Solution Shapiro

CHAPTER ONE\nGETTING STARTED:\nINTRODUCTORY\nCONCEPTS AND\nDEFINITIONS PROBLEM 1.1\nmass enters when intake valve is open\nmass exits when exhaust valve is open\nhot surfaces interact with surroundings\npiston exerts force on gas during compression, gas exerts force on piston during expansion\nAir in\nFuel in\nExhaust gas out\nshaft torque drives fan\nshaft torque transmits force to transmission PROBLEM 1.2\nMotor as system:\nelectric current flows\ntorque transmitted through shaft\n- speed of rotation of armature can change\n- temperature of motor parts can change\nwarm surface of motor interacts with surroundings\nEnlarged system:\n- chemical changes occur within the battery\n- speed of rotation of armature can change\n- temperature of motor parts and battery can change\n- mass is raised within system\nCOMMENT : The shaft torque and current flow interactions become internal to the enlarged system. PROBLEM 1.3\nA control volume encloses the solar collector.\n• Cool water enters the collector at 0, and hot water exits at 1.\n• Solar radiation impinges on the front of the collector.\n• Warm surfaces of the collector interact with the surroundings.\n• Some of the incoming radiation is reflected away and some is absorbed in the collector surface.\n\nA control volume encloses the solar collector, the tank, and the interconnected piping.\n• Cold water enters the tank at 0, and hot water exits at 1.\n• Warm surfaces of the collector, storage tank, and interconnected piping interact with the surroundings.\n• Solar radiation impinges on the front of the collector; some is reflected and some is absorbed.\n• The temperature of the water in the storage tank changes with time. PROBLEM 1.4\nA control volume encloses the valve and turbine.\n• Steam enters at 0 and exits at 1.\n• A torque is transmitted through the rotating shaft.\n• Warm surfaces of the turbine interact with the surroundings.\n\nWithin the control volume, steam flows across the valve and into the turbine blades.\nWhen the generator is included in the control volume,\n• Steam enters at 0 and exits at 1.\n• Warm surfaces of the turbine and the generator interact with the surroundings.\n• Electric current flows from the generator.\n\nNote that the transmitted torque does not cross the boundary of the enlarged control volume.\n\nPROBLEM 1.5\nA control volume encloses the engine-driven pump.\n• Water enters at 0 and exits at 1.\n• Air for combustion of the onboard fuel enters, and combustion gases exit.\n• Warm surfaces of the pump interact with the surroundings.\n\nWithin the pump, a piston is kept in motion within a cylinder during combustion of the onboard fuel. The piston motion is harnessed to pump the liquid. The amount of fuel within the system decreases with time.\n\nWhen the hose and nozzle are included, a high-speed water jet exits the extended control volume as the nozzle exits.\n1-4 PROBLEM 1.6\nsystem boundary\n• Two phases are present (liquid and gas).\n• not a pure substance because composition is different in each phase.\n\nsystem boundary\n• three phases are present (solid, liquid, and gas).\n• not a pure substance because composition of gas phases is different than that of the solid and liquid phases.\n\nPROBLEM 1.7\ntwo phases\nsystem boundary\nThe system is a pure substance. Although the liquid is vaporized, the system remains fixed in chemical composition and is chemically homogeneous.\n\nPROBLEM 1.8\nliquid water → liquid water & ice\nThe system is a pure substance. Although the phases change, the system remains of fixed chemical composition and is chemically homogeneous.\n\nPROBLEM 1.9\ndish of water\nThe system is not a pure substance during the process since the composition of the gas phase changes as water evaporates into the air.\nOnce all of the water evaporates, the gas phase comes to equilibrium and the composition becomes homogeneous. At this point, the gas phase can be treated as a pure substance.\n1-5 PROBLEM 1.10*\nm = 10 lb\ng = 31.0 ft/s²\nFgrav = mg = (10 lb)(31.0 ft/s²) | 1 lb | 1 lb | 32.2 lb ft/s²\n= 9.63 lb_f <-- Fgrav\n\nPROBLEM 1.11*\ng = 9.8 m/s²\nFgrav = 25 kN\nm = Fgrav / g = (25 kN) | (1000 N) | (1 kg m/s²) | (9.8 m/s²) | 1 N\n= 2551 kg\n\nPROBLEM 1.12\nm = 10 kg\n(a) g_local = Fgrav / m = (95 N) / (10 kg) | 1 kg m/s² | 1 N\n= 9.5 m/s²\n(b) Mass is unchanged.\nFgrav = mg = (10 kg)(9.81 m/s²) | 1 N | 1 kg m/s²\n= 98.1 N\n\nPROBLEM 1.13\nm = 10 lb\n(a) g_local = Fgrav / m = (9.6 lb) / (10 lb) | 32.2 lb ft/s² | 1 lb\n= 30.9 ft/s²\n(b) mass is unchanged.\nFgrav = mg = (10 lb)(32.2 ft/s²) | 1 lb | 32.2 lb ft/s²\n= 10 lb_f\n1-6 PROBLEM 1.14\nV = 25 ft³\nFgrav = 3.511 lb\ng_moon = 5.47 ft/s²\ngmars = 12.86 ft/s²\nAccordingly,\n(Fgrav)_mars = (g mars / g moon) (Fgrav)_moon = (12.86 ft/s²) / (5.47 ft/s²)(3.5 lb) = 8.23 lb_f <--\nThe density is \\rho = m / V\nApplying g_mars with data on mars:\nm = (8.23 lb_f) | (32.2 lb ft/s²) | (1 lb) / (12.86 ft/s²)\n= 20.61 lb\nThen\n\\rho = 20.61 lb / 25 ft³ = 0.824 lb / ft³\n1-7 PROBLEM 1.15*\nEg.1.10 is used on both parts: n = m / M, where M is from Tables A-1.\n\n(a) m = M/n, n = 10 kmol\nAr: m = (28.97 kg/kmol)(10 kmol) = 289.7 kg\nH₂O: m = (18.02 kg/kmol)(10 kmol) = 180.2 kg\nCur: m = (63.54 kg/kmol)(10 kmol) = 635.4 kg\nSO₂: m = (64.06 kg/kmol)(10 kmol) = 640.6 kg\n(b) n = m / M, m = 20 lb\nA.F: n = (20 lb) / (39.948 lb/mol) = 0.501 lb mol\nH₂: n = (20 lb) / (2.016 lb/mol) = 9.921 lb mol\nN₂: n = (20 lb) / (28.01 lb/mol) = 0.714 lb mol\nC: n = (20 lb) / (12.0116 lb/mol) = 1.665 lb mol\n1-7 PROBLEM 1.18\n\nFor a linear spring, Fspring = K(Δx), where Δx is the spring extension. Since Fspring = Fg = mg, we have K(Δx) = mg. Since m and K are independent of location, the local acceleration of gravity is proportional to the deflection. Thus\n\nmars:\n gmars / geart = (Δx)mars / (Δx)earth = 0.116 in / 0.291 in => gmars = (0.116)(32.174 ft/s²) = 12.825 ft/s²\n\nmoon:\n gmoon / geart = (Δx)moon / (Δx)earth = (5.471 / 32.174)(0.291 in) = 0.049 in\n\nPROBLEM 1.19\n\nDeceleration occurs from 5 mi/h to rest in 0.5 s. The average acceleration magnitude is\n\n|a|avg = Δv / Δt = (5 mi/h - 0) / (0.5 s) (5280 ft / 1 mi) (1 h / 3600 s)\n = 73.33 ft/s²\n\nor, in g's\n\n|a|avg = (73.33 ft/s² / 32.2 ft/s²) = 2.28 g's\n\ndeceleration (in g's)\n\nThus, the magnitude of the average force applied is\n\n|F|avg = m |a|avg = (50 lb)(73.33 ft/s²) / (32.2 lb-ft/s²) = 113.9 lbf\n