As the product of the base δ v i v i 1 v i i 1 n 1

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as the product of the base, Δ V i = V i +1 V i , i = 1 , . . . , N 1 , (4.69) and the average pressure of trapezoid i , P ave i = P i + P i +1 2 , i = 1 , . . . , N 1 . (4.70) We can summarize the calculations in Table 4.3. We see 1 W 2 = integraldisplay 2 1 PdV N 1 summationdisplay i =1 P ave i Δ V i = 11404 . 1 bar cm 3 . (4.71) Let us convert to kJ : ( 11404 . 1 bar cm 3 ) parenleftbigg 100 kPa bar parenrightbiggparenleftbigg m 3 (100 cm ) 3 parenrightbiggparenleftbigg kJ kPa m 3 parenrightbigg = 1 . 14 kJ. (4.72) A sketch of the process is given in Fig. 4.9. CC BY-NC-ND. 2011, J. M. Powers.
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4.2. WORK 91 Table 4.3: Tabular calculation of work. i P V P ave i Δ V i P ave i Δ V i bar cm 3 bar cm 3 barcm 3 1 20.0 454 18.05 86 1552.3 2 16.1 540 14.15 128 1811.20 3 12.2 668 11.05 112 1237.6 4 9.9 780 7.95 395 3140.25 5 6.0 1175 4.55 805 3662.75 6 3.1 1980 - - - 11404.1 0 500 1000 1500 2000 5 10 15 20 25 P (bar) V (cm 3 ) Figure 4.9: Sketch of P V diagram. Example 4.12 We are given air in the spring-restrained piston-cylinder arrangement of Fig. 4.10 with P 1 = 100 kPa , V 1 = 0 . 002 m 3 , x 1 = 0 m , no force on the piston at state 1, P atm = 100 kPa , and A = 0 . 018 m 2 . The air expands until V 2 = 0 . 003 m 3 . We know the spring is linear with F spring = kx with k = 16 . 2 kN/m . Find the final pressure of the air and the work done by the air on the piston. First note here that x is distance and not quality! The free body diagram is sketched in Fig. 4.10. For the piston to be in mechanical equilibrium, we require PA = P atm A + kx, (4.73) P = P atm + k A x. (4.74) CC BY-NC-ND. 2011, J. M. Powers.
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92 CHAPTER 4. WORK AND HEAT air A k P, V P atm F air = PA F spring = kx F atm = P atm A Figure 4.10: Sketch of piston-spring problem. This gives us P ( x ). So at the initial state, where x 1 = 0, we have P 1 = P atm = 100 kPa. We also need V ( x ): V ( x ) = V 1 + Ax. (4.75) Let us eliminate x . From Eq. (4.75), we get x = V V 1 A . (4.76) Substitute Eq. (4.76) into Eq. (4.74) to get P = P atm + k A parenleftbigg V V 1 A parenrightbigg , (4.77) P = P atm + k A 2 ( V V 1 ) . (4.78) Note, when V = V 1 , we find P = P atm . Now to get the work, we take 1 W 2 = integraldisplay 2 1 PdV = integraldisplay V 2 V 1 parenleftbigg P atm + k A 2 ( V V 1 ) parenrightbigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = P dV. (4.79) We integrate this to find 1 W 2 = bracketleftbigg P atm V + k 2 A 2 ( V V 1 ) 2 bracketrightbigg V 2 V 1 , (4.80) = P atm ( V 2 V 1 ) + k 2 A 2 ( V 2 V 1 ) 2 ( V 1 V 1 ) 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =0 , (4.81) = P atm ( V 2 V 1 ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright work done on atmosphere + 1 2 k parenleftbigg V 2 V 1 A parenrightbigg 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright work done on spring (4.82) The P V diagram is sketched in Fig. 4.11. Let us calculate the numerical values. P 2 = (100 kPa ) + parenleftbigg 16 . 2 kN m parenrightbigg 1 (0 . 018 m 2 ) 2 ( (0 . 003 m 3 ) (0 . 002 m 3 ) ) = 150 kPa. (4.83) CC BY-NC-ND. 2011, J. M. Powers.
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4.2. WORK 93 P V V 1 V 2 P 1 =P atm 1 2 1 k/A 2 1 W 2 = ∫ PdV 1 2 Figure 4.11: Sketch of P V diagram in piston-linear spring problem.
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