solns to practice set 2a 2010

# solns to practice set 2a 2010 - 3.66 3.69 A 1 3 tank is...

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Unformatted text preview: 3 .66 3 .69 A 1- 3 tank is ﬁlled with a gas at room temperature 20°C and pressure 100 kPa. How much mass is there if the gas is a) air, b) neon or c) propane '3 Solution: Use Table A2 to compare T and P to the critical T and P with T=20°C=293.15K; P=100kPa€cZ PC forall Air: T b“; TQM; Tom = 154.6 K so ideal gas; R: 0.287 kJJ'kg K Neon: T >> Tc = 44.4 K so ideal gas; R = 0.41195 11ng K Propane: T =1 Ta = 3'10 K, but P <41 PC = 4.25 MPa so gas R = 0.18855 kJa’kg K All states are ideal gas states so the ideal gas law applies PV = mRT PV 100><1 ‘9 mZEZozsa x 293.15 2 1'1” kg 100 x 1 0.41195 x 293.15 100 x] 0.18855 X29115 b) m = = 0.328 kg c) In = = 1.809 kg Is it reasonable to assume that at the given states the substance behaves as an ideal gas? Solution: a) Oxygen, 02 at 30°C, 3 MPa Ideal Gas (T >> To = 155 K from 131.2) b) Methane, CH4 at 30°C, 3 MPa Ideal Gas ( T )} To = 190 K from A.2) c) Water, H20 at 30°C, 3 MPa N0 compressed liquid P > PS,“ (B11) (1) R—134a at 30°C, 3 MPa N0 compressed liquid P > PS,“ (B51) e) R—134a at 30°C, 100 kPa Ideal Gas P is low <1. Pm (35.1) 3.?2 3.88 A spherical helium balloon of 10 m in diameter is at ambient T and P, 15°C and 100 kPa. How much helium does it contain? It can lift a total mass that equals the mass of displaced atmospheric air. How much mass of the balloon fabric and cage can then be lifted? We need to ﬁnd the masses and the balloon volume _E 3_E 3_ 3 V—ﬁD —610 —523.6m _ V_ PV_ 100 kPa x 523.6 m3 1? RT 2.0771 kakgK x 288 K mHE = pV = 87.5 kg m _Pv 10I01cPax523.ﬁm3 air RT = 0.28? kakgK x 288 K = 633 kg mliﬁ = mar—mHe = 633-815 = 545.5 kg What is the percent error in speciﬁc volume if the ideal gas model is used to represent the behavior of superheated ammonia at 40°C, 500 kPa'? What if the generalized compressibility chart, Fig. D. 1, is used instead? Solution: NH3 T =40°C = 313.15 K, Tc =405.5 K, PC = 11.35 MPa ﬁ'om TableAJ Table 13.2.2: v = 0.2923 m3mg RT _ 0.43819 X 313 Ideal gas: v = ? — 500 = 0.3056 msfkg 2:) 4.5% error _ _ 313.15 0.5 FigureDJ. Tr: 405-5 =0.7'72, PI=—11-35=0.044 3 2:11.97 ZRT 3 v = T = 0.2964 In fkg :b 1.4% error 3.92 Find the volume of 2 kg of ethylene at 270 K, 2500 kPa using Z ﬁ'om Fig. D. 1 Ethylene Table A2: Tc = 282.4 K, PC = 5.04 MPa Table AS: R = 0.2964 kag K The reduced temperature and pressure are: T 270 P 2.5 Tr=T =282_4=0.956, Pr=P =5_04=0.496 C C Enter the chart with these coordinates and read: Z = 0.?6 anRT 2 k x 0.?6 x 0.2964 kak -K x 270 K _ _ _ 3 V — P — 2500 kPa — 0.0487 n1 3.97 A new reﬁ‘igerant R-125 is stored as a liquid at -20 “C with a small amount of vapor. For a total of 1.5 kg R-125 ﬁnd the pressure and the volume. Solution: As there is no section B table use compressibility chart. Table A2: R-125 TC = 339.2 K PC = 3.62 MP3. Tr: TI'TC = 253.15 J“ 339.2 =0.746 We can read ﬁ‘om Figure D. 1 or a little more accurately interpolate from table D.4 entries: Pr sat : 0-16; Zg : 0-86; 2f: 0.029 P=P P =0.16x3620kPa=579kPa rsat C qu = meﬁq RTFP = 0.029 xl.5 kg x 0.06927:Ir kakgK x 253.15 KI 579 kPa = 0.0013 1113 Z sat vapor ...
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