Fundamentals of Thermodynamics SI Version

Transcription

1 Solution Manual Chapter 2 Fundamentals of hermodynamics SI Version Borgnakke Sonntag 8e Updated July 2013

2 2 Borgnakke and Sonntag CONEN CHAPER 2 SUBSECION PROB NO. Concept problems 1-15 Phase diagrams, triple and critical points General tables Ideal gas Compressibility factor Equations of state Reiew problems Linear interpolation Computer tables

3 3 Borgnakke and Sonntag In-ext Concept Questions

4 4 Borgnakke and Sonntag 2.a If the pressure is smaller than the smallest P sat at a gien, what is the phase? Refer to the phase diagrams in Figures 2.4 and 2.5. For a lower P you are below the aporization cure (or the sublimation cure) and that is the superheated apor region. You hae the gas phase. ln P S S L Critical Point Vapor 2.b An external water tap has the ale actiated by a long spindle so the closing mechanism is located well inside the wall. Why is that? By haing the spindle inside the wall the coldest location with water when the ale is closed is kept at a temperature aboe the freezing point. If the ale spindle was outside there would be some amount of water that could freeze while it is trapped inside the pipe section potentially rupturing the pipe.

5 5 Borgnakke and Sonntag 2.c What is the lowest temperature (approximately) at which water can be liquid? Look at the phase diagram in Fig At the border between ice I, ice III and the liquid region is a triple point which is the lowest where you can hae liquid. From the figure it is estimated to be about 255 K i.e. at -18 o C. 255 K - 18 C ln P S L lowest liquid CR.P. V

6 6 Borgnakke and Sonntag 2.d Some tools should be cleaned in water at a least 150 o C. How high a P is needed? If I need liquid water at 150 o C I must hae a pressure that is at least the saturation pressure for this temperature. able B.1.1: 150 o C, P sat = kpa.

7 7 Borgnakke and Sonntag 2.e Water at 200 kpa has a quality of 50%. Is the olume fraction V g /V tot < 50% or > 50%? his is a two-phase state at a gien pressure and without looking in the table we know that f is much smaller than g. From the definition of quality we get the masses from total mass, m, as m f = (1 x) m, he olumes are V f = m f f = (1 x) m f, m g = x m V g = m g g = x m g So when half the mass is liquid and the other half is apor the liquid olume is much smaller that the apor olume. he apor olume is thus much more than 50% of the total olume. Only right at the critical point is f = g for all other states g > f and the difference is larger for smaller pressures.

8 8 Borgnakke and Sonntag 2.f Why are most of the compressed liquid or solid regions not included in the printed tables? For the compressed liquid and the solid phases the specific olume and thus density is nearly constant. hese surfaces are ery steep nearly constant and there is then no reason to fill up a table with the same alue of for different P and. 2.g Why is it not typical to find tables for Ar, He, Ne or air like an Appendix B table? he temperature at which these substances are close to the two-phase region is ery low. For technical applications with temperatures around atmospheric or higher they are ideal gases. Look in able A.2 and we can see the critical temperatures as Ar : K He: 5.19 K Ne: 44.4 K It requires a special refrigerator in a laboratory to bring a substance down to these cryogenic temperatures.

9 9 Borgnakke and Sonntag 2.h What is the percent change in olume as liquid water freezes? Mention some effects the olume change can hae in nature and in our households. he density of water in the different phases can be found in ables A.3 and A.4 and in able B.1. From able B.1.1 f = m 3 /kg From able B.1.5 i = m 3 /kg Percent change: 100 i f = 100 f = 9.1 % increase Liquid water that seeps into cracks or other confined spaces and then freezes will expand and widen the cracks. his is what destroys any porous material exposed to the weather on buildings, roads and mountains. It can burst water pipes and crack engine blocks (that is why you put anti-freeze in it).

10 10 Borgnakke and Sonntag 2.i How accurate is it to assume that methane is an ideal gas at room conditions? From able A.2: c = K, P c = 4.60 MPa So at room conditions we hae much higher > c and P << P c so this is the ideal gas region. o confirm look in able B kpa, 300 K, = m 3 /kg Find the compressibility factor (R from able A.5) as Z = P/R = 100 kpa m3 /kg kj/kg-k 300 K = so Z is 1 with an accuracy of 0.2% better than most measurements can be done.

11 11 Borgnakke and Sonntag 2.j I want to determine a state of some substance, and I know that P = 200 kpa; is it helpful to write PV = mr to find the second property? NO. You need a second property. Notice that two properties are needed to determine a state. he EOS can gie you a third property if you know two, like (P,) gies just as you would get by entering a table with a set (P,). his EOS substitutes for a table when it is applicable.

12 12 Borgnakke and Sonntag 2.k A bottle at 298 K should hae liquid propane; how high a pressure is needed? (use Fig. D.1) o hae a liquid the pressure must be higher than or equal to the saturation pressure. here is no printed propane table so we use the compressibility chart and able A.2 Propane able A.2: c = K, P c = 4.25 MPa he reduced temperature is: r = c = = 0.806, for which we find in Fig. D.1: P r sat = 0.25 P = P r sat P c = MPa = 1.06 MPa

13 13 Borgnakke and Sonntag 2.l A bottle at 298 K should hae liquid propane; how high a pressure is needed? (use the software) o hae a liquid the pressure must be higher than or equal to the saturation pressure. here is no printed propane table but the software has propane included Start CA3, select cryogenic substances, propane select calculator, select case 4 (, x) = (25 o C, 0) P = MPa

14 14 Borgnakke and Sonntag Concept Problems

15 15 Borgnakke and Sonntag 2.1 Are the pressures in the tables absolute or gauge pressures? he behaior of a pure substance depends on the absolute pressure, so P in the tables is absolute.

16 16 Borgnakke and Sonntag 2.2 What is the minimum pressure for which I can hae liquid carbon dioxide? Look at the phase diagram in Fig he minimum P in the liquid phase is at the triple point. From able 2.2 this is at 520 kpa (a similar alue around kpa is seen in Fig. 2.5). ln P he 100 kpa is below the triple point. 100 kpa S L V

17 17 Borgnakke and Sonntag 2.3 When you skate on ice, a thin liquid film forms under the skate. How can that be? he ice is at some temperature below the freezing temperature for the atmospheric pressure of 100 kpa = 0.1 MPa and thus to the left of the fusion line in the solid ice I region of Fig As the skate comes oer the ice the pressure is increased dramatically right under the blade so it brings the state straight up in the diagram crossing the fusion line and brings it into a liquid state at same temperature. he ery thin liquid film under the skate changes the friction to be iscous rather than a solid to solid contact friction. Friction is thus significantly reduced. Comment: he latest research has shown that the pressure may not be enough to generate the liquid, but that such a liquid layer always exist on an ice surface, maybe only a few molecules thick (dependent upon temperature). At really low say -40 o C no such liquid layer exists which is why your finger can stick to such a surface.

18 18 Borgnakke and Sonntag 2.4 At a higher eleation like in mountains the pressure is lower, what effect does that hae for cooking food? A lower pressure means that water will boil at a lower temperature, see the aporization line in Fig. 2.4, or in able B.1.2 showing the saturated temperature as a function of the pressure. You therefore must increase the cooking time a little.

19 19 Borgnakke and Sonntag 2.5 Water at room temperature and room pressure has 1 10 n m 3 /kg what is n? See able B.1.1 or B.1.2 to determine it is in the liquid phase (you should know this already). able A.4 or from B1.1 at 20 o C: n = 3 ( = m 3 /kg) 2.6 Can a apor exist below the triple point temperature? Look at the phase diagrams in Figs 2.4 and 2.5. Below the triple point the sublimation cure has ery small pressures, but not zero. So for pressures below the saturation pressure the substance is a apor. If the phase diagram is plotted in linear coordinates the small apor region is nearly not isible. 2.7 In Example 2.1 b is there any mass at the indicated specific olume? Explain. his state is a two-phase mixture of liquid and apor. here is no mass at the indicated state, the alue is an aerage for all the mass, so there is some mass at the saturated apor state (fraction is the quality x) and the remainder of the mass is saturated liquid (fraction 1-x).

20 20 Borgnakke and Sonntag 2.8 Sketch two constant-pressure cures (500 kpa and kpa) in a - diagram and indicate on the cures where in the water tables you see the properties. MPa P B 1 4 B.1.3 C.P. B.1.3 B MPa B.1.3 B 500 kpa 1 4 B.1.1 B.1.3 B.1.5 B.1.5 he 30 MPa line in able B.1.4 starts at 0 o C and table ends at 380 o C, the line is continued in able B.1.3 starting at 375 o C and table ends at 1300 o C. he 500 kpa line in able B.1.4 starts at 0.01 o C and table ends at the saturated liquid state ( o C). he line is continued in able B.1.3 starting at the saturated apor state ( o C) continuing up to 1300 o C.

21 21 Borgnakke and Sonntag 2.9 If I hae 1 L of R-410A at 1 MPa, 20 o C how much mass is that? R-410A ables B.4: B.4.1 P sat = kpa at 20 o C so superheated apor. B.4.2 = m 3 /kg under subheading 1000 kpa m = V = m m 3 = kg = 35.2 g /kg P C.P. C.P kpa 20 C kpa 7.25 C 7.25 he P- (log-log) diagram from CA3, P in MPa and in m 3 /kg.

22 22 Borgnakke and Sonntag 2.10 Locate the state of ammonia at 200 kpa, -10 o C. Indicate in both the P- and the - diagrams the location of the nearest states listed in the printed able B.2 P C.P. C.P C C kpa 150 kpa

23 23 Borgnakke and Sonntag 2.11 Why are most of the compressed liquid or solid regions not included in the printed tables? For the compressed liquid and the solid phases the specific olume and thus density is nearly constant. hese surfaces are ery steep nearly constant and there is then no reason to fill up a table with the same alue of for different P and.

24 24 Borgnakke and Sonntag 2.12 How does a constant -process look like for an ideal gas in a P- diagram? For an ideal gas: P = R so then P = (R/) Constant is a straight line with slope (R/) in the P- diagram P Critical point S Liquid Vapor P = R/

25 25 Borgnakke and Sonntag 2.13 If = R/P for an ideal gas what is the similar equation for a liquid? he equation for a liquid is: = Constant = o If you include that increases a little with then: = o + C ( o ) where C is a small constant with units m 3 /kg-k.

26 26 Borgnakke and Sonntag 2.14 o sole for gien (P, ) in Eq.2.14, what is the mathematical problem? o sole for in Eq multiply with the de-numerators so we get the form P f 1 () = R f 2 () a ( b) his is a 3 rd order polynomial (a cubic) function in. he problem then is to find the roots or zero points in this cubic equation. he mathematical subject to study is to find zero points of functions (or roots). ypically you will do it by some iteration technique. Successie substitutions, bi-section, Newton-Raphson are some of the methods you should learn.

27 27 Borgnakke and Sonntag 2.15 As the pressure of a gas becomes larger, Z becomes larger than 1. What does that imply? P = Z R So for a gien P, the specific olume is then larger than predicted by the ideal gas law. he molecules are pressed so close together that they hae repulsie forces between them (the electron clouds are getting closer). he ideal gas law assumes the atoms (molecules) are point masses with no interactions between them and thus has a limit of zero specific olume as P goes to infinity. Real molecules occupy some olume and the outer shell has a number of electrons with negatie charges which can interact with one another if they are close enough.

28 28 Borgnakke and Sonntag Phase Diagrams, riple and Critical Points

29 29 Borgnakke and Sonntag 2.16 Carbon dioxide at 280 K can be in different phases. Indicate the pressure range you hae for each of the three phases (apor, liquid and solid). Look at the P- phase diagram in Fig. 2.5 at 280 K: P < 4000 kpa apor 4000 kpa < P < 400 MPa liquid 400 MPa < P solid ln P S L V 280 K

30 30 Borgnakke and Sonntag 2.17 Modern extraction techniques can be based on dissoling material in supercritical fluids such as carbon dioxide. How high are pressure and density of carbon dioxide when the pressure and temperature are around the critical point? Repeat for ethyl alcohol. CO 2 : able A.2: P c = 7.38 MPa, c = 304 K, c = m 3 /kg c = 1/ c = 1/ = 472 kg/m 3 C 2 H 5 OH: able A.2: P c = 6.14 MPa, c = 514 K, c = m 3 /kg c = 1/ c = 1/ = 275 kg/m 3

31 31 Borgnakke and Sonntag 2.18 he ice cap on the North Pole could be 1000 m thick with a density of 920 kg/m 3. Find the pressure at the bottom and the corresponding melting temperature. ICE = 920 kg/m 3 P = gh = 920 kg/m m/s m = Pa P = P o + P = = 9123 kpa See figure 3.7 liquid solid interphase => LS = 1 C

32 32 Borgnakke and Sonntag 2.19 Find the lowest temperature at which it is possible to hae water in the liquid phase. At what pressure must the liquid exist? here is no liquid at lower temperatures than on the fusion line, see Fig. 3.7, saturated ice III to liquid phase boundary is at 263K - 10 C and P 210 MPa ln P S L lowest liquid CR.P. V

33 33 Borgnakke and Sonntag 2.20 Water at 27 C can exist in different phases dependent upon the pressure. Gie the approximate pressure range in kpa for water being in each one of the three phases apor, liquid or solid. he phases can be seen in Fig. 2.4, a sketch of which is shown to the right. = 27 C = 300 From Fig. 2.4: P VL 3 MPa = 4 kpa, P LS = 10 3 MPa ln P S L S V CR.P. 0 < P < 4 kpa VAPOR MPa < P < 1000 MPa LIQUID P > 1000 MPa SOLID (ICE)

34 34 Borgnakke and Sonntag 2.21 Find the lowest temperature in Kelin for which you can see metal as a liquid if the metal is a. mercury b. zinc Assume the two substances hae a phase diagram similar to Fig. 2.5, then the triple point is the lowest with liquid possible the data is from able 2.1 a = -39 o C = 234 K b = 419 o C = 692 K

35 35 Borgnakke and Sonntag 2.22 A substance is at 2 MPa, 17 C in a rigid tank. Using only the critical properties can the phase of the mass be determined if the substance is oxygen, water or propane? Find state relatie to critical point properties which are from able A.2: a) Oxygen O 2 : 5.04 MPa K b) Water H 2 O : MPa K c) Propane C 3 H 8 : 4.25 MPa K State is at 17 C = 290 K and 2 MPa < P c for all cases: O 2 : >> c Superheated apor P < Pc H 2 O : << c ; P << P c you cannot say. C 3 H 8 : < c ; P < P c you cannot say ln P Liquid b Cr.P. c Vapor a

36 36 Borgnakke and Sonntag 2.23 Gie the phase for the following states. a. CO 2 40 C P 0.5 MPa able A.2 > c => also P << P c superheated apor assume ideal gas able A.5 b. Air 20 C P 200 kpa able A.2 superheated apor assume ideal gas able A.5 c. NH C P 600 kpa able B.2.2 or A.2 > c => superheated apor P C.P. a,b,c a, b, c P = const.

37 37 Borgnakke and Sonntag General ables

38 38 Borgnakke and Sonntag 2.24 Gie the phase for the following states. a. H 2 O 260 C P 5 MPa able B.1.1 or B.1.2 B.1.1 For gien read: P sat = MPa P > P sat => compressed liquid B.1.2 For gien P read: sat = 264 C < sat => compressed liquid b. H 2 O 2 C P 100 kpa able B.1.1 < triple point able B.1.5 at 2 C read: P sat = kpa since P > P sat => compressed solid P C.P. a a C.P. Note state b in P-, see the 3-D figure, is up on the solid face. b P = const. b P S b a L C.P. V

39 39 Borgnakke and Sonntag 2.25 Determine the phase of the substance at the gien state using Appendix B tables a) Water 100 C, 500 kpa b) Ammonia -10 C, 150 kpa c) R-410A 0 C, 350 kpa a) From able B.1.1 P sat (100 C) = kpa 500 kpa > P sat then it is compressed liquid OR from able B.1.2 b) Ammonia NH 3 : sat (500 kpa) = 152 C 100 C < sat then it is subcooled liquid = compressed liquid able B.2.1: P < P sat (-10 C) = 291 kpa Superheated apor c) R-410A able B.4.1: P < P sat (0 C) = 799 kpa Superheated apor. he S-L fusion line goes slightly to the left for water. It tilts slightly to the right for most other substances. ln P S L a b,c Vapor Cr.P.

40 40 Borgnakke and Sonntag 2.26 Gie the missing property of P-- and x for water at a. P = 10 MPa, = m 3 /kg b. 1 MPa, 190 C c. 400 C, 0.19 m 3 /kg d. 10 kpa, 10 C For all states start search in table B.1.1 (if gien) or B.1.2 (if P gien) a. P = 10 MPa, = m 3 /kg so look in B.1.2 at 10 MPa < f = m 3 /kg, so compressed liquid B.1.4: = = C b. 1 MPa, 190 C : Only one of the two look-ups is needed B.1.1: P < P sat = kpa so it is superheated apor B.1.2: > sat = C so it is superheated apor B.1.3: = ( ) = m3 /kg c. 400 C, 0.19 m 3 /kg: look in B.1.1: > c superheated apor B.1.3: 1 MPa = m 3 /kg, so look at higher P => P = 1600 kpa d. 10 kpa, 10 C : Only one of the two look-ups is needed From B.1.1: P > P g = kpa so compressed liquid From B.1.2: < sat = 45.8 C so compressed liquid From B.1.1: = f = m 3 /kg (at gien, not gien P) States shown are placed relatie to the two-phase region, not to each other. P C.P. a d c b a d C.P. c b P = const.

41 41 Borgnakke and Sonntag 2.27 For water at 200 kpa with a quality of 10%, find the olume fraction of apor. his is a two-phase state at a gien pressure: able B.1.2: f = m 3 /kg, g = m 3 /kg From the definition of quality we get the masses from total mass, m, as m f = (1 x) m, m g = x m he olumes are V f = m f f = (1 x) m f, V g = m g g = x m g So the olume fraction of apor is Fraction = V g V = V g x m g V g + V = f x m g + (1 x)m f = = = Notice that the liquid olume is only about 1% of the total. We could also hae found the oerall = f + x fg and then V = m.

42 42 Borgnakke and Sonntag 2.28 Determine whether refrigerant R-410A in each of the following states is a compressed liquid, a superheated apor, or a mixture of saturated liquid and apor. All cases are seen in able B.4.1 a. 50 C, 0.05 m 3 /kg From table B.4.1 at 50 C g = m 3 /kg since > g we hae superheated apor b. 1.0 MPa, 20 C From table B.4.1 at 20 C P g = kpa since P < P g we hae superheated apor c. 0.1 MPa, 0.1 m 3 /kg From table B.4.1 at 0.1 MPa (use 101 kpa) f = and g = m 3 /kg as f < < g we hae a mixture of liquid & apor d 20 C, 200 kpa superheated apor, P < P g = 400 kpa at -20 C 2.29 Show the states in Problem 2.28 in a sketch of the P- diagram. P C.P. C.P. States shown are placed relatie to the two-phase region, not to each other. c a, b d c P = const. a, b d

43 43 Borgnakke and Sonntag 2.30 How great is the change in liquid specific olume for water at 20 o C as you moe up from state i towards state j in Fig. 2.14, reaching kpa? State i, here a, is saturated liquid and up is then compressed liquid states a able B.1.1: f = m 3 /kg at 2.34 kpa b able B.1.4: f = m 3 /kg at 500 kpa c able B.1.4: f = m 3 /kg at 2000 kpa d able B.1.4: f = m 3 /kg at 5000 kpa e able B.1.4: f = m 3 /kg at kpa f able B.1.4: f = m 3 /kg at kpa Notice how small the changes in are for ery large changes in P. P e f d c b a = 20 o C f-a P f L C.P. S a V

44 44 Borgnakke and Sonntag 2.31 Fill out the following table for substance ammonia: P [kpa] [ o C] [m 3 /kg] x a) b) a) B.2.1 f < < g => two-phase mix Look in B.2.1 => x = ( - f )/ fg = ( )/ = b) B.2.1 P = P sat = kpa = f + x fg = = m 3 /kg 2.32 Place the two states a-b listed in Problem 2.31 as labeled dots in a sketch of the P- and - diagrams. P C.P. C.P b a b P = const. a

45 45 Borgnakke and Sonntag 2.33 Gie the missing property of P,, and x for R-410A at a. = -20 o C, P = 450 kpa b. P = 300 kpa, = m 3 /kg a) B.4.1 P > P sat = kpa compressed liquid ~ f = m 3 /kg x = undefined b) B.4.2 > g at 300 kpa superheated apor = 20 + [0 ( 20)] x = undefined ) = C P C.P. a b a b P = const.

46 46 Borgnakke and Sonntag 2.34 Determine the specific olume for R-410A at these states: a. 10 o C, 500 kpa b. 20 o C, 1500 kpa c. 20 o C, quality 15% a) able B.4.1: P < P sat = kpa, so superheated apor (-13.89) B.4.2 = ( ) 0 - (-13.89) = m 3 /kg b) able B.4.1: P > P sat = 1444 kpa, so compressed liquid f = m 3 /kg c) able B.4.1: f = m 3 /kg, fg = m 3 /kg so = f + x fg = = m 3 /kg States shown are placed relatie to the two-phase region, not to each other. P C.P. b c a b C.P. c P = const. a

47 47 Borgnakke and Sonntag 2.35 Gie the missing property of P,, and x for CH 4 at: a. = 155 K, = 0.04 m 3 /kg b. = 350 K, = 0.25 m 3 /kg a) B.7.1 < g = m 3 /kg 2-phase x = - f = fg = P = P sat = 1296 kpa b) B.7.1 > c and >> c superheated apor B.7.2 located between 600 & 800 kpa P = = 734 kpa P C.P. b a b a P = const.

48 48 Borgnakke and Sonntag 2.36 Gie the specific olume of carbon-dioxide at -20 o C for 2000 kpa and repeat for 1400 kpa. able B.3.1: -20 o C P sat = 1969 kpa, at 2000 kpa state is compressed liquid: at 1400 kpa state is superheated apor: = f = m 3 /kg = m 3 /kg he 2000 kpa is aboe and the 1400 kpa is below the aporization line. ln P S L a b V

49 49 Borgnakke and Sonntag 2.37 Calculate the following specific olumes a. CO 2 10 C, 80% quality b. Water 2 MPa, 45% quality c. Nitrogen 120 K, 60% quality All states are two-phase with quality gien. he oerall specific olume is gien by Eq.2.1 or 2.2 = f + x fg = (1 - x) f + x g a. CO 2 10 C, 80% quality in able B.3.1 = x = m 3 /kg b. Water 2 MPa, 45% quality in able B.1.2 = (1 - x) + x = m 3 /kg c. Nitrogen 120 K, 60% quality in able B.6.1 = x = m 3 /kg

50 50 Borgnakke and Sonntag 2.38 You want a pot of water to boil at 105 o C. How heay a lid should you put on the 15 cm diameter pot when P atm = 101 kpa? able B.1.1 at 105 o C : P sat = kpa A = 4 D2 = = m 2 F net = (P sat P atm ) A = ( ) kpa m 2 = kn = 350 N F net = m lid g m lid = F net /g = N 2 = 35.7 kg m/s Some lids are clamped on, the problem deals with one that stays on due to its weight.

51 51 Borgnakke and Sonntag 2.39 Water at 600 kpa with a quality of 25% has its pressure raised 50 kpa in a constant olume process. What is the new quality and temperature? State 1 from able B.1.2 at 600 kpa = f + x fg = = m 3 /kg State 2 has same at P = 650 kpa also from able B.1.2 x = f = fg = = sat = C P C.P. C.P. 162 C C 159

52 52 Borgnakke and Sonntag 2.40 A sealed rigid essel has olume of 1 m 3 and contains 2 kg of water at 100 C. he essel is now heated. If a safety pressure ale is installed, at what pressure should the ale be set to hae a maximum temperature of 200 C? Process: = V/m = constant State 1: 1 = 1/2 = 0.5 m 3 /kg from able B.1.1 it is 2-phase C.P. 500 kpa 400 kpa State 2: 200 C, 0.5 m 3 /kg able B.1.3 between 400 and 500 kpa so interpolate 100 C P ( ) = kpa

53 53 Borgnakke and Sonntag 2.41 Saturated water apor at 200 kpa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. How much is this distance and the temperature if the water is cooled to occupy half the original olume? State 1: B = g (200 kpa) = m 3 /kg, 1 = C Process: P = constant = 200 kpa State 2: P, 2 = 1 /2 = m 3 /kg able B < g so two phase 2 = sat = C Height is proportional to olume h 2 = h 1 2 / 1 = 0.1 m 0.5 = 0.05m P C.P. C.P P = 200 kpa 1

54 54 Borgnakke and Sonntag 2.42 Saturated liquid water at 60 C is put under pressure to decrease the olume by 1% keeping the temperature constant. o what pressure should it be compressed? State 1: = 60 C, x = 0.0; able B.1.1: = m 3 /kg Process: = constant = 60 C State 2:, = 0.99 f (60 C) = = m 3 /kg Between 20 & 30 MPa in able B.1.4, P 23.8 MPa P 2 C.P. 30 MPa 20 MPa 1 2 1

55 55 Borgnakke and Sonntag 2.43 In your refrigerator the working substance eaporates from liquid to apor at -20 o C inside a pipe around the cold section. Outside (on the back or below) is a black grille inside which the working substance condenses from apor to liquid at +45 o C. For each location find the pressure and the change in specific olume () if the substance is ammonia. he properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature. Substance ABLE P sat, kpa = fg, m 3 /kg Ammonia B o C Ammonia B o C P C.P. C.P C

56 56 Borgnakke and Sonntag 2.44 Repeat the preious problem with the substances a) R-134a b) R-410A In your refrigerator, the working substance eaporates from liquid to apor at -20 o C inside a pipe around the cold section. Outside (on the back or below) is a black grille inside which the working substance condenses from apor to liquid at +45 o C. For each location find the pressure and the change in specific olume (). he properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature. Substance ABLE P sat, kpa = fg, m 3 /kg R-134a B o C R-134a B o C R-410A B o C R-410A B o C P C.P. C.P C

57 57 Borgnakke and Sonntag 2.45 A glass jar is filled with saturated water at 500 kpa, quality 15%, and a tight lid is put on. Now it is cooled to 10 C. What is the mass fraction of solid at this temperature? Constant olume and mass 1 = 2 = V/m From able B.1.2: 1 = = m 3 /kg From able B.1.5: 2 = x 2 = 1 = m 3 /kg x 2 = mass fraction apor x solid = 1 - x 2 = or % P C.P. C.P P L C.P. S 1 V 2

58 58 Borgnakke and Sonntag 2.46 wo tanks are connected as shown in Fig. P2.46, both containing water. ank A is at 200 kpa, 0.5 m 3 /kg, V A 1 m 3 and tank B contains 3.5 kg at 0.5 MPa, 400 C. he ale is now opened and the two come to a uniform state. Find the final specific olume. Control olume: both tanks. Constant total olume and mass process. A B sup. apor State A1: (P, ) m A = V A / A = 1/0.5 = 2 kg State B1: (P, ) able B.1.3 B = m 3 /kg V B = m B B = 3.5 kg m 3 /kg = m 3 Final state: m tot = m A + m B = 5.5 kg V tot = V A + V B = m 3 2 = V tot /m tot = m 3 /kg

59 59 Borgnakke and Sonntag 2.47 Saturated apor R-410A at 60 o C changes olume at constant temperature. Find the new pressure, and quality if saturated, if the olume doubles. Repeat the question for the case the olume is reduced to half the original olume. 1: (, x) B.4.1: 1 = g = m 3 /kg, P 1 = P sat = kpa 2: 2 = 2 1 = m 3 /kg superheated apor Interpolate between 2000 kpa and 3000 kpa P 2 = = 2799 kpa : 3 = 1 /2 = m 3 /kg < g : two phase x 3 = 3 - f = fg = P 3 = P sat = kpa P C.P. C.P P = 3837 kpa 1 2

60 60 Borgnakke and Sonntag 2.48 A steel tank contains 6 kg of propane (liquid apor) at 20 C with a olume of m 3. he tank is now slowly heated. Will the liquid leel inside eentually rise to the top or drop to the bottom of the tank? What if the initial mass is 1 kg instead of 6 kg? Constant olume and mass 2 = 1 = V m = m3 6 kg = m 3 /kg Liq. C.P. Vapor A.2: c = m 3 /kg > 1 eentually reaches sat. liquid. leel rises to top a b 20 C If m = 1 kg 1 = m 3 /kg > c then it will reach saturated apor. leel falls c

61 61 Borgnakke and Sonntag 2.49 Saturated water apor at 60 C has its pressure decreased to increase the olume by 10% keeping the temperature constant. o what pressure should it be expanded? Initial state: Final state: = m 3 /kg from table B.1.1 = 1.10 g = = m 3 /kg Interpolate at 60 C between saturated (P = kpa) and superheated apor P = 10 kpa in ables B.1.1 and B.1.3 P ( ) = 18.9 kpa P C.P. C.P. 60 o C 10 kpa P = 10 kpa Comment:, P = 18 kpa (software) is not linear in P, more like 1/P, so the linear interpolation in P is not ery accurate.

62 62 Borgnakke and Sonntag 2.50 Ammonia at 20 o C with a quality of 50% and total mass 2 kg is in a rigid tank with an outlet ale at the top. How much apor (mass) can you take out through the ale assuming the temperature stays constant? he top has saturated apor at 20 o C as long as there is a two phase inside. When there is no more liquid and apor will be taken out pressure will drop for the remaining apor so we can take it out until we reach P 0. V = m 1 1 = 2 kg 0.5 ( ) m 3 /kg = m 3 2 = m 3 /kg (from able B.2.2 at 100 kpa) m 2 = V/ 2 = m 3 / m 3 /kg = kg m = m 1 m 2 = = kg P C.P. C.P. 20 o C P = 857 kpa

63 63 Borgnakke and Sonntag 2.51 A sealed rigid essel of 2 m 3 contains a saturated mixture of liquid and apor R- 134a at 10 C. If it is heated to 50 C, the liquid phase disappears. Find the pressure at 50 C and the initial mass of the liquid. P 2 1 Process: constant olume and constant mass. State 2 is saturated apor, from table B.5.1 P 2 = P sat (50 C) = MPa State 1: same specific olume as state 2 1 = 2 = m 3 /kg 1 = x x 1 = m = V/ 1 = 2 m 3 / m 3 /kg = kg; m liq = (1 - x 1 ) m = kg

64 64 Borgnakke and Sonntag 2.52 A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up by 5 C per hour due to a failure in the refrigeration system. How long time will it take before the methane becomes single phase and what is the pressure then? Use able B.7.1 Assume rigid tank = constant = 1 1 = P 2 = m 3 /kg We then also see that 1 > c = m 3 /kg All single phase when = g => 145 K 1 t = (5 C/h) = 5 hours; P = P sat = 824 kpa

65 65 Borgnakke and Sonntag 2.53 A 400-m 3 storage tank is being constructed to hold LNG, liquified natural gas, which may be assumed to be essentially pure methane. If the tank is to contain 98% liquid and 2% apor, by olume, at 100 kpa, what mass of LNG (kg) will the tank hold? What is the quality in the tank? CH 4 is in the section B tables. From able B.7.1: f m 3 /kg, (interpolated) From able B.7.2: g m 3 /kg (first entry 100 kpa) m liq = V liq = f = kg; m ap = V ap = g = kg m tot = kg, x = m ap / m tot =

66 66 Borgnakke and Sonntag 2.54 A piston/cylinder arrangement is loaded with a linear spring and the outside atmosphere. It contains water at 5 MPa, 400 C with the olume being 0.1 m 3. If the piston is at the bottom, the spring exerts a force such that P lift 200 kpa. he system now cools until the pressure reaches 1200 kpa. Find the mass of water, the final state ( 2, 2 ) and plot the P diagram for the process. P a 1 2? : able B = m 3 /kg m = V/ 1 = 0.1/ = 1.73 kg Straight line: P = P a + C 2 = 1 P 2 - P a P 1 - P a = m 3 /kg 2 < g (1200 kpa) so two-phase 2 = 188 C x 2 = ( )/ =

67 67 Borgnakke and Sonntag 2.55 A pressure cooker (closed tank) contains water at 100 C with the liquid olume being 1/20 of the apor olume. It is heated until the pressure reaches 2.0 MPa. Find the final temperature. Has the final state more or less apor than the initial state? State 1: V f = m f f = V g /20 = m g g /20 ; x 1 = m g m g + m f = able B.1.1: f = m 3 /kg, g = m 3 /kg 20 m f f / g m f + 20 m f f / g = 20 f 20 f + g = 1 = = m 3 /kg = State 2: 2 = 1 = m 3 /kg < g (2MPa) from B.1.2 so two-phase P 2 1 At state 2: 2 = f + x 2 fg = x => x 2 = More apor at final state 2 = sat (2MPa) = C

68 68 Borgnakke and Sonntag 2.56 A pressure cooker has the lid screwed on tight. A small opening with A = 5 mm 2 is coered with a petcock that can be lifted to let steam escape. How much mass should the petcock hae to allow boiling at 120 o C with an outside atmosphere at kpa? able B.1.1.: P sat = kpa F = mg = P A m = P A/g = ( ) N ms -2 = kg = 50 g

69 69 Borgnakke and Sonntag Ideal Gas Law

70 70 Borgnakke and Sonntag 2.57 What is the relatie (%) change in P if we double the absolute temperature of an ideal gas keeping mass and olume constant? What will it be if we double V haing m, constant. Ideal gas law: PV = mr State 2: P 2 V = mr 2 = mr2 1 = 2P 1 V P 2 = 2P 1 Relatie change = P/P 1 = P 1 /P 1 = 1 = 100% State 3: P 3 V 3 = mr 1 = P 1 V 1 P 3 = P 1 V 1 /V 3 = P 1 /2 Relatie change = P/P 1 = -P 1 /2P 1 = -0.5 = -50% P V V

71 71 Borgnakke and Sonntag 2.58 A 1-m 3 tank is filled with a gas at room temperature 20 C and pressure 450 kpa. How much mass is there if the gas is a) air, b) neon or c) propane? Use able A.2 to compare and P to the critical and P with = 20 C = K ; P = 450 kpa << P c for all Air : >> C,N2 ; C,O2 = K so ideal gas; R= kj/kg K Neon: >> c = 44.4 K so ideal gas; R = kj/kg K Propane: < c = 370 K, but P << P c = 4.25 MPa so gas R = kj/kg K All states are ideal gas states so the ideal gas law applies PV = mr a) m = PV R = 450 kpa m 3 = kg kj/kgk K b) m = PV R = 450 kpa m kj/kgk K c) m = PV R = 450 kpa m kj/kgk K = kg = kg

72 72 Borgnakke and Sonntag 2.59 Calculate the ideal gas constant for argon and hydrogen based on table A.2 and erify the alue with able A.5 argon: hydrogen: R = R _ R = R _ M = M = = kj/kgk same as able A.5 = kj/kgk same as able A.5

73 73 Borgnakke and Sonntag 2.60 A pneumatic cylinder (a piston cylinder with air) must close a door with a force of 500 N. he cylinder cross-sectional area is 5 cm 2. With V = 50 cm 3, = 20 C, what is the air pressure and its mass? F = PA P o A P = P o + F/A = 100 kpa N m 2 = 1100 kpa 1000 N/kN m = P 1 V kpa m3 R = = kg = 0.65 g kj/kgk 293 K Comment: Dependent upon your understanding of the problem you could also hae neglected the atmospheric pressure to get 1000 kpa and 0.59 g for answers.

74 74 Borgnakke and Sonntag 2.61 Is it reasonable to assume that at the gien states the substance behaes as an ideal gas? a) Oxygen, O 2 at 30 C, 3 MPa Ideal Gas (» c = 155 K from A.2) b) Methane, CH 4 at 30 C, 3 MPa Ideal Gas (» c = 190 K from A.2) c) Water, H 2 O at 30 C, 3 MPa NO compressed liquid P > P sat (B.1.1) d) R-134a at 30 C, 3 MPa NO compressed liquid P > P sat (B.5.1) e) R-134a at 30 C, 100 kpa Ideal Gas P is low < P sat (B.5.1) ln P Liq. c, d e Cr.P. Vapor a, b

75 75 Borgnakke and Sonntag 2.62 Helium in a steel tank is at 250 kpa, 300 K with a olume of 0.1 m 3. It is used to fill a balloon and when the pressure drops to 125 kpa, the flow of helium stops by itself. If all the helium still is at 300 K, how big a balloon did I get? State 1: m = V/ assume ideal gas so cb m = P 1 V kpa m R = kj/kgk 300 K = kg State 2: Same mass so then ( 2 = 1 ) c i r c u s t h e r m o V 2 = mr 2 P 2 = P 1 V 1 R 2 R 1 P 2 = V 1 P 1 P 2 = 0.1 m = 0.2 m3 he balloon olume is V balloon = V 2 V 1 = = 0.1 m 3

76 76 Borgnakke and Sonntag 2.63 A spherical helium balloon of 10 m in diameter is at ambient and P, 20 o 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 olume V = 6 D3 = = m 3 m He = V = V = PV R = 100 kpa m 3 = 86.0 kg kj/kgk 293 K m air = PV 100 kpa m3 R = kj/kgk 293 K = kg m lift = m air m He = = kg

77 77 Borgnakke and Sonntag 2.64 A glass is cleaned in 45 o C hot water and placed on the table bottom up. he room air at 20 o C that was trapped in the glass gets heated up to 40 o C and some of it leaks out so the net resulting pressure inside is 2 kpa aboe ambient pressure of 101 kpa. Now the glass and the air inside cools down to room temperature. What is the pressure inside the glass? 1 air: 40 o C, 103 kpa 2 air: 20 o C,? Constant Volume: V 1 = V 2, AIR Slight amount of liquid water seals to table top Constant Mass m 1 = m 2 Ideal Gas P 1 V 1 = m 1 R 1 and P 2 V 2 = m 1 R 2 ake Ratio P 2 = P = 103 kpa = 96.4 kpa his is a acuum relatie to atm pressure so the glass is pressed against table.

78 78 Borgnakke and Sonntag 2.65 Air in an automobile tire is initially at 10 C and 230 kpa. After the automobile is drien awhile, the temperature gets up to 10 C. Find the new pressure. You must make one assumption on your own. Assume constant olume V 2 = V 1 and that air is an ideal gas P 1 V 1 = mr 1 and P 2 V 2 = mr 2 so P 2 = P 1 2 / 1 = 230 kpa = kpa

79 79 Borgnakke and Sonntag 2.66 A rigid tank of 1 m 3 contains nitrogen gas at 600 kpa, 400 K. By mistake someone lets 0.5 kg flow out. If the final temperature is 375 K what is the final pressure? m = PV R = 600 kpa 1 m 3 = kg kj/kgk K m 2 = m 0.5 = kg P 2 = m 2 R 2 V = kg kj/kgk 375 K 1 m 3 = kpa

80 80 Borgnakke and Sonntag 2.67 Assume we hae 3 states of saturated apor R-134a at +40 o C, 0 o C and 40 o C. Calculate the specific olume at the set of temperatures and corresponding saturated pressure assuming ideal gas behaior. Find the percent relatie error = 100( g )/ g with g from the saturated R-134a table. R-134a. able alues from able B.5.1 P sat, g () Ideal gas constant from able A.5: R R-134a = kj/kg K P sat, kpa g, m 3 /kg ID.G. = R / P sat error % -40 o C o C o C P

81 81 Borgnakke and Sonntag 2.68 Do Problem 2.67 for R-410A. R-410A. able alues from able B.4.1 P sat, g () Ideal gas constant from able A.5: R R-410a = kj/kg K P sat, kpa g, m 3 /kg ID.G. = R / P sat error % -40 o C o C o C P

82 82 Borgnakke and Sonntag 2.69 A 1 m 3 rigid tank has propane at 125 kpa, 300 K and connected by a ale to another tank of 0.5 m 3 with propane at 250 kpa, 400 K. he ale is opened and the two tanks come to a uniform state at 325 K. What is the final pressure? Propane is an ideal gas (P << P c ) with R = kj/kgk from bl. A.5 m A = P AV A R A = m B = P BV B R B = V 2 = V A + V B = 1.5 m 3 m 2 = m A + m B = kg P 2 = m 2 R 2 V 2 = 125 kpa 1 m 3 = kg kj/kgk K 250 kpa m 3 = kg kj/kgk K kg kj/kgk 325 K 1.5 m 3 = 158 kpa

83 83 Borgnakke and Sonntag 2.70 A 1-m 3 rigid tank with air at 1 MPa, 400 K is connected to an air line as shown in Fig. P2.70. he ale is opened and air flows into the tank until the pressure reaches 5 MPa, at which point the ale is closed and the temperature inside is 450K. a. What is the mass of air in the tank before and after the process? b. he tank eentually cools to room temperature, 300 K. What is the pressure inside the tank then? P, known at both states and assume the air behaes as an ideal gas. P 1 V 1000 kpa 1 m m air1 = R = 3 = kg kj/kgk K m air2 = P 2 V 5000 kpa 1 m = 3 = kg R kj/kgk K Process 2 3 is constant V, constant mass cooling to 3 P 3 = P 2 3 / 2 ) = 5000 kpa (300/450) = 3.33 MPa

84 84 Borgnakke and Sonntag 2.71 A cylindrical gas tank 1 m long, inside diameter of 30 cm, is eacuated and then filled with carbon dioxide gas at 20 C. o what pressure should it be charged if there should be 2 kg of carbon dioxide? Assume CO 2 is an ideal gas, table A.5: R = kj/kg K V cyl = A L = 4 (0.3)2 1 = m 3 P V = mr => P = mr V P = 2 kg kj/kg ( ) K m 3 = 1567 kpa

85 85 Borgnakke and Sonntag Compressibility Factor

86 86 Borgnakke and Sonntag 2.72 Find the compressibility factor (Z) for saturated apor ammonia at 100 kpa and at 2000 kpa. able B.2.2: 1 = m 3 /kg, 1 = o C, P 1 = 100 kpa able A.5: 2 = m 3 /kg, 2 = o C, P 2 = 2000 kpa R = kj/kg K Extended gas law: P = ZR so we can calculate Z from this Z 1 = P kpa m 3 /kg R = kj/kg-k ) K = Z 2 = P kpa m 3 /kg R = kj/kg-k ) K = So state 1 is close to ideal gas and state 2 is not so close. Z 1 2 = 0.7 r = 0.7 r ln P r = 2.0 r = 1.2 r

87 87 Borgnakke and Sonntag 2.73 Find the compressibility for nitrogen at a kpa, 120 K ; b kpa, 300 K ; c. 120 K, = m 3 /kg able B.6 has the properties for nitrogen. a) B.6.2: = m 3 /kg Z = P/R = 2000 kpa m3 /kg kj/kg-k 120 K = b) B.6.2: = m 3 /kg Z = P/R = 2000 kpa m3 /kg kj/kg-k 300 K = c) B.6.1: f < < g so this is a two-phase state. Z = P/R = 2000 kpa m3 /kg kj/kg-k 120 K = 0.28 he liquid in this state is incompressible with a low olume and the apor is ery close to the critical point. If you calculate a quality it is x = he compressibility for the saturated apor alone is 0.44.

88 88 Borgnakke and Sonntag 2.74 Find the compressibility for carbon dioxide at 40 o C and 10 MPa using Fig. D.1 able A.2 CO2: c = K P c = 7.38 MPa r = / c = 313/304.1 = P r = P/P c = 10/7.38 = From Figure D.1: Z 0.3 (difficult to interpolate) Compare with table B.3.2: = m 3 /kg Z = P/R = kpa m3 /kg kj/kg-k 313 K = 0.27 Z = 2.0 r = 0.7 r = 1.1 r = 0.7 r ln P r

89 89 Borgnakke and Sonntag 2.75 What is the percent error in specific olume if the ideal gas model is used to represent the behaior of superheated ammonia at 40 C, 500 kpa? What if the generalized compressibility chart, Fig. D.1, is used instead? NH 3 = 40 C = K, c = K, P c = MPa from able A.1 able B.2.2: Ideal gas: = R P = m 3 /kg = kj/kgk 313 K 500 kpa = m 3 /kg 4.5% error Figure D.1: r = = 0.772, P r = = Z = 0.97 = ZR P = m3 /kg 1.4% error

90 90 Borgnakke and Sonntag 2.76 A cylinder fitted with a frictionless piston contains butane at 25 C, 500 kpa. Can the butane reasonably be assumed to behae as an ideal gas at this state? Solution Butane 25 C, 500 kpa, able A.2: c = 425 K; P c = 3.8 MPa r = = 0.701; P r = = 0.13 Look at generalized chart in Figure D.1 Actual P r > P r, sat = 0.1 => liquid!! not a gas he pressure should be less than 380 kpa to hae a gas at that.

91 91 Borgnakke and Sonntag 2.77 Estimate the saturation pressure of chlorine at 300 K. We do not hae a table in the B section for Chlorine so we must use the generalized chart. able A.2: P c = 7.98 MPa, c = K r = / c = 300 / = Figure D.1: P r sat = 0.13 (same estimation from able D.4) P = P c P r sat = = 1.04 MPa If you use the CA3 program then you will find P r sat = and P = 973 kpa

92 92 Borgnakke and Sonntag 2.78 A bottle with a olume of 0.1 m 3 contains butane with a quality of 75% and a temperature of 300 K. Estimate the total butane mass in the bottle using the generalized compressibility chart. We need to find the property the mass is: m = V/ so find gien 1 and x as : = f + x fg able A.2: Butane c = K P c = 3.8 MPa = 3800 kpa r = 300/425.2 = => From Fig. D.1 or table D.4: Z f 0.02; Z g 0.9; P r sat = 0.1 Z g f = 0.7 r ln P r = 2.0 r = 0.7 r P = P sat = P r sat P c = MPa 1000 kpa/mpa = 380 kpa f = Z f R/P = kj/kgk 300 K/380 kpa = m 3 /kg g = Z g R/P = kj/kgk 300 K/380 kpa = m 3 /kg = ( ) = m 3 /kg m = V = = kg

93 93 Borgnakke and Sonntag 2.79 Find the olume of 2 kg of ethylene at 270 K, 2500 kpa using Z from Fig. D.1 Ethylene able A.2: able A.5: c = K, P c = 5.04 MPa R = kj/kg K he reduced temperature and pressure are: r = c = = 0.956, P r = P P c = = Enter the chart with these coordinates and read: Z = 0.76 V = mzr P Z = = 0.7 r 2 kg kj/kg-k 270 K 2500 kpa ln P r = 2.0 r = 1.2 r r = 0.96 r = 0.7 = m 3

94 94 Borgnakke and Sonntag 2.80 For r = 0.7, what is the ratio g / f using Fig. D.1 compared to able. D.3 For the saturated states we can use Fig. D.1 with the estimates Z f = 0.02, Z g = 0.9 so g / f = (ZR/P) g / (ZR/P) f = Z g /Z f = = able D.3 list the entries more accurately than we can read the figure Z f = 0.017, Z g = so g / f = (ZR/P) g / (ZR/P) f = Z g /Z f = = Z = 2.0 r = 0.7 r = 0.7 r = 1.2 r ln P r

95 95 Borgnakke and Sonntag 2.81 Refrigerant R-32 is at -10 o C with a quality of 15%. Find the pressure and specific olume. For R-32 there is no section B table printed. We will use compressibility chart. From able A.2: c = K ; P c = 5.78 MPa ; From able A.5: R = kj/kg K r = / c = 263/351.3 = From able D.4 or Figure D.1, Z f ; Z g 0.86 ; P r sat 0.16 P = P r sat P c = = 925 kpa = f + x fg = (Z f + x Z fg ) R/P = [ ( )] kj/kg-k 263 K/ 925 kpa = m 3 /kg Z = 2.0 r = 0.7 r = 0.7 r ln P r

96 96 Borgnakke and Sonntag 2.82 o plan a commercial refrigeration system using R-123 we would like to know how much more olume saturated apor R-123 occupies per kg at -30 o C compared to the saturated liquid state. For R-123 there is no section B table printed. We will use compressibility chart. From able A.2 c = K ; P c = 3.66 MPa ; M = r = / c = 243/456.9 = 0.53 R = R _ /M = kj/kmol-k / kg/kmol = kj/kgk he alue of r is below the range in Fig. D.1 so use the table D.4 able D.4, Z g = Z f = P = P r P c = 42.5 kpa Z fg = = ; P r = P r sat = fg = Z fg R/P = kj/kgk 243 K / 42.5 kpa = m 3 /kg Comment: If you check with the software the solution is off by a factor of 6. he linear interpolation is poor and so is the approximation for P r sat so the real saturation pressure should be 6.75 kpa. Also the ery small alue of Z f is inaccurate by itself, minute changes in the cure gies large relatie ariations.

97 97 Borgnakke and Sonntag 2.83 A new refrigerant R-125 is stored as a liquid at -20 o C with a small amount of apor. For a total of 1.5 kg R-125 find the pressure and the olume. As there is no section B table use compressibility chart. able A.2: R-125 c = K P c = 3.62 MPa r = / c = / = We can read from Figure D.1 or a little more accurately interpolate from table D.4 entries: P r sat = 0.16 ; Z g = 0.86 ; Z f = P = P r sat P c = kpa = 579 kpa V liq = Z f m liq R/P = kg kj/kgk K/ 579 kpa = m 3 Z sat apor = 0.7 r sat liq ln P r = 2.0 r = 0.7 r

98 98 Borgnakke and Sonntag Equations of State For these problems see appendix D for the equation of state (EOS) and chapter 12.

99 99 Borgnakke and Sonntag 2.84 Determine the pressure of nitrogen at 160 K, = m 3 /kg using ideal gas, an der Waal Equation of State and the nitrogen table. Nitrogen from table A.2: c = K, P c = 3390 kpa, Ideal gas: P = R = kj/kg-k K m 3 /kg = kpa For an der Waal equation of state from able D.1 we hae b = 1 8 R c = = m P c /kg, a = 27 b 2 P c = 27 ( ) = kpa (m 3 /kg) 2 he EOS is: P = R b a 2 = able B.6.2: P = kpa = kpa

100 100 Borgnakke and Sonntag 2.85 Determine the pressure of nitrogen at 160 K, = m 3 /kg using Redlich- Kwong Equation of State and the nitrogen table. Nitrogen from table A.2: c = K, P c = 3390 kpa, r = / c = 160/126.2 = For Redlich-Kwong EOS we hae the parameters from able D.1 b = R c P = c 3390 = m 3 /kg, a = /2 r R2 2 c P c = / = kpa (m3 /kg) 2 he equation is: P = R b a 2 + b = = kpa able B.6.2: P = kpa.

101 101 Borgnakke and Sonntag 2.86 Determine the pressure of nitrogen at 160 K, = m 3 /kg using Soae Equation of State and the nitrogen table. Nitrogen from table A.2: c = K, P c = 3390 kpa, r = / c = 160/126.2 = For Soae EOS see Appendix D (ery close to Redlich-Kwong) P = R b a 2 + b where the parameters are from able D.1 and D.4 = f = = a o = [1 + f (1-1/2 r ) ] 2 = b = R c P = c 3390 = m 3 /kg, a = R2 2 c P = c 3390 = kpa (m 3 /kg) 2 P = R b a 2 + b = = kpa Nitrogen able B.6.2: P = kpa.