Thermodynamics: Understanding the Behavior of Pure Substances - Chapter III - Prof. Arland, Study notes of Thermodynamics

Download Thermodynamics: Understanding the Behavior of Pure Substances - Chapter III - Prof. Arland and more Study notes Thermodynamics in PDF only on Docsity! ENSC 2213 THERMODYNAMICS 1 ENSC 2213 THERMODYNAMICS Chapter III Part 1 EVALUATING PROPERTIES AJ Johannes, Ph.D., P.E. Professor of Chemical Engineering School of Chemical Engineering Oklahoma State University Copies of Class Lecture Notes Portions of these notes were originally compiled by Professor Robert L. Robinson, Jr. ENSC 2213 THERMODYNAMICS 2 Chapter III GENERAL BEHAVIOR OF PURE SUBSTANCES We have seen in the previous chapter that to use the First Law requires us to know the properties (e.g., internal energy etc) of the substance under study A. Objectives 1. Understand the general (qualitative) behavior of pure substances. 2. Provide the basis for evaluating quantitative (numerical) values of the properties of pure substances. >>> Sections 3.1 – 3.2.2 of your text ENSC 2213 THERMODYNAMICS 5 QUALITATIVE BEHAVIOR OF PURE SUBSTANCES We can gain an understanding of the nature of the relation among pressure, volume and temperature for a pure fluid by considering the behavior of 1 lbm of the fluid confined in a piston—cylinder arrangement, as shown below: Piston 1 lbm of fluid under the piston piston PISTON 1 lbm of fluid ENSC 2213 THERMODYNAMICS 6 HEATING A FLUID AT CONSTANT PRESSURE Consider heating a pure substance (like water) at constant pressure from an initial liquid state (room temperature, 70°F, and pressure, 1 atm., for water). The process would proceed as shown: ENSC 2213 THERMODYNAMICS 7 Constant Temperature Expansion Consider expanding a fluid from an initial liquid state at high pressure while keeping the fluid temperature constant. The process would, proceed as shown below: ENSC 2213 THERMODYNAMICS 10 B. Tabulated Properties The pVT Surface ENSC 2213 THERMODYNAMICS Pressure ‘Critical Liquid point Pressure v Triple point Vapor Pressure Triple line — Solid-vapor Temperature (b) 1 Specific volume (c) ENSC 2213 THERMODYNAMICS p,.= 22.09 MPa (3204 Ibf/in.2) 10 MPa T+ 5 Liquid Vapor a 1.014 bar (14.7 Ibffin.2) 5 20°C (68°F) Specific volume 12 ENSC 2213 THERMODYNAMICS 15 1. Saturated Property Region Example: Water Table A-2 (Temperature Table) Table A-3 (Pressure Table) ENSC 2213 THERMODYNAMICS fable A-2 Properties of Saturated Water {Liquid-Vapor): Temperature Table Tables in $1! Units iprsgsave Genueratn Specific Volume Enthalpy Entnpy bar = es) nike If Livkig * K, ee Bat, Sut, Sut Set Si sa] Sat. Temp. Press. | Liquid | Vaper Liquid Liquid] evap. | Vapor |Liguid | vaper c bar fo Koy ty iy hy Ss ty ol dort | 14002- | 266,136 0.c0 25003 -}2501.4 lanoon forsee | o1 4 0.00813, | 1.001 16.77 | 2380.9 | 16.78 ]2491.9 ||2508.7, (0.0610: }.USId f+ 5 olos'ré-|-.1.800E . | 147.120 3382.3, (o.a76) [9n2s7 f- 5. 6 0.00935 | -1enin) » | 137.734 33.6 ouye2 j9.o003 6 8 voier2 | Leone | 120.917 2386.4 0.1212. }8.98CL “f' 8 10 0.01228 | a.0uua | 106.379 2389.2 xvocs | 10 I Ouisl2 | 7.0004 | 99887 16.20 | 2300.5 0.1658 [8.8765 | t1 12 thnso2 93.784 23919 9.1808 |g.g24 | (2 W 0.61497 2303.3 0.1953 ]8.8285 | 13 4 a.cisys | 1.0008 2304.7 u2ney sane | 1 15 .or7as | t.ego9 2396.1 g78ta 1 1s 16 01813]. 1.0011 2307.4 8.7582 7/16 17 o.cisaa | 1.0012 2398.8 i751. 17 Ww 1.0084 2400.2 8.7123 - 18 w 1.0036 2016 8.6897 19 1.0018 2402.9 0.2966 [8.6672 20 v.c24u7 | 1.0020 86450 4) 0.82644 | 1.0092 gen 22 0.c2810 | 1,0024 BoUH 13 0.62985 | 1.0027 85704 j Ud 25 0.03169 | 1.09 209.8 | 104.bY 8.5580 | 25 26 0.03363. | 1.0032 241i ¢ 109.07 xs3e7. | 26 27, 0.03567; | 1.0035 24125 | 113.95 8.5156 || 27 26 0.03782}, 1.0037 tia} 17.43 4946 |} 28 29 8.04008, | -1.0040 215.2. 4-121.61 ea739 -|-24 a vues | 1.0043 | M166 0 31 06 | OBE | Misa a at 9.04759 i Be n u.5034 3 3 0.08324 M 33 0.05628 |. Jog60 146.67 2418.6 35 36 F 9.08847 2 L008a 180.85 5 MIG. 36 38 4.06633, jo 1.0071 (59.20. | 24274. |. 159.28 |24ib.s 38 au a.o7a84.| 1.0078 17s [24M |: 167.57. -|2406,7 40 a 0.09593 "| 1.0g99 Iga, | 2496.8 2704.8 arity] as, 16 9. ENSC 2213 THERMODYNAMICS Table A-3 Properties of Saturated Water (Liquid~Vapor): Pressure Table Specific Volume | Internal Energy Enthalpy Entropy mikg kivkg kitkg iékg - K Sat, Sat. Sat. Sat. Sat. Sat. Sat. Sat, Temp. | Liquid | Vapor | Liquid | Vapor | Liquid | Bvap. | Vapor | Liquid | Vapor clu x 10 bg ue & Ss 28,96: "L.0040 |" 34.800 12.45: 08226 | 8.4746 36.16 |. 40064 + 23.939 05210 | 83304 | AL St | 1.0084 | ta.i03 05926 | 82387 45:81 | 10102} 14674 016493 |- 8.1502 60.06 |-4.0172 |. 7.689 0.8320 | 7.9085 69.10] 1.0223) 5.229 0.9439 | 7.7686 75.87| 1.0265) 3.993 1.0259 | 7.6700 81.33] 1.0300 | 3.240 1.0910 ] 7.5939 85.94) 10331 | 2.732 1.1453] 7.5320 39.95} 1.0360] 2365 1.1919] 7.4797 93.50, 16380] 2.087 “19339 | 7.4346 96.71 | Logie |. 1369 1.2695 | 7.3049 99.63 | 150492") 4.694 13026} 7.35944. 411.4] 1.0528 | 1.159 #4336 | 7.2233 120.2: | Lb6d5.| 0.8857 45301) 7.1271 274 | 1.06721 0.7187 1.6072 | 7.0527 133.6 | 1.0732] 0.6088 16718] 6.9919 138.9 | 1.0786 | 0.5243 1.7275 | 6.9405 143.6 | 1.0836 | 0.4625 1.7766 | 6.8959 147.9 | 1.0882] 0.4140. 6.8565 1519-1 tose | a374d 68213 158.9 03157 6.7606 165.0 0.2729 6.7680 1704 0.2404 6.6628 175.4 0.2150 6.6226 1799 0.1944 6.5863 198.3 0.1318 6.4448 0 | 2124 0.09963 6.3409 250 | 224.0 0.07998 6.2575 0.06668 0.02737 5.8133 80.0 | 295.1 0.02352 5.7432 | 80.0 90.0 | 303.4 0.02048 5.6772]. 90.0 100, | 3111 0.01803, 5.6141 | 100. 10. | 318.2 0.01599 5.5527 | 110. 17 ENSC 2213 THERMODYNAMICS 20 Single Interpolation (cont’d.) T2 T1 v1 v2v T(?) p = constant Known o 0.1850 0.1757T 500 (600 500) 0.1996 0.1757 T 538.9 C ⎛ ⎞− = + −⎜ ⎟−⎝ ⎠ = ENSC 2213 THERMODYNAMICS 21 Example page 90 Determine the specific volume of water vapor at a state where p = 10 bar and T = 215°C. From Table A-4 Dr. AJ’s Way ( ) ( )( ) 240 200 215 200 215 200 240 200 o o o o C C C C v v v v − = + − − 3 215 0.2141o C mv kg = 215 200o oC C dvv v T dT = + Δ 1 1 2 1 2 1 v v T T v v T T − − = − − v .2060 215 200 .2275 .2060 240 200 − − = − − ENSC 2213 THERMODYNAMICS 22 2. Saturation Property Interpolation If asked to find a saturation property (e.g., psat @ Tgiven or hg @ pgiven), just interpolate linearly along the saturation line. Example Problem: Find uf for water at ps = 82 bars. From Table A-3 ps (bar) uf (kJ/kg) 80 1305.6 82 ? 90 1350.5 1 1 2 1 2 1 f f f p p u u p p u u u 1305.682 80 u @82bars 90 80 1350.5 1305.6 u 1314.6 kJ/kg − − = − − −− = ⇒ − − = ENSC 2213 THERMODYNAMICS 25 T p = 4.0 p = 4.2 p = 6.0 350 0.06645 0.04331 380 0.07065 0.06812 0.04535 400 0.07341 0.04739 Double Interpolation (cont’d.) Substance: Water Given: T = 380°C p = 4.2 MPa Find: v = ?? Step 3 Interpolate at T = 380°C (or at 4.2 MPa) as usual to get v Thus, v = 0.06812 m3/kg Volumes ENSC 2213 THERMODYNAMICS 26 C. The State Principle State Principle: Any two independent properties are sufficient to fix the state of a simple compressible system in equilibrium. As you know by now, the variables that describe the state of equilibrium (at this stage of our study) are: T, p, v, u This means u = f (T,p) p = f (T,v) v = f (T,p) u = f (T,v) T = f (p,v) ... and so on ?? Are T and p always independent ?? ENSC 2213 THERMODYNAMICS 27 D. Behavior in the Liquid-Vapor Mixture Region Consider a vaporization process at constant pressure, p = p1 (and, thus, constant temperature T = T1) (T and p constant) Note that although the total volume, V, changes, both specific volumes, vf and vg, remains constant (for fixed values of p, T). vf vgvf vg vg vf ENSC 2213 THERMODYNAMICS 30 Example 2: Given: Two kg of H2O are confined in a rigid volume of 0.1 m3 at 205 °C. V = 0.1 m3 m = 2 kg Solution: System: Water 1. Determine the State of the System at 205°C; the properties of water are as shown in the figure 0.001164 = vf vg = 0.11521 m3/kg v 1.723 MPa P 205oC ENSC 2213 THERMODYNAMICS 31 total volume vapor volume liquid volume Now v = V/m = 0.1/2 = 0.05 m3/kg So vf < v < vg ⇒ 2 - phase 2. Calculate the System Pressure 3. Calculate the System Quality 4. Calculate Volume % Vapor sat'd p p 1.723 MPa.= = f g f v v 0.05 0.001164x v v 0.11521 0.001164 − − = = − − ( ) 0.428 42.8%x or Mass fraction vapor = g g f fV m v m v= + % 100g g m v Volume Vapor V = × 100g g m v mv = × % ( / ) 100gVolume Vapor x v v x= ENSC 2213 THERMODYNAMICS 32 Thus Vol. % Vapor = Vol. % Liquid = Thus Vol. % Vapor = 100g x v v × ( )1 100f x v v − × ( )( )0.428 0.1152 100 0.05 x = 98.7%