CHAPTER 2: THE NATURE OF ENERGY

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1 Principles of Engineering Thermodynamics st Edition Reisel Solutions Manual Full Download: Chapter : The Nature of Energy CHAPTER : THE NATURE OF ENERGY.) A 5.00-kg light fiture is suspended above a theater s stage 5.0 meters above the ground. The local acceleration due to gravity is 9.75 m/s. Determine the potential energy of the light fiture. Given: m = 5.00 kg, z = 5.0 m; g = 9.75 m/s PE = mgz = (5.00 kg)(9.75 m/s )(5.0 m ) = 73 J.) In 908, Charles Street of the Washington, D.C., baseball team caught a baseball thrown from the top of the Washington Monument. If the height of the ball before it was thrown was 65 m, and the mass of the baseball was 0.45 kg, what was the potential energy of the baseball before it was thrown? Assume that the acceleration due to gravity was 9.8 m/s. Given: m = 0.45 kg, z = 65 m; g = 9.8 m/s PE = mgz = (0.45 kg)(9.8 m/s )(65 m ) = 35 J.3) 0.0 kg of water is about to fall over a cliff in a waterfall. The height of the cliff is 5 m. Determine the potential energy of the mass of water considering standard gravity on Earth. Given: m = 0.0 kg; g = 9.8 m/s ; z = 5 m PE = mgz = (0.0 kg)(9.8 m/s )(5 m ) =,80 J =.3 kj.4) A jet airplane is flying at a height of 0,300 m at a velocity of 40 m/s. If the mass of the airplane is 85,000 kg, and if the acceleration due to gravity is 9.70 m/s, determine the potential energy and the kinetic energy of the airplane. Given: m = 85,000 kg; g = 9.70 m/s ; z = 0,300 m; V = 40 m/s PE = mgz = (85,000 kg)(9.70 m/s )(0,300 m ) = J = 8,490 MJ 3 Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com

2 KE = ½ mv = ½ (85,000 kg)(40 m/s) = J =,450 MJ.5) A piece of space debris with a mass of 5 lbm is falling through the atmosphere towards earth at a height of 5,000 ft with a velocity of 750 ft/s. Determine the potential and kinetic energy of the object. Given: m = 5.0 lbm; z = 5,000 ft; V = 750 ft/s Assume: g = 3. ft/s (Acceleration due to gravity at 5,000 ft) PE = mgz/gc = (5.0 lbm)(3. ft/s )(5,000 ft) /(3.74 lbm-ft/lbf-s ) = 44,500 ft-lbf = 88 Btu KE = ½ mv /gc = ½ (5.0 lbm)(750 ft/s) / (3.74 lbm-ft/lbf-s ) = 3,000 ft-lbf = 69 Btu.6) A baseball with a mass of 0.45 kg is thrown by a pitcher at a velocity of 4.0 m/s. Determine the kinetic energy of the baseball. Given: m = 0.45 kg; V = 4.0 m/s KE = ½ mv = ½ (0.45 kg) (4.0 m/s) = 8 J.7) A brick with a mass of.50 kg, which was dropped from the roof of a building, is about to hit the ground at a velocity of 7.0 m/s. Determine the kinetic energy of the brick. Given: m =.50 kg; V = 7.0 m/s KE = ½ mv = ½ (.50 kg) (7.0 m/s) = 9J.8) An 80.0-kg rock that was hurled by a catapult is about to hit a wall while traveling at 7.50 m/s. What is the kinetic energy of the rock just before contacting the wall? Given: m = 80.0 kg; V = 7.50 m/s 4

3 KE = ½ mv = ½ (80.0 kg) (7.50 m/s) = 50 J =.5 kj.9) Steam flows through a pipe located 4 m above the ground. The velocity of the steam in the pipe is 80.0 m/s. If the specific internal energy of the steam is 765 kj/kg, determine the total internal energy, the kinetic energy, and the potential energy of.50 kg of steam in the pipe. Given: m =.50 kg; z = 4 m; V = 80.0 m/s; u = 765 kj/kg Assume: g = 9.8 m/s U = mu = (.50 kg) (765 kj/kg) = 4,50 kj KE = ½ mv = ½ (.50 kg)(80.0 m/s) = 4,800 J = 4.80 kj PE = mgz = (.50 kg)(9.8 m/s )(4 m) = 58.9 J = kj.0) Now, for problem.9 replace the steam with liquid water. If the specific internal energy of the liquid water is 0 kj/kg, determine the total internal energy, the kinetic energy, and the potential energy of.50 kg of water in the pipe. Given: m =.50 kg; z = 4 m; V = 80.0 m/s; u = 0 kj/kg Assume: g = 9.8 m/s U = mu = (.50 kg) (0 kj/kg) = 80 kj KE = ½ mv = ½ (.50 kg)(80.0 m/s) = 4,800 J = 4.80 kj PE = mgz = (.50 kg)(9.8 m/s )(4 m) = 58.9 J = kj.) Steam, with a specific internal energy of 05 Btu/lbm, is flowing through a pipe located 5 ft above the ground. The velocity of the steam in the pipe is 00 ft/s. Determine the total internal energy, the kinetic energy, and the potential energy of.50 lbm of steam in the pipe. Given: m =.50 lbm; z = 5ft; V = 00 ft/s; u = 05 Btu/lbm Assume: g = 3.74 ft/s U = mu = (.50 lbm) (05 Btu/lbm) = 330 Btu KE = ½ mv /gc = ½ (.50 lbm)(00 ft/s) /(3.74 lbm-ft/lbf-s ) = 550 ft-lbf =.00 Btu 5

4 PE = mgz/gc = (.50 lbm)(3.74 ft/s )(5 ft)/(3.74 lbm-ft/lbf-s ) = 37.5 ft-lbf = Btu.) Replace the steam in Problem. with liquid water. If the specific internal energy of the liquid qater is 45 Btu/lbm, determine the total internal energy, the kinetic energy, and the potential energy of.50 lbm of water in the pipe. Given: m =.50 lbm; z = 5ft; V = 00 ft/s; u = 45 Btu/lbm Assume: g = 3.74 ft/s U = mu = (.50 lbm) (45 Btu/lbm) = 3 Btu KE = ½ mv /gc = ½ (.50 lbm)(00 ft/s) /(3.74 lbm-ft/lbf-s ) = 550 ft-lbf =.00 Btu PE = mgz/gc = (.50 lbm)(3.74 ft/s )(5 ft)/(3.74 lbm-ft/lbf-s ) = 37.5 ft-lbf = Btu.3) Determine the specific energy of air at 5 o C, traveling at 35.0 m/s at a height of 0.0 m. Consider the specific internal energy of the air to be.6 kj/kg. Given: T= 5 o C ; z = 0.0 m; V = 35.0 m/s; u =.6 kj/kg Assume: g = 9.8 m/s e = u + ½ V = gz =.6 kj/kg + ½ (35.0 m/s) /(000 J/kJ) + (0.0 m)(9.8 m/s ) / (000 J/kJ) e = 3.3 kj/kg.4) Determine the total energy of water vapor with a mass of.50 kg, a specific internal energy of 780 kj/kg, traveling at a velocity of 56.0 m/s at a height of 3.50 m. Given: m =.50 kg ; z = 3.50 m; V = 56.0 m/s; u = 780 kj/kg Assume: g = 9.8 m/s E = mu + ½ mv + mgz = (.50 kg)[780 kj/kg + ½ (56.0 m/s) /(000 J/kJ) + (9.8 m/s )(3.50 m)/(000 J/kJ)] E = 6950 kj (6954 kj) 6

5 .5) A house has outside walls made of brick with a thermal conductivity of.0 W/m-K. The wall thickness is 0.0 m. (a) If the inside temperature of the wall is 0.0 o C and the outside temperature of the wall is -0.0 o C, determine the heat transfer rate per unit area from conduction (the conductive heat flu) for the wall. (b) For an inside wall temperature of 0.0 o C, plot the conductive heat flu through the wall for outside wall temperatures ranging from 30.0 o C to -5.0 o C. (c) For an outside wall temperature of o C, plot the conductive heat flu for inside wall temperatures ranging from 5.0 o C to 5.0 o C. Discuss how this relates to gaining cost savings by heating a home to a lower temperature in the winter. Given: =.0 W/m-K; = 0.0 m (a) To = -0.0 o C, Ti = 0.0 o C dt Q cond A d dt Ti To For the wall, approimate: d Q cond =-(.0 W/m-K)(50 K/m) = -80 W/m A 50 o C/m = 50 K/m (b) Heat Flu (W/m ) T o ( o C) 7

6 (c) Heat Flu (W/m ) T i ( o C).6) To reduce heat transfer losses through a glass window, you consider replacing the window with one of three alternatives: a sheet of tin 0.50-cm thick, a layer of brick 8.0- cm thick, and a combination of wood and insulation 4.0-cm thick. The original thickness of the glass is.0-cm. Consider the thermal conductivities of each to be the following: glass.40 W/m-K; tin 66.6 W/m-K; brick.0 W/m-K; wood/insulation combination 0.09 W/m-K. Considering the inside temperature of the surface to be 0.0 o C, and the outside temperature to be -5.0 o C, determine the heat conduction rate per unit area for the glass and the three alternatives, and discuss the relative merits of the three alternatives (ignoring material and installation costs). Given: Ti = 0.0 o C; To = -5.0 o C dt Ti To For the each material, approimate: d For Glass: =.40 W/m-K; =.0 cm = 0.0 m Q cond A = T i T o = -3,500 W/m Tin: = 66.6 W/m-K; = 0.5 cm = m Q cond A = T i T o = -66,400 W/m Brick: =.0 W/m-K; = 8.0 cm = 0.08 m Q cond A = T i T o = -375 W/m 8

7 Wood/Insulation: = 0.09 W/m-K; = 4.0 cm = 0.04 m Q cond A = T i T o = W/m The wood/insulation combination will offer the most resistance to heat flow, followed by the brick. The tin is a worse option than the original glass..7) A factory has a sheet metal wall with a thermal conductivity of 05 Btu/h-ft- o F. The wall thickness is inch. (a) If the inside temperature of the wall is 80.0 o F and the outside temperature of the wall is 0.0 o F, determine the heat transfer rate per unit area from conduction (the conductive heat flu) for the wall. (b) For an inside wall temperature of 80.0 o F, plot the conductive heat flu through the wall for outside wall temperatures ranging from 90.0 o F to 0.0 o F. (c) For an outside wall temperature of 0.0 o F, plot the conductive heat flu for inside wall temperatures ranging from 50.0 o F to 85.0 o F. Discuss how this relates to gaining cost savings by heating a building to a lower temperature in the winter. Given: = 05 Btu/h-ft- o F; = in = ft (a) To = 0.0 o F, Ti = 80.0 o F dt Q cond A d dt Ti To For the wall, approimate: o F/ft d Q cond =-(05 Btu/hr-ft- o F)(840.3 o F/ft) = -88,00 Btu/hr-ft A 9

8 (b) Heat Flu (Btu/hr-ft ) (c) T o ( o F) Heat Flu (Btu/hr-ft ) T i ( o F).8) You are stoking a fire in a furnace, and leave the stoker inside the furnace for several minutes. The end of the stoker inside the furnace reaches a temperature of 300 o C, while the other end of the stoker in the air is cooled sufficiently to maintain a temperature of 60 o C. The length of the stoker between these two ends is.0 m. If the cross-sectional area of the stoker is m, determine the rate of heat transfer that occurs if (a) the stoker is made of aluminum ( = 37 W/m-K), (b) iron ( = 80. W/m-K), and (c) granite ( =.79 W/m-K). Given: T = 300 o C; T = 60 o C; =.0 m; A = m For the each material, approimate: dt d T T 30

9 dt T T Q cond A A d (a) Aluminum: = 37 W/m-K Q cond = -(37 W/m-K)(0.000 m )(60 o C-300 o C)/.0m = 8.4 W (b) Iron: = 80. W/m-K Q cond = - (80. W/m-K)(0.000 m )(60 o C-300 o C)/.0m = 9.6 W (c) Granite: =.79 W/m-K Q cond = - (.79 W/m-K)(0.000 m )(60 o C-300 o C)/.0m = W.9) A conductive heat transfer of 5 W is applied to a metal bar whose length is m. The hot end of the bar is at 80 o C. Determine the temperature at the other end of the bar for (a) a copper bar ( = 40 W/m-K) with a cross-sectional area of m, (b) a copper bar ( = 40 W/m-K) with a cross-sectional area of m, and (c) a zinc bar ( = 6 W/m-K) with a cross-sectional area of m. Given: Q cond=50 W; = 0.50 m; T = 80 o C From dt T T A A d Q cond T = Q cond + T ka (a) Copper, = 40 W/m-K, A = m 5W T = (0.50m) + ( 40W m K )(0.0005m ) 80o C = 4.6 o C (b) Copper, = 40 W/m-K, A = m T = 76.3 o C (c) Zinc, = 6 W/m-K, A = m T = 54. o C Note: A material with a higher thermal conductivity will allow the same amount of heat flow with a smaller temperature difference than a material with a lower thermal conductivity. 3

10 .0) A wind blows over the face of a person on a cold winter day. The temperature of the air is -5.0 o C, while the person s skin temperature is 35.0 o C. If the convective heat transfer coefficient is 0.0 W/m -K, and the eposed surface area of the face is m, determine the rate of heat loss from the skin via convection. Given: Tf = -5.0 o C; Ts = 35.0 o C; h = 0.0 W/m -K; A= m Q conv ha( T Ts ) f = (0.0 W/m -K)(0.008 m )(-5 o C o C) = -3. W The heat transfer is from the person to the air..) Cooling water flows over a hot metal plate with a surface area of 0.5 m in a manufacturing process. The temperature of the cooling water is 5 o C, while the metal plate s surface temperature is maintained at 00 o C. If the convective heat transfer rate is 68.0 W/m -K, determine the rate of convective heat transfer from the plate. Given: A = 0.5 m ; Tf = 5 o C; Ts = 00 o C; h = 68.0 W/m -K Q conv ha( T Ts ) f = (68.0 W/m -K)(0.5 m )(5 o C - 00 o C) = -6,90 W = -6.9 kw The heat is transferred from the plate to the water..) Air passes over a hot metal bar which has a surface area of.5 ft. The temperature of the air is 70 o F, while the bar has a temperature maintained at 80 o F. If the convective heat transfer coefficient is 4 Btu/h-ft - o F, determine the rate of convective heat transfer from the plate. Given: A =.5 ft ; Tf = 70 o C; Ts = 80 o C; h = 4 Btu/h-ft - o F Q conv ha( T Ts ) f = (4 Btu/h-ft - o F)(.5 ft )(70 o F - 80 o F) = -,00 Btu/h The heat is transferred from the bar to the air..3) A cool early spring breeze passes over the roof of a poorly-insulated home. The surface area of the roof is 50 m, and the temperature of the roof is maintained at 0 o C from heat escaping the house. The air temperature is 5.0 o C, and the convective heat transfer coefficient is.0 W/m -K. Determine the rate of convective heat transfer from the roof. 3

11 Given: A = 50 m ; Tf = 5.0 o C; Ts = 0 o C; h =.0 W/m -K Q conv ha( T Ts ) f = (.0 W/m -K)(50 m )(5 o C - 0 o C) = -45,000 W = kw Heat is lost from the roof to the air..4) An electric space heater has a metal coil at a temperature of 50 o C, and is used to heat a space with an air temperature of 5 o C. If the surface area of the space heater is 0.0 m, and the emissivity of the heating element is 0.95, what is the rate of radiation heat transfer between the heater and the air? Given: Ts = 50 o C = 53 K; Tsurr = 5 o C = 88 K; A = 0.0 m ; = 0.95 rad 4 s 4 surr Q A T T = - (0.95)( W/m-K4 )(0.0m)[(53 K) 4 (88 K) 4 ] = W Heat is transferred from the heater to the air..5) An electrical resistance heater is placed inside a hollow cylinder. The heater has a surface temperature of 500 o F, while the inside of the cylinder is maintained at 80 o F. If the surface area of the heater is.5 ft, and the emissivity of the heater is 0.90, what is the net rate of radiation heat transfer from the heater? Given: Ts = 500 o F = 960 R; Tsurr = 80 o F = 540 R; A =.5 ft ; = 0.90 rad 4 s 4 surr Q A T T = - (0.90)( Btu/h-ft-R4 )(.5ft)[(960 R) 4 (540 R) 4 ] = Btu/h.6) A baker has developed a marvelous cookie recipe that works best if the cookie is cooled in a vacuum chamber through radiation heat transfer alone. The cookie is placed inside the chamber with a surface temperature of 5 o C. Assume that the cookie is thin enough so that the temperature is uniform throughout during the cooling process. The walls of the chamber are maintained at 0.0 o C. If the surface area of the cookie is m, and if the emissivity of the cookie is 0.80, determine the initial radiation heat transfer rate from the cookie to the walls of the chamber. Given: Ts = 5 o C = 398 K; Tsurr = 0 o C = 83 K; A = m ; =

12 4 4 Q rad A Ts T = - (0.80)( W/m-K4 )(0.005m)[(398 K) 4 (83 K) 4 ] = - surr 4.4 W Heat is transferred from the cookie to the walls..7) A building is losing heat to the surrounding air. The inside walls of the building are maintained at.0 o C, and the walls have a thermal conductivity of 0.50 W/m-K. The wall thickness is 0.0 m, and the outside wall temperature is held at.0 o C. The air temperature is -0.0 o C. Consider the emissivity of the outside of the walls to be Calculate the conductive heat flu and the radiative heat flu. Considering that the heat flu entering the outside of the wall from the inside must balance the heat flu leaving the outside of the wall via convection and radiation, determine the necessary convective heat transfer coefficient of the air passing over the outside wall. Given: To = Ts = o C = 75 K; Tsurr = -0 o C = 63 K; Ti =.0 o C = 95 K; = 0.0 m; = 0.85; = 0.50 W/m-K Conduction: Q cond = T i T o = - (0.50W/m-K)(95 K 75 K)/(0.0 m) = -00 W/m A Radiation: Q rad 4 4 T s Tsurr = - (0.85)( W/m -K 4 )[(75 K) 4 (63 K) 4 ] = A W/m Convection: The difference between the conduction and radiation is the convective heat flu: Q conv A = Q cond A Q rad A = -00 W/m (-45. W/m ) = W/m.8) A long cylindrical rod with a diameter of 3.0 cm is placed in the air. The rod is heated by an electrical current so that the surface temperature is maintained at 00 o C. The air temperature is o C. Air flows over the rod with a convective heat transfer coefficient of 5.0 W/m -K. Consider the emissivity of the rod to be 0.9. Ignoring the end effects of the rod, determine the heat transfer rate per unit length of the rod (a) via convection and (b) via radiation. Given: Ts = 00 o C = 473 K; Tf = Tsurr = o C = 94 K; h = 5.0 W/m -K; =0.9; D = 3.0 cm = 0.03 m 34

13 The surface area of the cylinder is A = DL = L m, where L is the length of the rod. Convection: Q conv A h ( Tf Ts ) = (5.0 W/m -K)( m)( o C - 00 o C) = W/m L L Radiation: Q rad A 4 4 T s Tsurr = -(0.9)( m)( W/m -K 4 )[(473 K) 4 (94 L L K) 4 ] = -09 W/m.9) A weight is attached to a horizontal load through a frictionless pulley. The weight has a mass of 5.0 kg, and the acceleration due to gravity is 9.80 m/s. The weight is allowed to fall.35 m. What is the work done by the falling weight as it pulls the load on a horizontal plane? Given: m = 5.0 kg; g = 9.80 m/s ; =.35 m W F d = W = (mg) = (5.0 kg)(9.80 m/s ) (.35 m) = 576 J.30) How much work is needed to lift a rock with a mass of 58.0 kg a distance of 5.0 m, where the acceleration due to gravity is 9.8 m/s? Given: m = 58.0 kg; = 5.0 m; g = 9.8 m/s W F d = W = (mg) = (58.0 kg)(9.8 m/s ) (5.0 m) = 8,530 J = 8.53 kj.3) How much work is needed to lift a 30 lbm rock a distance of 5.0 ft, where the acceleration due to gravity is 3.6 ft/s? Given: m = 30 lbm; = 5.0 ft; g = 3.6 ft/s 35

14 W F d = W = [(mg)/gc] = [(30 lbm)(3.6 ft/s )/(3.74 lbm-ft/lbf-s )] (5.0 ft) = 7,750 ft-lbf.3) A piston-cylinder device is filled with air. Initially, the piston is stabilized such that the length of the cylinder beneath the piston is 0.5 m. The piston has a diameter of 0.0 m. A weight with mass of 7.5 kg is placed on top of the piston, and the piston moves m, such that the new length of the air-filled cylinder beneath the piston is 0. m. Determine the work done by the newly-added mass. Given: = 0.5 m; = 0. m; D = 0.0 m; m = 7.5 kg Assume: g = 9.8 m/s The force eerted by the mass equals the weight of the mass: W= mg = (7.5 kg)(9.8 m/s ) = 7.7 N W F d = F ( ) = (7.7 N) (0. m 0.5 m ) = J The negative sign indicates that this is work that is adding energy to the system the air under the piston in the cylinder..33) 50 kpa of pressure is applied to a piston in a piston-cylinder device. This pressure causes the piston to move 0.05 m. The diameter of the piston is 0.0 m. Determine the work done on the gas inside the piston. Given: = m; P = 50 kpa; D = 0.0 m W mb V P dv = P V V V = (D /4) = m 3 Wmb = (50 kpa) ( m 3 ) = kn-m = kj = -96 J 36

15 .34) A piston-cylinder device is filled with 5 kg of liquid water at 50 o C. The specific volume of the liquid water is m 3 /kg. Heat is added to the water until some of the water boils, giving a liquid-vapor miture with a specific volume of 0.0 m 3 /kg. The pressure of the water is kpa. How much work was done by the epanding water vapor? Given: m = 5 kg; v = m 3 /kg; v = 0.0 m 3 /kg; P = kpa The water boils at constant pressure W mb V P dv = mp (v v) = (5 kg)(475.8 kpa)(0.0 m 3 /kg m 3 /kg) V = 83 kj In this case, work is being done on the surroundings by the epanding water, so the work represents energy out of the system..35) Air at 00 psi and 75 o F fills a piston-cylinder assembly to a volume of 0.55 ft 3. The air epands, in a constant-temperature process, until the pressure is 30 psi. (The constant-temperature process with a gas can be modeled as a polytropic process with n=.) Determine the work done by the air as it epands. Given: P = 00 psi; V = 0.55 ft 3 ; P = 30 psi PV = constant, so PV = PV V/V = P/P W mb PV = PVln (P/P) = (00 lbf/in )(44 in /ft ) (0.55 ft 3 ) ln ((00 psi/30 psi)) ln V V = 9,540 ft-lbf.36) Air at 450 kpa and 0 o C fills a piston-cylinder assembly to a volume of m 3. The air epands, in a constant-temperature process, until the pressure is 50 kpa. (The constant-temperature process with a gas can be modeled as a polytropic process with n=.) Determine the work done by the air as it epands. Given: P = 450 kpa; V = m 3 ; P = 50 kpa PV = constant, so PV = PV V/V = P/P 37

16 W mb PV = PVln (P/P) = (450 kpa) (0.075 m 3 ) ln ((450 kpa/50 kpa)) = 37. kj ln V V.37).5 kg of water vapor at 500 kpa fills a balloon. The specific volume of the water vapor is m 3 /kg. The water vapor condenses at constant pressure until a liquidvapor miture with a specific volume of m 3 /kg is present. Determine the work done in the process. Given: m =.5 kg; P = 500 kpa; v = m 3 /kg; v = m 3 /kg The water condenses at constant pressure W mb V P dv = mp (v v) = (.5 kg)(500 kpa)( m 3 /kg m 3 /kg) V = - kj.38) Air at 00 kpa and 50 o C fills a balloon with a volume of.85 m 3. The balloon cools and epands until the pressure is 400 kpa. The pressure and volume follow a relationship given by PV.3 = constant. Determine the work done by the air as it epands, and the final temperature of the air. Given: P = 00 kpa; T = 50 o C = 53 K; V =.85 m 3 ; P = 400 kpa; PV.3 = constant From the ideal gas law, m = PV/RT For air, R = 0.87 kj/kg-k, so m =.78 kg.3 V = V P = m 3 P The moving boundary work for a polytropic process with n : 3 3 PV PV (400kPa)(6.635m ) (00kPa)(.85m ) W mb = 550 kj n.3 From the ideal gas law: T = PV/mR = 406 K = 33 o C 38

17 .39) Oygen gas epands in a fleible container following a relationship of PV.5 = constant. The mass of the oygen is.75 lbm, and initially the pressure and temperature of the oygen is 50 psi and 50 o F, respectively. The epansion continues until the volume is double the original volume. Determine the final pressure and temperature of the oygen, and the work done by the oygen during the epansion. Given: m =.75 lbm; T = 50 o F = 60 R; P = 50 psi =,600 lbf/ft ; V = V; PV.5 = constant. P = P(V/V).5 = (,600 lbf/ft )(/).5 = 9,730 lbf/ft = 67.6 psi For O, R = 48.9 ft-lbf/lbm-r From the ideal gas law: V = mrt/p =.387 ft 3 Then, V = V = ft 3 From the ideal gas law: T = PV/mR = 550 R For the moving boundary work: 3 3 PV PV (9730 psf )(4.773 ft ) (,600 psf )(.387 ft ) W mb = 34,00 ft-lbf n.5.40) Nitrogen gas is compressed in a fleible container following a relationship of PV. = constant. The mass of the nitrogen is.5 kg, and initially the pressure and temperature of the nitrogen is 0 kpa and 5 o C, respectively. The compression continues until the volume reaches 0.0 m 3. Determine the final pressure and temperature of the nitrogen, and the work done on the nitrogen in the process. Given: m =.5 kg; P =0 kpa; T = 5 o C = 88 K; PV. = constant; V = 0.0 m 3 For N, R = kj/kg-k Using the ideal gas law: V = mrt/p =.068 m 3 For a polytropic process with n =.: PV. = PV. So, P = 060 kpa From the ideal gas law, T = PV/mR = 46 K = 89 o C For the moving boundary work: 3 3 PV PV (058kPa)(0.m ) (0kPa)(.068m ) W mb = -388 kj n.0 39

18 .4) Air epands in an isothermal process from a volume of 0.5 m 3 and a pressure of 850 kpa, to a volume of. m 3. The temperature of the air is 5 o C. Determine the work done by the air in this epansion process. Given: T = T = 5 o C ; V = 0.5 m 3 ; P = 850 kpa; V =. m 3 An isothermal process involving an ideal gas can be modeled as a polytropic process with n =. Therefore, W mb PV V ln V = (850 kpa)(0.5 m 3 ) ln (. m 3 / 0.5 m 3 ) = 37 kj.4) 3.0 kg of air is initially at 00 kpa and 0 o C. The air is compressed in a polytropic process, following the relationship PV n = constant. The air is compressed until the volume is 0.40 m 3. Determine the work done and the final temperature of the air for values of n of.0,.,.,.3, and.4, and plot the work as a function of n. Given: m = 3.0 kg; P = 00 kpa; T = 0 o C = 83 K; V = 0.40 m 3 For air, R = 0.87 kj/kg-k From the ideal gas law, V = mrt/p =.8 m 3 For a polytropic process with n = : W mb PV V ln V = - 7 kj PV PV For a polytropic process with n : W mb n To solve for P, use P = P (V/V) n The following values can be found from these two equations: n P (kpa) Wmb (kj)

19 W (kj) Chapter : The Nature of Energy n.43) A torque of 50 N-m is applied to a rotating shaft. Determine the work delivered for one revolution of the shaft. Given: T = 50 N-m revolution involves a rotation through radians. So, the rotating shaft work is W rs T d = T = 57 J =.57 kj.44) A torque of 50 N-m is applied to a rotating shaft. Determine the power used if the shaft rotates at 500 revolutions per minute. Given: T = 50 N-m, N = 500 rpm = 5 rps The rotating shaft power can be found from T d =T N = 80,00 J/s = 80. kw W rs.45) An engine delivers 75 hp of power to a rotating shaft. If the shaft rotates at 500 revolutions per minute, determine the torque eerted on the shaft by the engine. Given: W rs = 75 hp = 4,50 ft lbf ; N = 500 rpm = 4.67 rps s 4

20 From Problem.44, T = W rs πn = 58 ft-lbf.46) A steam turbine operates at 800 revolutions per minute (rpm). The turbine delivers 65.0 MW of power to the shaft of an electrical generator. Determine the torque on the steam turbine s shaft. Given: W rs = 65.0 MW = 65,000,000 W; N = 800 rpm = 30 rps From Problem.44, T = W rs πn = 345,000 N-m = 345 kn-m.47) A room fan operates on 0-volts and draws a current of.5 amps. What is the power used by the fan? Given: E = 0 V; I =.5 A W e I = -80 W.48) A 3-Watt CFL bulb is plugged into a 0-volt source. What is the current drawn by the bulb? Given: W e = -3 W; E = 0 V I = W e E = 0.9 A.49) An air compressor is plugged into a 08-volt outlet, and requires 0 kw of power to complete its compression process. What is the current drawn by the air compressor? Given: W e = -0 kw = -0,000 W; E = 08 V 4

21 I = W e E = 48. A.50) You are placed in an unfamiliar environment, with many odd pieces of electrical equipment using non-standard plugs. You need to determine the voltage of a particular outlet. The information on the machine plugged into the outlet reveals that the machine uses.50 kw of power and draws a current of 0.8 amps. What is the voltage of the outlet? Given: W e = -.50 kw = -,500 W; I = 0.8 A E = W e = 0 V I.5) A linear elastic spring with a spring constant of 50 N/m is compressed from a length of 0.5 m to a length of 0.7 m. Determine the work done in this compression process. Given: k = 50 N/m; = 0.5m ; = 0.7 m For a spring with a constant spring constant: W spring F d k d k( ) = ½ (50 N/m) [(0.7 m) (0.5 m) ] = - 4. J.5) A linear elastic spring has a spring constant of lbf/ft is compressed from a length of.5 ft to a length of.3 ft. Determine the work done in this compression process. Given: k = lbf/ft; =.5 ft ; =.3 ft For a spring with a constant spring constant: W spring F d k d k( ) = ½ ( lbf/ft) [(.3 ft) (.5 ft) ] = ft-lbf 43

22 .53) A linear elastic spring with a spring constant of.5 kn/m is initially at a length of 0.5 m. The spring epands, doing 40 J of work in the process. What is the final length of the spring? Given: k =,50 N/m; = 0.5m ; Wspring = 40 J For a spring with a constant spring constant: W spring F d = W spring k k d k( + = m ).54) Air, with a mass flow rate of 8.0 kg/s, enters a gas turbine. The enthalpy of the air entering the turbine is 85 kj/kg, the velocity of the air is 35 m/s, and the height of the air above the ground is.5 m. The acceleration due to gravity is 9.8 m/s. Determine the rate at which energy is being transferred into the gas turbine via mass flow by this air. Given: m = 8.0 kg/s; h = 85 kj/kg; V = 35 m/s; z =.5 m; g = 9.8 m/s E massflow m h V gz = (8.0 kg/s)[85 kj/kg + ½ (35 m/s) /(000 J/kJ) + (9.8 m/s )(.5 m)/(000 J/kJ)] = 7,03 kw = 7.0 MW.55) A jet of liquid water with a velocity of 4 m/s and an enthalpy of 6 kj/kg enters a system at a mass flow rate of 0 kg/s. The potential energy of the water is negligible. What is the rate of energy transfer to the system via mass flow for the water jet? Given: m = 0 kg/s; h = 6 kj/kg; V = 4 m/s; z = 0 m; E massflow m h V gz = (0 kg/s)[6 kj/kg + ½ (4 m/s) /(000 J/kJ) + 0] = 3,0 kw = 3. MW 44

23 .56) Steam enters a steam turbine at a volumetric flow rate of 0 ft 3 /s. The enthalpy of the steam is 30 Btu/lbm, and the specific volume of the steam is.43 ft 3 /lbm. The velocity of the steam as it enters the steam turbine is 50 ft/s, and the height of the entrance above the ground is 0 ft. The acceleration due to gravity is 3.7 ft/s. Determine the rate at which energy is being transferred via mass flow for the steam. Given: V = 0 ft 3 /s; h = 30 Btu/lbm; v =.43 ft 3 /lbm; V = 50 ft/s; z = 0 ft; g = 3.7 ft/s First, finding the mass flow: m = V v = 45.7 lbm/s gz E massflow m h V gc gc =(45.7 lbm/s)[30 Btu/lbm + ½ [(50 ft/s) /(3.74 lbm-ft/lbf-s )][ Btu/ ftlbf] + [(3.7 ft/s )(0 ft)/( 3.74 lbm-ft/lbf-s )][ Btu/ ft-lbf]] = 59,80 Btu/s = 6.6 Btu/h.57) You have steam available to add energy to a system via mass flow. The enthalpy of the steam is 750 kj/kg, and the velocity of the steam is 0 m/s. The potential energy of the steam is negligible. If you must add 33. MW of energy to the system with this steam, what is the required mass flow rate of steam entering the system? Given: E massflow = 33,00 kw; h = 750 kj/kg; V = 0 m/s; z = 0 m; m h V m = E massflow E massflow gz, so h+ = 33,00 kw V +gz 750 kj kg + ( 0 m ) s (000 J +0 kj ) =.0 kg/s 45

24 Principles of Engineering Thermodynamics st Edition Reisel Solutions Manual Full Download: Chapter : The Nature of Energy 46 Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com