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ENGG 3260 (1)

Home-assignment-2-2011(ENGG3260) - answer.pdf
Home-assignment-2-2011(ENGG3260) - answer.pdf

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University of Guelph
ENGG 3260
Linda Gerber

ENGG 3260: Thermodynamics Home Assignment 2 (Chapter 2) 1. A person gets into an elevator at the lobby level of a hotel together with his 30-kg th suitcase, and gets out at the 10 floor 35 m above. Determine the amount of energy consumed by the motor of the elevator that is now stored in the suitcase. Answer: 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. 2  1kJ/kg  ∆ Esuitcase∆PE =mg z = (30kg)(9.81m/s )(35m)  2 2 =10.3kJ 1000m /s  The suitcase on 10 floor has 10.3 kJ more energy compared to an identical suitcase on the lobby level. c. Discussion: Noting that 1 kWh = 3600 kJ, the energy transferredto the suitcase is 10.3/3600 = 0.0029 kWh, which is very small. 2. 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 3500 kg/s steadily. Determine the power generation potential. Answer: a. Assumptions: A) The elevation of the reservoir remains constant. B) The mechanical energy of water at the turbine exit is negligible. 160 m b. The total mechanical energy water in a reservoir possesses is equivalent to the potential energy of water at the free surface, and it Turbine Generato can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and mgz for a given mass flow rate.  1kJ/kg  emech= pe = gz = (9.81m/s )(160m) =1.574kJ/kg 1000m /s 2  The power generation potential becomes W&max = Emech = memech= (3500kg/s)(1.574kJ/kg) 1kW =5509kW 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. 3. At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical 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 1.25 kg/m . 3 Answer: 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 = 1.25 kg/m . 3 10 60 m c. Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, 2 which is V /2 per unit mass, and mV 2 / 2for a given mass flow rate: 2 2 V (10 m/s)  1kJ/kg  emech = ke = 2 = 2  2 2 = 0.050 kJ/kg 1000m /s  πD 2 2 m = VA = V ρ = (1.25 kg/m )(10 m/s)(60 m) = 35,340 kg/s 4 4 W&max= E&mech= me mech= (35,340 kg/s)(0.050 kJ/kg) = 1770 kW 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 3 m/s at a rate of 500m /s at a location 90 m above the lake surface. Determine the local mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location. (figure) Answer: 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 mechanical energy of water at the turbine exit is negligible. b. Properties: We take the density of 3 water to be ρ = 1000 kg/m . c. Noting that the sum of the flow River 3 m/s energy and the potential energy is constant for a given fluid body, we can take the elevation of the 90 m entire river water to be the elevation of the free surface, and ignore the flow energy. Then the total mechanical energy of the river water per unit mass becomes V 2  2 (3m/s)2  1kJ/kg  emech = pe+ ke = gh + = (9.81m/s )(90m) +  2 2 = 0.887kJ/kg 2  2 1000m /s  The power generation potential of the river water is obtained by multiplying the total mechanical energy by the mass flow rate, m = ρV = (1000 kg/m )(500 m /s) = 500,000 kg/s & & W max = Emech = memech = (500,000 kg/s)(0.887 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, (2) the car engine, (3) the car wheels, (4) the road, and (5) the air surrounding the car. Answer: ① The car's radiator transfers heat from the hot engine cooling fluid to the cooler air. No work interaction occurs in the radiator. ② The hot engine transfers heat to cooling fluid and ambient air while delivering work to the transmission. ③ 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 mechanical power. What is this power in (1) N, m, and s units; and (2) kg, m, and s units? Answer: Using appropriate conversion factors, we obtain: (1) W& = (5W) 1J/s1N⋅m =5N⋅m/s 1W  1J  & 1J/s1N⋅m 1kg ⋅/s 2  2 3 (2) W = (5W)    =5kg⋅m /s 1W  1J  1N  7. How much work, in kJ, can a spring whose spring constant is 3 kN/cm produce after it has been compressed 3 cm from its unloaded length? Answer: Since there is no preload, F = kx. Substituting this into the work expression gives: 2 2 2 k 2 2 W = ∫Fds = ∫ kxdx = ∫ xdx = ( 2 x 1) F 1 1 1 2 300kN/m = [(0.03m) −0 2 ] x 2 = 0.135kN⋅m = (0.135kN⋅m)  1kJ = 0.135 kJ 1kN⋅m  8. A damaged 1200-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, (2) for constant velocity of 50 km/h on a 30°(from horizontal) uphill road, and (3) to accelerate on a level road from stop to 90 km/h in 12 s. Answer: 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 totalW +a g (1) Zero. & (2) W a 0 . & & ∆z o W totalW =gmg(z − z2)/ 1t = mg = mgV z = mgV sin 30 ∆t 50,000 m  1 kJ/kg  = (1200 kg)(9.81m/s )   (0.5) = 81.7 kW  3600 s 1000 m /s 2 & (3) W g 0 .  2  & & 1 2 2 1  90,000 m   1 kJ/kg  W totalW a = 2 m ( 2 −V1 ) /∆ = 2 (1200 kg) 3600 s  −0  2 2 /(12 s)= 31.3 kW   1000 m /s  9. A classroom that normally contains 40 people is to be air-conditioned with window air- conditioning units of 5-kW cooling capacity. A person at rest may be assumed to dissipate heat at a rate of about 360 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 21℃, determine the number of window air-conditioning units required. Answer: 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: Qcooling Q&lights &peopleQ &heat gain where Q& =10×100 W =1kW lights Qpeople 40×360 kJ / h = 4 kW Q&heat gain,000 kJ / h = 4.17 kW Room · Substituting, 15,000 kJ/h Q cool Q&cooling+4+4.17 = 9.17 kW 40 people The number of air-conditioning units required is: 9.17 kW =1.83→2 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 365 days a year. The storage room is actually used for an average of 3 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 $32 and it takes 1 hour to install it at a cost of $40. Answer: a. Assumption: The electrical energy consumed by the ballasts is negligible. b. The plant operates 12 hours a day, and thus currently the lights are on for the entire 12 hour period. The motion sensors installed will keep the lights on for 3 hours, and off for the remaining 9 hours every day. This corresponds to a total of 9×365 = 3285 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) = (24 lamps)(60 W/lamp )(3285 hours/year) = 4730 kWh/year 2) Cost Savings = (Energy Savings)(Unit cost of energy) = (4730 kWh/year)($0.08/kWh) = $378/year The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor. Implementation Cost = Material + Labor = $32 + $40 = $72 This gives a simple payback period of Implementation cost $72 Simple payback period= = = 0.19 year (2.3months) Annualcostsavings $378/ year The motion sensor will pay for itself in about 2 months. 11. A university campus has 200 classrooms and 400 faculty offices. The classrooms are required with 12 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 240 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 $0.082/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. Answer: a. The total electric power consumed by the lights in the classrooms and faculty offices is: E = (Power consumed per lamp)×(No.of lamps) = (200×12×110 W) = 264,000 = 264 kW lighting, classroom Elighting, officeser consumed per lamp)×(No.of lamps) = (400×6×110 W) = 264,000 = 264 kW Elighting, totlighting, classrolighting, offices 264 = 528kW Noting that the campus is open 240 days a year, the total number of unoccupied work hours per year is Unoccupied hou
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