At v 1 0 02 m 3 t 1 50 c when p 100 kpa the piston

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at V 1 = 0 . 02 m 3 , T 1 = 50 C . When P = 100 kPa , the piston leaves the stops. The water is heated from its initial state to a final state of 200 C . Find diagrams for the process in the P T , T v , and P v planes and the work done by the water. At state 1, we have v 1 = V 1 m = 0 . 02 m 3 2 kg = 0 . 01 m 3 kg . (4.61) Go to the saturated water tables, temperature entry. At T 1 = 50 C , we find v f = 0 . 001012 m 3 /kg and v g = 12 . 0318 m 3 /kg. This gives v fg = 12 . 0308 m 3 /kg. Since v f < v 1 < v g , we have a two-phase mixture at the initial state. This fixes P 1 = 12 . 350 kPa . Now we have x 1 = v 1 v f v fg = parenleftBig 0 . 01 m 3 kg parenrightBig parenleftBig 0 . 001012 m 3 kg parenrightBig 12 . 0308 m 3 kg = 0 . 000747083 . (4.62) Next, heat at constant volume until the piston leaves the stops at P = 100 kPa , or the fluid becomes saturated. We search the saturated water tables and examine the state at P = 100 kPa . We see at that pressure that v f = 0 . 001043 m 3 /kg and v g = 1 . 69296 m 3 /kg . So when P reaches 100 kPa , we still have v f < v < v g , so the water is still a two-phase mixture. CC BY-NC-ND. 2011, J. M. Powers.
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4.2. WORK 89 v v T T P P 1 2 3 1 2 3 1 2 3 Figure 4.8: Sketch of T v , P v , and P T diagrams. We define then state 2 as the state where P 2 = 100 kPa with v 2 = v 1 = 0 . 01 m 3 /kg . From here on the process is isobaric. It is useful at this stage to consider a sketch of the processes given in Fig. 4.8. Now at P 2 = 100 kPa , we find that T 2 = 99 . 62 C . And we have v 2 = v 1 = 0 . 01 m 3 /kg . Using v f and v g at P = 100 kPa , we find x 2 = v 2 v f v fg = parenleftBig 0 . 01 m 3 kg parenrightBig parenleftBig 0 . 001043 m 3 kg parenrightBig 1 . 69296 m 3 kg = 0 . 00529073 . (4.63) Now we heat isobarically until T 3 = 200 C , with P 3 = P 2 = 100 kPa . This gives us two properties, so we know the state. The superheat tables give us v 3 = 2 . 17225 m 3 /kg . Now the final volume is V 3 = mv 3 = (2 kg ) parenleftbigg 2 . 17226 m 3 kg parenrightbigg = 4 . 34452 m 3 . (4.64) Let us get the work. 1 W 3 = 1 W 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =0 + 2 W 3 . (4.65) But 1 W 2 = 0 since this is an isochoric process. So 1 W 3 = 2 W 3 = integraldisplay 3 2 PdV = P 2 integraldisplay 3 2 dV = P 2 ( V 3 V 2 ) = P 2 m ( v 3 v 2 ) . (4.66) Substituting numbers, we find 1 W 3 = (100 kPa )(2 kg ) parenleftbiggparenleftbigg 2 . 17226 m 3 kg parenrightbigg parenleftbigg 0 . 01 m 3 kg parenrightbiggparenrightbigg = 432 . 452 kJ. (4.67) A summary table is given in Table 4.1. Example 4.11 Measured P V data for an internal combustion engine is obtained. Estimate the work. CC BY-NC-ND. 2011, J. M. Powers.
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90 CHAPTER 4. WORK AND HEAT Table 4.1: Numerical values for steam heating example variable units state 1 state 2 state 3 P kPa 12 . 350 100 151 . 665 v m 3 kg 0.01 0.01 2.17255 T C 50 99.62 200 x - 0 . 000747083 0 . 00529073 - V m 3 0.02 0.02 4.34452 Table 4.2: Values for P and V in an expansion process. P ( bar ) V ( cm 3 ) 20.0 454 16.1 540 12.2 668 9.9 780 6.0 1175 3.1 1980 The data is given in Table 4.2. Here we have N = 6 points. The best way to address this problem is via numerical integration. We could use a variety of methods like Simpson’s rule. Let us take what amounts to the trapezoidal method. We will estimate 1 W 2 = integraldisplay 2 1 PdV N 1 summationdisplay i =1 P ave i Δ V i . (4.68) Here we are taking the area of trapezoid i
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