ENGG 60: Thermodynamics Home Assignment (Chapter ) 1. A person gets into an elevator at the lobby level of a hotel together with his 0-kg suitcase, and gets out at the 10 th floor 5 m above. Determine the amount of energy consumed by the motor of the elevator that is now stored in the suitcase. a. Assumption: The vibrational effects in the elevator are negligible. b. The energy stored in the suitcase is stored in the form of potential energy, which is mgz. E suitcase PE mg z (0 kg)(9.81m/s 1kJ/kg )(5 m) 1000 m /s 10. kj The suitcase on 10 th floor has 10. kj more energy compared to an identical suitcase on the lobby level. c. Discussion: Noting that 1 kwh 600 kj, the energy transferred to the suitcase is 10./600 0.009 kwh, which is very small.. Electric power is to be generated by installing a hydraulic turbine-generator at a site 160 m below the free surface of a large water reservoir that can supply water at a rate of 500 kg/s steadily. Determine the power generation potential. a. Assumptions: A) The elevation of the reservoir remains constant. B) The anical energy of water at the turbine exit is negligible. b. The total anical energy water in a reservoir possesses is equivalent to the potential energy of water at the free surface, and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. e 1kJ/kg pe gz (9.81m/s )(160 m) 1000 m /s 1.574 kj/kg 160 m Turbine Generato
The power generation potential becomes W & max E& m& e 1kW (500 kg/s)(1.574 kj/kg) 5509 kw 1kJ/s The reservoir has the potential to generate 1766 kw of power. c. Discussion: This problem can also be solved by considering a point at the turbine inlet, and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir.. At a certain location, wind is blowing steadily at 10 m/s. Determine the anical energy of air per unit mass and the power generation potential of a wind turbine with 60-m-diameter blades at that location. Take the air density to be 1.5 kg/m. a. Assumption: The wind is blowing steadily at a constant uniform velocity. Wind b. Properties: The density of air is Wind given to be ρ 1.5 kg/m. c. Kinetic energy is the only form of anical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V / per unit mass, and m& V / for a given mass flow rate: e V (10 m/s) 1kJ/kg ke 1000 m /s 0.050 kj/kg πd π (60 m) m& ρva ρv (1.5 kg/m )(10 m/s) 5,40 kg/s 4 4 W & E& m& e (5,40 kg/s)(0.0 50 kj/kg) 1770 kw max 10 60 m 1770 kw of actual power can be generated by this wind turbine at the stated conditions.
d. Discussion: The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions. 4. Consider a river flowing toward a lake at an average velocity of m/s at a rate of 500m /s at a location 90 m above the lake surface. Determine the local anical energy of the river water per unit mass and the power generation potential of the entire river at that location. (figure) a. Assumptions: A) The elevation given is the elevation of the free surface of the river. B) The velocity given is the average velocity. C) The anical energy of water at the turbine exit is negligible. b. Properties: We take the density of water to be ρ 1000 kg/m. c. Noting that the sum of the flow energy and the potential energy is constant for a given fluid body, we can take the elevation of the entire river water to be the elevation of the free surface, and ignore the flow energy. Then the total anical energy of the river water per unit mass becomes e V pe + ke gh + (9.81m/s ( m/s) )(90 m) + 1kJ/kg 1000 m /s 0.887 kj/kg The power generation potential of the river water is obtained by multiplying the total anical energy by the mass flow rate, m & ρv & (1000 kg/m )(500 m /s) 500,000 kg/s River m/s 90 m W & max E& m& e (500,000 kg/s)(0.88 7 kj/kg) 444,000 kw 444 MW 444 MW of power can be generated from this river as it discharges into the lake if its power potential can be recovered completely. d. Discussion: Note that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored in the analysis. Also, the power output of an actual turbine will be less than 444 MW because of losses and inefficiencies.
5. Consider an automobile traveling at a constant speed along a road. Determine the direction of the heat and work interactions, taking the following as the system: (1) the car radiator, () the car engine, () the car wheels, (4) the road, and (5) the air surrounding the car. 1 The car's radiator transfers heat from the hot engine cooling fluid to the cooler air. No work interaction occurs in the radiator. The hot engine transfers heat to cooling fluid and ambient air while delivering work to the transmission. The warm tires transfer heat to the cooler air and to some degree to the cooler road while no work is produced. No work is produced since there is no motion of the forces acting at the interface between the tire and road. 4 There is minor amount of heat transfer between the tires and road. Presuming that the tires are hotter than the road, the heat transfer is from the tires to the road. There is no work exchange associated with the road since it cannot move. 5 Heat is being added to the atmospheric air by the hotter components of the car. Work is being done on the air as it passes over and through the car. 6. A small electrical motor produces 5 W of anical power. What is this power in (1) N, m, and s units; and () kg, m, and s units? Using appropriate conversion factors, we obtain: 1J/s 1 N m (1) W & (5 W) 5 N m/s 1 W 1J 1J/s 1 N m () 1kg m/s W & (5 W) 5 kg m /s 1W 1J 1 N 7. How much work, in kj, can a spring whose spring constant is kn/cm produce after it has been compressed cm from its unloaded length? Since there is no preload, F kx. Substituting this into the work expression gives: k W Fds kxdx k xdx ( x 1 1 [(0.0 m) 0 ] 1 x ) 00 kn/m 0.15 kn m 1kJ (0.15 kn m) 0.15 kj 1kN m 1 F x
8. A damaged 100-kg car is being towed by a truck. Neglecting the friction, air drag, and rolling resistance, determine the extra power required (1) for constant velocity on a level road, () for constant velocity of 50 km/h on a 0 (from horizontal) uphill road, and () to accelerate on a level road from stop to 90 km/h in 1 s. a. Assumption: Air drag, friction, and rolling resistance are negligible. b. The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, W & W& a + W& g total (1) Zero. () &W a 0. W& total & z o Wg mg( z z1) / t mg mgvz mgv sin 0 t 1 kj/kg 50,000 m (100 kg)(9.81m/s ) (0.5) 81.7 kw 600 s 1000 m /s () & W g 0. W& total 1 1 90,000 m 1 kj/kg W& a m( V V1 ) / t (100 kg) 1. kw 0 /(1 s) 600 s 1000 m /s 9. A classroom that normally contains 40 people is to be air-conditioned with window airconditioning units of 5-kW cooling capacity. A person at rest may be assumed to dissipate heat at a rate of about 60 kj/h. There are 10 light-bulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kj/h. If the room air is to be maintained at a constant temperature of 1, determine the number of window air-conditioning units required. a. Assumption: There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. b. The total cooling load of the room is determined from: Q& Q& + Q& + Q& cooling lights people heat gain where
Q& Q& Q& lights people heat gain Substituting, 10 100 W 1 kw 40 60 kj / h 4 kw 15,000 kj / h 4.17 kw 15,000 kj/h Room Q cool & 1+ 4 + 417. 9.17 kw Q cooling The number of air-conditioning units required is: 9.17 kw 1.8 units 5 kw/unit 10. The lighting needs of a storage room are being met by 6 fluorescent light fixtures, each fixture containing four lamps rated at 60 W each. All the lamps are on during operating hours of the facility, which are 6 AM to 6 PM 65 days a year. The storage room is actually used for an average of h a day. If the price of electricity is $0.08/kWh, determine the amount of energy and money that will be saved as a result of installing motion sensors. Also, determine the simple payback period if the purchase price of the sensor is $ and it takes 1 hour to install it at a cost of $40. a. Assumption: The electrical energy consumed by the ballasts is negligible. b. The plant operates 1 hours a day, and thus currently the lights are on for the entire 1 hour period. The motion sensors installed will keep the lights on for hours, and off for the remaining 9 hours every day. This corresponds to a total of 9 65 85 off hours per year. Disregarding the ballast factor, the annual energy and cost savings become: 1) Energy Savings (Number of lamps)(lamp wattage)(reduction of annual operating hours) (4 lamps)(60 W/lamp )(85 hours/year) 470 kwh/year ) Cost Savings (Energy Savings)(Unit cost of energy) (470 kwh/year)($0.08/kwh) $78/year The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor. Implementation Cost Material + Labor $ + $40 $7
This gives a simple payback period of Simple payback period Implementation cost Annual cost savings $7 $78 / year 0.19 year (. months) The motion sensor will pay for itself in about months. 11. A university campus has 00 classrooms and 400 faculty offices. The classrooms are required with 1 fluorescent tubes, each consuming 110 W, including the electricity used by the ballasts. The faculty offices, on average, have half as many tubes. The campus is open 40 days a year. The classrooms and faculty offices are not occupied an average of 4 h a day, but the lights are kept on. If the unit cost of electricity is $0.08/kWh, determine how much the campus will save a year it the lights in the classrooms and faculty offices are turned off during unoccupied periods. a. The total electric power consumed by the lights in the classrooms and faculty offices is: E& E& E& lighting, classroom lighting, offices lighting, total (Power consumed per lamp) (No. of lamps) (00 1 110 W) 64,000 64 kw (Power consumed per lamp) (No. of lamps) (400 6 110 W) 64,000 64 kw E& lighting, classroom + E& lighting, offices 64 + 64 58 kw Noting that the campus is open 40 days a year, the total number of unoccupied work hours per year is Unoccupied hours (4 hours/day)(40 days/year) 960 h/yr Then the amount of electrical energy consumed per year during unoccupied work period and its cost are Energy savings ( E & lighting, total )(Unoccupied hours) (58 kw)(960 h/yr) 506,880 kwh Cost savings (Energy savings)(unit cost of energy) (506,880 kwh/yr)($0.08/kwh) $41,564/yr b. Discussion: Note that simple conservation measures can result in significant energy and cost savings. 1. Consider a room that is initially at the outdoor temperature of 0. The room contains a 100-W lightbulb, a 110-W TV set, a 00-W refrigerator, and a 1000-W iron. Assuming no heat transfer through the walls, determine the rate of increase of the energy content of the room when all of these electric devices are on.
a. Assumptions: A) The room is well sealed, and heat loss from the room is negligible. B) All the appliances are kept on. b. Taking the room as the system, the rate form of the energy balance can be written as: E 1 & 44 & in Eout desystem / dt de dt E& room / in 144 Rate of net energy transfer by heat, work,and mass Rate of changein internal, kinetic, potential, etc. energies Since no energy is leaving the room in any form, and thus E & 0. Also, E & in E& + E& + E& + E& lights 100 + 110 + 00 + 1000 W 1410 W TV refrig Substituting, the rate of increase in the energy content of the room becomes de dt & room / E in 1410 W iron c. Discussion: Note that some appliances such as refrigerators and irons operate intermittently, switching on and off as controlled by a thermostat. Therefore, the rate of energy transfer to the room, in general, will be less. 1. A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m /s. Determine the minimum power that must be supplied to the fan. Take the density of air to be 1.18 kg/ m. a. Assumption: The fan operates steadily. b. Properties: The density of air is given to be ρ 1.18 kg/m. c. A fan transmits the anical energy of the shaft (shaft power) to anical energy of air (kinetic energy). For a control volume that encloses the fan, the energy balance can be written as: E 1 & 44 & 0 (steady) in Eout 0 E in Eout Rate of net energy transfer by heat, work,and mass W & m& ke m& sh, in Where: air out desystem / dt 1444 444 Rate of changein internal, kinetic, potential, etc.energies air V out m & air ρv & (1.18 kg/m )(9 m /s) 10.6 kg/s & & out Electricity ROOM - Lights - TV - Refrig
Substituting, the minimum power input required is determined to be W & sh, in Vout (8 m/s) 1J/kg m& air (10.6 kg/s) 40 J/s 40 W 1m /s d. Discussion: The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of anical shaft energy to kinetic energy of air. 14. An escalator in a shopping center is designed to move 0 people, 75 kg each, at a constant of 0.8 m/s at 45 slope. Determine the minimum power input needed to drive this escalator. What would your answer be if the escalator velocity were to be doubled? a. Assumptions: A) Air drag and friction are negligible. B) The average mass of each person is 75 kg. C) The escalator operates steadily, with no acceleration or breaking. D) The mass of escalator itself is negligible. b. At design conditions, the total mass moved by the escalator at any given time is Mass (0 persons)(75 kg/person) 50 kg The vertical component of escalator velocity is V vert V sin 45 (0.8 m/s)sin45 Under stated assumptions, the power supplied is used to increase the potential energy of people. Taking the people on elevator as the closed system, the energy balance in the rate form can be written as: E 1 & 44 & in Eout desystem / dt 144 Rate of net energy transfer by heat, work,and mass Rate of changein internal, kinetic, potential, etc. energies E E& sys in desys / dt t PE mg z W & t t in mgv vert That is, under stated assumptions, the power input to the escalator must be equal to the rate of increase of the potential energy of people. Substituting, the required power input becomes: W & 1 kj/kg (50 kg)(9.81m/s )(0.8 m/s)sin45 1.5 kj/s 1.5 kw 1000 m /s in mgv vert
When the escalator velocity is doubled to V 1.6 m/s, the power needed to drive the escalator becomes: W & 1 kj/kg (50 kg)(9.81m/s )(1.6 m/s)sin45 5.0 kj/s 5.0 kw 1000 m /s in mgv vert c. Discussion: Note that the power needed to drive an escalator is proportional to the escalator velocity. 15. Can the combined turbine-generator efficiency be greater than either the turbine efficiency or the generator efficiency? Explain. No, the combined pump-motor efficiency cannot be greater that either of the pump efficiency of the motor efficiency. Explain: This is because η pump -motor η pumpη motor, and both η pump and η motor are less than one, and a number gets smaller when multiplied by a number smaller than one. 16. A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is replaced by a high-efficiency 75-hp motor that has an efficiency of 95.4 percent. Determine the reduction in the heat gain of the room due to higher efficiency under fullload conditions. a. Assumptions: A) The motor and the equipment driven by the motor are in the same room. B) The motor operates at full load so that f load 1. b. The heat generated by a motor is due to its inefficiency, and the difference between the heat generated by two motors that deliver the same shaft power is simply the difference between the electric power drawn by the motors, W& W& in, electric, standard in, electric, efficient W& W& shaft shaft / η / η motor motor (75 746 W)/0.91 61,484 W (75 746 W)/0.954 58,648 W The reduction in heat generation becomes Q & reduction W& W& in, electric, standard in, electric, efficient 61,484 58,648 86 W 17. An exercise room has eight weight-lifting machines that have no motors and four treadmills each equipped with a.5-hp (shaft output) motor. The motors operate at an average load factor of 0.7, at which their efficiency is 0.77. During peak evening hours, all 1 pieces of exercising equipment are used continuously, and there are also two people doing light exercises while waiting in line for one piece of the equipment.
Assuming the average rate of heat dissipation from people in an exercise room is 55 W, determine the rate of heat gain of the exercise room from people and the equipment at peak load conditions. a. Assumption: The average rate of heat dissipated by people in an exercise room is 55 W. b. The 8 weight lifting machines do not have any motors, and thus they do not contribute to the internal heat gain directly. The usage factors of the motors of the treadmills are taken to be unity since they are used constantly during peak periods. Noting that 1 hp 746 W, the total heat generated by the motors is Q & motors ( No.of motors) W& motor f load f usage / ηmotor 4 (.5 746 W) 0.70 1.0/0.77 678 W The heat gain from 14 people is Q& ( No. of people ) Q& 14 ( 55 W) 750 W people person The total rate of heat gain of the exercise room during peak period becomes: Q & total Q& + Q& 678 + 750 14,1 W motors people 18. At a certain location, wind is blowing steadily at 7 m/s. Determine the anical energy of air per unit mass and the power generation potential of a wind turbine with 80-m-diameter blades at that location. Also determine the actual electric power generation assuming an overall efficiency of 0 percent. Take the air density to be 1.5 kg/ m. a. Assumptions: A) The wind is blowing steadily at a constant uniform velocity. Wind Wind B) The efficiency of the wind turbine is independent of the wind speed. b. Properties: The density of air is given to be ρ 1.5 kg/m. 7 m/s 80 m c. Kinetic energy is the only form of anical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V / per unit mass, and m& V / for a given mass flow rate: e V (7 m/s) ke 1kJ/kg 1000 m /s 0.045 kj/kg
πd π (80 m) m& ρva ρv (1.5 kg/m )(7 m/s) 4,98 kg/s 4 4 W & E& m& e (4,98 kg/s)(0.045 kj/kg) 1078 kw max The actual electric power generation is determined by multiplying the power generation potential by the efficiency, W & η W& (0.0)(1078 kw) kw elect wind turbine max kw of actual power can be generated by this wind turbine at the stated conditions. d. Discussion: The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions. 19. Water is pumped from a lake to a storage tank 0 m above at a rate of 70 L/s while consuming 0.4 kw of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump-motor unit and (b) the pressure difference between the inlet and the exit of the pump. (figure) a. Assumptions: A) The elevations of the tank and the lake remain constant. B) Frictional losses in the pipes are negligible. C) The changes in kinetic energy are negligible. D) The elevation difference across the pump is negligible. b. Properties: We take the density of water to be ρ 1000 kg/m. c. (a) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point. We also take the lake surface as the reference level (z 1 0), and thus the potential energy at points 1 and are pe 1 0 and pe gz. The flow energy at both points is zero since both 1 and are open to the atmosphere (P 1 P P atm ). Further, the kinetic energy at both points is zero (ke 1 ke 0) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point are 0 m 1 Pump Storage m & ρv & (1000 kg/m )(0.070 m /s) 70 kg/s
1kJ/kg pe gz (9.81m/s )(0 m) 1000 m /s 0.196 kj/kg The rate of increase of the anical energy of water becomes: E &, fluid m& ( e,out e,in ) m& ( pe 0) m& pe (70 kg/s)(0.196 kj/kg) 1.7 kw The overall efficiency of the combined pump-motor unit is determined from its definition, η E& W 1.7 kw 0.4 kw,fluid pump -motor & elect,in 0.67 or 67.% (b) Now we consider the pump. The change in the anical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful anical energy supplied by the pump, which is 1.7 kw: E&, fluid m& ( e,out e,in ) Solving for P and substituting, &E,fluid P V & 1.7 kj/s 1kPa m 0.070 m /s 1kJ P P1 m& V & P ρ 196 kpa The pump must boost the pressure of water by 196 kpa in order to raise its elevation by 0 m. d. Discussion: Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the anical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.