Exam (elaborations) TEST BANK FOR Classical Mechanics and Thermodynamics Chapters 1-9 By Sonntag, Borgnakke and Van Wylen. (Solution Manual)

Exam (elaborations) TEST BANK FOR Classical Mechanics and Thermodynamics Chapters 1-9 By Sonntag, Borgnakke and Van Wylen. (Solution Manual) correspondence between the problem set in this fifth edition versus the problem set in the 4'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 4 mod 21 13 41E 33E mod 2 new 22 14 42E 34E mod 3 new 23 15 43E 35E 4 7 mod 24 17 44E 36E 5 2 mod 25 18 45E 37E 6 new 26 new 46E 38E 7 new 27 19 47E 39E 8 new 28 20 48E 40E 9 5 mod 29 21 49E 41E 10 6 30 22 11 8 mod 31 23 12 new 32 24 13 9 mod 33 new 14 10 mod 34 25 mod mod 16 new 36 27 mod 17 new 37 28 18 16 mod 38 29 19 new 39E 31E mod 20 12 40E 32E 2-2 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 = å F = F - mg F = mg = 2 ´ 9.80665 = 19.613 N 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 = å F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2 2.3 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. Solution: Acceleration is the time rate of change of velocity. ma = å F ; a = dV / dt = (60 ´ 1000) / (3600 ´ 5) = 3.33 m/s2 Fnet = ma = 1075 ´ 3.333 = 3583 N 2.4 A washing machine has 2 kg of clothes spinning at a rate that generates an acceleration of 24 m/s2. What is the force needed to hold the clothes? Solution: F = ma = 2 kg ´ 24 m/s2 = 48 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 / dt => Dt = dV/a = [ ( 75 - 20 ) / 4 ] ´ ( 1000 / 3600 ) Dt = 3.82 sec ; F = ma = 1200 ´ 4 = 4800 N 2-3 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. a = dV / dt => ò dV = ò a dt => DV = a Dt = 3 ´ 10 = 30 m/s V = 30 m/s ; F = ma = 950 ´ 3 = 2850 N 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: ma = å F Þ a = å F / m m = msteel + mpropane = 15 + (1.75 ´ 44.094) = 92.165 kg a = 2000 / 92.165 = 21.7 m/s2 2.8 A rope hangs over a pulley with the two equally long ends down. On one end you attach a mass of 5 kg and on the other end you attach 10 kg. Assuming standard gravitation and no friction in the pulley what is the acceleration of the 10 kg mass when released? Solution: Do the equation of motion for the mass m2 along the downwards direction, in that case the mass m1 moves up (i.e. has -a for the acceleration) m2 a = m2 g - m1 g - m1a (m1 + m2 ) a = (m2 - m1 )g This is net force in motion direction a = (10 - 5) g / (10 + 5) = g / 3 = 3.27 m/s2 g 1 2 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 Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N 2-4 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 m m m Beam Balance Reading is 5 kg This is mass comparison Spring Balance Reading is in kg units length µ F µ g Reading will be 5 6 kg This is force comparison 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: v = V/m = 0.5/1 = 0.5 m3/kg v = V/n = V m/M = Mv = 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 = r V = rair ( Vtot - mgranite / r ) = 1.15 [ 5 - (900 / 2400) ] = 1.15 ´ 4.625 = 5.32 kg v = V / m = 5 / (900 + 5.32) = 0.00552 m3/kg 2.13 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 2-5 2.14 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. Solution: Force balance: F­ = PA = F¯ = P0A + mpg ; P0 = 1 bar = 100 kPa A = (p/4) D2 = (p/4) ´ 0.1252 = 0.01227 m2 mp = (P-P0)A/g = ( 1500 - 100 ) ´ 1000 ´ 0.01227 / 9.80665 = 1752 kg 2.15 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 : Dl = 725 mm = 0.725 m; r = 13550 kg/m3 P = r gDl = 13550 ´ 9.80665 ´ 0.725 ´ 10-3 = 96.34 kPa 2.16 A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gunpowder 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 ) p [ ( 0.152 /4 )/5 ] = 24.38 m/s2 2.17 Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relative to the horizontal direction. Solution: ma = F = ( P1 - P0 ) A - mg sin 400 ma = ( 7000 - 101 ) ´ p ´ ( 0.152 / 4 ) - 5 ´ 9.80665 ´ 0.6428 = 121.9 - 31.52 = 90.4 N a = 90.4 / 5 = 18.08 m/s2 2-6 2.18 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. P2.18. With an outside atmospheric pressure of 100 kPa, what should the water pressure be to lift the piston? Solution: Force balance: F­ = F¯ = PA = mpg + P0A P = P0 + mpg/A = 100 kPa + (100 ´ 9.80665) / (0.01 ´ 1000) = 100 kPa + 98.07 = 198 kPa 2.19 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: F¯ = ma = mg = 740 ´ 9.80665 = 7256.9 N Force balance: F­ = ( P - P0 ) A = F¯ => P = P0 + F¯ / A A = p D2 (1 / 4) = 0. m2 P = 101 + 7256.9 / (0. ´ 1000) = 332 kPa 2.20 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. Solution: Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPa P = Pgauge + P0 = 1250 + 96 = 1346 kPa 2.21 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: DP = P0 - Ptank = rgDl Dl = ( P0 - Ptank ) / rg = [(97 - 85 ) ´ 1000 ] / (13550 ´ 9.80665) = 0.090 m = 90 mm 2-7 2.22 A 5-kg piston in a cylinder with diameter of 100 mm is loaded with a linear spring and the outside atmospheric pressure of 100 kPa. The spring exerts no force on the piston when it is at the bottom of the cylinder and for the state shown, the pressure is 400 kPa with volume 0.4 L. The valve is opened to let some air in, causing the piston to rise 2 cm. Find the new pressure. Solution: A linear spring has a force linear proportional to displacement. F = k x, so the equilibrium pressure then varies linearly with volume: P = a + bV, with an intersect a and a slope b = dP/dV. Look at the balancing pressure at zero volume (V -> 0) when there is no spring force F = PA = PoA + mpg and the initial state. These two points determine the straight line shown in the P-V diagram. Piston area = AP = (p/4) ´ 0.12 = 0.00785 m2 400 106.2 2 1 0 0.4 P V 0.557 P 2 a = P0 + mpg Ap = 100 + 5 ´ 9.80665 0.00785 = 106.2 kPa intersect for zero volume. V2 = 0.4 + 0.00785 ´ 20 = 0.557 L P2 = P1 + dP dV DV = 400 + (400-106.2) 0.4 - 0 (0.557 - 0.4) = 515.3 kPa 2.23 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. P2.23, what should the length of the column in the tilted tube be relative to the U-tube? Solution: h H 30° DP = F/A = mg/A = Vrg/A = hrg = 0.25 ´ 1000 ´ 9.807 = 2452.5 Pa = 2.45 kPa h = H ´ sin 30° Þ H = h/sin 30° = 2h = 50 cm 2-8 2.24 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: DP = r1gh1 = 900 ´ 9.807 ´ 0.2 = 1765.26 Pa = 1.77 kPa hhg = DP/ (rhg g) = (r1 gh1) / (rhg g) = 900 13600 ´0.2 = 0.0132 m= 13.2 mm 2.25 Two reservoirs, A and B, open to the atmosphere, are connected with a mercury manometer. Reservoir A is moved up/down so the two top surfaces are level at h3 as shown in Fig. P2.25. Assuming that you know rA, rHg and measure the heights h1, h2 , and h3, find the density rB. Solution: Balance forces on each side: P0 + rAg(h3 - h2) + rHggh2 = P0 + rBg(h3 - h1) + rHggh1 Þ rB = rAè ç æ ø ÷ ö h3 - h2 h3 - h1 + rHgè ç æ ø ÷ ö h2 - h1 h3 - h1 2.26 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 4m. 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 ´ pD2 ´ (1 / 4) = 10 ´ p ´ 22 ´ ( 1 / 4) = 31.416 m3 VB = H ´ pD2 ´ (1 / 4) = 2.5 ´ p ´ 42 ´ ( 1 / 4) = 31.416 m3 Tanks have the same volume, so same mass of water F = mg = r V g = 1000 ´ 31.416 ´ 9.80665 = N Tanks have same net force up (holds same m in gravitation field) Pbot = P0 + r H g Pbot,A = 101 + (1000 ´ 10 ´ 9.80665 / 1000) = 199 kPa Pbot,B = 101 + (1000 ´ 2.5 ´ 9.80665 / 1000) = 125.5 kPa 2-9 2.27 The density of mercury changes approximately linearly with temperature as rHg = 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: DP = rgh Þ h = DP/rg ; rsu = 13507.5 ; rw = 13632.5 hsu = 100´103/(13507.5 ´ 9.807) = 0.7549 m hw = 100´103/(13632.5 ´ 9.807) = 0.7480 m Dh = hsu - hw = 0.0069 m = 6.9 mm 2.28 Liquid water with density r is filled on top of a thin piston in a cylinder with crosssectional 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 H h P 0 Piston: F­ = F¯ PA = P0A + mH2Og P = P0 + mH2Og/A P = P0 + (H-h)rg h, V air P P 0 2.29 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: P 0 g Force balance: F­ = F¯ = P0A + mpg = PA P = P0 + mpg/A = 101.325 + 5 ´ 25 / (1000 ´ 0.0015) = 184.7 kPa 2-10 2.30 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: DP = rgh ; rHg = 13600 DP = 13600 ´ 9.5 ´ 0.2 = 25840 Pa = 25.84 kPa 2.31 Repeat the previous problem if the flow inside the apparatus is liquid water, r @ 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. Solution: Balance forces in the manometer: P1 P2 · · h h 1 2 H (H - h2) - (H - h1) = DhHg = h1 - h2 P1A + rH2Oh1gA + rHg(H - h1)gA = P2A + rH2Oh2gA + rHg(H - h2)gA Þ P1 - P2 = rH2O(h2 - h1)g + rHg(h1 - h2)g P1 - P2 = rHgDhHgg - rH2ODhHgg = 13600 ´ 0.2 ´ 9.5 - 1000 ´ 0.2 ´ 9.5 = = 23940 Pa = 23.94 kPa 2.32 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: A B P P 0 0 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 2-11 2.33 Two hydraulic piston/cylinders are of same size and setup as in Problem 2.32, but with neglible piston masses. A single point force of 250 N presses down on piston A. Find the needed extra force on piston B so that none of the pistons have to move. Solution: No motion in connecting pipe: PA = PB & Forces on pistons balance AA = 75 cm2 ; AB = 25 cm2 PA = P0 + FA / AA = PB = P0 + FB / AB FB = FA AB / AA = 250 ´ 25 / 75 = 83.33 N 2.34 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: DP = rgh Pocean= P0 + DP = 1025 ´ 100 + 1000 ´ 9.81 ´ 15 = 2.4965 ´ 105 Pa = 250 kPa Phill = P0 - DP = 1025 ´ 100 - 1.18 ´ 9.81 ´ 250 = 0.99606 ´ 105 Pa = 99.61 kPa 2.35 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. P2.35. 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 + rgl = 125 + 1000 ´ 9.807 ´ 25 ´ 10-3 = 370 kPa 2-12 2.36 Two cylinders are connected by a piston as shown in Fig. P2.36. Cylinder A is used as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg and there is standard gravity. What is the gas pressure in cylinder B? Solution: Force balance for the piston: PBAB + mpg + P0(AA - AB) = PAAA AA = (p/4)0.12 = 0.00785 m2; AB = (p/4)0.0252 = 0. m2 PBAB = PAAA - mpg - P0(AA - AB) = 500´ 0.00785 - (25 ´ 9.807/1000) - 100 (0.00785 - 0.) = 2.944 kN PB = 2.944/0. = 5996 kPa = 6.0 MPa 2.37 Two cylinders are filled with liquid water, r = 1000 kg/m3, and connected by a line with a closed valve. A has 100 kg and B has 500 kg of water, their cross-sectional areas are AA = 0.1 m2 and AB = 0.25 m2 and the height h is 1 m. Find the pressure on each side of the valve. The valve is opened and water flows to an equilibrium. Find the final pressure at the valve location. Solution: VA = vH2OmA = mA/r = 0.1 = AAhA => hA = 1 m VB = vH2OmB = mB/r = 0.5 = ABhB => hB = 2 m PVB = P0 + rg(hB+H) = + 1000 ´ 9.81 ´ 3 = 130 755 Pa PVA = P0 + rghA = + 1000 ´ 9.81 ´ 1 = 111 135 Pa Equilibrium: same height over valve in both Vtot = VA + VB = h2AA + (h2 - H)AB Þ h2 = hAAA + (hB+H)AB AA + AB = 2.43 m PV2 = P0 + rgh2 = 101.325 + (1000 ´ 9.81 ´ 2.43)/1000 = 125.2 kPa 2.38 Using the freezing and boiling point temperatures for water in both Celsius and Fahrenheit scales, develop a conversion formula between the scales. Find the conversion formula between Kelvin and Rankine temperature scales. Solution: TFreezing = 0 oC = 32 F; TBoiling = 100 oC = 212 F DT = 100 oC = 180 F Þ ToC = (TF - 32)/1.8 or TF = 1.8 ToC + 32 For the absolute K & R scales both are zero at absolute zero. TR = 1.8 ´ TK 2-13 English Unit Problems 2.39E A 2500-lbm car moving at 15 mi/h is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/h. What are the force and total time required? Solution: a = dV dt = DV / D t Þ D t = DV / a D t = (50 -15) ´ 1609.34 ´ 3.28084/(3600 ´ 15) = 3.42 sec F = ma = 2500 ´ 15 / 32.174 lbf= 1165 lbf 2.40E Two pound moles of diatomic oxygen gas are enclosed in a 20-lbm steel container. A force of 2000 lbf now accelerates this system. What is the acceleration? Solution: mO2 = nO2 MO2 = 2 ´ 32 = 64 lbm mtot = mO2 + msteel = 64 + 20 = 84 lbm a = Fgc mtot = (2000 ´ 32.174) / 84 = 766 ft/s2 2.41E A bucket of concrete of total mass 400 lbm is raised by a crane with an acceleration of 6 ft/s2 relative to the ground at a location where the local gravitational acceleration is 31 ft/s2. Find the required force. Solution: F = ma = Fup - mg Fup = ma + mg = 400 ´ ( 6 + 31 ) / 32.174 = 460 lbf 2.42E One pound-mass of diatomic oxygen (O2 molecular weight 32) is contained in a 100-gal tank. Find the specific volume on both a mass and mole basis (v and v ). Solution: v = V/m = 15/1 = 15 ft3/lbm v¯ = V/n = V m/M = Mv = 32 ´15 = 480 ft3/lbmol 2-14 2.43E A 30-lbm steel gas tank holds 10 ft3 of liquid gasoline, having a density of 50 lbm/ft3. What force is needed to accelerate this combined system at a rate of 15 ft/s2? Solution: m = mtank + mgasoline = 30 + 10 ´ 50 = 530 lbm F = ma gC = (530 ´ 15) / 32.174 = 247.1 lbf 2.44E A differential pressure gauge mounted on a vessel shows 185 lbf/in.2 and a local barometer gives atmospheric pressure as 0.96 atm. Find the absolute pressure inside the vessel. Solution: P = Pgauge + P0 = 185 + 0.96 ´ 14.696 = 199.1 lbf/in2 2.45E A U-tube manometer filled with water, density 62.3 lbm/ft3, shows a height difference of 10 in. What is the gauge pressure? If the right branch is tilted to make an angle of 30° with the horizontal, as shown in Fig. P2.23, what should the length of the column in the tilted tube be relative to the U-tube? Solution: h H 30° DP = F/A = mg/AgC = hrg/gC = [(10/12) ´ 62.3 ´ 32.174] / 32.174 ´144 = Pgauge = 0.36 lbf/in2 h = H ´ sin 30° Þ H = h/sin 30° = 2h = 20 in = 0.833 ft 2.46E A piston/cylinder with cross-sectional area of 0.1 ft2 has a piston mass of 200 lbm resting on the stops, as shown in Fig. P2.18. With an outside atmospheric pressure of 1 atm, what should the water pressure be to lift the piston? Solution: P = P0 + mpg/Agc = 14.696 + (200´32.174) / (0.1 ´ 144 ´ 32.174) = 14.696 + 13.88 = 28.58 lbf/in2 2-15 2.47E The density of mercury changes approximately linearly with temperature as rHg = 851.5 - 0.086 T lbm/ft3 T in degrees Fahrenheit so the same pressure difference will result in a manometer reading that is influenced by temperature. If a pressure difference of 14.7 lbf/in.2 is measured in the summer at 95 F and in the winter at 5 F, what is the difference in column height between the two measurements? Solution: DP = rgh/gc Þ h = DPgc/rg rsu = 843.33 lbm/ft3; rw = 851.07 lbm/ft3 hsu = 14.7 ´ 144 ´ 32.174 843.33 ´ 32.174 = 2.51 ft = 30.12 in hw = 14.7 ´ 144 ´ 32.174 851.07 ´ 32.174 = 2.487 ft = 29.84 in Dh = hsu - hw = 0.023 ft = 0.28 in 2.48E A piston, mp = 10 lbm, is fitted in a cylinder, A = 2.5 in.2, that contains a gas. The setup is in a centrifuge that creates an acceleration of 75 ft/s2. Assuming standard atmospheric pressure outside the cylinder, find the gas pressure. Solution: P 0 g F¯ = F­ = P0A + mpg = PA P = P0 + mpg/Agc = 14.696 + 10 ´ 75 2.5 ´ 32.174 = 14.696 + 9.324 = 24.02 lbf/in2 2.49E At the beach, atmospheric pressure is 1025 mbar. You dive 30 ft down in the ocean and you later climb a hill up to 300 ft elevation. Assume the density of water is about 62.3 lbm/ft3 and the density of air is 0.0735 lbm/ft3. What pressure do you feel at each place? Solution: DP = rgh; P0 = (1.025/1.01325) ´14.696 = 14.866 lbf/in2 Pocean = P0 + DP = 14.866 + 62.3 ´ 30 ´ g gc ´ 144 = 27.84 lbf/in2 Phill = P0 -DP = 14.866 - 0.0735 ´ 300 ´ g gc ´ 144 = 14.71 lbf/in2 3-1 CHAPTER 3 The SI set of problems are revised from the 4th edition as: New Old New Old New Old 1 new 21 new 41 33 2 new 22 13 mod 42 34 3 new 23 16 mod 43 35 4 new mod 5 new 25 new 45 37 mod 6 new 26 18 mod 46 38 mod 7 7 mod 40 41 42 mod 43 12 new 45 14 8 mod 34 27 mod 54 46 47 48 mod 57 49 18 new 38 30 mod 58 50 19 15 mod 20 new 40 32 mod 60 52 The english unit problem set is revised from the 4th edition as: New Old New Old New Old 61 new mod 78 70 71 72 65 new 73 65 81 new 66 new 67 59 mod mod mod indicates a modification from the previous problem that changes the solution but otherwise is the same type problem. 3-2 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 K From Fig. 3.6: PVL » 4 ´ 10-3 MPa = 4 kPa, PLS = 103 MPa ln P T V L S CR.P. P < 4 kPa VAPOR P > 1000 MPa SOLID(ICE) 0.004 MPa < P < 1000 MPa LIQUID 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 T V L S CR.P. lowest T liquid 3.3 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: rICE = 920 kg/m3 DP = rgH = 920 ´ 9.80665 ´ 1000 = Pa P = Po + DP = 101.325 + 9022 = 9123 kPa See figure 3.6 liquid solid interphase => TLS = -1°C 3-3 3.4 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: 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 ln P T Vapor Liquid Cr.P. a c b 3.5 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 Tr = ( 25 + 273 ) / 425 = 0.701; Pr = 0.5/3.8 = 0.13 Look at generalized chart in Figure D.1 Actual Pr > Pr, sat => liquid!! not a gas 3.6 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: Table A.2 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 Neon : T >> Tc = 44.4 K so ideal gas; R = 0.41195 Propane: T < Tc = 370 K, but P << Pc = 4.25 MPa so gas R = 0.18855 a) m = PV/RT = 100 ´1 / 0.287 ´ 293.15 = 1.189 kg b) m = 100 ´ 1/ 0.41195 ´ 293.15 = 0.828 kg c) m = 100 ´ 1 / 0.18855 ´ 293.15 = 1.809 kg 3-4 3.7 A cylinder has a thick piston initially held by a pin as shown in Fig. P3.7. The cylinder contains carbon dioxide at 200 kPa and ambient temperature of 290 K. The metal piston has a density of 8000 kg/m3 and the atmospheric pressure is 101 kPa. The pin is now removed, allowing the piston to move and after a while the gas returns to ambient temperature. Is the piston against the stops? Solution: Force balance on piston determines equlibrium float pressure. Piston mp = Ap ´ l ´ r rpiston = 8000 kg/m3 Pext on CO2 = P0 + mpg Ap = 101 + Ap ´ 0.1 ´ 9.807 ´ 8000 Ap ´ 1000 = 108.8 kPa Pin released, as P1 > Pfloat piston moves up, T2 = To & if piston at stops, then V2 = V1 ´ 150 / 100 Þ P2 = P1 ´ V1 / V2 = 200 ´ 100 150 = 133 kPa > Pext Þ piston is at stops, and P2 = 133 kPa 3.8 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: P = mRT/V Vcyl = A ´ L = p 4 (0.2)2 ´ 1 = 0. m3 Þ P = 1.2 ´ 0.18892 (273.15 + 25)/0. = 2152 kPa 3-5 3.9 A 1-m3 rigid tank with air at 1 MPa, 400 K is connected to an air line as shown in Fig. P3.9. The valve is opened and air flows into the tank until the pressure reaches 5 MPa, at which point the valve is closed and the temperature inside is 450K. a. What is the mass of air in the tank before and after the process? b. The tank eventually cools to room temperature, 300 K. What is the pressure inside the tank then? Solution: P, T known at both states and assume the air behaves as an ideal gas. mair1 = P1V RT1 = 1000 ´ 1 0.287 ´ 400 = 8.711 kg mair2 = P2V RT2 = 5000 ´ 1 0.287 ´ 450 = 38.715 kg Process 2 ® 3 is constant V, constant mass cooling to T3 P3 = P2 ´ (T3/T2) = 5000 ´ (300/450) = 3.33 MPa 3.10 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 = p 6 (0.15)3 = 0. m3 M = _ mRT PV = 0.0025 ´ 8.3145 ´ 298.2 875 ´ 0. = 4.009 » MHe => Helium Gas 3-6 3.11 A piston/cylinder arrangement, shown in Fig. P3.11, contains air at 250 kPa, 300°C. The 50-kg piston has a diameter of 0.1 m and initially pushes against the stops. The atmosphere is at 100 kPa and 20°C. The cylinder now cools as heat is transferred to the ambient. a. At what temperature does the piston begin to move down? b. How far has the piston dropped when the temperature reaches ambient? Solution: Piston Ap = p 4 ´ 0.12 = 0.00785 m2 Balance forces when piston floats: Pfloat = Po + mpg Ap = 100 + 50 ´ 9.807 0.00785 ´ 1000 = 162.5 kPa = P2 = P3 To find temperature at 2 assume ideal gas: 2 1 P V P2 Vstop 3 T2 = T1 ´ P2 P1 = 573.15 ´ 162.5 250 = 372.5 K b) Process 2 -> 3 is constant pressure as piston floats to T3 = To = 293.15 K V2 = V1 = Ap ´ H = 0.00785 ´ 0.25 = 0.00196 m3 = 1.96 L Ideal gas and P2 = P3 => V3 = V2 ´ T3 T2 = 1.96 ´ 293.15 372.5 = 1.54 L DH = (V2 -V3)/A = (1.96-1.54) ´ 0.001/0.00785 = 0.053 m = 5.3 cm 3.12 Air in a tank is at 1 MPa and room temperature of 20°C. It is used to fill an initially empty balloon to a pressure of 200 kPa, at which point the diameter is 2 m and the temperature is 20°C. Assume the pressure in the balloon is linearly proportional to its diameter and that the air in the tank also remains at 20°C throughout the process. Find the mass of air in the balloon and the minimum required volume of the tank. Solution: Assume air is an ideal gas. Balloon final state: V2 = (4/3) p r3 = (4/3) p 23 = 33.51 m3 m2bal = P2 V2 / RT2 = 200´ 33.51 / 0.287 ´ 293.15 = 79.66 kg Tank must have P2 ³ 200 kPa => m2 tank ³ P2 VTANK /RT2 Initial mass must be enough: m1 = m2bal + m2 tank = P1V1 / R T1 P1VTANK / R T1 = m2bal + P2VTANK / RT2 => VTANK = RTm2bal / (P1 – P2) = 0.287 ´ 293.15 ´ 79.66/ (1000 – 200 ) = 8.377 m3 3-7 3.13 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 30- min period? Solution: Use ideal gas P << lowest P in steam tables. R is from table A.5 m = m . Dt with mass flow rate as: m . = V . /v = PV . /RT (ideal gas) Þ m = PV . Dt/RT = 0.1 ´ 0.5 ´ 30´60 (0.46152 ´ 323.15) = 0.603 kg 3.14 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 P = P0 + C è ç æ ø ÷ ö 1 - D1 D D1 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 P = 400 = P0 + C ( 1 – 1 4 ) 1 4 = 100 + C ´ 3/16 => C = 1600 The pressure is: P = 100 + 1600 ( 1 – X –1) X –1; X = D / D1 Differentiate to find max: dP dD = C ( - X –2 + 2 X –3 ) / D1 = 0 => - X –2 + 2 X –3 = 0 => X = 2 at max P => D = 2D1 = 2 m; V = p 6 D3 = 4.18 m3 Pmax = 100 + 1600 ( 1 - 1 2 ) 1 2 = 500 kPa Helium is ideal gas A.5: m = PV / RT = 500 ´ 4.189 2.0771 ´ 293.15 = 3.44 kg mTANK, 1 = PV/RT = 2000 ´ 12/(2.0771 ´ 293.15) = 39.416 kg mTANK, 2 = 39.416 – 3.44 = 35.976 kg PT2 = mTANK, 2 RT/V = ( mTANK, 1 / mTANK, 2 )´ P1 = 1825.5 kPa 3-8 3.15 The helium balloon described in Problem 3.14 is released into the atmosphere and rises to an elevation of 5000 m, with a local ambient pressure of Po = 50 kPa and temperature of -20°C. What is then the diameter of the balloon? Solution: Balloon of Problem 3.14, where now after filling D = 4 m, we have : m1 = P1V1/RT1 = 400 (p/6) 43 /2.077 ´ 293.15 = 22.015 kg P1 = 400 = 100 + C(1 - 0.25)0.25 => C = 1600 For final state we have : P0 = 50 kPa, T2 = T0 = -20°C = 253.15 K State 2: T2 and on process line for balloon, i.e. the P-V relation: P = 50 + 1600 ( D* -1 - D* -2 ), D* = D/D1 ; V = (p/6) D3 P2V2 = m R T2 = 22.015 ´ 2.077 ´ 253.15 = 11575 or PD+3 = 11575 ´ 6/ p = 22107 substitute P into the P-V relation 22107 D* -3 = 50 + 1600 ( D* -1 - D* -2 ) Divide by 1600 13.8169 D* -3 - 0.03125 - D* -1 + D* -2 = 0 Multiply by D*3 13.8169 - 0.03125 D* 3 - D* 2 + D* 1 = 0 Qubic equation. By trial and error D* = 3.98 so D = D*D1 = 3.98 m 3.16 A cylinder is fitted with a 10-cm-diameter piston that is restrained by a linear spring (force proportional to distance) as shown in Fig. P3.16. The spring force constant is 80 kN/m and the piston initially rests on the stops, with a cylinder volume of 1 L. The valve to the air line is opened and the piston begins to rise when the cylinder pressure is 150 kPa. When the valve is closed, the cylinder volume is 1.5 L and the temperature is 80°C. What mass of air is inside the cylinder? Solution: Fs = ksDx = ks DV/Ap ; V1 = 1 L = 0.001 m3, Ap = p 4 0.12 = 0. m2 State 2: V3 = 1.5 L = 0.0015 m3; T3 = 80°C = 353.15 K The pressure varies linearly with volume seen from a force balance as: PAp = P0 Ap + mp g + ks(V - V0)/Ap Between the states 1 and 2 only volume varies so: P3 = P2 + ks(V3-V2) Ap 2 = 150 + 80´103(0.0015 - 0.001) 0. ´ 1000 = 798.5 kPa m = P3V3 RT3 = 798.5 ´ 0.0015 0.287 ´ 353.15 = 0.012 kg P v 2 3 1 3-9 3.17 Air in a tire is initially at -10°C, 190 kPa. After driving awhile, the temperature goes 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 = 190 ´ 283.15/263.15 = 204.4 kPa 3.18 A substance is at 2 MPa, 17°C in a 0.25-m3 rigid tank. Estimate the mass from the compressibility factor if the substance is a) air, b) butane or c) propane. Solution: Figure D.1 for compressibility Z and table A.2 for critical properties. Nitrogen Pr = 2/3.39 = 0.59; Tr = 290/126.2 = 2.3; Z » 0.98 m = PV/ZRT = 2000 ´ 0.25/(0.98 ´ 0.2968 ´ 290) = 5.928 kg Butane Pr = 2/3.80 = 0.526; Tr = 290/425.2 = 0.682; Z » 0.085 m = PV/ZRT = 2000 ´ 0.25/(0.085 ´ 0.14304 ´ 290) = 141.8 kg Propane Pr = 2/4.25 = 0.47; Tr = 290/369.8 = 0.784; Z » 0.08 m = PV/ZRT = 2000 ´ 0.25/(0.08 ´ 0.18855 ´ 290) = 114.3 kg ln Pr Z Tr = 2.0 a c b Tr = 0.7 Tr = 0.7 0.1 1 3.19 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 => Z @ 0.96 m = PV ZRT = 3000 ´ 5 0.96 ´ 0.2081 ´ 243.2 = 308.75 kg Ideal gas Z = 1 m = PV/RT = 296.4 kg 4% error 3-10 3.20 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: m = V/v so find v given T1 and x as : v = vf + x vfg Tr = 300/425.2 = 0.705 => Fig. D.1 Zf » 0.02; Zg » 0.9 P = Psat = Prsat ´ 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. m3/kg m = 0.1/0. = 1.303 kg 3.21 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; R= 0.3193 v = V/m = 0.045/2 = 0.0225 m3/kg State given by (P, v) v = ZRT/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: Zg = 0.59; Tr = 0.94 vg = 0.59 ´ 0.3193 ´ 0.94 ´ 308.3/4300 = 0.0127 too small Tr = 1 => Z = 0.7 => v = 0.7 ´ 0.3193 ´ 1 ´ 308.3/4300 = 0.016 Tr = 1.2 => Z = 0.86 => v = 0.86 ´ 0.3193 ´ 1.2 ´ 308.3/4300 = 0.0236 Interpolate to get: Tr » 1.17 T » 361 K 3.22 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 30°C, 3 MPa Ideal Gas ( T » Tc = 190 K from A.2) c) Water, H2O at 30°C, 3 MPa NO compressed liquid P > Psat (B.1.1) 3-11 d) R-134a at 30°C, 3 MPa NO compressed liquid P > Psat (B.5.1) e) R-134a at 30°C, 100 kPa Ideal Gas P is low < Psat (B.5.1) ln P T Vapor Liq. Cr.P. a, b c, d e 3.23 Determine whether water at each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor. Solution: All states start in table B.1.1 (if T given) or B.1.2 (if P given) a. 10 MPa, 0.003 m3/kg vf = 0.; vg = 0.01803 m3/kg, so mixture of liquid and vapor. b. 1 MPa, 190°C : T > Tsat = 179.91oC so it is superheated vapor c. 200°C, 0.1 m3/kg: v < vg = 0.12736 m3/kg, so it is two-phase d. 10 kPa, 10°C : P > Pg = 1.2276 kPa so compressed liquid e. 130°C, 200 kPa: P < Pg = 270.1 kPa so superheated vapor f. 70°C, 1 m3/kg vf = 0.; vg = 5.042 m3/kg, so mixture of liquid and vapor States shown are placed relative to the two-phase region, not to each other. P C.P. v T C.P. v a T d e c b e d a c b P = const. f f 3-12 3.24 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 superheated vapor, v > vg =0.01167 at 50°C b. 1.0 MPa, 20°C compressed liquid, P > Pg = 909.9 kPa at 20°C c. 0.1 MPa, 0.1 m3/kg mixture liq. & vapor, vf < v < vg at 0.1 MPa d. 50°C, 0.3 m3/kg superheated vapor, v > vg = 0.01167 at 50°C e -20°C, 200 kPa superheated vapor, P < Pg = 244.8 kPa at -20°C f. 2 MPa, 0.012 m3/kg superheated vapor, v > vg = 0.01132 at 2 MPa States shown are placed relative to the two-phase region, not to each other. P C.P. v T C.P. v T a d e c b e c a d b P = const. f f 3.25 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 % v2 = 0.19444 m3/kg so error = 7.49 % 3-13 3.26 Determine the quality (if saturated) or temperature (if superheated) of the following substances at the given two states: Solution: a) Water, H2O, use Table B.1.1 or B.1.2 1) 120°C, 1 m3/kg => v > vg superheated vapor, T = 120 °C 2) 10 MPa, 0.01 m3/kg => two-phase v < vg x = ( 0.01 – 0. ) / 0.01657 = 0.516 b) Nitrogen, N2, table B.6 1) 1 MPa, 0.03 m3/kg => superheated vapor since v > vg Interpolate between sat. vapor and superheated vapor B.6.2: T @ 103.73 + (0.03-0.02416)´(120-103.73)/(0.03117-0.02416) = 117 K 2) 100 K, 0.03 m3/kg => sat. liquid + vapor as two-phase v < vg v = 0.03 = 0. + x ´ 0. Þ x = 0.959 c) Ammonia, NH3, table B.2 1) 400 kPa, 0.327 m3/kg => v > vg = 0.3094 m3/kg at 400 kPa Table B.2.2 superheated vapor T @ 10 °C 2) 1 MPa, 0.1 m3/kg => v < vg 2-phase roughly at 25 °C x = ( 0.1 – 0. ) / 0. = 0.7776 d) R-22, table B.4 1) 130 kPa, 0.1 m3/kg => sat. liquid + vapor as v < vg vf @ 0. m3/kg, vg @ 0.1684 m3/kg v = 0.1 = 0. + x ´ 0.16768 Þ x = 0.592 2) 150 kPa, 0.17 m3/kg => v > vg superheated vapor, T @ 0°C 3.27 Calculate the following specific volumes Solution: a. R-134a: 50°C, 80% quality in Table B.5.1 v = 0. + x ´ 0. = 0.01228 m3/kg b. Water 4 MPa, 90% quality in Table B.1.2 v = 0.(1-x) + x ´ 0.04978 = 0.04493 m3/kg c. Methane 140 K, 60% quality in Table B.7.1 v = 0.00265 + x ´ 0.09574 = 0.06009 m3/kg d. Ammonia 60°C, 25% quality in Table B.2.1 v = 0. + x ´ 0.04697 = 0.01358 m3/kg 3-14 3.28 Give the phase and the specific volume. Solution: a. H2O T = 275°C P = 5 MPa Table B.1.1 or B.1.2 Psat = 5.94 MPa => superheated vapor v = 0.04141 m3/kg b. H2O T = -2°C P = 100 kPa Table B.1.5 Psat = 0.518 kPa =>compressed solid v @ vi = 0. m3/kg c. CO2 T = 267°C P = 0.5 MPa Table A.5 sup. vap. assume ideal gas v = RT P = 0.18892 ´ 540 500 = 0.204 m3/kg d. Air T = 20°C P = 200 kPa Table A.5 sup. vap. assume ideal gas v = RT P = 0.287 ´ 293 200 = 0.420 m3/kg e. NH3 T = 170°C P = 600 kPa Table B.2.2 T > Tc => sup. vap. v = (0.34699 + 0.36389)/2 = 0.3554m3/kg States shown are placed relative to the two-phase region, not to each other. P C.P. v T C.P. v T a c, d, e b a c, d, e b P = const. 3.29 Give the phase and the specific volume. Solution: a. R-22 T = -25°C P = 100 kPa => Table B.4.1 Psat = 201 kPa sup. vap. B.4.2 v @ (0.22675 + 0.23706)/2 = 0.2319 m3/kg b. R-22 T = -25°C P = 300 kPa => Table B.4.1 Psat = 201 kPa compr. liq. as P > Psat v @ vf = 0. m3/kg c. R-12 T = 5°C P = 300 kPa => Table B.3.1 Psat = 362.6 kPa sup. vap. B.3.2 v @ (0.0569 + 0.05715)/2 = 0.05703 m3/kg d. Ar T = 200°C P = 200 kPa Table A.5 ideal gas v = RT P = 0.20813 ´ 473 200 = 0.4922 m3/kg 3-15 e. NH3 T = 20°C P = 100 kPa => Table B.2.1 Psat = 847.5 kPa sup. vap. B.2.2 v = 1.4153 m3/kg States shown are placed relative to the two-phase region, not to each other. P C.P. v T C.P. v T a, c, e b b a, c, e P = const. d d 3.30 Find the phase, quality x if applicable and the missing property P or T. Solution: a. H2O T = 120°C v = 0.5 m3/kg < vg Table B.1.1 sat. liq. + vap. P = 198.5 kPa, x = (0.5 - 0.00106)/0.8908 = 0.56 b. H2O P = 100 kPa v = 1.8 m3/kg Table B.1.2 v > vg sup. vap., interpolate in Table B.1.3 T = 1.8 - 1.694 1.93636 - 1.694 (150 – 99.62) + 99.62 = 121.65 °C c. H2O T = 263 K v = 200 m3/kg Table B.1.5 sat. solid + vap., P = 0.26 kPa, x = (200-0.001)/466.756 = 0.4285 d. Ne P = 750 kPa v = 0.2 m3/kg; Table A.5 ideal gas, T = Pv R = 750 ´ 0.2 0.41195 = 364.1 K e. NH3 T = 20°C v = 0.1 m3/kg Table B.2.1 sat. liq. + vap. , P = 857.5 kPa, x = (0.1-0.00164)/0.14758 = 0.666 States shown are placed relative to the two-phase region, not to each other. P C.P. v T C.P. v T b a, e b P = const. d d a, e c c 3-16 3.31 Give the phase and the missing properties of P, T, v and x. Solution: a. R-22 T = 10°C v = 0.01 m3/kg Table B.4.1 sat. liq. + vap. P = 680.7 kPa, x = (0.01-0.0008)/0.03391 = 0.2713 b. H2O T = 350°C v = 0.2 m3/kg Table B.1.1 v > vg sup. vap. P @ 1.40 MPa, x = undefined c. CO2 T = 800 K P = 200 kPa Table A.5 ideal gas v = RT P = 0.18892 ´ 800 200 = 0.756 m3/kg d. N2 T = 200 K P = 100 kPa Table B.6.2 T > Tc sup. vap. v = 0.592 m3/kg e. CH4 T = 190 K x = 0.75 Table B.7.1 P = 4520 kPa sat. liq + vap. v = 0.00497 + x ´ 0.003 = 0.00722 m3/kg States shown are placed relative to the two-phase region, not to each other. P C.P. v T C.P. v T b a, e b P = const. c, d a, e c, d 3.32 Give the phase and the missing properties of P, T, v and x. These may be a little more difficult if the appendix tables are used instead of the software. Solution: a) R-22 at T = 10°C, v = 0.036 m3/kg: Table B.4.1 v > vg at 10°C => sup. vap. Table B.4.2 interpolate between sat. and sup. both at 10°C P = 680.7 + (0.036-0.03471)(600-680.7)/(0.04018-0.03471) = 661.7 kPa b) H2O v = 0.2 m3/kg , x = 0.5: Table B.1.1 sat. liq. + vap. v = (1-x) vf + x vg => vf + vg = 0.4 m3/kg since vf is so small we find it approximately where vg = 0.4 m3/kg. vf + vg = 0.39387 at 150°C, vf + vg = 0.4474 at 145°C. An iterpolation gives T @ 149.4°C, P @ 468.2 kPa 3-17 c) H2O T = 60°C, v = 0. m3/kg: Table B.1.1 v < vf = 0. => compr. liq. see Table B.1.4 v = 0. at 5 MPa so P @ 0.5(5000 + 19.9) = 2.51 MPa d) NH3 T = 30°C, P = 60 kPa : Table B.2.1 P < Psat => sup. vapor interpolate in Table B.2.2 v = 2.94578 + (60-50)(1.95906-2.94578)/(75-50) = 2.551 m3/kg v is not linearly proportional to P (more like 1/P) so the computer table gives a more accurate value of 2.45 e) R-134a v = 0.005m3/kg , x = 0.5: sat. liq. + vap. Table B.5.1 v = (1-x) vf + x vg => vf + vg = 0.01 m3/kg vf + vg = 0. at 65°C, vf + vg = 0. at 70°C. An iterpolation gives: T @ 68.7°C, P = 2.06 MPa States shown are placed relative to the two-phase region, not to each other. P C.P. v T C.P. v T a b, e a P = const. c b, e d c d 3.33 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 Table B.2.2: v = 0.2923 m3/kg Ideal gas: v = RT P = 0.48819 ´ 313 500 = 0.3056 m3/kg Þ 4.5% error Figure D.1: Tr = 313.15/405.5 = 0.772, Pr = 0.5/11.35 = 0.044 Þ Z = 0.97 v = ZRT/P = 0.2964 m3/kg Þ 1.4% error 3-18 3.34 What is the percent error in pressure if the ideal gas model is used to represent the behavior of superheated vapor R-22 at 50°C, 0.03082 m3/kg? What if the generalized compressibility chart, Fig. D.1, is used instead (iterations needed)? Solution: Real gas behavior: P = 900 kPa from Table B.4.2 Ideal gas constant: R = R _ /M = 8.31451/86.47 = 0. P = RT/v = 0. ´ (273.15 + 50) / 0.03082 = 1008 kPa which is 12% too high Generalized chart Fig D.1 and critical properties from A.2: Tr = 323.2/363.3 = 0.875; Pc = 4970 kPa Assume P = 900 kPa => Pr = 0.181 => Z @ 0.905 v = ZRT/P = 0.905 ´ 0. ´ 323.15 / 900 = 0.03125 too high Assume P = 950 kPa => Pr = 0.191 => Z @ 0.9 v = ZRT/P = 0.9 ´ 0. ´ 323.15 / 950 = 0. too low P @ 900 + ( 950 - 900 ) ´ 0.03082 - 0. 0.03125 - 0. = 938 kPa 4.2 % high 3.35 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: The methane Table B.7.2 linear interpolation between 225 and 250 K. Þ v @ 0.03333 + 243.15-225 250-225 ´(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 m = V/v = 2/0.042 = 47.62 kg Error: 5.83 kg 10.9% too small 3.36 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. m3/kg ; DP = gh vf = 9.807 ´ 8 0. = 74578 Pa = 74.578 kPa Pbottom = Ptop + DP = 143.3 + 74.578 = 217.88 kPa 3-19 3.37 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 v1 = 1/2 = 0.5 m3/kg 2-phase 200°C, 0.5 m3/kg seen in Table B.1.3 to be between 400 and 500 kPa so interpolate P @ 400 + 0.5-0.53422 0.42492-0.53422 ´ (500-400) = 431.3 kPa T C.P. v 100 C 500 kPa 400 kPa 3.38 A 500-L tank stores 100 kg of nitrogen gas at 150 K. To design the tank the pressure must be estimated and three different methods are suggested. Which is the most accurate, and how different in percent are the other two? a. Nitrogen tables, Table B.6 b. Ideal gas c. Generalized compressibility chart, Fig. D.1 Solution: State 1: 150 K, v = V/m = 0.5/100 = 0.005 m3/kg a) Table B.6, interpolate between 3 & 6 MPa with both at 150 K: 3 MPa : v = 0.01194 6 MPa : v = 0. P= 3 + (0.005-0.01194)´(6-3)/(0.-0.01194) = 5.707 MPa b) Ideal gas table A.5: P = RT v = 0.2968 ´ 150 0.005 = 8.904 MPa c) Table A.2 Tc = 126.2 K, Pc = 3.39 MPa so Tr = 150/126.2 = 1.189 Z is a function of P so it becomes trial and error. Start with P = 5.7 MPa Pr @ 1.68 Þ Z = 0.60 Þ P = ZRT v = 5342 kPa Þ Pr = 1.58 Þ Z = 0.62 Þ P = 5520 kPa OK ANSWER: a) is the most accurate with others off by b) 60% c) 1% 3-20 3.39 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 at P = 100 kPa from Table B.7.1 by interpolation. mliq = Vliq vf = 0.9 ´ 400 0.00236 = kg; mvap = Vvap vg = 0.1 ´ 400 0.5726 = 69.9 kg mtot = 152 612 kg, x = mvap / mtot = 4.58´10-4 (If you use computer table, vf @ 0., vg @ 0.5567) 3.40 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 = const = v1 = 0. + 0.25´0.30367 = 0. All single phase when v = vg => T @145 K Dt = DT/5°C @ (145 – 120 ) / 5 = 5 hours P = Psat= 824 kPa 3.41 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: H2O T = 60°C , x = 0.0; Table B.1.1 v = 0.99 ´ vf (60°C) = 0.99´0. = 0. m3/kg Between 20 & 30 MPa in Table B.1.4, P @ 23.8 MPa 3.42 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: From initial state: 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 P @ 19.941 + (8.4378 - 7.6707)(10-19.941)/(15.3345-7.6707) = 18.9 kPa Comment: T,v Þ P = 18 kPa (software) v is not linear in P, more like 1/P, so the linear interpolation in P is not very accurate. 3-21 3.43 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: At 240°C, 20 MPa: v = 0. m3/kg (from B.1.4) m . = V . /v = 0.05/0. = 41.5 kg/s vf (240°C) = 0. Þ m . = 40.68 kg/s error 2% vf (20 MPa) = 0. Þ m . = 24.56 kg/s error 41% 3.44 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 Þ v1 = v2 From Table B.1.2 and B.1.5: v1 = 0. + 0.25 ´ 0.3738 = 0. = v2 = 0. + x2 ´ 446.756 Þ x2 = 0.0002 mass fraction vapor xsolid =1-x2 = 0.9998 or 99.98 % 3.45 A cylinder/piston arrangement contains water at 105°C, 85% quality with a volume of 1 L. The system is heated, causing the piston to rise and encounter a linear spring as shown in Fig. P3.45. At this point the volume is 1.5 L, piston diameter is 150 mm, and the spring constant is 100 N/mm. The heating continues, so the piston compresses the spring. What is the cylinder temperature when the pressure reaches 200 kPa? Solution: P1 = 120.8 kPa, v1 = vf + x vfg = 0. + 0.85*1.41831 = 1.20661 m = V1/ v1 = 0.001 1.20661 = 8.288´10-4 kg v2 = v1 (V2 / V1) = 1.20661´ 1.5 = 1.8099 & P = P1 = 120.8 kPa ( T2 = 203.5°C ) P3 = P2 + (ks/Ap 2) m(v3-v2) linear spring P v 1 2 200 3 1 1.5 liters Ap = (p/4) ´ 0.152 = 0.01767 m2 ; ks = 100 kN/m (matches P in kPa) 200 = 120.8 + (100/0.01767 2 ) ´ 8.288´10-4(v3-1.8099) 200 = 120.8 + 265.446 (v3 – 1.8099)=> v3 = 2.1083 m3/kg T3 @ 600 + 100 ´ (2.1083 – 2.01297)/(2.2443-2.01297) @ 641°C 3-22 3.46 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, v = C From table B.2.1: v1 = v2 = 0. = 0. + x1 ´ 0.04697 => x1 = 0.01515 liquid = 1 - x1 = 0.948 T v 60 C Crit. point 1 3.47 For a certain experiment, R-22 vapor is contained in a sealed glass tube at 20°C. It is desired to know the pressure at this condition, but there is no means of measuring it, since the tube is sealed. However, if the tube is cooled to -20°C small droplets of liquid are observed on the glass walls. What is the initial pressure? Solution: R-22 fixed volume (V) & mass (m) at 20°C cool to -20°C ~ sat. vapor 1 1 2 20 C o 20 C - o P T T T v v = const = vg at -20°C = 0. m3/kg State 1: 20°C, 0. m3/kg interpolate between 250 and 300 kPa in Table B.4.2 => P = 291 kPa 3-23 3.48 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 v2 = v1 = V/m = 0.0025 m3/kg v T 20°C C.P. V c vc = 0.203/44.094 = 0. > v1 eventually reaches sat. liq. Þ level rises to top If m = 1 kg Þ v1 = 0.015 > vc then it will reach sat. vap. Þ level falls 3.49 A cylinder containing ammonia is fitted with a piston restrained by an external force that is proportional to cylinder volume squared. Initial conditions are 10°C, 90% quality and a volume of 5 L. A valve on the cylinder is opened and additional ammonia flows into the cylinder until the mass inside has doubled. If at this point the pressure is 1.2 MPa, what is the final temperature? Solution: State 1 Table B.2.1: v1 = 0.0016 + 0.9(0. - 0.0016) = 0.18513 m3/kg P1 = 615 kPa; V1 = 5 L = 0.005 m3 m1 = V/v = 0.005/0.18513 = 0.027 kg State 2: P2 = 1.2 MPa, Flow in so: m2 = 2 m1 = 0.054 kg Process: Piston Fext = KV2 = PA => P = CV2 => P2 = P1 (V2/V1)2 From the process equation we then get: V2 = V1 (P2/P1) 1/2 = 0.005 ( 1200 615 ) 1/2 = 0. m3 v2 = V/m = 0. 0.054 = 0.12934 m3/kg