For a cpig k c p c v c p c v constant 593 example 53

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For a CPIG, k = c P c v = c P c v = constant. (5.93) Example 5.3 Given Ar at P 1 = 140 kPa , T 1 = 10 C , V 1 = 200 which undergoes a polytropic expansion to P 2 = 700 kPa , T 2 = 180 C , find 1 Q 2 CC BY-NC-ND. 2011, J. M. Powers.
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118 CHAPTER 5. THE FIRST LAW OF THERMODYNAMICS To calculate 1 Q 2 , we will need to invoke the first law: U 2 U 1 = 1 Q 2 1 W 2 . Now for the CPIG, we will be able to calculate Δ U from knowledge of Δ T , and we can easily compute 1 W 2 from its definition as integraltext 2 1 PdV . This will let us calculate the heat transfer. Ar is a noble gas. Because it is noble, it is well modeled as a CPIG over a wide range of temperatures. From Table A.5 in BS, we find the data M = 39 . 948 kg kmole , R = 0 . 2081 kJ kg K , c v = 0 . 312 kJ kg K . (5.94) Note that c v R = 0 . 312 kJ kg K 0 . 2081 kJ kg K = 1 . 49928 3 2 . (5.95) This result will be seen to be valid for all monatomic gases, such as He , Ne , H , O , etc. Also note that MR = parenleftbigg 39 . 948 kg kmole parenrightbiggparenleftbigg 0 . 2081 kJ kg K parenrightbigg = 8 . 31318 kJ kmole K = R. (5.96) Now we need to have absolute temperatures. Adding 273 . 15 to both temperatures, we get T 1 = 283 . 15 K, T 2 = 453 . 15 K. (5.97) Also V 1 = (200 ) 0 . 001 m 3 = 0 . 2 m 3 . (5.98) Let us compute the mass m . From one incarnation of the ideal gas law, we have m = P 1 V 1 RT 1 = (140 kPa )(0 . 2 m 3 ) parenleftBig 0 . 2081 kJ kg K parenrightBig (283 . 15 K ) = 0 . 475192 kg. (5.99) Now at state 2, we have P 2 V 2 = mRT 2 , (5.100) V 2 = mRT 2 P 2 = (0 . 475192 kg ) parenleftBig 0 . 2081 kJ kg K parenrightBig (453 . 15 K ) 700 kPa = 0 . 0640155 m 3 . (5.101) Now for the polytropic process, we have P 1 V n 1 = P 2 V n 2 , (5.102) P 1 P 2 = parenleftbigg V 2 V 1 parenrightbigg n , (5.103) ln parenleftbigg P 1 P 2 parenrightbigg = ln parenleftbigg V 2 V 1 parenrightbigg n , (5.104) ln parenleftbigg P 1 P 2 parenrightbigg = n ln parenleftbigg V 2 V 1 parenrightbigg , (5.105) n = ln parenleftBig P 1 P 2 parenrightBig ln parenleftBig V 2 V 1 parenrightBig , (5.106) = ln ( 140 kP a 700 kP a ) ln ( 0 . 0640155 m 3 0 . 2 m 3 ) = 1 . 41279 . (5.107) CC BY-NC-ND. 2011, J. M. Powers.
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5.5. CALORIC EQUATIONS OF STATE. 119 We have calculated the work for a polytropic process before. Repeating, we find 1 W 2 = integraldisplay 2 1 PdV = P 1 V n 1 integraldisplay 2 1 dV V n = P 1 V n 1 bracketleftbigg V 1 n 1 n bracketrightbigg V 2 V 1 = P 1 V n 1 1 n ( V 1 n 2 V 1 n 1 ) . (5.108) But since P 1 V n 1 = P 2 V n 2 , the work reduces to simply 1 W 2 = P 2 V 2 P 1 V 1 1 n = (700 kPa )(0 . 0640155 m 3 ) (140 kPa )(0 . 2 m 3 ) 1 1 . 41279 = 40 . 7251 kJ. (5.109) The work is negative, so the gas was worked upon in compression. Now the first law tells us U 2 U 1 = 1 Q 2 1 W 2 , (5.110) 1 Q 2 = U 2 U 1 + 1 W 2 , (5.111) = mc v ( T 2 T 1 ) + 1 W 2 , (5.112) = (0 . 475192 kg ) parenleftbigg 0 . 312 kJ kg K parenrightbigg ((453 . 15 K ) (283 . 15 K )) + ( 40 . 7251 kJ ) , (5.113) = 15 . 5209 kJ. (5.114) The heat transfer is negative, so heat was lost from the system, even though the temperature went up . The reason the temperature went up is that the internal energy was raised more by work than it lost by heat transfer. A plot of this process is given in Fig. 5.10. 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0 200 400 600 800 1000 1 2 T 2 = 453.15 K T 1 = 283.15 K P (kPa) v (m 3 /kg) Figure 5.10: Plot of polytropic compression of Ar .
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