mode, Eq. 2.17 applies: That is, W = ∫(p dV) = ∫[(A/V) + B] dV = A ln(V2/V1) + B(V2-V1) = (0.06 bar.m^3) [ ln(0.03 m^3 / 0.01 m^3) ] + (3.0 bar) [(0.03 - 0.01) m^3] = [0.65 * 9 + 0.06 * 3 bar.m^3 * 10^5 N.m/1 bar] = 10^5 N / 103 N.m = 12.94 kJ