# Solution Manual For Fundamentals of Engineering Thermodynamics, 9th Edition

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Problem 2.1 A baseball has a mass of 0.3 lb. What is the kinetic energy relative to home plate of a 94 mile per hour fastball, in Btu? KE = ยฝ m V2 relative to home plate = ยฝ (0.3 lb){(44 = 0.114 Btu mi 5280 ft h 1 mi )| || 1h 2 |} = 2851.1 ftโlb/s2 | 60 s 1 lbf 32.2 ftโlb/s2 || 1 Btu 778 ftโlbf | Problem 2.2 Determine the gravitational potential energy, in kJ, of 2 m3 of liquid water at an elevation of 30 m above the surface of Earth. The acceleration of gravity is constant at 9.7 m/s2 and the density of the water is uniform at 1000 kg/m3. Determine the change in gravitational potential energy as the elevation decreased by 15 m. KNOWN: The elevation of a known quantity of water is decreased from a given initial value by a given amount. FIND: Determine the initial gravitational potential energy and the change in gravitational potential energy. V = 2 m3 g = 9.7 m/s2 ฯ = 1000 kg/m3 SCHEMATIC AND GIVEN DATA: ENGINERING MODEL: (1) The water is a closed system. (2) The acceleration of gravity is constant. (3) The density of water is uniform. ฮz = – 15 m z1 = 30 m ANALYSIS: The initial gravitational potential energy is ??1 = ???1 = (?V)??1 = (1000 kg m3 m 1N ? 1 kgโ 2 ) (2 m3 ) (9.7 2 ) (30 m) | m s || 1 kJ = 582 kJ The change in potential energy is โ?? = ??(?2 โ ?1 ) = ??โ? = (1000 kg m3 m 1N ? 1 kgโ 2 ) (9.7 2 ) (โ15 m) | = โ145.5 kJ m s || | 103 Nโm 1 kJ 103 Nโm | Problem 2.3 An object whose weight is 100 lbf experiences a decrease in kinetic energy of 500 ftโlbf and an increase in potential energy of 1500 ftโlbf. The initial velocity and elevation of the object, each relative to the surface of the earth, are 40 ft/s and 30 ft, respectively. If g = 32.2 ft/s2, determine (a) the final velocity, in ft/s. (b) the final elevation, in ft. KNOWN: An object experiences specified changes in kinetic and potential energy. The initial velocity and elevation are known. FIND: Determine the final velocity and the final elevation. SCHEMATIC AND GIVEN DATA: ฮKE = โ 500 ftโlbf ฮPE = + 1500 ftโlbf V1 = 40 ft/s z1 = 30 ft z ENGINEERING MODEL: (1) The object is a closed system. (2) The acceleration of gravity is constant; g = 32.2 ft/s2. (3) Velocity and elevation are measured relative to the surface of the earth. Fgrav = mg = 100 lbf ANALYSIS: (a) The change in kinetic energy is: ฮKE = 1/2 m(V22 โ V12). Solving for V2 1/2 2โKE V2 = [( ? ) + V12 ] The mass is m= Fgrav ? = (1) 100 lbf 32.2 ftโlb/s2 | | = 100 lb 32.2 ftโs2 1 lbf So, inserting values into (1) and converting units 2(โ500 ftโlbf) 32.2 ftโlb/s2 V2 = [ (100 lb) | ? lbf ft 1/2 | + (40 )2 ] s = 35.75 ft/s (b) The change in potential energy is: ฮPE = mg(z2 โ z1). With mg = Fgrav, z2 = ฮPE/Fgrav + z1 = (1500 ftโlbf)/(100 lbf) + (30 ft) = 45 ft Problem 2.4 A construction crane weighing 12,000 lbf fell from a height of 400 ft to the street below during a severe storm. For g = 32.05 ft/s2, determine mass, in lb, and the change in gravitational potential energy of the crane, in ftโlbf. KNOWN: A crane of known weight falls from a known elevation to the street below. FIND: Determine the change in gravitational potential energy of the crane. SCHMATIC AND GIVEN DATA: Fgrav = 12,000 lbf g = 32.05 ft/s2 z2 = 0 z1 = 400 ft ENGINEERING MODEL: (1) The crane is the closed system. (2) The acceleration of gravity is constant. ANALYSIS: To get the mass, note that Fgrrav = mg. Thus ?= Fgrav ? = 12000 lbf 32.174 lbโft/s2 32.05 ft/s 1 lbf | 2 | = 12,046 lb The change in gravitational potential energy is โ?? = ??(?2 โ ?1 ) = Fgrav โ? = (12000 lbf)(โ400 ft) = 4.8 x 106 ft โ lbf Problem 2.5 An automobile weighing 2500-lbf increases its gravitational potential energy by 2.25 x 104 Btu in going from an elevation of 5,183 ft in Denver to the highest elevation on Trail Ridge road in the Rocky Mountains. What is the elevation at the high point of the road, in ft? KNOWN: An automobile of known weight increases its gravitational potential energy by a given amount. The initial elevation is known. FIND: Determine the final elevation. Fgrav = 2500 lbf ฮPE = 2.25 x 104 Btu z2 = ? z1 = 5183 ft ENGINEERING MODEL: (1) The automobile is the closed system. (2) The acceleration of gravity is constant. (2) Velocity and elevation are each measured relative to a stationary observer on the surface of the earth. ANALYSIS: The change in gravitational potential energy is: ฮPE = mg(z2 โ z1). With Fgrav = mg, we get ฮPE = Fgrav(z2 โ z1) Solving for z2 ?2 = โPE F???? + ?1 = (2.25 ? 104 Btu) 778 ftโlbf (2500 lbf) | 1 Btu | + 5183 ft = 12,185 ft Problem 2.6 An object of mass 15 kg is at an elevation of 100 m relative to the surface of the Earth. What is the potential energy of the object, in kJ? If the object were initially at rest, to what velocity, in m/s, would you have to accelerate it for the kinetic energy to have the same value as the potential energy you calculated above? The acceleration of gravity is 9.8 m/s2. KNOWN: An object has known mass and elevation and is initially at rest. The acceleration of gravity is given. FIND: Determine the initial potential energy and the final velocity if it were accelerated to have kinetic energy equal to the initial potential energy. SCHEMATIC AND GIVEN DATA: m = 15 kg V2 = ? V1 = 0 g = 9.8 m/s2 ENGINEERING MODEL: (1) The object is a closed system. (2) The acceleration of gravity is constant: g = 9.8 m/s2. (3) The final kinetic energy is equal to the initial gravitational potential energy. (4) Velocity and elevation are each measured relative to a stationary observer on the surface of the earth. m = 15 kg z1 = 100 m ANALYSIS: First, evaluate the initial potential energy. PE1 = mgz1 = (15 kg)(9.8 m/s2)(100 m) | For KE2 = PE1 โ 1N 1 kgโm/s2 || 1 kJ 103 Nโm | = 14.7 kJ ยฝ mV22 = PE1 So V2 = โ 2PE1 ? =โ 2(14.7 kJ) 103 Nโm (15 kg) | 1 kJ || 1 kgโm/s2 1N | = 44.27 m/s 2.7 An automobile having a mass of 900 kg initially moves along a level highway at 100 km/h relative to the highway. It then climbs a hill whose crest is 50 m above the level highway and parks at a rest area located there. For the automobile, determine its changes in kinetic and potential energy, each in kJ. For each quantity, kinetic and potential energy, specify your choice of datum and reference value at that datum. Let g = 9.81 m/s2. KNOWN: Data are provided for an automobile on the open road. FIND: Determine the changes in kinetic and potential energy for the automobile and specify an appropriate datum for each. SCHEMATIC AND GIVEN DATA: V2 = 0 stationary observer on level road V=0 z=0 z2 = 50m m = 900 kg V1= 100 km/h ENGINEERING MODEL: (1) The automobile is the closed system. (2) The acceleration of gravity is constant; g = 9.81 m/s2. (3) Velocities and elevations are measured relative to the stationary observer on the level road. ANALYSIS: The change in kinetic energy is ฮKE = ยฝ m (V22 โ V12) = ยฝ (900 kg)[0 โ (100 km/h)2| 1h 2 103 m 2 1N 1 km 1 kgโm/s2 | | 3600 s | | || 1 kJ 103 Nโm | = โ 347.2 kJ (decrease) The change in potential energy is ฮPE = mg(z2 โ z1) = (900 kg)(9.81 m/s2)(50 m โ 0) | 1N 1 kgโm/s2 || 1 kJ 103 Nโm | = +441.5 kJ (increase) 2.8 Vehicle crumple zones are designed to absorb energy during an impact by deforming to reduce transfer of energy to occupants. How much kinetic energy, in Btu, must a crumple zone absorb to fully protect occupants in a 3000 lb vehicle that suddenly decelerates from 10 mph to 0 mph? 2.9 In a recent airline disaster, an airliner flying at 30,000 ft, 550 mi/h, lost power and fell to Earth. The mass of the aircraft was 255,000 lb. If the magnitude of the work done by drag force on the plane during the fall was 2.96 x 106 Btu, estimate the velocity of the aircraft at the time of impact, in mi/h. Let g = 32.08 ft/s2. KNOWN: An airliner falls from a known elevation and velocity to the surface of the Earth. The magnitude of the work done against drag force is specified, and the average acceleration of gravity is given. FIND: Estimate the velocity at the time of impact with the Earth. SCHEMATIC AND GIVEN DATA: 0 m = 255,000 lb g = 32.08 ft/s2 V1 = 550 mi/h 5280 ft 1h 1 mi 3600 s z2 = 0 z1 = 100 m ?drag = 2.96 x 106 Btu z ANALYSIS: The work done against drag is ? Wdrag = โซ? 2 ?drag โ ?? = 806.7 ft/s ENGINEERING MODEL: (1) The airliner is a closed system. (2) The acceleration of gravity is constant: g = 32.08 ft/s2. (3) The only forces acting on the airliner as it falls are the drag force opposing motion and the force of gravity. Fdrag โ opposite to direction of motion 1 Fdrag โds is negative. So ds – direction of motion Wdrag = โ 2.96 x 106 Btu The work done against drag equals the change in kinetic energy plus the change in potential energy ? Wdrag = โซ? 2 ?drag โ ?? = ยฝ m(V22 โ V12) + mg(z2 โ z1) 1 Problem 2.9 (Continued) Solving for V2 2?drag V2 = โ ? + 2??1 + v12 lb = 2 2(โ2.96 x 106 Btu) 778.17 ftโlbf 32.174 ftโs2 ft 2 ) ft โ (30,000 | | + 2 (32.08 ) ft) + (806.7 (255,000 lb) 1 Btu 1 lbf s2 s2 = 1412 ft/s = 1412 ft/s 1 mi 3600 s 5280 ft 1h = 962.7 mi/h COMMENTS: The work done against drag is reduces the final kinetic energy, so it is negative in this case. Also, be careful in applying the English system units in this problem. Problem 2.10 Two objects having different masses are propelled vertically from the surface of Earth, each with the same initial velocities. Assuming the objects are acted upon only by the force of gravity, show that they reach zero velocity at the same height. KNOWN: Two objects are propelled upward from the surface of Earth with the same initial velocities and are acted upon only by the force of gravity. FIND: Show that they reach zero velocity at the same height. V2 = 0 SCHEMATIC AND GIVEN DATA: ENGINEERING MODEL: (1) Each object is a closed system. (2) The acceleration of gravity is constant. (3) The only force acting is the force of gravity. ANALYSIS: For an object moving vertically under the influence of gravity only, Eq. 2.11 applies 1 2 z1 = 0 z2 V1 V1 m1 m2 ??(V22 โ V12 ) + ??(?2 โ ?1 ) = 0 For V2 = 0 and z1 = 0 1 โ ?V12 + ???2 = 0 2 Thus ?2 = V12 /2? Since the final height doesnโt depend on mass, both objects will reach zero velocity at the same final height. Problem 2.11 An object whose mass is 100 lb falls freely under the influence of gravity from an initial elevation of 600 ft above the surface of Earth. The initial velocity is downward with a magnitude of 50 ft/s. The effect of air resistance is negligible. Determine the velocity, in ft/s, of the object just before it strikes Earth. Assume g = 31.5 ft/s2. KNOWN: An object of known mass falls freely from a known elevation and with a given initial velocity. The only force acting is the force of gravity. m = 100 lb g = 31.5 ft/s2 FIND: Determine the velocity of the object just before it strikes Earth. SCHEMATIC AND GIVEN DATA: ENGINEERING MODEL: (1) The object is a closed system. (2) The acceleration of gravity is constant: g = 51.5 ft/s2. (3) The only force acting on the object is the force of gravity. ANALYSIS: Since the only force acting on the object is the force of gravity, Eq. 2.11 applies. Thus 0 1 2 2) 1 ?(V2 โ V1 + ??(?2 โ ?1 ) = 0 V1 = 50 ft/s z1 = 600 ft z2 = 0 V2 2 Solving for V2 V2 = โV12 + 2??1 Inserting values ft2 ft s s V2 = โ502 2 + 2 (31.5 2 ) (600 ft) = 200.7 ft/s 1. Note that the mass cancels out. Any object falling freely under the influence of gravity, with no effects of air resistance, would reach the same final velocity. 2.12 During the packaging process, a can of soda of mass 0.4 kg moves down a surface inclined at 20o relative to the horizontal, as shown in Fig. P2.12. The can is acted upon by a constant force R parallel to the incline and by the force of gravity. The magnitude of R is 0.05 N. Ignoring friction between the can and the inclined surface, determine the canโs change in kinetic energy, in J, and whither it is increasing or decreasing. If friction between the can and the inclined surface were significant, what would effect would that have on the value of the change in kinetic energy? Let g = 9.8 m/s2. Problem 2.12 (Continued) Problem 2.13 Jack, who weighs 150 lbf, runs 5 miles in 43 minutes on a treadmill set at a one-degree incline (Fig. P2.13). The treadmill display shows he has burned 620 kcal. For jack to break even calorie-wise, how much vanilla ice cream, in cups, may he have after his workout? ๏ท ๏ท Exercise value = 620 kcal Calorific value; 1 cup of vanilla ice cream = 264 kcal To break even calorie-wise, Jack may have 620 kcal = 2.35 cups 264 kcal/cup Problem 2.14 An object initially at an elevation of 5 m relative to Earthโs surface and with a velocity of 50 m/s is acted on by an applied force R and moves along a path. Its final elevation is 20 m and its velocity is 100 m/s. The acceleration of gravity is 9.81 m/s2. Determine the work done on the object by the applied force, in kJ. KNOWN: An object moves along a path due to the action of an applied force. The elevation and velocities are known initially and finally. V2 = 100 m/s FIND: Determine the work of the applied force. R SCHEMATIC AND GIVEN DATA: m = 50 kg V1 = 50 m/s ENGINEERING MODEL: (1) The object is a closed system. (2) R is the only force acting on the object other z than the force of gravity. (3) g = 9.81 m/s2 and is constant. z2 = 20 m z1 = 5 m ANALYSIS: To find the work of force R we use 2 1 Work = โซ1 ? โ ?? = ?(V22 โ V12 ) + ??(?2 โ ?1 ) 2 Inserting values and converting units 1 m2 m 1N 1 kJ Work = { (50 kg)(1002 โ 502 ) 2 + (50 kg) (9.81 2 ) (20 โ 5)m} | || 3 | 2 2 s s 1 kg โ m/s 10 N โ m = 187.5 + 7.36 = 194.9 kJ Problem 2.15 An object of mass 10 kg, initially at rest, experiences a constant horizontal acceleration of 4 m/s2 due to the action of a resultant force applied for 20 s. Determine the total amount of energy transfer by work, in kJ. KNOWN: A system of known mass experiences a constant horizontal acceleration due to an applied force for a specified length of time. m = 10 kg ax = 4 m/s2 FIND: Determine the amount of energy transfer by work. Fx ฮt = 20 s V1 = 0 SCHEMATIC AND GIVEN DATA: ENGINEERING MODEL: (1) The object is a closed system. (2) The horizontal acceleration is constant. x ANALYSIS: The work of the resultant force is determined using Eq. 2.6 0 ?2 1 2 2 โซ? F? ?? = ?(V2 โ V1 ) 2 1 To find V2, use the fact that the acceleration is constant ax = ?V โ dV = axdt ?? โ V ? 1 1 2 2 โซV ?V = โซ? a? ?? 0 or (V2 โ V1) = ax(t2 โ t1) = axฮt Thus V2 = (4 m/s2) (20 s) = 80 m/s Finally, the work of the resultant force is ? 1 2 โซ? F? ?? = 2 ?V22 1 1 m2 1N 2 s 1 kgโm/s2 = (10 kg)(802 ) 2 | || 1 kJ | = 32 kJ 103 Nโm Problem 2.16 An object with an initial horizontal velocity of 20 ft/s experiences a constant horizontal acceleration due to the action of a resultant force applied for 10 s. The work of the resultant force is 10 Btu. The mass of the object is 55 lb. Determine the constant horizontal acceleration, in ft/s2. KNOWN: Data are known for an object accelerating horizontally under the action of a constant resultant force for a specified amount of time. The work of the force is given. FIND: Determine the constant horizontal acceleration. SCHEMATIC AND GIVEN DATA: V1 = 20 ft/s F V2 m = 55 lb Finally ฮt = 10 s Work of force F is 10 Btu Initially ENGINEERING MODEL: (1) The object is a closed system. (2) The resultant force is constant over the time interval. (3) All forces and motions are horizontal. ANALYSIS: The constant horizontal acceleration is a = dV/dt = (V2 โ V1)/ฮt. To find V2 = use Eq. 2.6 ?2 2 2 โซ? ? โ ?? = ยฝ m(V2 โ V1 ) 1 Work, W, of the resultant force F So 2? V2 = โ ? 2(10 Btu) 778 ftโlbf + V12 = โ (55 lb) | 1 Btu 32.2 ftโlb/s2 || 1 lbf ft | + (20 )2 = 97.52 ft/s The acceleration is a = (V2 โ V1)/ฮt = [(97.52 โ 20)ft/s](10s) = 7.752 ft/s2 s Problem 2.17 A gas in a piston-cylinder assembly undergoes a process for which the relationship between pressure and volume is pV2 = constant. The initial pressure is 1 bar, the initial volume is 0.1 m3, and the final pressure is 9 bar. Determine (a) the final volume, in m3, and (b) the work for the process, in kJ. KNOWN: A gas in a piston-cylinder assembly undergoes a process during which pV2 = constant. State data are provided. FIND: Determine the final volume occupied by the gas and the work for the process. SCHEMATIC AND GIVEN DATA: p (bar) 9 Gas .2 ENGINEERING MODEL: (1) The gas is a closed system. (2) Volume change is the only work mode. (3) the process of the obeys pV2 = constant. pV2 = constant p1 = 1 bar V1 = 0.1 m3 p2 = 9 bar .1 1 0.1 V (m3) ANALYSIS: (a) We have pV2 = constant. Thus p1V12 = p2V22. Solving for V2 ?1 1/2 V2 = [ ] ?2 1 1/2 V1 = [ ] 9 (0.1 m3) = 0.0333 m3 (b) Using Eq. 2.17 to determine the work 2 W = โซ1 ??? = ?2 ?2 โ?1 ?1 (1โ2) (See Example 2.1(a) for the integration) Inserting values and converting units W= (9 bar)(0.033 m3 )โ(1 bar)(0.1 m3 ) 105 N/m2 (โ1) | 1 bar || 1 kJ 103 Nโm | = โ 20 kJ Negative sign denotes energy transfer to the gas by work during the compression process. Problem 2.18 Carbon dioxide (CO2) gas within a piston-cylinder assembly undergoes a process from a state where p1 = 5 lbf/in.2, V1 = 2.5 ft3 to a state where p2 = 20 lbf/in.2, V2 = 0.5 ft3. The relationship between pressure and volume during the process is given by p = 23.75 โ 7.5V, where V is in ft3 and p is in lbf/in.2 Determine the work for the process, in Btu. KNOWN: CO2 gas within a piston-cylinder assembly undergoes a process where the p-V relation is given. The initial and final states are specified. FIND: Determine the work for the process. SCHEMATIC AND GIVEN DATA: 25 p1 = 5 lbf/in.2 V1 = 2.5 ft3 p (lbf/in^2) p2 = 20 lbf/in. V2 = 0.5 ft3 CO2 2 p = 23/75 โ 7.5 V . 2 20 15 10 . 1 W 5 ENGINEERING MODEL: (1) The CO2 is the closed system. (2) The p-V relation during the process is linear. (3) Volume change is the only work mode. 0 0 0.5 1 1.5 2 2.5 3 V (ft^3) ANALYSIS: The given p-V relation can be used with Eq. 2.17 as follows: ?2 ? ?2 7.5? 2 2 ? = โซ ??? = โซ [23.75 โ 7.5?]?? = [23.75? โ ] 2 ? ?1 ?1 1 7.5 2 [? โ ?12 ] = 23.75[?2 โ ?1 ] โ 2 2 lbf 144 in.2 7.5 lbfโin.2 144 in.2 3 [0.5 ? = (23.75 2 ) | | โ 2.5]ft โ ( )| | [0.52 โ 2.52 ](ft 3 )2 in. 1 ft 2 2 ft 3 1 ft 2 = -3600 ftโlbf = (โ3600 ft โ lbf) | 1 Btu | = โ4.63 Btu (negative sign denotes energy transfer in.) 778 ftโlbf Alternative Solution Since the p-V relation is linear, W can also be evaluated geometrically as the area under the process line: ? = ?ave (?2 โ ?1 ) = ( = -4.63 Btu ?1 + ?2 20 + 5 lbf 144 in2 1 Btu | (0.5 โ 2.5)ft 3 | | ) (?2 โ ?1 ) = ( ) 2| 2 2 2 in 1 ft 778 ft โ lbf PROBLEM 2.19 Problem 2.20 Nitrogen (N2) gas within a piston-cylinder assembly undergoes a process from p1 = 20 bar, V1 = 0.5 m3 to a state where V2 = 2.75 m3. The relationship between pressure and volume during the process is pV1.35 = constant. For the N2, determine (a) the pressure at state 2, in bar, and (b) the work, in kJ. KNOWN: N2 gas within a piston-cylinder assembly undergoes a process where the p-V relation is pV1.35 = constant. Data are given at the initial and final states. FIND: Determine the pressure at the final state and the work. ENGINEERING MODEL: (1) The N2 is the closed system. (2) The p-V relation is specified for the process. (3) Volume change is the only work mode. SCHEMATIC AND GIVEN DATA: pV 1.35= constant p1 = 20 bar, V1 = 0.5 m3 V2 = 2.75 m3 N2 ANALYSIS: (a) ?1 ?1? = ?2 ?2? 0.5 m3 ?2 = (20 bar) ( 2.75 m 1.35 3) โ ? ? ?2 = ?1 ( 1 ) ; n = 1.35. Thus ?2 = 2 bar (b) Since volume change is the only work mode, Eq. 2.17 applies. Following the procedure of part (a) of Example 2.1, we have W= ?2 ?2 โ?1 ?1 1โ? = = 1285.7 kJ (2 bar)(2.75 m3) โ(20)(0.5) 105 N/m2 1โ1.35 | 1 bar || 1 kJ | 103 Nโm Problem 2.21 Air is compressed slowly in a piston-cylinder assembly from an initial state where p1 = 1.4 bar, V1 = 4.25 m3, to a final state where p2 = 6.8 bar. During the process, the relation between pressure and volume follows pV = constant. For the air as the closed system, determine the work, in Btu. KNOWN: Air is compressed in a piston-cylinder assembly from a known initial state to a known final pressure. The pressure-volume relation during the process is specified. FIND: Determine the work for the air as the closed system. SCHEMATIC AND GIVEN DATA: p1 = 1.4 V1 = 4.25 m3 Air W=? p2 = 6.8 bar p 6.8 bar .2 pV = constant ENGINEERING MODEL: (1) The air is a closed system. (2) The process is polytropic, with pV = constant. .1 1.4 bar V 3 4.25 m V2 = ? ANALYSIS: The work for the polytropic process can be determined by integrating ? W = โซ? 2 ??? . The pressure-volume relation is pV = constant. 1 ? ?2 ???????? 1 1 W = โซ? 2 ??? = โซ? ? dV = (constant) ln (V2/V1) The constant can be evaluated from the given data as constant = p1V1, and V2 can be determined using p1V1 = p2V2. Thus V2 = (p1/p2)V1 = (1.4/6.8)(4.25 m3) = 0.875 m3 And W = (p1V1) ln (V2/V1) W = (1.4 bar)(4.25 m3) ln (0.875/4.25)| 105 N/m2 1 bar || 1 kJ 103 Nโm | = โ 940.4 kJ Negative โ energy transfer by work is in, as expected Problem 2.22 Air contained within a piston-cylinder assembly is slowly compressed. As shown in Fig P2.32, during this first process the pressure first varies linearly with volume and then remains constant. Determine the total work, in kJ. KNOWN: Air within a piston-cylinder assembly undergoes two processes in series. FIND: Determine the total work. SCEMATIC AND GIVEN DATA: ENGINEERING MODEL: (1) The air within the piston-cylinder assembly is the closed system. (2) The two-step p-V relation is specified graphically. (3) Volume change is the only work mode. 3 150 2 + + 1 p (kPa) + 100 1 2 50 0.015 0.055 0.07 V (m3) ANALYSIS: Since volume change is the work mode, Eq. 2.17 applies. Furthermore, the integral can be evaluated geometrically in terms of the total area under process lines: 1 2 ?2 ?1 + ?2 ? = โซ ??? = ?ave (?2 โ ?1 ) + ?2 (?3 โ ?2 ) = ( ) (?2 โ ?1 ) + ?2 (?3 โ ?2 ) 2 ?1 = [( 100+150 103 Nโm2 2 1 kPa ) kPa(0.055 โ 0.07)m3 + (150)(0.015 โ 0.055)] | = (-1.875 kJ) + (-6 kJ) = -7.875 kJ (in) || 1 kJ | 103 Nโm Problem 2.23 A gas contained within a piston-cylinder assembly undergoes three processes in series: Process 1-2: Constant volume from p1 = 1 bar, V1 = 4 m3 to state 2, where p2 = 2 bar. Process 2-3: Compression to V3 = 2 m3, during which the pressure-volume relationship is pV = constant. Process 3-4: Constant pressure to state 4, where V4 = 1 m3. Sketch the processes in series on p-V coordinates and evaluate the work for each process, in kJ. KNOWN: A gas contained within a piston-cylinder assembly undergoes three processes in series. State data are provided. FIND: Sketch the processes on pโV coordinates and evaluate the work for each process. SCHEMATIC AND GIVEN DATA: 5 4 p (bar) Gas . . 4 3 pV = constant Note: p2V2 = p3V3 So p3 = (V2/V3)p2 = (4/3)(2 bar) = 4 bar 3 .2 .1 2 1 V (m3) 1 2 3 4 ENGINEERING MODEL: (1) The gas in the piston-cylinder assembly is a closed system. (2) The gas undergoes three processes in series, as illustrated. (3) Volume change is the only work mode. ANALYSIS: The work is evaluated using Eq. 2.17: W = โซ pdV Process 1-2: The volume is constant, so W12 = 0. ? ???????? Process 2-3: W23 = โซ? 3 2 ? ?? = constant ln (V3/V2) = (p2V2) ln (V3/V2) = (2 bar)(4 m3) ln(2/4) | 105 N/m2 1 bar || 1 kJ | = โ 554.5 kJ (in) 103 Nโm Process 3-4: For the constant-pressure process W34 = p(V4 โ V3) = (4 bar)(1 โ 2)m3 | 105 N/m2 1 bar || 1 kJ 103 Nโm | = โ 400 k J (in) Problem 2.24 Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state (a cycle). Process 1-2: Expansion from State 1 where p1 = 10 bar, V1 = 1 m3, to State 2 where V2 = 4 m3. During the process, pressure and volume are related by pV1.5 = constant. Process 2-3: Constant volume heating to State 3 where p3 = 10 bar. Process 3-1: Constant pressure compression to State 1. Sketch the processes on p-V coordinates and evaluate the work for each process, in kJ. What is net work for the cycle, in kJ? KNOWN: Carbon dioxide gas undergoes a cycle in a piston-cylinder assembly. Data are provided for the key states and the different process that make up the cycle. FIND: Sketch the processes on p-V coordinates, and determine the work for each process and the net work for the cycle overall. SCHEMATIC AND GIVEN DATA: CO2 ENGINEERING MODEL: (1) The CO2 is a closed system. (2) Processes 1-2 and 3-1 are quasi-equilibrium processes. (3) Process 2-3 is at constant volume and Process 3-1 is at constant pressure. Process 1-2: Expansion from State 1 where p1 = 10 bar, V1 = 1 m3, to State 2 where V2 = 4 m3. During the process, pressure and volume are related by pV1.5 = constant. Process 2-3: Constant volume heating to State 3 where p3 = 10 bar. Process 3-1: Constant pressure compression to State 1. ANALYSIS: First, for Process 1-2, pV1.5 = constant. Thus p1V11.5 = p2V21.5 โ p2 = (V1/V2)1.5 p1 = (1/4)1.5 (10 bar) = 1.25 bar With p3 = p2 = 10 bar, and V3 = V2 = 4 m3, all states are known and the p-V diagram is . . p (bar) . 1 2 3 4 V (m3) Problem 2.24 (Continued) ? Process 1-2 The work is W = โซ? 2 ??? . For a polytropic process with pV1.5 = constant, the 1 integral becomes W12 = ?2 ?2 โ?1 ?1 1โ1.5 (1.25 bar)(4 m3 )โ(10 bar)(1 m3 ) 105 N/m2 (1โ1.5) 1 bar =[ ]| || 1 kJ | 103 Nโm = 1000 kJ (out) Process 2-3 Since there is no volume change, W23 = 0 ? Process 3-1 The work is W31 = โซ? 1 ??? . Since the pressure is constant, the integral becomes 3 W31 = p3(V1 โ V3) = (10 bar) (1 โ 4) m3 | 105 N/m2 1 bar || 1 kJ 103 Nโm | = โ 3000 kJ (in) The net work for the cycle is: Wcycle = W12 + W23 + W31 = (1000) + (0) + (โ 3000) = โ 2000 kJ (in) Problem 2.25 Air contained within a piston-cylinder assembly undergoes three processes in series: Process 1-2: Compression during which the pressure-volume relationship is pV = constant from p1 = 10 lbf/in.2, V1 = 4 ft3 to p2 = 50 lbf/in.2 Process 2-3: Constant volume from state 2 to state 3 where p = 10 lbf/in.2 Process 3-1: Constant pressure expansion to the initial state. Sketch the processes in series on a p-V diagram. Evaluate (a) the volume at state 2, in ft3, and (b) the work for each process, in Btu. KNOWN: Air within a piston-cylinder assembly undergoes three processes in series. FIND: Sketch the processes in series on a p-V diagram. Evaluate (a) the volume at state 2, and (b) the work for each process. SCHEMATIC AND GIVEN DATA: p 2 Air pV = constant Process 1-2: Compression during which the pressure-volume relationship is pV = constant from p1 = 10 lbf/in.2, V1 = 4 ft3 to p2 = 50 lbf/in.2 Process 2-3: Constant volume from state 2 to state 3 where p = 10 lbf/in.2 Process 3-1: Constant pressure expansion to the initial state. 3 V ENGINEERING MODEL: (1) The gas is the closed system. (2) Volume change is the only work mode. (3) Each of the three processes is specified. ANALYSIS: (a) For process 1-2; pV = constant. Thus p1V1 = p2V2, and 10lbfโin.2 ) (4 ft 3 ) = 0.8 ft3 50lbfโin.2 ? ?2 = ( 1 ) ?1 = ( ?2 1 (b) Since volume change is the only work mode, Eq. 2.17 applies. Process 1-2: For process 1-2, pV = constant = p1V1. Thus ? ? ? ? ? 1 1 ? ?1 ?1 ?12 = โซ? 2 ??? = โซ? 2 ?? = ?ln ( 2) = (?1 ?1 )ln ( 2 ) Problem 2.25 (Continued) Inserting values and converting units ?12 = (10 lbf in.2 0.8 ft3 ) (4 ft 3 ) ln ( 4 ft3 144 in.2 )| 1 ft 2 || 1 Btu 778 ftโlbf | = -11.92 Btu (in) Process 2-3: Constant volume (piston does not move). Thus W23 = 0 ? Process 3-1: Constant pressure processes (p3 = p1): ?31 = โซ? 1 ??? = ?1 (?1 โ ?3 ) 3 Noting that V3 = V2 ?31 = (10 lbf 144 in.2 ) (4 โ 0.8)ft 3 | in2 . 1 ft 2 || 1 Btu 778 ftโlbf | = 5.92 Btu (out) 1. The net work for the three process is Wnet = W12 + W23 + W31 = (-11.92) + 0 + (5.92)) = – 6 kJ (net work is negative – in) PROBLEM 2.26 A 0.15-m-diameter pulley turns a belt rotating the driveshaft of a power plant pump. The torque applied by the belt on the pulley is 200 N โ m, and the power transmitted is 7 kW. Determine the net force applied by the belt on the pulley, in kN, and the rotational speed of the driveshaft, in RPM. Problem 2.27 A 10-V battery supplies a constant current of 0.5 amp to a resistance for 30 min. (a) Determine the resistance, in ohms. (b) For the battery, determine the amount of energy transfer by work, in kJ. KNOWN: Operating data are given for a 10-V battery providing current to a resistance. FIND: Determine the resistance and the amount of energy transfer by work. SCHEMATIC AND GIVEN DATA: โ + ENGINEERING MODEL: (1) The resistor is the closed system. (2) The current is constant with time. Welec 10-V battery resistance i = 0.5 amp ฮt = 30 min ANALYSIS: For current flow through a resistor: Voltage = Resistance * Current (Ohmโs Law) Thus Resistance = Voltage Current = 10 amps 1 ohm | = 20 ohm 0.5 volts 1 voltโamp | With Eq. 2.21applied to the battery which is discharging โ 1 watt amp |?ฬelec | = (voltage)(current) = (10 volt)(0.5 amp)| | = 5 watt 1 volt So, for 30 min of continuous operation, the energy transfer by work to the resistor is ? Welec = โซ? 2|?ฬelec |?? = |?ฬelec |ฮt 1 = (5 watt)(30 min)| 60 s 1 min || 1 J/s || 1 watt 1 kJ | = 9 kJ 103 J Problem 2.28 An electric heater draws a constant current of 6 amp, with an applied voltage of 220 V, for 24 h. Determine the instantaneous electric power provided to the heater, in kW, and the total amount of energy supplied to the heater by electrical work, in kWโh. If electric power is valued at \$0.08/kWโh, determine the cost of operation for one day. KNOWN: An electric heater draws a constant current at a specified voltage for a given length of time. The cost of electricity is specified. FIND: Determine the instantaneous power provided to the heater and the total amount of energy supplied by electrical work. Determine the cost of operation for one day. SHEMATIC AND GIVEN DATA: i = 6 amp E = 220 V ฮt = 10 h ENGINEERING MODEL: (1) The heater is a closed system. (2) The current and voltage are constant. ANALYSIS: The constant power input to the heater is given by Eq. 2.21 1 Wโamp 1 kW ?ฬin = E I = (220 V)(6 amp) | | | 3 | = 1.320 kW 1V 10 W Thus, the total energy input is ? ?in = โซ? 2 ?ฬin ?? = ?ฬin โ? = (1.320 kW)(24 h) = 31.68 kW โ h 1 Using the specified cost of electricity Cost per day = (31.68 kWโh) (\$0.08/kWโh) = \$2.53 ?ฬin Problem 2.29 Problem 2.30 Problem 2.31 Figure P2.31 shows an object whose mass is 5 lb attached to a rope wound around a pulley. The radius of the pulley is 3 in. If the mass falls at a constant velocity of 5 ft/s, determine the power transmitted to the pulley, in hp, and the rotational speed of the shaft, in revolutions per minute (RPM). The acceleration of gravity is 32.2 ft/ s2. KNOWN: An object attached to a rope wound around a pulley falls at a constant velocity. FIND: Find the power transmitted to the pulley and the rotational speed. SCHEMATIC AND GIVEN DATA: ENGINEERING MODEL: (1) The object falls at a constant speed. (2) The acceleration of gravity is constant. ANALYSIS: The power is obtained using Eq. 2.13 ?ฬ = ? โ ? = (??)V ft = (5 lb) (32.2 s2 ft 1 lbf s 32.2 lbโftโ? 2 ) (5 ) | | = 25 ftโlb/s Converting to horsepower lbf ?ฬ = (25 ft โ ) | s 1 hp | = 0.0455 hp 550 ftโlbfโs The rotational speed of the pulley is related to the velocity of the object and the radius by V = Rฯ. Thus ?= V R 5 ftโs =( โ )| 3 12 ft 1 rev 60 s 2? 1 min || | = 191 rev/min Problem 2.32 Problem 2.32 (Continued) Problem 2.33 Problem 2.34 A fan forces air over a computer circuit board with a surface area of 70 cm2 to avoid overheating. The air temperature is 300 K while the circuit board surface temperature is 340 K. Using data from Table 2.1, determine the largest and smallest heat transfer rates, in W, that might be encountered for this forced convection. Problem 2.35 A 6-in. insulated frame wall of a house has an average thermal conductivity of 0.04 Btu/hโftโoR. The inner surface of the wall is at 68oF and the outer surface is at 40oF. Determine at steady state the rate of heat transfer through the wall, in Btu/h. If the wall is 20 ft x 10 ft, determine the total amount of energy transfer in 10 hours, in Btu. KNOWN: The inner and outer temperatures, thermal conductivity, and thickness of a frame wall are specified. The wall area is also known. FIND: Determine the rate of energy transfer by heat through the wall and the total amount of energy transferred in 10 hours. SCHEMATIC AND GIVEN DATA: 20 ft . Tout = 40oF 10 ft Tin = 68oF ENGINEERING MODEL: (1) The wall is a closed system at steady state. (2) Conduction follows Fourierโs law. (3) The temperature profile through the wall is linear. ฮบ = 0.04 Btu/hโftโoR . Assume a linear temperature profile. L = 6 in. ANALYSIS: For conduction through a plane wall with constant thermal conductivity and assuming a linear temperature profile Note: ฮT (oR) = ฮT (0F) ?ฬ = ?A(?in โ?out ) ? (0.04 = Btu )(20 ? 10 ft2 )(68โ40)o R hโftโo R (6โ12 ft) = 448 Btu/h The total amount of energy transfer by heat is ฬ = ?ฬโ? = (448 Btu/h)(10 h) = 4480 Btu Q = โซ ??? COMMENTS: Note that the temperature difference has the same value in oF and oR. Also, since the heat transfer rate is constant for steady state conduction, the integral reduces to ?ฬโ?. Problem 2.36 As shown in Fig. P2.36, an oven wall consists of a 0.635-cm-thick layer of steel (ฮบs = 15.1 W/mโK) and a layer of brick (ฮบb = 0.72 W/mโK). At steady state, a temperature decrease of 0.7oC occurs over the steel layer. The inner temperature of the steel layer is 300 K. If the temperature of the outer surface of the brick must be no greater than 40oC, determine the thickness of brick, in cm, that ensures this limit is met. What is the rate of conduction, in kW per m2 of wall surface area? KNOWN: Steady-state data are provided for a composite wall formed from a steel layer and a brick layer. FIND: Determine the minimum thickness of the brick layer to keep the outer surface temperature of the brick at or below a specified value. SCHEMATIC AND GIVEN DATA: Ti = 300 K Tm o ENGINEERING MODEL: (1) The wall is ฮT = -0.7 C the system at steady state. (2) The temperature varies linearly through each layer. ?ฬ ? โ?i = โ?s [ m A steel ?s ] and ฮบs = 15.1 W/mโK ฮบb = 0.72 W/mโK Ls = 0.635 cm ANALYSIS: Using Eq. 2.31 together with model assumption 2 ( ) To โค 40oC ?ฬ ( ) A brick ? โ?m = โ?b [ o ?b ] where Tm denotes the temperature at the steel-brick interface. At steady state, the rate of conduction to the interface through the steel must equal the rate of conduction from the interface through the brick: (?ฬโA)steel = (?ฬโA)brick . Thus ? โ?i โ?s [ m ?s ? โ?m ] = โ?b [ o ?b ] (300 โ 0.7) = 299.3 oC And solving for Lb we get ?b = ?b ?o โ?m [ ?s ?m โ?? ] ?s = โ 0.7 oC Problem 2.36 (Continued) ?b = ( 0.72 WโmโK 299.3โ ?o )[ ] (0.635 cm) 15.1 WโmโK 0.7 Since To โค 40 oC ?b โฅ ( 0.72 15.1 299.3โ 40 )[ 0.7 ] (0.635 cm) Lb โฅ 11.22 cm The rate of conduction is ?ฬ ( ) A steel ? โ?i = โ?s [ m ?s 299.3โ300 ] = โ(15.1 Wโm โ K) [ 0.635 cm ]| 100 cm 1m || 1 kW | = 1.665 kW/m2 103 W or ?ฬ ( ) A brick ? โ?m = โ?s [ o ?b 40โ299.3 100 cm 11.22 cm 1m ] = โ(0.72 Wโm โ K) [ The slight difference is due to round-off. ]| || 1 kW | = 1.664 kW/m2 103 W Problem 2.37 A composite plane wall consists of a 3-in.-thick layer of insulation (ฮบi = 0.029 Btu/hโftโoR) and a 0.75-in.-thick layer of siding (ฮบs = 0.058 Btu/hโftโoR). The inner temperature of the insulation is 67oF. The outer temperature of the siding is -8oF. Determine at steady state (a) the temperature at the interface of the two layers, in oF, and (b) the rate of heat transfer through the wall in Btu per ft2 of surface area. KNOWN: Energy transfer by conduction occurs through a composite wall consisting of two layers. FIND: Determine the temperature at the interface between the two layers and the rate of heat transfer per unit area through the wall. SCHEMATIC AND GIVEN DATA: T1 = 67 oF T3 = -8 oF ENGINEERING MODEL: (1) The wall is the system at steady state. (2) The temperature varies linearly through each layer. T2 = ? ฮบi = 0.029 Btu/hโftโoR ANALYSIS: With Eq. 2.17, and recognizing that at steady state the rates of energy conduction must be equal through each layer insulation ?ฬ Li = 3 in. ? ? โ?1 = โ?i [ 2 ?i ? โ?2 ] = โ?s [ 3 ?s ] (*) siding ฮบs = 0.058 Btu/hโftโoR Ls = 0.75 in. Solving for T2 ? ?i ? ?s ( i ?1 + s ?3 ) ?2 = (? โ )+(? โ i ?i ?i = ?i s ?s ) 0.029 BtuโhโftโR 12 in. | 3 in. Thus ?2 = 1 ft ?s | = 0.116 Btu/hโoR (0.116)(527)+(0.928)(452) (0.116)+(0.928) = ?s 0.058 BtuโhโftโR 12 .75 in. = 460.3 oR = 0.33 oF Thus, using Eq. (*) ?ฬ ? ?ฬ ? ? โ?1 = โ?i [ 2 ?i ] = (โ0.029 ? โ?2 = โ?s [ 3 ?2 ] = (โ0.058 Btu (0.33โ67)R )[ hโftโR Btu 3 12 )[ hโftโR ft ] = 7.73 Btu/ft2 (โ8โ0.33)R 0.75 12 ft ] = 7.73 Btu/ft2 | | = 0.928 Btu/hโoR 1 Problem 2.38 Complete the following exercise using heat transfer relations: (a) Referring to Fig. 2.12, determine the rate of conduction heat transfer, in W, for ฮบ = 0.07 W/mโK, A = 0.125 m2, T1 = 298 K, T2 = 273 K. (b) Referring to Fig. 2.14, determine the rate of convection heat transfer from the surface to the air, in W, for h = 10 W/m2, A = 0.125 m2, Tb = 305 K, Tf = 298 K. (a) Referring to Fig. 2.12, determine the rate of conduction heat transfer, in W, for ฮบ = 0.07 W/mโK, A = 0.125 m2, T1 = 298 K, T2 = 273 K. Using Eq. 2.31 and noting that the temperature varies linearly through the wall ?2 โ ?1 ?ฬ? = โ?A [ ] ? = โ (0.07 W mโK (273โ298)K ) (0.125 m2 ) [ (0.127 m) ] = 1.722 W (b) Referring to Fig. 2.14, determine the rate of convection heat transfer from the surface to the air, in W, for h = 10 W/m2, A = 0.125 m2, Tb = 305 K, Tf = 298 K. Using Eq. 2.34 ?ฬc = hA[Tb โ Tf] = (10 W m2 ) (0.125 m2 )[305 โ 298]K = 8.75 W Problem 2.39 At steady state, a spherical interplanetary electronics-laden probe having a diameter of 0.5 m transfers energy by radiation from its outer surface at a rate of 150 W. If the probe does not receive radiation from the sun of deep space, what is the surface temperature, in K? Let ฮต = 0.8. KNOWN: Steady-state operating data are provided for a spherical interplanetary probe. FIND: Determine the surface temperature of the sphere. SCHEMATIC AND GIVEN DATA: ?ฬe = 150 W Ts ENGINEERING MODEL: (1) The probe is at steady state. (2) The probe emits but does not receive radiation. D = 0.5 m ฮต = 0.8 ANALYSIS: In this case, Eq. 2.32 applies: ?ฬe = ??A? 4 , where A = ฯ d 2 = 0.7854 m2 The Stefan-Boltzmann constant is ฯ = 5.67 x 10-8 W/m2โK4. Thus Ts = [ 1/4 ?ฬe ??A ] =[ 150 W 1/4 ] 10โ8 W )(0.7854 m2 ) m2 K4 (0.8)(5.67 x = 254.7 K Problem 2.40 A body whose surface area is 0.5 m2, emissivity is 0.8, and temperature is 150oC is placed in a large, evacuated chamber whose walls are at 25oC. What is the rate at which radiation is emitted by the surface, in kW? What is the net rate at which radiation is exchanged between the body and the walls, in kW? KNOWN: Data are provided for a body placed in a large, evacuated chamber. FIND: Determine the rate at which radiation is emitted from the surface and the net rate at which radiation is exchanged between the body and the chamber. SCHEMATIC AND GIVEN DATA: ?ฬe Body at surface temperature Tb ENGINEERING MODEL: (1) The area of body is much less than that of the chamber walls. (2) The chamber is evacuated. A = 0.5 m2 ฮต = 0.8 Tb = 150oC = 423 K Ts = 25oC = 298 K Chamber surface at Ts ANALYSIS: The rate radiation is emitted from the surface is given by Eq. 2.32, where ฯ = 5.67 x 10-8 W/m2โK4 is the Stefan-Boltzmann constant. That is ?ฬe = ฮตฯATb4 = (0.8)(5.67 x 10-8 W/m2โK4 )(0.5 m2)(423 K)4 = 726 W With the assumptions in the Engineering Model, the net rate at which energy is exchanged by radiation between the body and the chamber walls is given by Eq. 2.33. Thus (?ฬe )net = ฮตฯA(Tb4 โ Ts4) = (0.8)(5.67 x 10-8 W/m2โK4 )(0.5 m2)[(423 K)4 โ (298 K)4] = 547 W Problem 2.41 The outer surface of the grill shown in Fig. P2.41 is at 47oC and the emissivity is 0.93. The heat transfer coefficient for convection between the hood and the surroundings at 27oC is 10 W/m2 โ K. Determine the rate of heat transfer between the grill hood and the surroundings by convection and radiation, in kW per m2 of surface area. Problem 2.42 Each line of the following table gives data for a process of a closed system. Each entry has the same energy units. Determine the missing entries. Process a b c d e Process a b c d e Q +50 E1 -20 W +20 -60 -40 +50 Q +50 +50 -40 -90 +50 E2 +50 +60 +40 +50 0 +150 W -20 +20 -60 -90 +150 E1 -20 +20 +40 +50 +20 Process a: W = Q – ฮE = +50 โ (+ 70) = -20 ฮE = E2 โ E1 E2 = ฮE + E1 = +70 + (-20) = +50 Process b: Q = ฮE + W = +30 + (+20) = +50 ฮE = E2 โ E1 E1 = E2 โ ฮE = +50 โ (+30) = +20 Process c: ฮE = E2 โ E1 = +60 โ (+40) = +20 Q = ฮE + W = +20 + (-60) = -40 Process d: W = Q โ ฮE = (-90) โ 0 = -90 ฮE = E2 โ E1 E2 = ฮE + E1 = 0 +50 = +50 Process e: ฮE = Q โ W = +50 โ (+150) = -100 E1 = E2 โ ฮE = (-80) โ (-100) = +20 E2 +50 +50 +60 +50 -80 -80 ฮE +70 +30 +20 0 -100 ฮE +70 +30 ๏ฝ Problem 2.43 Each line of the following table gives data for a process of a closed system. Each entry has the same energy units. Determine the missing entries. Process a b c d e Process a b c d e Q E1 +15 +7 +6 W โ 25 +27 โ4 +10 โ 10 E2 +30 +15 +10 โ8 +3 Q โ 10 +27 โ4 +10 +3 W โ 25 โ 12 +10 โ 10 +3 E1 +15 +7 +6 โ 10 โ8 Process a: ฮE = E2 โ E1 = + 30 โ (+15) = +15 Q = ฮE + W = +15 + (โ 25) = -10 Process b: W = ฮE โ Q = +15 โ (+27) = โ 12 ฮE = E2 โ E1 E2 = ฮE + E1 = +15 + (+7) = +22 Process c: ฮE = Q โ W = (โ 4) โ (+10) = โ 14 ฮE = E2 โ E1 E2 = ฮE + E1 = (โ 14) + (+6) = โ 8 Process d: Q = ฮE + W = (+20) + (โ 10) = +10 ฮE = E2 โ E1 E1 = E2 โ ฮE = (+10) โ (+20) = โ 10 Process e: W = Q โ ฮE = (+3) โ (0) = +3 ฮE = E2 โ E1 E1 = E2 โ ฮE = (โ 8) โ (0) = โ 8 E2 + 30 +22 โ8 +10 -8 ฮE +15 +15 โ 14 +20 0 ฮE ๏ฝ +20 0 Problem 2.44 A closed system of mass of 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per kg, an elevation decrease of 50 m, and an increase in velocity from 15 m/s to 30 m/s. The specific internal energy decreases by 5 kJ/kg and the acceleration of gravity is constant at 9.7 m/s2. Determine the heat transfer for the process, in kJ. KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity. FIND: Determine the heat transfer for the process. V1 = 15 m/s SCHEMATIC AND GIVEN DATA: ฮu = -5 kJ/kg W/m = + 0.147 kJ/kg 50 m V2 = 30 m/s ENGINEERING MODEL: (1) The system is a closed system. (2) The acceleration of gravity is constant. z m = 10 kg g = 9.7 m/s2 ANALYSIS: ฮU + ฮPE + ฮKE = Q – W โ Q = ฮU + ฮPE + ฮKE – W W = m [W/m] = 10 kg [-0.147 kJ/kg] = -1.47 kJ ฮU = mฮu = 10 kg [- 5 kJ/kg] = -50 kJ ฮKE = ? 2 (V22 โ V12 ) = 10 kg 2 [(30 m 2 m 2 s s ) โ (15 ฮPE = mg(z2 โ z1) = (10 kg) (9.7 m/s2)(-50 m)| 1N 1 kJ | | | = +3.38 kJ 2 1 kgโmโs 103 Nโm ) ]| 1N 1 kgโmโs2 Q = (-50) + (-4.85) + (3.38) โ (-1.47) = -50 kJ (out) || 1 kJ 103 Nโm | = – 4.85 kJ Problem 2.45 Problem 2.46 Problem 2.47 An electric motor operating at steady state draws a current of 20 amp at a voltage of 110 V. The output shaft rotates at a constant speed of 2000 RPM and exerts a torque of 9.07 Nโm. Determine (a) the magnitude of the power input, in W. (b) the output power, in W. (c) the cost of 24 hours of operation if electricity is valued at \$0.09 per kWโh. KNOWN: An electric motor operates at steady state. The input current and voltage are known and the rotational speed and torque of the exit shaft are specified. The cost of electricity is also specified. FIND: Determine (a) the magnitude of the power input, (b) the output power, and (c) the cost of 24 hours of operation. SCHEMCATIC AND GIVEN DATA: 20 amps T = 9.07 Nโm 110 V 2000 RPM ENGINEERING MODEL: (1) The motor operates at steady state. ANALYSIS: (a) The magnitude of the input power is |?ฬelec | = (20 amp) (110 V) | 1W || 1 voltโamp 1 kW 103 W | = 2.2 kW (input) (b) The shaft power is |?ฬshaft | = T x ฯ = (9.07 Nโm)(2000 rev min ) (2ฯ 1 rev 1 min 1 kJ 60 s 103 Nโm )| || || 1 kW | = 1.90 kW (out) 1 kJ/s (c) For steady operation (constant with time) the total amount of energy transferred in by electricity is Welec = โซ|?ฬelec |?? = |?ฬelec | ฮt = (2.2 Kw)(24 h) = 52.8 kWโh Cost = (52.8 kWโh)(\$0.09/kWโh) = \$4.75 (per day) COMMENT: Note, for steady state operation the inputs and outputs are constant with time. Problem 2.48 An electric motor draws a current of 10 amp with a voltage of 110 V, as shown in Fig. P2.48. The output shaft develops a torque of 9.7 Nโm and a rotational speed of 1000 RPM. For operation at steady state, determine for the motor (a) the electric power required, in kW. (b) the power developed by the output shaft, in kW. (c) the average surface temperature, Ts, in oC, if heat transfer occurs by convection to the surroundings at Tf = 21oC. KNOWN: Operating data are provided for an electric motor at steady state. FIND: Determine (a) the electric power required, (b) the power developed by the output shaft, and (c) average the surface temperature. Tf = 21oC Ts T = 9.7 Nโm 1000 RPM 10 amp ?ฬ = hA(?f โ ?s ) hA = 3.9 W/K 110 V ENGINEERING MODEL: (1) The motor is the closed system. (2) The system is at steady state. ANALYSIS: (a) Using Eq. 2.21 1 Wโamp 1 kW ?ฬelectric = – (voltage) (current) = – (110 V)(10 amp)| | | 3 | = -1.1 kW (in) 1V 10 W (b) Using Eq. 2.20 ?ฬshaft = (torque) (angular velocity) = (9.7 N โ m) (1000 rev 2ฯ rad 1 min rev 60 s )| min || || 1 kW | = 1.016 kW (out) 103 Nโmโs (c) To determine the surface temperature, first find the rate of energy transfer by heat using the energy balance 0 ?? = ?ฬ โ ?ฬ = ?ฬ โ (?ฬelectric + ?ฬshaft ) ?? ?ฬ = (?ฬelectric + ?ฬshaft ) = (-1.1 kW) + (1.016 kW) = -0.084 kW The surface temperature of the motor is 103 W ?s = (?ฬโhA) + ?? = (-0.084 kW)/(3.9 W/K)| | + 294 K 1 kW = 315.5 K = 42.5 oC Problem 2.49 A gas is contained is a vertical piston-cylinder assembly by a piston with a face area of 40 in.2 and a weight of 100 lbf. The atmosphere exerts a pressure of 14.7 lbf/in.2 on top of the piston. A paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the piston increases. The change in internal energy of the gas is 2.12 Btu. The piston and cylinder are poor thermal conductors, and friction between them can be neglected. Determine the elevation increase of the piston, in ft. KNOWN: A rotating shaft transfers a specified amount of energy to a gas contained in a vertical piston-cylinder assembly and the piton elevation increases. Data are provided for the piston and the change in internal energy of the gas. FIND: Determine the elevation increase of the piston. SCHEMATIC AND GIVEN DATA: State 1 patm = 14.7 lbf/in.2 State 2 ฮzpist Apist = 40 in.2 Fgrav, pist = 100 lbf Gas Gas ฮUgas = 2.12 Btu Q=0 Wpw = โ 3 Btu Process ENGINEERING MODEL: (1) The gas and the piston are the closed system. (2) Energy transfer by heat is negligible; Q = 0. (3) Kinetic energy effects are negligible. (4) The potential energy change of the gas is negligible, but the potential energy change of the piston in considered. (5) The change of internal energy of the piston is negligible. (6) Friction between the piton and the cylinder wall is neglected. ANALYSIS: The change in elevation of the piston is related to its change in potential energy by ฮPEpist = (mpist g) ฮzpist = (Fgrav, piston) ฮzpist (*) To evaluate the change in internal energy of the piston, we apply the energy balance to the system consisting of the gas and the piston: ฮEgas + ฮEpist = Q โ W Problem 2.49 (Continued) For the piston: ฮEpist = ฮUpist + ฮKEpist + ฮPEpist For the gas: ฮEgas = ฮUgas + ฮKEgas + ฮPEgas There are two mechanisms for mechanical work for the system chosen: the paddle wheel work and the work of the piston against the atmospheric pressure: W = Wpw + Watm Since the atmospheric pressure is constant: Watm = patmApist ฮzpist Combining these results ฮUgas + (Fgrav, piston) ฮzpist = Q โ (Wpw + patmApist ฮzpist) Solving for ฮzpist ฮzpist = โ โ?gas โ?pw Fgrav,piston+patm Apist Inserting values and converting units ฮzpist = โ (2.12 Btu)โ(โ3 Btu) 778 Btu | | = 0.995 ft (100 lbf)+(40 in.2 )(14.7 lbfโin.2 ) 1 ftโlbf COMMENTS: Note that the gas and the piston can be considered individually as closed systems. With this approach, the work done by the gas on the piston must be considered, which involves evaluating the change in volume of the gas and the gas pressure. The approaches are algebraically equivalent, but the approach chosen leads to a simpler analysis. Problem 2.50 A gas undergoes a process in a piston-cylinder assembly during which the pressure-specific volume relation is pv1.2 = constant. The mass of the gas is 0.4 lb and the following data are known: p1 = 160 lbf/in.2, V1 = 1 ft3, and p2 = 390 lbf/in.2 During the process, heat transfer from the gas is 2.1 Btu. Kinetic and potential energy effects are negligible. Determine the change in specific internal energy of the gas, in Btu/lb. KNOWN: A gas is compressed in a piston-cylinder assembly. The pressure-specific volume relation is specified. p FIND: Determine the change in specific internal energy. 2 . pv1.2 = constant SCHEMATIC AND GIVEN DATA: Q = -2.1 Btu gas m = 0.4 lb . p1 = 160 lbf/in.2 V1 = 1 ft3 p2 = 390 lbf/in.2 W 1 v ENGINEERING MODEL: (1) The gas is a closed system. (2) The process follows pv1.2 = constant. (3) Kinetic and potential energy effects are negligible. ANALYSIS: The change in specific internal energy will be found from an energy balance. First, determine the work. Since volume change is the only work mode, Eq. 2.17 applies: ? ?2 ????? 1 ? 1.2 W = โซ? 2 ??? = = โซ? 1 ?? = (?2 ?2 โ?1 ?1 ) 1โ1.2 Evaluating V2 ?1 1โ 1.2 V2 = ( ) ?2 1โ 1.2 160 lbfโin.2 ?1 = ( ) (1 ft 3 ) = 0.4759 ft3 390 lbfโin.2 Thus (390 lbfโin.2 )(0.4759 ft3 )โ(160)(1) 144 in.2 1โ1.2 1 ft2 W=[ ]| || 1 Btu | = -23.69 Btu (in) 778 ftโlbf Now, writing the energy balance: โ?? + โ?? + โ? = ? โ ? With ฮU = mฮu 1 ฮu = ?โ? ? = (โ2.1 Btu)โ (โ23.69 Btu) 0.4 lb = 54.0 Btu 1. The amount of energy transfer in by work exceeds the amount of energy transfer out by heat, resulting in a net increase in internal energy. Problem 2.51 Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle wheel that transfers energy to the CO at a constant rate of 14 W for 1 h. During the process, the specific internal energy of the carbon monoxide increases by 10 kJ/kg. If no overall changes in kinetic or potential energy occur, determine (a) the specific volume at the final state, in m3/kg. (b)the energy transfer by work, in kJ. (c) the energy transfer by heat, in kJ, and the direction of the heat transfer. Problem 2.52 Steam in a piston-cylinder assembly undergoes a polytropic process. Data for the initial and final states are given in the accompanying table. Kinetic and potential energy effects are negligible. For the process, determine the work and heat transfer, each in Btu per lb of steam. p (lbf/in.2) 100 40 State 1 2 v (ft3/lb) 4.934 11.04 u (Btu/lb) 1136.2 1124.2 KNOWN: Steam undergoes a polytropic process in a piston-cylinder assembly. Data are known at the initial and final states. FIND: Determine the work and heat transfer, each per unit mass of steam. p (lbf/in.2) 100 40 State 1 2 Steam v (ft3/lb) 4.934 11.04 p 1 . pvn = constant u (Btu/lb) 1136.2 1124.2 . 2 v ENGINEERING MODEL: (1) The steam is a closed system. (2) The process is polytropic, and volume change is the only work mode. (3) Kinetic and potential energy effects are negligible. ANALYSIS: Since the process is polytropic, Eq 2.17 applies for the work: ?2 ????? ? W/m = โซ? 2 ??? = โซ? 1 ?? 1 ?? = (?2 ?2 โ?1 ?1 ) 1โ? The pressures and specific volumes are known at each state, but n is unknown. To find n, pvn = constant , as follows: ?1 ?1? = ?2 ?2? โ ?1 ?2 ? ? = ( 2) ?1 โ ?= ln(?1 โ?2 ) ln(100โ40) = = 1.1377 ln(?2 โ?1 ) ln(11.04โ4.934) Thus W/m = (40 lbfโin.2 )(11.04 ft3 โlb)โ(100)(4.934) 144 in.2 1โ1.1377 | The heat transfer is obtained using the energy balance. 1 ft2 || 1 Btu | = 69.63 Btu/lb (out) 778 ftโlbf Problem 2.52 (Continued) ฮU + ฮKE + ฮPE = Q โ W โ Q = ฮU + W With ฮU = m ฮu = m(u2 โ u1) Q/m = (u2 โ u1) + (W/m) = (1124.2 โ 1136.2) Btu/lb + (69.63 Btu/lb) = 57.63 Btu/lb (in) Problem 2.53 Air expands adiabatically in a piston-cylinder assembly from an initial state where p1 = 100 lbf/in.2, v1 = 3.704 ft3/lb, and T1 = 1000 oR, to a final state where p2 = 50 lbf/in.2 The process is polytropic with n = 1.4. The change in specific internal energy, in Btu/lb, can be expressed in terms of temperature change as ฮu = (0.171)(T2 โ T1). Determine the final temperature, in oR. Kinetic and potential energy effects can be neglected. KNOWN: Air undergoes a polytropic process with known n in a piston-cylinder assembly. Data are known at the initial and final states, and the change in specific internal energy is expressed as a function of temperature change. FIND: Determine the final temperature. p 1 SCHEMATIC AND GIVEN DATA: p1 = 100 lbf/in.2 v1 = 3.704 ft3/lb Air T1 = 1000 oR p2 = 50 lbf/in.2 ฮu = (0.171Btu/lbโoR)(T2 โ T1) Q=0 . pv1.4 = constant .2 ENGINEERING MODEL: (1) The air is a closed system. (2) The process is polytropic with n = 1.4 and volume change is the only work mode. (3) The process is adiabatic: Q = 0. (4) Kinetic and potential energy effects are negligible. v ANALYSIS: To find the final temperature, we will use the energy balance with the given expression for change in specific internal energy as a function of temperature change. First, determine the work using Eq. 2.17 ? ? ????? 1 ? 1.4 W/m = โซ? 2 ??? = โซ? 2 1 ?? = (?2 ?2 โ?1 ?1 ) 1โ1.4 For the polytropic proess, ?1 ?11.4 = ?2 ?21.4 . Thus 1 ? 1.4 ?2 = ( 1 ) ?1 = ?2 1 100 lbfโin.2 1.4 ( ) (3.704 ft 3 /lb) = 6.077 ft3/lb 50 lbfโin.2 So, the work is W/m = (50 lbfโin.2 )(6.077 ft3 โlb)โ(100 )(3.704) 1โ1.4 | 144 in.2 1 ft2 || 1 Btu | = 30.794 Btu/lb 778 ftโlbf The energy balance is: ฮU + ฮKE + ฮPE = Q โ W. With ฮU = m(u2 โ u1) Problem 2.53 (Continued) (u2 โ u1) = โ W/m Inserting values (0.171 Btu/lbโoR)(T2 โ 1000 oR) = โ (30.794 Btu/lb) Solving; T2 = (-30.794 Btu/lb)/(0.171 Btu/lbโoR) + 1000 = 819.9 oR Problem 2.54 Problem 2.55 Process 1-2 2-3 3-4 4-1 ฮE -1200 400 Q 0 800 -200 W -200 400 Problem 2.56 A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2: Constant volume V1 = 2 m3, p1 = 1 bar, to p2 = 3 bar, U2 โ U1 = 400 kJ. Process 2-3: Constant pressure compression to V3 = 1 m3. Process 3-1: Adiabatic expansion, with W31 = 150 kJ. There are no significant changes in kinetic or potential energy. Determine the net work for the cycle, in kJ, and the heat transfers for Processes 1-2 and 2-3, in kJ. Is this a power cycle or refrigeration cycle? Explain. KNOWN: Data are provided for a gas undergoing a thermodynamic cycle consisting of three processes. FIND: Determine the net work for the cycle and the heat transfers for processes 1-2 and 2-3. Explain whether it is a power cycle or a refrigeration cycle. SCHEMATIC AND GIVEN DATA: p .3 3 Gas ENGINEERING MODEL: (1) The gas is a closed system. (2) Kinetic and potential energy effects are neglected. (3) Volume change is the only work mode. (4) Process 3-1 is adiabatic, so Q31 = 0. .2 .1 1 V 1 2 ANALYSIS: Beginning with Process 1-2, the volume is constant, so W12 = 0. The energy balance reduces to ฮU12 = Q12 โ W12. Thus Q12 = ฮU12 = 400 kJ (in) Process 2-3 is at constant pressure. So, with Eq. 2.17 105 N/m2 ? 1 kJ W23 = โซ? 3 ??? = p2(V3 โ V2) = (3 bar)(1 โ 2)m3| | | 3 | = โ 300 kJ (in) 1 bar 10 Nโm 2 The energy balance becomes ฮU23 = Q23 โ W23. So Q23 = ฮU23 + W23 To get ฮU23, letโs consider that for a cycle, ฮUcycle = 0. So ฮU12 + ฮU23 + ฮU31 = 0 โ ฮU23 = โ ฮU12 โ ฮU31 Problem 2.56 (Continued) Now, for Process 3-1: Q31 = 0, so ฮU31 = โ W31 = โ 150 kJ And ฮU23 = โ ฮU12 โ ฮU31 = โ (400 kJ) โ (โ 150 kJ) = โ 250 kJ Finally Q23 = ฮU23 + W23 = (โ 250 k J) + (- 300 kJ) = โ 550 kJ The net work for the cycle is Wcycle = W12 + W23 + W31 = 0 + (โ 300 k J) + (150 kJ) = โ 150 kJ (in) From the p-V diagram, we see that the cycle is executed in a counter clockwise fashion. And, the net work is in. So, the cycle is a refrigeration cycle. Problem 2.57 A gas undergoes a cycle in a piston-cylinder assembly consisting of the following three processes: Process 1-2: Constant pressure, p = 1.4 bar, V1 = 0.028 m3, W12 = 10.5 kJ Process 2-3: Compression with pV = constant, U3 = U2 Process 3-1: Constant volume, U1 โ U3 = -26.4 kJ There are no significant changes in kinetic or potential energy. (a) Sketch the cycle on a p-V diagram. (b) Calculate the net work for the cycle, in kJ. (c) Calculate the heat transfer for process 1-2, in kJ KNOWN: A gas undergoes a cycle consisting of three processes. FIND: Sketch the cycle on a p-V diagram and determine the net work for the cycle and the heat transfer for process 1-2. SCHEMATIC AND GIVEN DATA: Process 1-2: Constant pressure, p = 1.4 bar, V1 = 0.028 m3, W12 = 10.5 kJ Process 2-3: Compression with pV = constant, U3 = U2 Process 3-1: Constant volume, U1 โ U3 = -26.4 kJ Gas ENGINEERING MODEL: (1) The gas is a closed system. (2) Kinetic and potential energy effects are negligible. (3) The compression from state 2 to 3 is a polytropic process. ANALYSIS: (a) Since W12 > 0, the process is an expansion. Thus p 3 1 . . . 2 V Problem 2.57 (Continued) 0 (b) The net work for the cycle is Wcycle = W12 +W23 + W31. W12 = 10.5 kJ, so we need W23. ? ? ????? 2 2 W23 = โซ? 3 ??? = โซ? 3 ? ? ? ?2 ?2 ?? = (?2 ?2 )ln ( 3) = (?2 ?2 )ln ( 1) (*) where V3 = V1 has been incorporated. But, we still need to evaluate V2. For Process 1-2 at constant pressure ? W12 = โซ? 2 ??? = ?(?2 โ ?1 ) 1 or V2 = ?12 ? + ?1 = (10.5 kJ) 103 Nโm (1.4 bar) | 1 kJ || 1 bar | + 0.028 m3 = 0.103 m3 105 Nโm2 Thus, with Eq. (*) 0.028 W23 = (1.4 bar)(0.103 m3 )ln ( )| 0.103 105 Nโm2 1 kJ || 3 | = -18.78 kJ 1 bar 10 Nโm Thus Wcycle = 10.5 kJ + (-18.78 kJ) + 0 = -8.28 kJ 0 0 (c) To get Q12, we apply the energy balance to process 1-2: ฮKE + ฮPE + (U2 โ U1) = Q12 โ W12 With U2 = U3, Q12 = (U3 โ U1) + W12 = (+26.4 kJ) + (10.5 kJ) = 36.9 kJ Problem 2.58 The net work of a power cycle operating as in Fig. 2.17a is 10,000 kJ, and the thermal efficiency is 0.4. Determine the heat transfers Qin and Qout, each in kJ. ?= ?cycle ?in โ Qin = ?cycle ? Qin = (10,000 kJ) / (0.4) = 25,000 kJ Wcycle = 10,000 kJ ฮท = 0.4 Wcycle = Qcycle = Qin – Qout Thus Qout = Qin โ Wcycle = 25,000 โ 10,000 = 15,000 kJ Problem 2.59 For a power cycle operating as shown in Fig. 2.17a, the energy transfer by heat into the cycle, Qin, is 500 MJ. What is the net work developed, in MJ, if the cycle thermal efficiency is 30%? What is the value of Qout, in MJ? ฮท= ?cycle ?in Wcycle = ฮทQin = (0.3)(500 MJ) = 150 MJ Qin = 500 MJ Wcycle = Qcycle = Qin – Qout n Thus ฮท = 30% Qout = Qin โ wcycle = 500 MJ โ 150 MJ = 350 MJ Problem 2.60 For a power cycle operating as in fig. 2.17a, Qin = 17 x 106 Btu and Qout = 12 x 106 Btu. Determine Wcycle, in Btu, and ฮท. Wcycle = Qcycle = Qin โ Qout = (17 x 106) โ (12 x 106) = 5 x 106 Btu Qin = 17 x 106 Btu ฮท= ?cycle ?in = 5 x 106 Btu 17 x 106 Btu = 0.294 (29.4%) Alternatively Qout = 12 x 106 Btu ฮท=1โ ?out ?in =1โ 12 x 106 17 x 106 = 0.294 Problem 2.61 A system undergoing a power cycle develops a steady power output of 0.3 kW while receiving energy input by heat transfer of 2400 Btu/h. Determine the thermal efficiency and the total amount of energy developed by work, in kW โ h, for one full year of operation. The thermal efficiency, based on constant rates of energy transfer, is ฮท = ?ฬcycle /?ฬin ?ฬin = 2400 Btu/h ฮท = [(0.3 kW)/(2400 Btu/h)]| 3.413 Btu/h 1W || 103 W 1 kW = 0.427 (42.7 %) ?ฬcycle = 0.3 kW ?ฬout For one year of steady operation W = โซ ?ฬcycle ?? = ?ฬcycle ฮt = (0.3 k W) (365 days) (24 h/day) = 2628 kW โ h | Problem 2.62 ) Problem 2.63 A concentrating solar collector system, as shown in Fig. P2.63, provides energy by heat transfer to a power cycle at a rate of 2 MW. The cycle thermal efficiency is 36%. Determine the power developed by the cycle, in MW. What is the work output, in MWโh, for 4380 hours of steadystate operation? If the work is valued at \$0.08/kWโh, what is the total dollar value of the work output? Power Cycle ?ฬin = 2 MW Atmosphere ?ฬout ?ฬcycle The power developed is ?ฬcycle = ฮท?ฬin = (0.36) (2 MW) = 0.72 MW For 4380 hours of steady-state operation Wcycle = ?ฬcycle ฮt = (0.72 MW)(4380 h) = 3153.6 MWโh The total dollar value is \$ Value = (3153.6 MWโh)(\$0.08/kWโh)| 103 kW 1 MW | = \$252,300 Problem 2.64 Problem 2.65 Problem 2.66 Problem 2.67 Problem 2.68 For a refrigerator with automatic defrost and a top-mounted freezer, the electric power required is approximately 420 watts to operate. If the coefficient of performance is 2.9, determine the rate that energy is removed from its refrigerated space, in watts. Evaluating electricity at \$0.10/kW โ h, and assuming the units runs 60% of the time, estimate the cost of one monthโs operation, in \$. KNOWN: Data are provided for steady operation of a refrigerator-freezer. FIND: Determine the rate that energy is removed from the refrigerated space, and estimate the cost of one monthโs operation. SCHEMATIC AND GIVEN DATA: ?ฬ out Kitchen (Hot body) Refrigerator components ?ฬ in = ? ฮฒ = 2.9 ENGINEERING MODEL: (1) The refrigeration components execute a refrigeration cycle. (2) Data are given for steady operation. = 420 W Refrigerated space (Cold body) ?ฬin ANALYSIS: The coefficient of performance for steady operation is ฮฒ = ฬ . Thus ?cycle ?ฬin = ฮฒ ?ฬcycle = (2.9)(420 W) = 1218 W The total amount of electric energy input for one monthโs operation is determined for steady operation as Welec = ?ฬcycle ฮt Where ฮt = (0.6)(30 days/month)(24 h/day) = 432 h/month And Welec = (420 W)(432 h/month)| 1 kW | = 181.4 kW โ h/month 103 W Cost = (\$0.1/ kW โ h) (181.4 kW โ h/month) = \$18.14 / month Problem 2.69 A window-mounted room air conditioner removes energy by heat transfer from a room and rejects energy by heat transfer to the outside air. For steady operation, the air conditioner cycle requires a power input of 0.434 kW and has a coefficient of performance of 6.22. Determine the rate that energy is removed from the room air, in kW. If electricity is valued at \$0.1/kWโh, determine the cost of operation for 24 hours of operation. KNOWN: Steady-state operating data are provided for an air conditioner. FIND: Determine the rate energy is removed from the room and air the cost of 24 hours of operation. SCHEMATIC AND GIVEN DATA: Room air ?ฬin Outside air ?ฬout Air Conditioner ?ฬin = 0.434 kW ENGINEERING MODEL: (1) The system shown in the schematic undergoes a refrigeration cycle. (2) Energy transfers are positive in the directions of the arrows. (3) The cycle operates steadily for 24 hours. (4) Electricity is valued at \$0.1/kWโh. Refrigeration Cycle, ฮฒ = 6.22 Electric cost: \$0.1/kWโh ฮฒ = 6.22 ANALYSIS: Using Eq. 2.45 on a time rate basis ?ฬin ฮฒ= ฬ ?cycle โ ?ฬin = ??ฬcycle = (6.22)(0.434 kW) = 2.70 kW The total amount of electric energy input by work for 24 h of operation is Wcycle = ?ฬcycle โ? = (0.434 kW)(24 h) = 10.42 kWโh Thus, the total cost is Total cost = (10.42 kWโh)(\$0.1/kWโh) = \$1.04 (for 24 hours) Problem 2.70 Problem 2.71 A heat pump delivers energy by heat transfer to a dwelling at a rate of 11.7 kW. The coefficient of performance is 2.8. (a) Determine the power input to the cycle, in kW. (b) Evaluating electricity at \$0.1/kW โ h, determine the cost of electricity during the heating season when the heat pump operates for 1800 hours. KNOWN: Operating data are provided for a residential heat pump. FIND: Determine the power input to the cycle, and the seasonal operating cost. SCHEMATIC AND GIVEN DATA: Seasonal hours of operation = 1800 h ENGINEERING MODEL: (1) The closed system undergoes a heat pump cycle. (2) The cycle operates steadily for 1800 h during the heating season. (3) Electricity is valued at \$0.1/kW โ h. ?ฬ in ?ฬ cycle = ? ฮฒ = 2.8 ?ฬ out = 11.7 kW ANALYSIS: (a) The coefficient of performance for steady operation of the heat pump cycle is: ฮณ = ?ฬout /?ฬcycle . Thus ฬ ?out 11.7 kW ?ฬcycle = = = 4.179 kW ? 2.8 (b) Based upon modeling assumptions, the cost to operate the heat pump is estimated to be Cost = (4.179 kW) (1800 h/season) (\$0.1/kW โ h) = \$752.22/season

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