Exercícios de Termodinâmica Resolvidos Cap3

SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3

SONNTAG BORGNAKKE VAN WYLEN

FUNDAMENTALS

of

Thermodynamics

Sixth Edition

Sonntag, Borgnakke and van Wylen

CONTENT CHAPTER 3

SUBSECTION Correspondence table Study guide problems Phase diagrams, triple and critical points General tables Ideal gas Compressibility factor Review problems Linear interpolation Computer tables English unit problems PROB NO. 1-20 21-28 29-63 64-79 79-89 90-115 116-121 122-127 128-158

Sonntag, Borgnakke and van Wylen

Correspondence Table CHAPTER 3 6th edition

Sonntag/Borgnakke/Wylen

The set of problems have a correspondence to the 5th edition Fundamentals of Thermodynamics as: Problems 3.1 through 3.20 are all new New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 5th new 2 1 3 new 4 28 mod new 23 28 mod 24 new new new 29 new new 27 mod new 37 41 new new new new 36 new 58 35 42 new 43 new 40 44 New 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 5th 46 48 39 mod 57 51 new new 5 new 22 6 new 8 new 10 13 new 25 new new new 17 14 19 33 new new new new 20 new 21 18 26 mod 16 mod 30 mod New 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 5th 30 mod 31 mod 32 new 60 55 new 59 53 54 50 49 45 56 9 52 7 47 11 12 16 38 34 new new new new new new new new new new 86 87

Sonntag, Borgnakke and van Wylen

The English unit problem correspondence is New 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 5th Ed. new new new new new new 61E 68E a-c 68E d-f new 70E 73E 74E new 76E SI 5 7 9 11 17 23 27 30 30 40 36 47 41 44 51 New 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 5th Ed. 77E new 79E 62E new 69E c+d 70E d 72E 64E new 81E new 71E 80E 83E 65E 66E SI 53 62 58 69 65 81 113 74 49 99 95 61 106 89 -

The Computer, design and open-ended problem correspondence is New 159 160 161 162 5th new new 88 89 New 163 164 165 166 5th 90 91 92 93 New 167 168 5th 94 95

mod indicates a modification from the previous problem that changes the solution but otherwise is the same type problem.

Sonntag, Borgnakke and van Wylen

Concept-Study Guide Problems

Sonntag, Borgnakke and van Wylen

3.1 What is the lowest temperature (approximately) at which water can be liquid? ln P Look at the phase diagram in Fig. 3.7. At the border between ice I, ice III and the liquid region is a triple point which is the lowest T where you can have liquid. From the figure it is estimated to be about 255 K i.e. at -18oC. T 255 K - 18 C 3.2 What is the percent change in volume as liquid water freezes? Mention some effects in nature and for our households the volume change can have. The density of water in the different phases can be found in Tables A.3 and A.4 and in Table B.1. From Table B.1.1 vf = 0.00100 m3/kg From Table B.1.5 Percent change: 100 vi = 0.0010908 m3/kg vi vf 0.0010908 0.001 = 100 = 9.1 % increase 0.001 vf

lowest T liquid

L S V

CR.P.

T

Liquid water that seeps into cracks or other confined spaces and then freezes will expand and widen the cracks. This is what destroys any pourous material exposed to the weather on buildings, roads and mountains. 3.3 When you skate on ice a thin liquid film forms under the skate; how can that be? The 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. 3.7. As the skate comes over 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. The very thin liquid film under the skate changes the friction to be viscous rather than a solid to solid contact friction. Friction is thus significantly reduced.

Sonntag, Borgnakke and van Wylen

3.4 An external water tap has the valve activated by a long spindle so the closing mechanism is located well inside the wall. Why is that? Solution: By having the spindle inside the wall the coldest location with water when the valve is closed is kept at a temperature above the freezing point. If the valve 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. 3.5 Some tools should be cleaned in water at a least 150oC. How high a P is needed? Solution: If I need liquid water at 150oC I must have a pressure that is at least the saturation pressure for this temperature. Table B.1.1: 150oC Psat = 475.9 kPa.

3.6 Are the pressures in the tables absolute or gauge pressures? Solution: The behavior of a pure substance depends on the absolute pressure, so P in the tables is absolute. 3.7 If I have 1 L ammonia at room pressure and temperature (100 kPa, 20oC) how much mass is that? Ammonia Tables B.2: B.2.1 Psat = 857.5 kPa at 20oC so superheated vapor. B.2.2 v = 1.4153 m3/kg under subheading 100 kPa 3 V 0.001 m m= v = = 0.000 706 kg = 0.706 g 1.4153 m3/kg

Sonntag, Borgnakke and van Wylen

3.8 How much is the change in liquid specific volume for water at 20oC as you move up from state i towards state j in figure 3.12 reaching 15 000 kPa? State "i", here "a", is saturated liquid and up is then compressed liquid states a Table B.1.1: vf = 0.001 002 m3/kg at 2.34 kPa b c d e f Table B.1.4: Table B.1.4: Table B.1.4: Table B.1.4: Table B.1.4: vf = 0.001 002 m3/kg at 500 kPa vf = 0.001 001 m3/kg at 2000 kPa vf = 0.001 000 m3/kg at 5000 kPa vf = 0.000 995 m3/kg at 15 000 kPa vf = 0.000 980 m3/kg at 50 000 kPa

Notice how small the changes in v are for very large changes in P. P e d c b a f-a T = 20 C v v

o

f

T

P f T a L C.P. V

S

v

Sonntag, Borgnakke and van Wylen

3.9 For water at 100 kPa with a quality of 10% find the volume fraction of vapor. This is a two-phase state at a given pressure: Table B.1.2: vf = 0.001 043 m3/kg, vg = 1.6940 m3/kg From the definition of quality we get the masses from total mass, m, as mf = (1 x) m, mg = x m The volumes are Vf = mf vf = (1 x) m vf, Vg = mg vg = x m vg So the volume fraction of vapor is Vg Vg x m vg Fraction = V = V + V = x m v + (1 x)m v g f g f = 0.1694 0.1 1.694 = 0.17034 = 0.9945 0.1 1.694 + 0.9 0.001043

Notice that the liquid volume is only about 0.5% of the total. We could also have found the overall v = vf + xvfg and then V = m v.

Sonntag, Borgnakke and van Wylen

3.10 Sketch two constant-pressure curves (500 kPa and 30 000 kPa) in a T-v diagram and indicate on the curves where in the water tables you see the properties.

MPa P 30 B 1 4

B.1.3 C.P. B.1.3 B.1.2 B.1.5 v

T B 1 4

30 MPa B.1.3 500 kPa B.1.1 B.1.3

0.5

B.1.5

v

The 30 MPa line in Table B.1.4 starts at 0oC and table ends at 380oC, the line is continued in Table B.1.3 starting at 375oC and table ends at 1300oC. The 500 kPa line in Table B.1.4 starts at 0.01oC and table ends at the saturated liquid state (151.86oC). The line is continued in Table B.1.3 starting at the saturated vapor state (151.86oC) continuing up to 1300oC.

Sonntag, Borgnakke and van Wylen

3.11 Locate the state of ammonia at 200 kPa, -10oC. Indicate in both the P-v and the T-v diagrams the location of the nearest states listed in the printed table B.2

P T

C.P.

C.P. 200 kPa

290.9 200

-10 C -18.9 C

T

0 -10 -18.9

150 kPa

v

v

3.12 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 volume and thus density is nearly constant. These surfaces are very steep nearly constant v and there is then no reason to fill up a table with the same value of v for different P and T.

Sonntag, Borgnakke and van Wylen

3.13 Water at 120oC with a quality of 25% has its temperature raised 20oC in a constant volume process. What is the new quality and pressure? Solution: State 1 from Table B.1.1 at 120oC v = vf + x vfg = 0.001060 + 0.25 0.8908 = 0.22376 m3/kg State 2 has same v at 140oC also from Table B.1.1 v - vf 0.22376 - 0.00108 x= v = = 0.4385 0.50777 fg P = Psat = 361.3 kPa

P C.P.

140 C 120 C

T

C.P.

361.3 198.5

T v

140 120 v

Sonntag, Borgnakke and van Wylen

3.14 Water at 200 kPa with a quality of 25% has its temperature raised 20oC in a constant pressure process. What is the new quality and volume? Solution: State 1 from Table B.1.2 at 200 kPa v = vf + x vfg = 0.001061 + 0.25 0.88467 = 0.22223 m3/kg State 2 has same P from Table B.1.2 at 200 kPa T = T + 20 = 120.23 + 20 = 140.23oC

2 sat

so state is superheated vapor x = undefined 20 v = 0.88573 + (0.95964 0.88573)150 - 120.23 = 0.9354 m3/kg

P C.P.

140 C

T

C.P. 200 kPa

200

120.2 C

T v

140 120

v

3.15 Why is it not typical to find tables for Ar, He, Ne or air like an Appendix B table? The temperature at which these substances are close to the two-phase region is very low. For technical applications with temperatures around atmospheric or higher they are ideal gases. Look in Table A.2 and we can see the critical temperatures as Ar : 150.8 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.

Sonntag, Borgnakke and van Wylen

3.16 What is the relative (%) change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant? Repeat if we double V having m, T constant. Ideal gas law: State 2: PV = mRT P2V = mRT2 = mR2T1 = 2P1V P2 = 2P1 Relative change = P/P1 = P1/P1 = 1 = 100% P3V3 = mRT1 = P1V1 P3 = P1V1/V3 = P1/2 Relative change = P/P1 = -P1/2P1 = -0.5 = -50%

P 2 2 1 3 T2 T1 V 1 3 V T

State 3:

Sonntag, Borgnakke and van Wylen

3.17 Calculate the ideal gas constant for argon and hydrogen based on table A.2 and verify the value with Table A.5 The gas constant for a substance can be found from the universal gas constant from the front inside cover and the molecular weight from Table A.2 R 8.3145 Argon: R = M = 39.948 = 0.2081 kJ/kg K R 8.3145 Hydrogen: R = M = 2.016 = 4.1243 kJ/kg K

Sonntag, Borgnakke and van Wylen

3.18 How close to ideal gas behavior (find Z) is ammonia at saturated vapor, 100 kPa? How about saturated vapor at 2000 kPa? Table B.2.2: Table A.5: Extended gas law: v1 = 1.1381 m3/kg, R = 0.4882 kJ/kg K Pv = ZRT so we can calculate Z from this T1 = -33.6oC, P1 = 100 kPa P2 = 2000 kPa

v2 = 0.06444 m3/kg, T2 = 49.37oC,

P1v1 100 1.1381 Z1 = RT = = 0.973 1 0.4882 (273.15 - 33.6) P2v2 2000 0.06444 Z2 = RT = = 0.8185 2 0.4882 (273.15 + 49.37) So state 1 is close to ideal gas and state 2 is not so close. Z 1 2 Tr = 0.7 Tr= 2.0 Tr = 1.2 Tr = 0.7 0.1 1 ln Pr

Sonntag, Borgnakke and van Wylen

3.19 Find the volume of 2 kg of ethylene at 270 K, 2500 kPa using Z from Fig. D.1 Ethylene Table A.2: Table A.5: Tc = 282.4 K, Pc = 5.04 MPa R = 0.2964 kJ/kg K

The reduced temperature and pressure are: T 270 Tr = T = 282.4 = 0.956, c P 2.5 Pr = P = 5.04 = 0.496 c Z = 0.76

Enter the chart with these coordinates and read: V= Z Tr = 0.7

mZRT 2 0.76 0.2964 270 = 0.0487 m3 P = 2500

Tr= 2.0 Tr = 1.2 Tr = 0.96 Tr = 0.7 0.5 1 ln Pr

0.1

Sonntag, Borgnakke and van Wylen

3.20 With Tr = 0.85 and a quality of 0.6 find the compressibility factor using Fig. D.1 For the saturated states we will use Table D.4 instead of the figure. There we can see at Tr = 0.85 Zf = 0.062, Zg = 0.747 Z = (1 x) Zf + xZg = (1 0.6) 0.062 + 0.6 0.747 = 0.473

Sonntag, Borgnakke and van Wylen

Phase Diagrams, Triple and Critical Points

3.21 Modern extraction techniques can be based on dissolving 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. Solution: CO2 : Table A.2: Pc = 7.38 MPa, Tc = 304 K, vc = 0.00212 m3/kg c = 1/vc = 1/0.00212 = 472 kg/m3 C2H5OH: Table A.2: Pc = 6.14 MPa, Tc = 514 K, vc = 0.00363 m3/kg c = 1/vc = 1/0.00363 = 275 kg/m3

Sonntag, Borgnakke and van Wylen

3.22 Find the lowest temperature at which it is possible to have water in the liquid phase. At what pressure must the liquid exist? Solution: ln P There is no liquid at lower temperatures than on the fusion line, see Fig. 3.6, saturated ice III to liquid phase boundary is at T 263K - 10 C and P 2100 MPa

lowest T liquid

L S V

CR.P.

T

Sonntag, Borgnakke and van Wylen

3.23

Water at 27 C can exist in different phases dependent upon the pressure. Give the approximate pressure range in kPa for water being in each one of the three phases vapor, liquid or solid. Solution: ln P The phases can be seen in Fig. 3.6, a sketch of which is shown to the right. T = 27 C = 300 From Fig. 3.6: P 4 10-3 MPa = 4 kPa, PLS = 103 MPa

VL

S L CR.P.

S

V T

0 P 4 kPa 0.004 MPa P 1000 MPa P 1000 MPa

VAPOR LIQUID SOLID(ICE)

Sonntag, Borgnakke and van Wylen

3.24 What is the lowest temperature in Kelvins for which you can see metal as a liquid if the metal is a. silver b. copper Solution: Assume the two substances have a phase diagram similar to Fig. 3.6, then we can see the triple point data in Table 3.2 Ta = 961oC = 1234 K Tb = 1083oC = 1356 K

Sonntag, Borgnakke and van Wylen

3.25 If density of ice is 920 kg/m3, find the pressure at the bottom of a 1000 m thick ice cap on the north pole. What is the melting temperature at that pressure? Solution: ICE = 920 kg/m3 P = gH = 920 kg/m3 9.80665 m/s2 1000 = 9022 118 Pa P = Po + P = 101.325 + 9022 = 9123 kPa See figure 3.6 liquid solid interphase = TLS = -1 C

Sonntag, Borgnakke and van Wylen

3.26 Dry ice is the name of solid carbon dioxide. How cold must it be at atmospheric (100 kPa) pressure? If it is heated at 100 kPa what eventually happens? Solution: The phase boundaries are shown in Figure 3.6 At 100 kPa the carbon dioxide is solid if T 190 K It goes directly to a vapor state without becoming a liquid hence its name. ln P The 100 kPa is below the triple point. 100 kPa S L V T

Sonntag, Borgnakke and van Wylen

3.27

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 nitrogen, water or propane ? Solution: Find state relative to critical point properties which are from Table A.2: a) Nitrogen N2 : 3.39 MPa 126.2 K b) Water c) Propane H2O : 22.12 MPa 647.3 K 369.8 K ln P Liquid b Cr.P. a Vapor T C3H8 : 4.25 MPa

State is at 17 C = 290 K and 2 MPa Pc for all cases: N2 : T Tc Superheated vapor P Pc H2O : T Tc ; P Pc you cannot say. C3H8 : T Tc ; P Pc you cannot say

c

Sonntag, Borgnakke and van Wylen

3.28 Give the phase for the following states. Solution: a. CO2 b. Air c. NH3 T = 267 C P = 0.5 MPa Table A.2 superheated vapor assume ideal gas Table A.5 T = 20 C P = 200 kPa Table A.2 superheated vapor assume ideal gas Table A.5 T = 170 C P = 600 kPa Table B.2.2 or A.2 T Tc = superheated vapor

T a,b,c a, b, c

P C.P.

T v

P = const. v

Sonntag, Borgnakke and van Wylen

3.29 Determine the phase of the substance at the given state using Appendix B tables a) Water 100 C, 500 kPa b) Ammonia -10 C, 150 kPa c) R-12 0 C, 350 kPa Solution: a) From Table B.1.1 Psat(100 C) = 101.3 kPa Tsat(500 kPa) = 152 C 500 kPa Psat then it is compressed liquid OR from Table B.1.2 100 C Tsat then it is subcooled liquid = compressed liquid b) Ammonia NH3 : Table B.2.1: P Psat(-10 C) = 291 kPa Superheated vapor c) R-12 Table B.3.1: P Psat(0 C) = 309 kPa Compressed liquid. ln P The S-L fusion line goes slightly to the left for water. It tilts slightly to the right for most other substances. L a, c S b Vapor T Cr.P.

Sonntag, Borgnakke and van Wylen

3.30 Determine whether water at each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor. a. P = 10 MPa, v = 0.003 m3/kg b. 1 MPa, 190 C 3/kg c. 200 C, 0.1 m d. 10 kPa, 10 C Solution: For all states start search in table B.1.1 (if T given) or B.1.2 (if P given) a. P = 10 MPa, v = 0.003 m3/kg = v f v vg = so look in B.1.2 at 10 MPa so mixture of liquid and vapor.

vf = 0.001452; vg = 0.01803 m3/kg, b. 1 MPa, 190 C : Only one of the two look-ups is needed B.1.1: P Psat = 1254.4 kPa so it is superheated vapor B.1.2: T Tsat = 179.91 C so it is superheated vapor c. 200 C, 0.1 m3/kg: look in B.1.1 vf = 0.001156 m3/kg ; vg = 0.12736 m3/kg, = so mixture of liquid and vapor. = vf v vg d. 10 kPa, 10 C : From B.1.1: From B.1.2: Only one of the two look-ups is needed P Pg = 1.2276 kPa so compressed liquid T Tsat = 45.8 C so compressed liquid

P C.P. T C.P. P = const. d b a c T v d b a c v

States shown are placed relative to the two-phase region, not to each other.

Sonntag, Borgnakke and van Wylen

3.31 Give the phase for the following states. Solution: a. H2O T = 275 C P = 5 MPa Table B.1.1 or B.1.2 = superheated vapor = superheated vapor Table B.1.1 T Ttriple point B.1.1 Psat = 5.94 MPa B.1.2 Tsat = 264 C b. H2O T = -2 C P = 100 kPa

Table B.1.5 at -2 C Psat = 0.518 kPa since P Psat = compressed solid

P C.P. a T C.P.

States shown are placed relative to the two-phase region, not to each other. Note state b in P-v, see in 3-D figure, is up on the solid face.

a P = const.

b

T v b

v

P L S T b C.P. a V

v

Sonntag, Borgnakke and van Wylen

3.32 Determine whether refrigerant R-22 in each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor. Solution: All cases are seen in Table B.4.1 a. 50 C, 0.05 m3/kg b. 1.0 MPa, 20 C c. 0.1 MPa, 0.1 m3/kg From table B.4.1 at 50 C vg = 0.01167 m3/kg since v vg we have superheated vapor From table B.4.1 at 20 C Pg = 909.9 kPa since P Pg we have compressed liquid From table B.4.1 at 0.1 MPa (use 101 kPa) vf = 0.0007 and vg = 0.2126 m3/kg as vf v vg we have a mixture of liquid & vapor d -20 C, 200 kPa superheated vapor, P Pg = 244.8 kPa at -20 C

P C.P. T C.P. P = const. b c a d T v b c a d v

States shown are placed relative to the two-phase region, not to each other.

Sonntag, Borgnakke and van Wylen

General Tables

3.33 Fill out the following table for substance water: Solution: P [kPa] T [ oC] v [m3/kg] a) 500 20 0.001002 b) 500 151.86 0.20 c) 1400 200 0.14302 d) 8581 300 0.01762

x Undefined 0.532 Undefined 0.8

a) Table B.1.1 P Psat so it is compressed liquid = Table B.1.4 b) Table B.1.2 vf v vg so two phase L + V

v - vf x = v = (0.2 0.001093) / 0.3738 = 0.532 fg T = Tsat = 151.86oC c) Only one of the two look-up is needed Table B.1.1 200oC P Psat = Table B.1.2 Table B.1.3 d) Table B.1.1 1400 kPa

= superheated vapor

T Tsat = 195oC

subtable for 1400 kPa gives the state properties since quality is given it is two-phase

v = vf + x vfg = 0.001404 + 0.8 0.02027 = 0.01762 m3/kg 3.34 Place the four states a-d listed in Problem 3.33 as labeled dots in a sketch of the P-v and T-v diagrams. Solution:

8581 1400 500 P C.P. d a b v T c T 300 200 152 20 a

C.P. d c b v P = const.

Sonntag, Borgnakke and van Wylen

3.35 Determine the phase and the specific volume for ammonia at these states using the Appendix B table. a. 10oC, 150 kPa b. 20oC, 100 kPa c. 60oC, quality 25% Solution: Ammonia, NH3, properties from Table B.2 a) Table B.2.1: P Psat(-10 C) = 291 kPa Superheated vapor B.2.2 b) Table B.2.1 at given T: Superheated vapor B.2.2 c) Table B.2.1 enter with T (this is two-phase L + V) v = vf + x vfg = 0.001834 + x 0.04697 = 0.01358 m3/kg Psat = 847.5 kPa v = 1.4153 m3/kg so P Psat v = 0.8336 m3/kg

Sonntag, Borgnakke and van Wylen

3.36 Give the phase and the specific volume. Solution: a. R-22 T = -25 C P = 100 kPa Table B.4.1 at given T: Psat = 201 kPa sup. vap. B.4.2 b. R-22 T = -25 C P = 300 kPa Table B.4.1 at given T: Psat = 201 kPa

3

so

P Psat

=

v (0.22675 + 0.23706)/2 = 0.2319 m3/kg so

compr. liq. as P Psat

v vf = 0.000733 m /kg so P Psat

c. R-12 T = 5 C P = 200 kPa Table B.3.1 at given T: Psat = 362.6 kPa sup. vap. B.3.2

v (0.08861 + 0.09255)/2 = 0.09058 m3/kg

P C.P. T a, c T v b C.P. P = const. a, c

States shown are placed relative to the two-phase region, not to each other.

b

v

Sonntag, Borgnakke and van Wylen

3.37 Fill out the following table for substance ammonia: Solution: P [kPa] T [ oC] v [m3/kg] a) 1200 50 0.1185 b) 2033 50 0.0326 a) b) x Undefined 0.5

B.2.1 v vg = superheated vapor Look in B.2.2 B.2.1 P = Psat = 2033 kPa v = vf + x vfg = 0.001777 + 0.5 0.06159 = 0.0326 m3/kg

3.38 Place the two states a-b listed in Problem 3.37 as labeled dots in a sketch of the Pv and T-v diagrams. Solution:

P C.P. 2033 1200 b a T v 50 b T C.P. P = const. a

v

Sonntag, Borgnakke and van Wylen

3.39 Calculate the following specific volumes a. R-134a: 50 C, 80% quality b. Water c. Nitrogen Solution: All states are two-phase with quality given. The overall specific volume is given by Eq.3.1 or 3.2 v = vf + x vfg = (1-x)vf + x vg a. R-134a: 50 C, 80% quality in Table B.5.1 4 MPa, 90% quality 120 K, 60% quality

v = 0.000908 + x 0.01422 = 0.01228 m3/kg b. Water 4 MPa, 90% quality in Table B.1.2 v = 0.001252(1-x) + x 0.04978 = 0.04493 m3/kg c. Nitrogen 120 K, 60% quality in Table B.6.1 v = 0.001915 + x 0.00608 = 0.005563 m3/kg

Sonntag, Borgnakke and van Wylen

3.40 Give the phase and the missing property of P, T, v and x. a. R-134a T = -20oC, P = 150 kPa b. R-134a P = 300 kPa, v = 0.072 m3/kg c. CH4 T = 155 K, v = 0.04 m3/kg d. CH4 T = 350 K, v = 0.25 m3/kg

Solution: a) B.5.1

P Psat = 133.7 kPa compressed liquid v vf = 0.000738 m3/kg x = undefined

b) B.5.2

v vg at 300 kPa T = 10 + (20-10) x = undefined

(

superheated vapor 0.072 - 0.07111 0.07441 - 0.07111 = 12.7 C

)

c) B.7.1

v vg = 0.04892 m3/kg 2-phase v - vf 0.04-0.002877 x= v = = 0.806 0.04605 fg P = Psat = 1295.6 kPa T Tc and v vc superheated vapor B.7.2 located between 600 & 800 kPa 0.25-0.30067 P = 600 + 200 0.2251-0.30067 = 734 kPa

P d T v a v T

d) B.7.1

C.P. c

d b P = const.

c a

b

Sonntag, Borgnakke and van Wylen

3.41

A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100 C. The vessel is now heated. If a safety pressure valve is installed, at what pressure should the valve be set to have a maximum temperature of 200 C? Solution: Process: v = V/m = constant State 1: v1 = 1/2 = 0.5 m3/kg from Table B.1.1 it is 2-phase State 2: 200 C, 0.5 m3/kg Table B.1.3 between 400 and 500 kPa so interpolate 0.5-0.53422 P 400 + 0.42492-0.53422 (500-400) = 431.3 kPa T

C.P.

500 kPa 400 kPa 100 C v

Sonntag, Borgnakke and van Wylen

3.42 Saturated liquid water at 60 C is put under pressure to decrease the volume by 1% keeping the temperature constant. To what pressure should it be compressed? Solution: State 1: T = 60 C , x = 0.0; Table B.1.1: v = 0.001017 m3/kg Process: T = constant = 60 C State 2: T, v = 0.99 vf (60 C) = 0.99 0.001017 = 0.0010068 m3/kg Between 20 & 30 MPa in Table B.1.4,

P 2 C.P. T 2 1 1

P 23.8 MPa

30 MPa 20 MPa

v

v

Sonntag, Borgnakke and van Wylen

3.43 Saturated water vapor 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 if the temperature is changed to a) 200 oC and b) 100 oC. Solution: State 1: (200 kPa, x = 1) in B.1.2: State a: (200 kPa, 200 oC) B.1.3: State b: (200 kPa, 100 oC) B.1.1: v1 = vg (200 kPa) = 0.8857 m3/kg va = 1.083 m3/kg vb = 0.001044 m3/kg

As the piston height is proportional to the volume we get ha = h1 (va /v1) = 0.1 (1.0803 / 0.8857) = 0.12 m hb = h1 (vb / v1) = 0.1 (0.001044 / 0.8857) = 0.00011 m

P C.P. b 1 a T v 200 120 100 b 1 v T C.P. P = 200 kPa 200 a

Sonntag, Borgnakke and van Wylen

3.44 You want a pot of water to boil at 105oC. How heavy a lid should you put on the 15 cm diameter pot when Patm = 101 kPa? Solution: Table B.1.1 at 105oC : Psat = 120.8 kPa

A = 4 D2 = 4 0.152 = 0.01767 m2 Fnet = (Psat Patm) A = (120.8 - 101) kPa 0.01767 m2 = 0.3498 kN = 350 N Fnet = mlid g 350 mlid = Fnet/g = 9.807 = 35.7 kg

Some lids are clamped on, the problem deals with one that stays on due to its weight.

Sonntag, Borgnakke and van Wylen

3.45 In your refrigerator the working substance evaporates from liquid to vapor at -20 oC inside a pipe around the cold section. Outside (on the back or below) is a black grille inside which the working substance condenses from vapor to liquid at +40 oC. For each location find the pressure and the change in specific volume (v) if a) the substance is R-12 b) the substance is ammonia Solution: The properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature. v = vfg 0.017 0.108 0.0814 0.622

Substance R-12 R-12 Ammonia Ammonia

P C.P. 40C 4 1

TABLE B.3.1 B.3.1 B.2.1 B.2.1

T 40 oC -20 oC 40 oC -20 oC

Psat , kPa 961 151 1555 190

T 40 3 T 2 v -20

C.P. 4 1 3 2 v

Sonntag, Borgnakke and van Wylen

3.46 In your refrigerator the working substance evaporates from liquid to vapor at -20 oC inside a pipe around the cold section. Outside (on the back or below) is a black grille inside which the working substance condenses from vapor to liquid at +40 oC. For each location find the pressure and the change in specific volume (v) if: a) the substance is R-134a b) the substance is R-22 Solution: The properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature. v = vfg 0.019 0.146 0.0143 0.092

Substance R-134a R-134a R-22 R-22

P C.P. 40C 4 1

TABLE B.5.1 B.5.1 B.4.1 B.4.1

T 40 oC -20 oC 40 oC -20 oC

Psat , kPa 1017 134 1534 245

T 40 3 T 2 v -20

C.P. 4 1 3 2 v

Sonntag, Borgnakke and van Wylen

3.47 A water storage tank contains liquid and vapor in equilibrium at 110 C. The distance from the bottom of the tank to the liquid level is 8 m. What is the absolute pressure at the bottom of the tank? Solution: Saturated conditions from Table B.1.1: Psat = 143.3 kPa vf = 0.001052 m3/kg ; gh 9.807 8 P = v = 0.001052 = 74 578 Pa = 74.578 kPa f Pbottom = Ptop + P = 143.3 + 74.578 = 217.88 kPa

H

Sonntag, Borgnakke and van Wylen

3.48 Saturated water vapor 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 volume? Solution: State 1: Process: State 2: B 1.2 v1 = vg (200 kPa) = 0.8857 m3/kg, P = constant = 200 kPa P, v2 = v1/2 = 0.44285 m3/kg Table B.1.2 T1 = 120.2 C

v2 vg so two phase T2 = Tsat = 120.2 C Height is proportional to volume h2 = h1 v2/v1 = 0.1 0.5 = 0.05m

P C.P. 2 1 T v 120 2 T

C.P. P = 200 kPa 1 v

200

Sonntag, Borgnakke and van Wylen

3.49 Two tanks are connected as shown in Fig. P3.49, both containing water. Tank A is at 200 kPa, v = 0.5 m3/kg, V = 1 m3 and tank B contains 3.5 kg at 0.5 MPa, 400 C. A The valve is now opened and the two come to a uniform state. Find the final specific volume. Solution: Control volume: both tanks. Constant total volume and mass process.

A

B sup. vapor

State A1: (P, v)

mA = VA/vA = 1/0.5 = 2 kg vB = 0.6173 m3/kg

State B1: (P, T) Table B.1.3

VB = mBvB = 3.5 0.6173 = 2.1606 m3 Final state: mtot = mA + mB = 5.5 kg Vtot = VA + VB = 3.1606 m3 v2 = Vtot/mtot = 0.5746 m3/kg

Sonntag, Borgnakke and van Wylen

3.50

Determine the mass of methane gas stored in a 2 m3 tank at -30 C, 3 MPa. Estimate the percent error in the mass determination if the ideal gas model is used. Solution: Methane Table B.7.1 at -30 C = 243.15 K Tc = 190.6 K, so superheated vapor in Table B.7.2. Linear interpolation between 225 and 250 K. 243.15-225 v 0.03333 + 250-225 (0.03896 - 0.03333) = 0.03742 m3/kg m = V/v = 2/0.03742 = 53.45 kg Ideal gas assumption v = RT/P = 0.51835 243.15/3000 = 0.042 m3/kg m = V/v = 2/0.042 = 47.62 kg Error: 5.83 kg 10.9% too small

Sonntag, Borgnakke and van Wylen

3.51 Saturated water vapor at 60 C has its pressure decreased to increase the volume by 10% keeping the temperature constant. To what pressure should it be expanded? Solution: Initial state: Final state: v = 7.6707 m3/kg from table B.1.1 v = 1.10 vg = 1.1 7.6707 = 8.4378 m3/kg

Interpolate at 60 C between saturated (P = 19.94 kPa) and superheated vapor P = 10 kPa in Tables B.1.1 and B.1.3 8.4378 - 7.6707 P 19.941 + (10 - 19.941) = 18.9 kPa 15.3345 - 7.6707

P C.P. T C.P. P = 10 kPa T v v

60 C 10 kPa

o

Comment: T,v P = 18 kPa (software) v is not linear in P, more like 1/P, so the linear interpolation in P is not very accurate.

Sonntag, Borgnakke and van Wylen

3.52 Saturated water vapor 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 heated to occupy twice the original volume? Solution: From B.1.2, v1 = 0.8857 m3/kg P2 = P1, v2 = 2v1 = 2 0.8857 = 1.7714 m3/kg Since the cross sectional area is constant the height is proportional to volume h2 = h1 v2/v1 = 2h1 = 0.2 m 2: From B.1.3., Interpolate for the temperature 1.7714 1.5493 T2 = 400 + 100 1.78139 1.5493 496 C

Sonntag, Borgnakke and van Wylen

3.53 A boiler feed pump delivers 0.05 m3/s of water at 240 C, 20 MPa. What is the mass flowrate (kg/s)? What would be the percent error if the properties of saturated liquid at 240 C were used in the calculation? What if the properties of saturated liquid at 20 MPa were used? Solution: State 1: (T, P) compressed liquid seen in B.1.4: v = 0.001205 m3/kg . . m = V/v = 0.05/0.001205 = 41.5 kg/s . vf (240 C) = 0.001229 m3/kg m = 40.68 kg/s error 2% . vf (20 MPa) = 0.002036 m3/kg m = 24.56 kg/s error 41%

P C.P. 20 MPa 240 C

o

T 240

C.P. P = 20 MPa

v

v

The constant T line is nearly vertical for the liquid phase in the P-v diagram. The state is at so high P, T that the saturated liquid line is not extremely steep.

Sonntag, Borgnakke and van Wylen

3.54

Saturated vapor R-134a at 50oC changes volume at constant temperature. Find the new pressure, and quality if saturated, if the volume doubles. Repeat the question for the case the volume is reduced to half the original volume. Solution: 1: 2: (T, x) B.4.1: v1 = vg = 0.01512 m3/kg, P1 = Psat = 1318 Kpa

v2 = 2v1 = 0.03024 m3/kg superheated vapor Interpolate between 600 kPa and 800 kPa 0.03024 0.03974 P2 = 600 + 200 0.02861 0.03974 = 771 kPa v3 = v1/2 = 0.00756 m3/kg vg : two phase v3 - vf 0.00756 0.000908 = = 0.4678 x3 = v 0.01422 fg P3 = Psat = 1318 kPa

P C.P. 3 1 2 T v 50 T C.P. P = 1318 kPa 1318 3 1 2 v

3:

Sonntag, Borgnakke and van Wylen

3.55 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? Solution: Use Table B.7.1 Assume rigid tank v = constant = v1 v1 = 0.002439 + 0.25 0.30367 = 0.078366 m3/kg v1 vc = 0.00615 m3/kg All single phase when v = vg = T 145 K We then also see that t = T/(5 C/h) (145 120 ) / 5 = 5 hours P = Psat= 824 kPa

Sonntag, Borgnakke and van Wylen

3.56 A glass jar is filled with saturated water at 500 kPa, quality 25%, and a tight lid is put on. Now it is cooled to -10 C. What is the mass fraction of solid at this temperature? Solution: Constant volume and mass v1 = v2 = V/m From Table B.1.2: v1 = 0.001093 + 0.25 0.3738 = 0.094543 From Table B.1.5: v2 = 0.0010891 + x2 446.756 = v1 = 0.094543 x2 = 0.0002 mass fraction vapor or 99.98 % xsolid =1 - x2 = 0.9998

P C.P. T C.P.

1 T 2 v 1 2 v

P L T 1 S 2 v C.P. V

Sonntag, Borgnakke and van Wylen

3.57

Saturated (liquid + vapor) ammonia at 60 C is contained in a rigid steel tank. It is used in an experiment, where it should pass through the critical point when the system is heated. What should the initial mass fraction of liquid be? Solution: Process: Constant mass and volume, From table B.2.1: v2 = vc = 0.004255 m3/kg v1 = 0.001834 + x1 0.04697 = 0.004255 = x1 = 0.01515 liquid mass fraction = 1 - x1 = 0.948 1 60 C v T Crit. point

v=C

Sonntag, Borgnakke and van Wylen

3.58

A steel tank contains 6 kg of propane (liquid + vapor) at 20 C with a volume of 0.015 m3. The tank is now slowly heated. Will the liquid level inside eventually rise to the top or drop to the bottom of the tank? What if the initial mass is 1 kg instead of 6 kg? Solution: Constant volume and mass

T C.P.

V 0.015 m3 v2 = v1 = m = 6 kg = 0.0025 m3/kg A.2: vc = 0.00454 m3/kg v1 eventually reaches sat. liquid. level rises to top If m = 1 kg v1 = 0.015 m3/kg vc then it will reach saturated vapor. level falls

Liq.

Vapor

a

vc

b

20 C v

Sonntag, Borgnakke and van Wylen

3.59

A 400-m3 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 90% liquid and 10% vapor, by volume, at 100 kPa, what mass of LNG (kg) will the tank hold? What is the quality in the tank? Solution: CH4 is in the section B tables. From Table B.7.1: From Table B.7.2: vf 0.002366 m3/kg, vg 0.55665 m3/kg (interpolated) (first entry 100 kPa)

Vliq 0.9 400 Vvap 0.1 400 mliq = v = 0.002366 = 152 155.5 kg; mvap = v = 0.55665 = 71.86 kg f g mtot = 152 227 kg, x = mvap / mtot = 4.72 10-4

Sonntag, Borgnakke and van Wylen

3.60 A sealed rigid vessel of 2 m3 contains a saturated mixture of liquid and vapor R134a 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. Solution: Process: constant volume and constant mass. P 2 1 State 2 is saturated vapor, from table B.5.1 P2 = Psat(50 C) = 1.318 MPa State 1: same specific volume as state 2 v = v = 0.015124 m3/kg

1 2

v

v1 = 0.000794 + x1 0.048658 x1 = 0.2945

m = V/v1 = 2/0.015124 = 132.24 kg;

mliq = (1 - x1)m = 93.295 kg

Sonntag, Borgnakke and van Wylen

3.61 A pressure cooker (closed tank) contains water at 100 C with the liquid volume being 1/10 of the vapor volume. It is heated until the pressure reaches 2.0 MPa. Find the final temperature. Has the final state more or less vapor than the initial state? Solution: State 1: Vf = mf vf = Vg/10 = mgvg/10 ; vf = 0.001044 m3/kg, vg = 1.6729 m3/kg mg 10 mfvf / vg 10 vf 0.01044 x1 = m + m = m + 10 m v / v = 10 v + v = 0.01044 + 1.6729 = 0.0062 g f f f f g f g Table B.1.1: v1 = 0.001044 + 0.0062 1.67185 = 0.01141 m3/kg State 2: v2 = v1 = 0.01141 m3/kg vg(2MPa) from B.1.2 so two-phase P 2 At state 2: = v2 = vf + x2 vfg x2 = 0.104

0.01141 = 0.001177 + x2 0.09845 More vapor at final state T2 = Tsat(2MPa) = 212.4 C v

1

Sonntag, Borgnakke and van Wylen

3.62 A pressure cooker has the lid screwed on tight. A small opening with A = 5 mm2 is covered with a petcock that can be lifted to let steam escape. How much mass should the petcock have to allow boiling at 120oC with an outside atmosphere at 101.3 kPa? Table B.1.1.: Psat = 198.5 kPa

F = mg = P A m = P A/g (198.5-101.3) 1000 5 10-6 = 9.807 = 0.0496 kg = 50 g

Sonntag, Borgnakke and van Wylen

3.63

Ammonia at 10 oC and mass 0.1 kg is in a piston cylinder with an initial volume of 1 m3. The piston initially resting on the stops has a mass such that a pressure of 900 kPa will float it. Now the ammonia is slowly heated to 50oC. Find the final pressure and volume. Solution: C.V. Ammonia, constant mass. Process: V = constant unless P = Pfloat P V 1 State 1: T = 10 oC, v1 = m = 10 = 0.1 m3/kg From Table B.2.1 vf v vg P2 P1 1 V 1a 2

v - vf 0.1 - 0.0016 x1 = v = 0.20381 = 0.4828 fg

State 1a: P = 900 kPa, v = v1 = 0.1 m3/kg vg at 900 kPa This state is two-phase T1a = 21.52oC Since T2 T1a then v2 v1a State 2: 50oC and on line(s) means P2 = 900 kPa which is superheated vapor. Table B.2.2 : v2 = 0.16263 m3/kg V2 = mv2 = 1.6263 m3

Sonntag, Borgnakke and van Wylen

Ideal Gas Law

3.64 A cylinder fitted with a frictionless piston contains butane at 25 C, 500 kPa. Can the butane reasonably be assumed to behave as an ideal gas at this state ? Solution Butane 25 C, 500 kPa, Table A.2: Tc = 425 K; Pc = 3.8 MPa 25 + 273 0.5 Tr = 425 = 0.701; Pr = 3.8 = 0.13 Look at generalized chart in Figure D.1 Actual Pr Pr, sat = 0.1 = liquid! not a gas The pressure should be less than 380 kPa to have a gas at that T.

Sonntag, Borgnakke and van Wylen

3.65 A spherical helium balloon of 10 m in diameter is at ambient T and P, 15oC 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? V = 6 D3 = 6 103 = 523.6 m3 V PV mHe = V = v = RT 100 523.6 = = 87.5 kg 2.0771 288 PV 100 523.6 mair = RT = = 633 kg 0.287 288 mlift = mair mHe = 633-87.5 = 545.5 kg

Sonntag, Borgnakke and van Wylen

3.66 Is it reasonable to assume that at the given states the substance behaves as an ideal gas? Solution: a) Oxygen, O2 at 30 C, 3 MPa Ideal Gas ( T Tc = 155 K from A.2) b) Methane, CH4 at c) Water, H2O d) R-134a e) R-134a at 30 C, 3 MPa 30 C, 3 MPa Ideal Gas ( T Tc = 190 K from A.2) NO compressed liquid P Psat (B.1.1)

at 30 C, 3 MPa NO compressed liquid P Psat (B.5.1) at 30 C, 100 kPa Ideal Gas P is low Psat (B.5.1) ln P c, d Liq. e Cr.P. a, b Vapor T

Sonntag, Borgnakke and van Wylen

3.67

A 1-m3 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 ? Solution: Use Table A.2 to compare T and P to the critical T and P with T = 20 C = 293.15 K ; P = 100 kPa Pc for all Air : T TC,N2; TC,O2 = 154.6 K so ideal gas; R= 0.287 kJ/kg K Neon: T Tc = 44.4 K so ideal gas; R = 0.41195 kJ/kg K Propane: T Tc = 370 K, but P Pc = 4.25 MPa so gas R = 0.18855 kJ/kg K All states are ideal gas states so the ideal gas law applies PV = mRT PV a) m = RT = b) m = c) m = 100 1 = 1.189 kg 0.287 293.15

100 1 = 0.828 kg 0.41195 293.15 100 1 = 1.809 kg 0.18855 293.15

Sonntag, Borgnakke and van Wylen

3.68 A rigid tank of 1 m3 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 then the final pressure? Solution: PV m = RT = 600 1 = 5.054 kg 0.2968 400

m2 = m - 0.5 = 4.554 kg m2RT2 4.554 0.2968 375 P2 = V = = 506.9 kPa 1

Sonntag, Borgnakke and van Wylen

3.69 A cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then filled with carbon dioxide gas at 25 C. To what pressure should it be charged if there should be 1.2 kg of carbon dioxide? Solution: Assume CO2 is an ideal gas, table A.5: R = 0.1889 kJ/kg K Vcyl = A L = 4(0.2)2 1 = 0.031416 m3 P V = mRT P= = mRT P= V

1.2 kg 0.1889 kJ/kg (273.15 + 25) K = 2152 kPa 0.031416 m3

Sonntag, Borgnakke and van Wylen

3.70

A glass is cleaned in 45oC hot water and placed on the table bottom up. The room air at 20oC that was trapped in the glass gets heated up to 40oC and some of it leaks out so the net resulting pressure inside is 2 kPa above ambient pressure of 101 kPa. Now the glass and the air inside cools down to room temperature. What is the pressure inside the glass? Solution: 1 air: 40oC, 103 kPa 2 air: 20oC, ? Constant Volume: V1 = V2,

AIR

Slight amount of liquid water seals to table top

Constant Mass m1 = m2 Ideal Gas P1V1 = m1RT1 Take Ratio

and

P2V2 = m1RT2

T1 20 + 273 P2 = P1 T = 103 40 + 273 = 96.4 kPa

2

Sonntag, Borgnakke and van Wylen

3.71 A hollow metal sphere of 150-mm inside diameter is weighed on a precision beam balance when evacuated and again after being filled to 875 kPa with an unknown gas. The difference in mass is 0.0025 kg, and the temperature is 25 C. What is the gas, assuming it is a pure substance listed in Table A.5 ? Solution: Assume an ideal gas with total volume: V = 6(0.15)3 = 0.001767 m3 mRT 0.0025 8.3145 298.2 M = PV = = 4.009 MHe 875 0.001767 = Helium Gas

Sonntag, Borgnakke and van Wylen

3.72 A vacuum pump is used to evacuate a chamber where some specimens are dried at 50 C. The pump rate of volume displacement is 0.5 m3/s with an inlet pressure of 0.1 kPa and temperature 50 C. How much water vapor has been removed over a 30min period? Solution: Use ideal gas since P lowest P in steam tables. From table A.5 we get R = 0.46152 kJ/kg K . . . . m = m t with mass flow rate as: m= V/v = PV/RT . 0.1 0.5 30 60 m = PVt/RT = = 0.603 kg (0.46152 323.15)

(ideal gas)

Sonntag, Borgnakke and van Wylen

3.73 A 1 m3 rigid tank has propane at 100 kPa, 300 K and connected by a valve to another tank of 0.5 m3 with propane at 250 kPa, 400 K. The valve is opened and the two tanks come to a uniform state at 325 K. What is the final pressure? Solution: Propane is an ideal gas (P Pc) with R = 0.1886 kJ/kgK from Tbl. A.5 PAVA 100 1 mA = RT = = 1.7674 kg 0.1886 300 A PBVB 250 0.5 m = RT = = 1.6564 kg 0.1886 400 B V2 = VA + VB = 1.5 m3 m2 = mA + mB = 3.4243 kg m2RT2 3.4243 0.1886 325 = = 139.9 kPa P2 = V 1.5 2

Sonntag, Borgnakke and van Wylen

3.74 Verify the accuracy of the ideal gas model when it is used to calculate specific volume for saturated water vapor as shown in Fig. 3.9. Do the calculation for 10 kPa and 1 MPa. Solution: Look at the two states assuming ideal gas and then the steam tables. Ideal gas: v = RT/P = v1 = 0.46152 (45.81 + 273.15)/10 = 14.72 m3/kg v2 = 0.46152 (179.91 + 273.15)/1000 = 0.209 m3/kg Real gas: Table B.1.2: v1 = 14.647 m3/kg so error = 0.3 % 3/kg so error = 7.49 % v2 = 0.19444 m

Sonntag, Borgnakke and van Wylen

3.75

Assume we have 3 states of saturated vapor R-134a at +40 oC, 0 oC and -40 oC. Calculate the specific volume at the set of temperatures and corresponding saturated pressure assuming ideal gas behavior. Find the percent relative error = 100(v - vg)/vg with vg from the saturated R-134a table. Solution: R-134a. Table values from Table B.5.1 Psat, vg(T)

Ideal gas constant from Table A.5: RR-134a = 0.08149 kJ/kg K T -40 oC 0 oC 40 oC Psat , kPa 51.8 294 1017 vg 0.35696 0.06919 0.02002 vID.G. = RT / Psat 0.36678 0.07571 0.02509 error % 2.75 9.4 25.3

P 3 2 1 v

T 3 2 1 v

Sonntag, Borgnakke and van Wylen

3.76 Do Problem 3.75, but for the substance R-12. Solution: R-12.

Table values from Table B.3.1

Psat, vg(T)

Ideal gas constant from Table A.5: RR-12 = 0.08149 kJ/kg K T -40 oC 0 oC 40 oC Psat , kPa 64.2 308.6 960.7 vg 0.24191 0.05539 0.01817 vID.G. = RT / Psat 0.2497 0.06086 0.02241 error % 3.2 9.9 23.4

P 3 2 1 v

T 3 2 1 v

Sonntag, Borgnakke and van Wylen

3.77 Do Problem 3.75, but for the substance ammonia. Solution: NH3.

Table values from Table B.2.1

Psat, vg(T)

Ideal gas constant from Table A.5: Rammonia = 0.4882 kJ/kg K T -40 oC 0 oC 40 oC Psat , kPa 71.7 429.6 1555 vg 1.5526 0.28929 0.08313 vID.G. = RT / Psat 1.5875 0.3104 0.09832 error % 2.25 7.3 18.3

P 3 2 1 v

T 3 2 1 v

Sonntag, Borgnakke and van Wylen

3.78

Air in an automobile tire is initially at -10 C and 190 kPa. After the automobile is driven awhile, the temperature gets up to 10 C. Find the new pressure. You must make one assumption on your own. Solution:

Assume constant volume and that air is an ideal gas P2 = P1 T2/T1 283.15 = 190 263.15 = 204.4 kPa

Sonntag, Borgnakke and van Wylen

3.79 An initially deflated and flat balloon is connected by a valve to a 12 m3 storage tank containing helium gas at 2 MPa and ambient temperature, 20 C. The valve is opened and the balloon is inflated at constant pressure, Po = 100 kPa, equal to ambient pressure, until it becomes spherical at D1 = 1 m. If the balloon is larger than this, the balloon material is stretched giving a pressure inside as D1 D1 P = P0 + C 1 - D D The balloon is inflated to a final diameter of 4 m, at which point the pressure inside is 400 kPa. The temperature remains constant at 20 C. What is the maximum pressure inside the balloon at any time during this inflation process? What is the pressure inside the helium storage tank at this time? Solution: At the end of the process we have D = 4 m so we can get the constant C as 1 1 P = 400 = P0 + C ( 1 4 ) 4 = 100 + C 3/16 = C = 1600 1 1 X = D / D1 The pressure is: P = 100 + 1600 ( 1 X ) X ; Differentiate to find max: dP 2 3 + 2 X ) / D1 = 0 dD = C ( - X 2 3 X=2 = - X + 2 X = 0 = 3 3 at max P = D = 2D1 = 2 m; V = 6 D = 4.18 m 1 1 Pmax = 100 + 1600 ( 1 - 2 ) 2 = 500 kPa PV Helium is ideal gas A.5: m = RT = PV mTANK, 1 = RT = 500 4.189 = 3.44 kg 2.0771 293.15

2000 12 = 39.416 kg 2.0771 293.15 mTANK, 2 = 39.416 3.44 = 35.976 kg PT2 = mTANK, 2 RT/V = ( mTANK, 1 / mTANK, 2 ) P1 = 1825.5 kPa

Sonntag, Borgnakke and van Wylen

Compressibility Factor

3.80

Argon is kept in a rigid 5 m3 tank at -30 C, 3 MPa. Determine the mass using the compressibility factor. What is the error (%) if the ideal gas model is used? Solution: No Argon table, so we use generalized chart Fig. D.1 Tr = 243.15/150.8 = 1.612, Pr = 3000/4870 = 0.616 = PV m = ZRT = 3000 5 = 308.75 kg 0.96 0.2081 243.2 4% error

Z 0.96

Ideal gas Z = 1 PV m = RT = 296.4 kg

Sonntag, Borgnakke and van Wylen

3.81 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 from Table A.1 v = 0.2923 m3/kg RT 0.48819 313 = 0.3056 m3/kg 4.5% error Ideal gas: v = P = 500 313.15 0.5 Figure D.1: Tr = 405.5 = 0.772, Pr = 11.35 = 0.044 Z = 0.97 ZRT v = P = 0.2964 m3/kg 1.4% error Table B.2.2:

Sonntag, Borgnakke and van Wylen

3.82

A new refrigerant R-125 is stored as a liquid at -20 oC 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 A.2: R-125 Tc = 339.2 K Pc = 3.62 MPa Tr = T / Tc = 253.15 / 339.2 = 0.746 We can read from Figure D.1 or a little more accurately interpolate from table D.4 entries: Pr sat = 0.16 ; Zg = 0.86 ; Zf = 0.029 P = Pr sat Pc = 0.16 3620 = 579 kPa PVliq = Zf mliq RT = 0.029 1.5 0.06927 253.15 / 579 = 0.0013 m3 Z sat vapor Tr= 2.0 Tr = 0.7 sat liq. 0.1 1 ln Pr Tr = 0.7

Sonntag, Borgnakke and van Wylen

3.83 Many substances that normally do not mix well do so easily under supercritical pressures. A mass of 125 kg ethylene at 7.5 MPa, 296.5 K is stored for such a process. How much volume does it occupy? Solution: There is no section B table for ethylene so use compressibility chart. Table A.2: Ethylene Tc = 282.4 K Pc = 5.04 MPa Tr = T/Tc = 296.5 / 282.4 = 1.05 ; Z = 0.32 from Figure D.1 V = mZRT / P = 125 0.32 0.2964 296.5 / 7500 = 0.469 m3 Pr = P/Pc = 7.5 / 5.04 = 1.49

Z Tr = 0.7

Tr= 2.0 Tr = 1.05 Tr = 0.7 0.1 1 ln Pr

Sonntag, Borgnakke and van Wylen

3.84 Carbon dioxide at 330 K is pumped at a very high pressure, 10 MPa, into an oilwell. As it penetrates the rock/oil the oil viscosity is lowered so it flows out easily. For this process we need to know the density of the carbon dioxide being pumped. Solution: There is not a B section table so use compressibility chart Table A.2 CO2: Tc = 304.1 K Pc = 7.38 MPa Tr = T/Tc = 330/304.1 = 1.085 Pr = P/Pc = 10/7.38 = 1.355 From Figure D.1: Z 0.45 = 1/v = P / ZRT = 10000/(0.45 0.1889 330) = 356 kg/m3

Z Tr = 0.7

Tr= 2.0 Tr = 1.1 Tr = 0.7 0.1 1 ln Pr

Sonntag, Borgnakke and van Wylen

3.85 To plan a commercial refrigeration system using R-123 we would like to know how much more volume saturated vapor R-123 occupies per kg at -30 oC compared to the saturated liquid state. Solution: For R-123 there is no section B table printed. We will use compressibility chart. From Table A.2 Tc = 456.9 K ; Pc = 3.66 MPa ; M = 152.93 Tr = T/Tc = 243/456.9 = 0.53 R = R/M = 8.31451 / 152.93 = 0.0544 The value of Tr is below the range in Fig. D.1 so use the table D.4 Table D.4, Zg = 0.979 Zf = 0.00222 Pr = Pr sat = 0.0116 Zfg = 0.979 - 0.0022 = 0.9768; P = Pr Pc = 42.5 vfg = Zfg RT/P = 0.9768 0.0544 243 / 42.5 = 0.304 m3/kg

Sonntag, Borgnakke and van Wylen

3.86

A bottle with a volume of 0.1 m3 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. Solution: We need to find the property v the mass is: m = V/v so find v given T1 and x as : v = vf + x vfg Table A.2: Butane Tc = 425.2 K = Zf 0.02; Zg 0.9; Pr sat = 0.1 Pc = 3.8 MPa = 3800 kPa Tr = 300/425.2 = 0.705

From Fig. D.1 or table D.4:

Z

g Tr = 0.7 f 0.1 1

Tr= 2.0 Tr = 0.7

ln Pr

P = Psat = Pr sat Pc = 0.1 3.80 1000 = 380 kPa vf = ZfRT/P = 0.02 0.14304 300/380 = 0.00226 m3/kg vg = ZgRT/P = 0.9 0.14304 300/380 = 0.1016 m3/kg v = 0.00226 + 0.75 (0.1016 0.00226) = 0.076765 m3/kg V 0.1 m = v = 0.076765 = 1.303 kg

Sonntag, Borgnakke and van Wylen

3.87

Refrigerant R-32 is at -10 oC with a quality of 15%. Find the pressure and specific volume. Solution: For R-32 there is no section B table printed. We will use compressibility chart. From Table A.2: Tc = 351.3 K ; Pc = 5.78 MPa ; From Table A.5: R = 0.1598 kJ/kg K Tr = T/Tc = 263/351.3 = 0.749 From Table D.4 or Figure D.1, Zf 0.029 ; Zg 0.86 ; Pr sat 0.16 P = Pr sat Pc = 0.16 5780 = 925 kPa v = vf + x vfg = (Zf + x Zfg) RT/P = [0.029 + 0.15 (0.86 0.029)] 0.1598 263 / 925 = 0.007 m3/kg Z Tr = 0.7

Tr= 2.0 Tr = 0.7

0.1

1

ln Pr

Sonntag, Borgnakke and van Wylen

3.88

A mass of 2 kg of acetylene is in a 0.045 m3 rigid container at a pressure of 4.3 MPa. Use the generalized charts to estimate the temperature. (This becomes trial and error). Solution: Table A.2, A.5:

Pr = 4.3/6.14 = 0.70; Tc = 308.3 K;

R = 0.3193 kJ/kg K

v = V/m = 0.045/2 = 0.0225 m3/kg ZRT State given by (P, v) v= P Since Z is a function of the state Fig. D.1 and thus T, we have trial and error. Try sat. vapor at Pr = 0.7 = Fig. D.1

• Avaliações:
• 4
• 49
• Denuncie