(Solution Manual) Introduction to Engineering Thermodynamics

Sonntag,Borgnakke, 1

Engineering

Solution Manual

1st Edition

• $15.00 $50.00

---

Description

Solution Manual for Introduction to Engineering Thermodynamics, The downloadable file consists of 765 pages, and the total size of the file is 7.39 MB. Purchase now only for highly discounted price.

Author(s): Sonntag, Borgnakke

---

---

Study Note Preview

2-1 CHAPTER 2 The correspondence between the problem set in this first edition versus the problem set in the 5'th edition text. Problems that are new are marked new and those that are only slightly altered are marked as modified (mod). New Old New Old New Old 1 1 21 new 41 new 2 2 22 18 42 32 3 new 23 19 43 34 4 3 24 15 44 35 5 5 25 new 6 6 26 20 7 7 27 21 8 new 28 new 9 9 29 new 10 10 30 23 11 11 31 24 12 12 32 new 13 new 33 new 14 new 34 26 15 new 35 27 16 13 36 new 17 new 37 28mod 18 14 38 29 19 new 39 30 20 16 40 31 2-2 F m F m 2.1 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in this gravitational field ? How much mass can a force of 1 N support ? Solution: ma = 0 =  F = F - mg F = mg = 2  9.80665 = 19.613 N g F = mg => m = F/g = 1 / 9.80665 = 0.102 kg 2.2 A model car rolls down an incline with a slope so the gravitational “pull” in the direction of motion is one third of the standard gravitational force (see Problem 2.1). If the car has a mass of 0.45 kg find the acceleration. Solution: ma =  F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2 g This acceleration does not depend on the mass of the model car. 2.3 A force of 125 N is applied to a mass of 12 kg in addition to the standard gravitation. If the direction of the force is vertical up find the acceleration of the mass. Solution: x Fup = ma = F – mg a = ( F – mg ) / m = F / m – g = 125/12 – 9.807 g = 0.61 ms-2 2.4 A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5 seconds. If the total car and driver mass is 1075 kg find the necessary force. 2-3 = dt Solution: Acceleration is the time rate of change of velocity. dV 60  1000 dt 3600  5 = 3.33 m/s2 ma =  F ; Fnet = ma = 1075  3.333 = 3583 N 2.5 A 1200 -kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed of 75 km/h. What are the force and total time required? Solution: a = dV = V => t = V = (75 − 20) 1000 = 3.82 sec dt t a 3600  5 F = ma = 1200  4 = 4800 N 2.6 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What force is needed and what is the final velocity? Solution: Constant acceleration can be integrated to get velocity. dV =>  dV =  a dt => V = a t V = a t = 3  10 = 30 m/s => V = 30 m/s F F = ma = 950  3 = 2850 N a = a = 2-4  2.7 A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2 kN now accelerates this system. What is the acceleration? Solution: m = m steel + mpropane = 15 + (1.75  44.094) = 92.165 kg ma =  F  a =  F / m a = 2000 / 92.165 = 21.7 m/s2 2.8 A car of mass 1775 kg travels with a velocity of 100 km/h. Find the kinetic energy. How high should it be lifted in the standard gravitational field to have a potential energy that equals the kinetic energy? Solution: Standard kinetic energy of the mass is KIN = ½ m V2 = ½  1775 100  1000 2   3600  = ½  1775  27.778 = 684 800 J = 684.8 kJ Standard potential energy is POT = mgh h = ½ m V2 / mg = 684 800 = 39.3 m 1775 9.807 2.9 A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration of 2 m/s2 relative to the ground at a location where the local gravitational acceleration is 9.5 m/s2. Find the required force. Solution: F = ma = Fup − mg g Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N Fup 2-5 6 m/M 2.10 On the moon the gravitational acceleration is approximately one -sixth that on the surface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface on the moon. What is the expected reading? If this mass is weighed with a spring scale that reads correctly for standard gravity on earth (see Problem 2.1), what is the reading? Solution: Moon gravitation is: g = gearth/6 Beam Balance Reading is 5 kg Spring Balance Reading is in kg units This is mass comparison Force comparison length  F  g Reading will be 5 kg 2.11 One kilogram of diatomic oxygen (O2 molecular weight 32) is contained in a 500- L tank. Find the specific volume on both a mass and mole basis (v and v ). Solution: From the definition of the specific volume v = V/m = 0.5/1 = 0.5 m3/kg v = V/n = V = M v = 32  0.5 = 16 m3/kmol 2.12 A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the rest of the volume is air with density 1.15 kg/m3. Find the mass of air and the overall (average) specific volume. Solution: mair =  V = air ( Vtot − mgranite /  ) = 1.15 [ 5 - (900 / 2400) ] = 1.15  4.625 = 5.32 kg v = V / m = 5 / (900 + 5.32) = 0.00552 m3/kg m m m 2-6 2.13 A tank has two rooms separated by a membrane. Room A has 1 kg air and volume 0.5 m3, room B has 0.75 m3 air with density 0.8 kg/m3. The membrane is broken and the air comes to a uniform state. Find the final density of the air. Solution: Density is mass per unit volume m = mA + mB = mA + BVB = 1 + 0.8  0.75 = 1.6 kg V = VA + VB = 0.5 + 0.75 = 1.25 m3  = m = 1.6 = 1.28 kg/m3 V 1.25 2.14 A 1 m3 container is filled with 400 kg of granite stone, 200 kg dry sand and 0.2 m3 of liquid 25°C water. Use properties from tables A.3 and A.4. Find the average specific volume and density of the masses when you exclude air mass and volume. Solution: Specific volume and density are ratios of total mass and total volume. mliq = Vliq/vliq = Vliq liq = 0.2  997 = 199.4 kg mTOT = mstone + msand + mliq = 400 + 200 + 199.4 = 799.4 kg Vstone = mv = m/ = 400/ 2750 = 0.1455 m3 Vsand = mv = m/  = 200/ 1500 = 0.1333 m3 VTOT = Vstone + Vsand + Vliq = 0.1455 + 0.1333 + 0.2 = 0.4788 m3 v = VTOT / mTOT = 0.4788/799.4 = 0.000599 m3/kg  = 1/v = mTOT/VTOT = 799.4/0.4788 = 1669.6 kg/m3 2-7 mg P 850  9.807 500   1000 2.15 A 1 m3 container is filled with 400 kg of granite stone, 200 kg dry sand and 0.2 m3 of liquid 25°C water. Use properties from tables A.3 and A.4 and use air density of 1.1 kg/m3. Find the average specific volume and density of the 1 m3 volume. Solution: Specific volume and density are ratios of total mass and total volume. Vstone = mv = m/  = 400/ 2750 = 0.1455 m3 Vsand = mv = m/  = 200/ 1500 = 0.1333 m3 Vair = VTOT − Vstone − Vsand − Vliq = 1− 0.1455 − 0.1333 − 0.2 = 0.5212 m3 mair = Vair/vair = Vair air = 0.5212  1.1 = 0.573 kg mliq = Vliq/vliq = Vliq liq = 0.2  997 = 199.4 kg mTOT = mstone + msand + mliq + mair = 400 + 200 + 199.4 + 0.573  800 kg v = VTOT / mTOT = 1/800 = 0.00125 m3/kg  = 1/v = mTOT/VTOT = 800/1 = 800 kg/m3 2.16 A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800 kg/m3. If the system is decelerated with 6 m/s2 what is the needed force? Solution: m = mtank + mgasoline = 15 + 0.3  800 = 255 kg F = ma = 255  6 = 1530 N 2.17 A hydraulic lift has a maximum fluid pressure of 500 kPa. What should the piston - cylinder diameter be so it can lift a mass of 850 kg? Solution: F = P A = F = mg A =  r2 =  D2/4 PA = P  D2/4 = mg  D2 = 4mg / P  D = 2 = 2 = 0.146 m 2.18 A vertical hydraulic cylinder has a 125 -mm diameter piston with hydraulic fluid inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity, find the piston mass that will create a pressure inside of 1500 kPa. 2-8 Solution: Force balance: F = PA = F = P0A + mpg ; P0 = 1 bar = 100 kPa A = (/4) D2 = (/4)  0.1252 = 0.01227 m2 mp = (P−P ) A = ( 1500 − 100 )  1000  0.01227 = 1752 kg 0 g 9.80665 2.19 Blue manometer fluid of density 925 kg/m3 shows a column height difference of 6 cm vacuum with one end attached to a pipe and the other open to P0 = 101 kPa. What is the absolute pressure in the pipe? Solution: Since the manometer shows a vacuum we have PPIPE = P0 - P P = gh = 925  9.807  0.06 = 544.3 Pa = 0.544 kPa PPIPE = 101 – 0.544 = 100.46 kPa 2.20 A cannon -ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gun- powder is burned a pressure of 7 MPa is created in the gas behind the ball. What is the acceleration of the ball if the cylinder (cannon) is pointing horizontally? Solution: The cannon ball has 101 kPa on the side facing the atmosphere. ma = F = P1  A − P0  A a = (P1 − P0 )  A / m = ( 7000 − 101 )  [ ( 0.152 /4 )/5 ] = 24.38 m/s2 2-9 2.21 What is the pressure at the bottom of a 5 m tall column of fluid with atmospheric pressure 101 kPa on the top surface if the fluid is a) water at 20°C b) glycerine 25°C or c) light oil Solution: Table A.4: H2O = 997 ; Glyc = 1260 ; Oil = 910 kg/m3 P = gh P = Ptop + P a) P = gh = 997 9.807  5 = 48887.9 Pa P = 101 + 48.99 = 149.9 kPa b) P = gh = 1260  9.807  5 = 61784 Pa P = 101 + 61.8 = 162.8 kPa c) P = gh = 910 9.807  5 = 44622 Pa P = 101 + 44.6 = 145.6 kPa 2.22 A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg resting on the stops, as shown in Fig. P2.22. With an outside atmospheric pressure of 100 kPa, what should the water pressure be to lift the piston? Solution: The force acting down on the piston comes from gravitation and the outside atmospheric pressure acting over the top surface. Force balance: F = F = PA = mpg + P0A Now solve for P P = P0 + mpg/A = 100 kPa + (100  9.80665) / (0.01  1000) = 100 kPa + 98.07 = 198 kPa 2-10 2.23 The hydraulic lift in an auto-repair shop has a cylinder diameter of 0.2 m. To what pressure should the hydraulic fluid be pumped to lift 40 kg of piston/arms and 700 kg of a car? Solution: Force acting on the mass by the gravitational field F = ma = mg = 740  9.80665 = 7256.9 N Force balance: F = ( P - P0 ) A = F => P = P0 + F / A A =  D2 (1 / 4) = 0.031416 m2 P = 101 + 7256.9 / (0.031416  1000) = 332 kPa 2.24 A barometer to measure absolute pressure shows a mercury column height of 725 mm. The temperature is such that the density of the mercury is 13550 kg/m3. Find the ambient pressure. Solution: Hg : l = 725 mm = 0.725 m;  = 13550 kg/m3 The external pressure P balances the column of height l so from Fig.2.10 P =  gl = 13550  9.80665  0.725  10-3 = 96.34 kPa 2.25 The density of atmospheric air is about 1.15 kg/m3, which we assume is constant. How large an absolute pressure will a pilot see when flying 1500 m above ground level where the pressure is 101 kPa. Solution: Assume g and  are constant then the pressure difference to carry a column of height 1500 m is from Fig.2.10 P = gh = 1.15  9.807  1500 = 16917 Pa = 16.9 kPa The pressure on top of the column of air is then P = P0 – P = 101 – 16.9 = 84.1 kPa 2.26 A differential pressure gauge mounted on a vessel shows 1.25 MPa and a local barometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure inside the vessel. 2-11 Solution: Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPa P = Pgauge + P0 = 1250 + 96 = 1346 kPa 2.27 The absolute pressure in a tank is 85 kPa and the local ambient absolute pressure is 97 kPa. If a U -tube with mercury, density 13550 kg/m3, is attached to the tank to measure the vacuum, what column height difference would it show? Solution: P = P0 - Ptank = gl l = ( P0 - Ptank ) / g = [(97 - 85 )  1000 ] / (13550  9.80665) = 0.090 m = 90 mm 2.28 A submarine maintains 101 kPa inside it and it dives 240 m down in the ocean having an average density of 1030 kg/m3. What is the pressure difference between the inside and the outside of the submarine hull? Solution: Assume inside is 101 kPa then P is from the 240 m column water. P = Lg = (1030  240  9.807) / 1000 = 2424 kPa 2.29 Assume we use a pressure gauge to measure the air pressure at street level and at the roof of a tall building. If the pressure difference can be determined with an accuracy of 1 mbar (0.001 bar) what uncertainty in the height estimate does that corresponds to? Solution: air = 1.169 Kg/m3 from Table A.5 P = 0.001 bar = 100 Pa L = P/g = 100 / (1.169  9.807) = 8.72 m 2.30 A U-tube manometer filled with water, density 1000 kg/m3, shows a height difference of 25 cm. What is the gauge pressure? If the right branch is tilted to make an angle of 30° with the horizontal, as shown in Fig. P2.30, what should the length of the column in the tilted tube be relative to the U-tube? 2-12 Solution: Same height in the two sides in the direction of g. P = F/A = mg/A = Vg/A = hg = 0.25  1000  9.807 = 2452.5 Pa = 2.45 kPa h h = H  sin 30°  H = h/sin 30° = 2h = 50 cm 2.31 The difference in height between the columns of a manometer is 200 mm with a fluid of density 900 kg/m3. What is the pressure difference? What is the height difference if the same pressure difference is measured using mercury, density 13600 kg/ m3, as manometer fluid? Solution: P = 1gh1 = 900  9.807  0.2 = 1765.26 Pa = 1.77 kPa h = P/ ( g) = ( gh ) / ( g) = 900  0.2 = 0.0132 m= 13.2 mm hg hg 1 1 hg 13600 2.32 An exploration submarine should be able to go 4000 m down in the ocean. If the ocean density is 1020 kg/m3 what is the maximum pressure on the submarine hull? Solution : P = Lg = (1020  400 9.807) / 1000 = 40011 kPa  40 MPa 2.33 The main waterline into a tall building has a pressure of 600 kPa at 5 m elevation below ground level. How much extra pressure does a pump need to add to ensure a water line pressure of 200 kPa at the top floor 150 m above ground? Solution: The pump exit pressure must balance the top pressure plus the column P. The pump inlet pressure provides part of the absolute pressure. Pafter pump = Ptop + P P = gh = 997  9.807  (150 + 5) = 1515525 Pa = 1516 kPa Pafter pump = 200 + 1516 = 1716 kP Ppump = 1716 – 600 = 1116 kPa H 30o 2-13 2.34 Two vertical cylindrical storage tanks are full of liquid water, density 1000 kg/m3, the top open to the atmoshere. One is 10 m tall, 2 m diameter, the other is 2.5 m tall with diameter 4 m. What is the total force from the bottom of each tank to the water and what is the pressure at the bottom of each tank? Solution: VA = H  D2  (1 / 4) = 10    22  ( 1 / 4) = 31.416 m3 VB = H  D2  (1 / 4) = 2.5    42  ( 1 / 4) = 31.416 m3 Tanks have the same volume, so same mass of water F = mg =  V g = 1000  31.416  9.80665 = 308086 N Tanks have net force up as (holds same m in gravitation field) Fnet A = F + P0A = 308086 + 101325  3.1416 = 626 408 N Fnet B = F + P0A = 308086 + 101325  12.5664 = 1 581 374 N Pbot = P0 +  H g Pbot,A = 101 + (1000  10  9.80665 / 1000) = 199 kPa Pbot,B = 101 + (1000  2.5  9.80665 / 1000) = 125.5 kPa 2-14 2.35 The density of mercury changes approximately linearly with temperature as Hg = 13595 − 2.5 T kg/ m3 T in Celsius so the same pressure difference will result in a manometer reading that is influenced by temperature. If a pressure difference of 100 kPa is measured in the summer at 35°C and in the winter at −15°C, what is the difference in column height between the two measurements? Solution: P = gh  h = P/g ; su = 13507.5 ; w = 13632.5 kg/m3 hsu = 100103/(13507.5  9.807) = 0.7549 m hw = 100103/(13632.5  9.807) = 0.7480 m h = hsu - hw = 0.0069 m = 6.9 mm 2.36 A mercury thermometer measures temperature by measuring the volume expansion of a fixed mass of liquid Hg due to a change in the density, see problem 2.35. Find the ralative change (%) in volume for a change in temperature from 10°C to 20°C. Solution: From 10°C to 20°C At 10°C : Hg = 13595 – 2.5  10 = 13570 kg/m3 At 20°C : Hg = 13595 – 2.5  20 = 13545 kg/m3 V = m/ Relative Change = (V20 – V10) / V10 = (m/20 - m/10) / (m/10) = 10/20 – 1 = (13570/13545) – 1 = 0.0018 (0.18%) 2-15 2.37 Liquid water with density  is filled on top of a thin piston in a cylinder with cross - sectional area A and total height H. Air is let in under the piston so it pushes up, spilling the water over the edge. Deduce the formula for the air pressure as a function of the piston elevation from the bottom, h. Solution: Force balance P0 Piston: F = F PA = P0A + mH2Og H P = P0 + mH2Og/A h P = P0 + (H − h)g h, V air 2.38 A piston, mp= 5 kg, is fitted in a cylinder, A = 15 cm2, that contains a gas. The setup is in a centrifuge that creates an acceleration of 25 m/s2 in the direction of piston motion towards the gas. Assuming standard atmospheric pressure outside the cylinder, find the gas pressure. Solution: Force balance: F = F = P0A + mpg = PA g P = P0 + mpg/A = 101.325 + 5  25 / (1000  0.0015) = 184.7 kPa 2.39 A piece of experimental apparatus is located where g = 9.5 m/s2 and the temperature is 5°C. An air flow inside the apparatus is determined by measuring the pressure drop across an orifice with a mercury manometer (see Problem 2.27 for density) showing a height difference of 200 mm. What is the pressure drop in kPa? Solution: P = gh ; Hg = 13600 kg/m3 P = 13600  9.5  0.2 = 25840 Pa = 25.84 kPa 2.40 Repeat the previous problem if the flow inside the apparatus is liquid water,   1000 kg/m3, instead of air. Find the pressure difference between the two holes flush with the bottom of the channel. You cannot neglect the two unequal water columns. P P0 P0 2-16 . h1 h2 P1 P2 Solution: Balance forces in the manometer: (H - h2) - (H - h1) = hHg = h1 - h2 P A +  h gA +  (H - h )gA H 1 H2O 1 Hg 1 = P2A + H2Oh2gA + Hg(H - h2)gA  P1 - P2 = H2O(h2 - h1)g + Hg(h1 - h2)g P1 - P2 = HghHgg - H2OhHgg = 13600  0.2  9.5 - 1000  0.2  9.5 = 25840 - 1900 = 23940 Pa = 23.94 kPa 2.41 A dam retains a lake 6 m deep. To construct a gate in the dam we need to know the net horizontal force on a 5 m wide and 6 m tall port section that then replaces a 5 m section of the dam. Find the net horizontal force from the water on one side and air on the other side of the port. Solution: Pbot = P0 + P P = gh = 997 9.807  6 = 58665 Pa = 58.66 kPa Neglect P in air Fnet = Fright – Fleft = Pavg A - P0A Pavg = P0 + 0.5 P Since a linear pressure variation with depth. Fnet = (P0 + 0.5 P)A - P0A = 0.5 P A = 0.5  58.66  5  6 = 880 kN 2-17 2.42 Two piston/cylinder arrangements, A and B, have their gas chambers connected by a pipe. Cross -sectional areas are AA = 75 cm2 and AB = 25 cm2 with the piston mass in A being mA = 25 kg. Outside pressure is 100 kPa and standard gravitation. Find the mass mB so that none of the pistons have to rest on the bottom. Solution: Force balance for both pistons: F = F A: mPAg + P0AA = PAA B: mPBg + P0AB = PAB Same P in A and B gives no flow between them. mPAg AA + P0 = mPBg AB + P0 => mPB = mPA AA/ AB = 25  25/75 = 8.33 kg 2.43 At the beach, atmospheric pressure is 1025 mbar. You dive 15 m down in the ocean and you later climb a hill up to 250 m elevation. Assume the density of water is about 1000 kg/m3 and the density of air is 1.18 kg/m3. What pressure do you feel at each place? Solution: P = gh Pocean= P0 + P = 1025  100 + 1000  9.81  15 = 2.4965  105 Pa = 250 kPa Phill = P0 - P = 1025  100 - 1.18  9.81  250 = 0.99606  105 Pa = 99.61 kPa 2.44 In the city water tower, water is pumped up to a level 25 m above ground in a pressurized tank with air at 125 kPa over the water surface. This is illustrated in Fig. P2.44. Assuming the water density is 1000 kg/m3 and standard gravity, find the pressure required to pump more water in at ground level. Solution: Pbottom = Ptop + gl = 125 + 1000  9.807  25  10-3 = 370 kPa Po Po B A 3-1 CHAPTER 3 CONTENT SUBSECTION PROB NO. PAGE Correspondence table 3-2 Phase diagrams, triple and critical points 1-10 3-3 General tables 11-62 3-10 Compressibility factor 63-75 3-42 Linear interpolation 76-81 3-52 Computer tables 82-87 3-57 3-2 The SI set of problems have a correspondence to the 5th edition Fundamentals of Thermodynamics as: New Old New Old New Old 1 1 31 new 61 12 2 2 32 43 62 new 3 new 33 new 63 19 4 3 34 44 64 33 5 4 35 46 65 new 6 new 36 48 66 38 7 23 37 55 67 new 8 24 38 47 68 39 mod 9 new 39 52 69 new 10 28 mod 40 51 70 new 11 new 41 57 71 20 12 new 42 56 72 new 13 29 43 59 73 34 14 27 mod 44 5 74 21 15 26 45 8 75 18 16 new 46 22 76 new 17 new 47 new 77 new 18 30 mod 48 6 78 new 19 31 mod 49 7 79 new 20 new 50 10 80 new 21 new 51 new 81 new 22 58 52 9 82 new 23 new 53 13 83 new 24 36 54 25 84 new 25 37 55 11 85 new 26 41 56 new 86 86 27 new 57 new 87 87 28 new 58 new 29 60 59 17 30 42 60 16 mod indicates a modification from the previous problem that changes the solution but otherwise is the same type problem. 3-3 Phase diagrams, triple and critical points 3.1 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: 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: PVL    − MPa = 4 kPa, PLS = 103 MPa T 0 < P < 4 kPa VAPOR 0.04 MPa < P < 1000 MPa LIQUID P > 1000 MPa SOLID(ICE) 3.2 Find the lowest temperature at which it is possible to have water in the liquid phase. At what pressure must the liquid exist? Solution: 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 ln P L S lowest T liquid CR.P. V T ln P S L CR.P. S V 3-4 3.3 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 is 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/v c = 1/0.00212 = 472 kg/m3 C2H5OH: Table A.2: Pc = 6.14 MPa , Tc = 514 K, vc = 0.00363 m3/kg c = 1/v c = 1/0.00363 = 275 kg/m3 3.4 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  9.80665  1000 = 9022118 Pa P = Po + P = 101.325 + 9022 = 9123 kPa See figure 3.6 liquid solid interphase => TLS = −1C 3-5 ln P Liquid Cr.P. c a b Vapor 3.5 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 H2O : 22.12 MPa 647.3 K c) Propane C3H8 : 4.25 MPa 369.8 K 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 T 3.6 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. 3-6 3.7 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 d. 10 kPa, 10C b. 1 MPa, 190C e. 130C, 200 kPa c. 200C, 0.1 m3/kg f. 70C, v = 1 m3/kg 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 so look in B.1.2 at 10 MPa vf = 0.001452; vg = 0.01803 m3/kg, => vf < v < vg => so mixture of liquid and vapor. 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, => vf < v < vg => so mixture of liquid and vapor. d. 10 kPa, 10C : Only one of the two look -ups is needed From B.1.1: P > P g = 1.2276 kPa so compressed liquid From B.1.2: T < Tsat = 45.8 C so compressed liquid e. 130C, 200 kPa: Only one of the two look -ups is needed From B.1.1: P < Pg = 270.1 kPa so superheated vapor From B.1.2: T > Tsat = 120.23 C so superheated vapor f. 70C, v = 1 m3/kg so look in B.1.1 vf = 0.001023; vg = 5.042 m3/kg, => vf < v < vg => so mixture of liquid and vapor. T C.P. States shown are P = const. placed relative to the two-phase region, not to each other. b d a c e f v v P C.P. d b a c T f e