SOLUTIONS to Problem Set I - EEE 2010

Solution to problem 337 svna b 534cm3mol c 2620

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Unformatted text preview: Table F.2. Mol. Wt. = 18.015 gm/mol; V = 2252 cm3/mol. Ans. V = 2268cm3/mol Ans. V = 124.99 cm3. Mol. Wt. /gm k. Pc = 220.55 bar; Tc = 647.1 K For R/K; a = 1.427 ×108 bar K1/2/mol2; b = 21.1345 cm3/mol ; Rewrite as / Substitute and rearrange to get V = 26.4048, 191.9724 & 30398.11 Vl = 26.4048/18 = 1.466cm3/gm Vg = 30398.11/18 = 1688 cm3/gm Similarly for VDV, Vl = 2.167 cm3/gm; Vg = 1692.65cm3/gm Steam Table: Vl = 1.044cm3/gm, Vg = 1673 cm3/gm VDW slightly over predicts the values l. Solution to problem 3.37 (SVNA) B = -53.4cm3/mol; C = 2620 cm6/mol2; D = 5000 cm9/mol3; Given 1 ; f(P,V) = Find (V) -10 i=0-10; Pi = (10 + 20i) bar; Vi = RT/Pi (guess) , Zi = Eq. (3.12) 1 1 ; Eq (3.38); 2 ; T = 273.15K Eq (3.39) Plot Z, Zi, Z1i, Z2i vs. Pi for different values of i m. Solution to problem 3.45 (SVNA) (Abbott Correlation) T = 298.15K; Tc = 425.1 K1; Tr = T/Tc; Tr = 0.701 Pr = P/Pc; Pr = 0.064 P = 2.43bar; Pc = 37.96 bar; ω = 0.200; Vvap = 16m3; mol. Wt. = 58.123 gm/mol B0 = -0.661; B1 = -0.624 ; V = 9.469×103 cm3/mol mvap = Vvap / (V / mol. Wt.); mvap = 98.213 kg Ans. n. Solution to problem 3.46 (SVNA) (Lee-Kesler Correlation) Part (a) T = 333.15K; Tc = 305.3K; Tr=T/Tc; Tr = 1.901 P = 14000 kPa; Pc = 48.72 bar; Pr = P/Pc; Pr = 2.874 Ω=0.100; Vtotal = 0.15m3; mol. Wt. = 30.07 gm/mol Z1 = -0.037 From tables E3 and E4, Z0 = 0.463; Z = Z0 + ωZ1; Z = 0.459; V = ZRT / P; V = 90.87 cm3/mol methane = Vtotal / (V/mol.wt); methane = 49.64kg Ans. Part (b) V = Vtotal/40 kg; P = 20000kPa; PV=ZRT = ZRTcPc Or Tr = α/Z where α = PV / RTc; α = 29.548 mol/kg Whence Tr = 0.889/Z at Pr = P/Pc; Pr = 4.105 This equation giving Tr as a function of Z and Eq. 3.57 in conjunction with tables E3 and E4 are two relations in the same variables which must be satisfied at the given reduced pressure. The intersection of these two relations can be found by one means or another to occur at about: Tr = 1.283 and Z = 0.693 Whence T = Tr. Tc T = 391.7K or 118.5°C Ans. B. LIQUIDS a. Solution to problem 3.53 (SVNA) (Lyderson charts) For n-pentane Tc = 469.7K; Pc = 33.7 bar; ρ1 = 0.63 gm/cm3 T1 = 291.15 K1; P1 = 1 bar; T2 = 413.15K; P2 = 120 bar Tr1 = 0.62; Pr1 = 0.03; Tr2 = 0.88; Pr2 = 3.561 From Fig. 3.16 ρr1 = 2.69; ρr2 = 2.27 By Eq. 3.75 ρ2 = ρ1 (ρr2/ρr1); ρ2 = 0.532 gm/cm3 Ans. b. Solution to problem 3.54 (SVNA) (Lyderson charts) For ethanol Tc = 513.9 K; T = 453.15K; Tr = T/Tc; Tr = 0.882 Pc = 61.48 bar; P = 200bar; Pr = P/Pc; Pr = 3.253 Vc = 167 cm3/mol; mol. Wt. = 46.069 gm/mol From Fig. 3.16; ρr = 2.28; ρ = ρr.ρc = ρr / Vc ρ = ρr / (Vc/molwt); ρ = 0.629 gm/cm3 Ans. c. Solution to problem 3.55 (SVNA) (Rackett) For ammonia Tc = 405.7 K; T = 293.15K; Tr = T/Tc; Tr = 0.723 Pc = 112.8 bar; P=857kPa; Pr = P/Pc; Pr = 0.076 ω=0.253 Vc = 72.5 cm3/mol; Zc = 0.242; Eq. 3.72 B0 = -0.627; . ; Vliquid = 27.11 cm3/mol B1 = -0.534 ; Vvapor = 2616 cm3/mol ΔV = 2589 cm3/mol Ans. ΔV = Vvapor – Vliquid; d. Solution to problem 3.44 (SVNA) (Rackett, Liquid + Vapor) T = 320 K1; P = 16 bar; Tc = 369.8K; Pc = 42.48 bar 3 Ω = 0.152; Zc = 0.276; mol Wt. = 44.097 gm/mol Vc = 200cm /mol; Tr = T/Tc; Pr = P/Pc; Tr = 0.865; Pr = 0.377 . ; B0 = -0.449; mliq = 0.8Vtank / (Vliq/molwt); mliq = 127.594 kg Ans. B1 = -0.177 ; Vvapor = 1.318 ×103 cm3/mol mvap = 0.2Vtank / (Vvap/mol wt); e. mvap = 2.341 kg Ans. . ; Zc = 0.229; Vc = 55.9 cm3/mol; Tr = 373.15/647.1 = 0.5765 Tc = 647.1K . . Vsat = 55.9 0.229 sat 3 V = 17.646cm /mol or 0.9803 cm3/g Experimentally it is 1.044 cm3/gm which is very close....
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