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Solution Manual Renewable Energy 2nd Edition by Mehmet Kanoglu & Yunus Cengel

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Solution Manual Renewable Energy 2nd Edition by Mehmet Kanoglu & Yunus Cengel

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  1. 1-1 Email: smtb98@gmail.com Telegram: https://t.me/solumanu Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 SOLUTIONS MANUAL Chapter 1 INTRODUCTION TO RENEWABLE ENERGY PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGraw Hill and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw Hill. Email: smtb98@gmail.com Telegram: https://t.me/solumanu Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 complete document is available on https://solumanu.com/ *** contact me if site not loaded

  2. Email: smtb98@gmail.com Telegram: https://t.me/solumanu Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 smtb98@gmail.com 1-2 smtb98@gmail.com Why Renewable Energy? 1-1 The combustion of fossil fuels produces the following undesirable emissions and adverse effects: • Carbon dioxide (CO2), causes global warming • Nitrogen oxides (NOx) and hydrocarbons (HC), cause smog • Carbon monoxide (CO), toxic • Sulfur dioxide (SO2), causes acid rain • Particulate matter (PM): causes adverse health effects 1-2 Carbon dioxide is not an air pollutant. CO2 is a greenhouse gas causing global warming. It is a natural product of a fossil fuel combustion. Other emissions, on the other hand, are harmful air pollutants. 1-3 The concern over the depletion of fossil fuels and pollutant and greenhouse emissions associated by their combustion can be tackled by essentially two methods: (i) Using renewable energy sources to replace fossil fuels. (ii) Implementing energy efficiency practices in all aspects of energy production, distribution, and consumption so that less fuel is used while obtaining the same useful output. 1-4Energy efficiency is to reduce energy use to the minimum level, but to do so without reducing the standard of living, the production quality, and the profitability. Energy efficiency is an expression for the most effective use of the energy resources, and it result in energy conservation. Energy efficiency can only reduce the fossil fuel use while renewable energy can directly replace fossil fuels. 1-5 Main renewable energy sources include solar, wind, hydropower, geothermal, and biomass. Ocean, wave, and tidal energies are also renewable sources but they are currently not economical and the technologies are still in the experimental and developmental stage. 1-6 An energy source is called renewable if it can be renewed and sustained without any depletion and any significant effect on the environment. Coal, oil, and natural gas are not renewable since they are depleted by use and they emit harmful pollutants and greenhouse gases. 1-7 Although solar energy is sufficient to meet the entire energy needs of the world, currently it is not economical to do so because of the low concentration of solar energy on earth and relativelyhigh capital cost of harnessing it. 1-8Solar and wind are the fastest growing renewables. Hydropower represents the greatest amount of electricity production among renewable. complete document is available on https://solumanu.com/ *** contact me if site not loaded

  3. 1-3 1-9Wind and hydro (water) energies are converted to electricity only while solar, biomass, and geothermal can be converted to both electricity and thermal energy. 1-10 We do not agree. The electricity used by the electric cars is generated somewhere else mostly by burning fuel and thus emitting pollution. Therefore, each time an electric car consumes 1 kWh of electricity, it bears the responsibility for the pollutions emitted as 1 kWh of electricity (plus the conversion and transmission losses) is generated elsewhere. 1-11 The electric cars can be claimed to be zero emission vehicles only when the electricity they consume is generated by emission-free renewable resources such as hydroelectric, solar, wind, and geothermal energy. 1-12 Most energy in the world is consumed by the _______ sector. (a) residential (b) commercial (c) industrial (d) transportation Answer: (c) industrial 1-13 The emission from fossil fuel combustion that is not an air pollutant is (a) CO (b) CO2 (c) NOx (d) SO2 Answer: (b) CO2 1-14 Which emission causes acid rain? (a) CO (b) CO2 (c) NOx (d) SO2 Answer: (d) SO2 1-15 Which source should not be considered as a main renewable energy source? (a) Wind (b) Hydro (c) Tidal (d) Biomass Answer: (c) Tidal 1-16 The fastest growing renewable energy sources in the world are (a) Wind and solar (b) Hydro and biomass (c) Solar and hydro (e) Geothermal and biomass Answer: (a) Wind and solar 1-17 Which renewable energy sources are only used for electricity generation? (a) Wind and solar (b) Hydro and solar (c) Solar and geothermal (e) Hydro and geothermal Answer: (d) Wind and hydro (e) service (e) PM (e) PM (e) Geothermal (d) Biomass and wave (d) Wind and hydro complete document is available on https://solumanu.com/ *** contact me if site not loaded

  4. 1-4 smtb98@gmail.com smtb98@gmail.com 1-18 Which renewable energy source should not be considered as the manifestation of solar energy in different forms? (a) Wind (b) Hydro (c) Wave (d) Biomass Answer: (e) Geothermal 1-19 Reducing energy use to the minimum level, but to do so without reducing the standard of living, the production quality, and the profitability is called (a) Energy renewability (b) Energy efficiency (e) Fuel conservation Answer: (b) Energy efficiency Dimensions and Units 1-20 The SI is a simple and logical system based on a decimal relationship between the various units. The English system, however, has no apparent systematic numerical base, and various units in this system are related to each other rather arbitrarily 1-21 In SI, the units of mass, length, time, and energy are the kilogram (kg), meter (m), second (s), Joule (J), and Watt (W), respectively. The respective units in the English system are the pound-mass (lbm), foot (ft), second (s), British thermal units (Btu), and Btu/h. 1-22 The unit kW (or kJ/s) is for the rate of energy, which is the amount of energy per unit time. On the other hand, kWh is a unit for the amount of energy. We show that 1 kWh = 3600 kJ, as follows: s 3600 h s     1-23 We show that 1 kJ = 1 kPam3, as follows: kN m kPa 2   1-24 We show that 1 kJ/kg = 1000 m2/s2, as follows: (e) Geothermal (c) Energy minimization (d) Fuel savings     kJ = =     kWh 1 1 3600 kJ h 1   3 3  = =  =   ( m ) kN m kJ m 2   J N m kg m 1 m = = = ( m ) 2 2 kg kg kg s s Therefore, 1 J/kg = 1 m2/s2 1 kJ/kg = 1000 m2/s2 complete document is available on https://solumanu.com/ *** contact me if site not loaded

  5. 1-5 1-25 Noting that 1 Quad = 11015 Btu, 1 Btu = 1.055 kJ, and 1 kWh = 3412 Btu, we express oil and renewable consumptions in Btu, GJ, and kWh, as follows: 15 1 10 Btu 1 Quad 17 1.89 10 Btu Oil Consumption (188.8 Quad) 15 1 10 Btu 1.055 kJ 1 Quad 1 GJ 11 1.99 10 GJ Oil Consumption (188.8 Quad) 6 1 Btu 1 10 kJ 15 1 10 Btu 1 Quad 1 kWh 3412 Btu 13 5.53 10 kWh Oil Consumption (188.8 Quad) 15 1 10 Btu 1 Quad 16 1.94 10 Btu Renewable Consumption (19.4 Quad) 15 1 10 Btu 1.055 kJ 1 Quad 1 GJ 10 2.05 10 GJ Renewable Consumption (19.4 Quad) 6 1 Btu 1 10 kJ 15 1 10 Btu 1 Quad 1 kWh 3412 Btu 12 5.69 10 kWh Renewable Consumption (19.4 Quad) EES (Engineering Equation Solver) SOLUTION EnergyConsumption_Oil=188.8 [Quad] EnergyConsumption_Oil_Btu=188.8E15 [Btu] EnergyConsumption_Oil_GJ=188.8E15 [Btu]*convert(Btu,GJ) EnergyConsumption_Oil_kWh=188.8E15 [Btu]*convert(Btu,kWh) EnergyConsumption_Renew=19.4 [Quad] EnergyConsumption_Renew_Btu=19.4E15 [Btu] EnergyConsumption_Renew_GJ=19.4E15 [Btu]*convert(Btu,GJ) EnergyConsumption_Renew_kWh=19.4E15 [Btu]*convert(Btu,kWh) complete document is available on https://solumanu.com/ *** contact me if site not loaded

  6. 1-6 smtb98@gmail.com 1-26 Noting that 1 Quad = 11015 Btu, 1 Btu = 1.055 kJ, and 1 toe = 41.868 GJ, we express total world delivered energy consumptions by fuel and by end-use sector, in toe, as follows: smtb98@gmail.com 15 1 10 Btu 1.055 kJ 1 Quad 1 GJ 1 toe Energy Consumption by fuel (523.9 Quad) 6 1 Btu 41.868 GJ 1 10 kJ 10 1.32 10 toe 15 1 10 Btu 1.055 kJ 1 Quad 1 GJ 1 toe Energy Consumption by sector (382.0 Quad) 6 1 Btu 41.868 GJ 1 10 kJ 9 9.63 10 toe EES SOLUTION Energy_fuel=495.2 [Quad] Energy_sector=359.0 [Quad] Energy_fuel_toe=Energy_fuel*convert(Quad, GJ)/41.868 Energy_sector_toe=Energy_sector*Convert(Quad, GJ)/41.868 "1 toe = 41.868 GJ" 1-27 Since 21.7 percent of total electricity is produced from natural gas-burning power plants, 9 10 kWh 1 TWh 12 Electricity produced (0.217)(23,332 TWh) 5.063 10 kWh The heat input by the natural gas consumption is 15 1 10 Btu 1 Quad 1 kWh 3412 Btu 13 Fuel energy consumed (45 Quad) 1.319 10 kWh The overall thermal efficiency is 12 Electricity produced Fuel energy consumed 5.063 10 kWh 1.319 10 kWh 38.4% Overall thermal efficiency 0.384 13

  7. 1-7 1-28 Noting that 1 TWh =1109 kWh, the amount of electricity produced in OECD countries is 9 1 10 kWh 1 tWh 13 Electricity produced (0.553)(19,028 tWh) 1.052 10 kWh Noting that the average thermal efficiency is 38%, the amount of energy consumed to produce this much electricity is 13 Electricity produced Thermal efficiency 1.052 10 kWh 0.38 13 Energy consumed 2.769 10 kWh Now, we express this in tWh, Quad, and toe units: 1 tWh 1 10 kWh 13 27,691 tWh Energy consumed (2.769 10 kWh) 9 3412 Btu 1 kWh 1 Quad 1 10 Btu 13 94.48 Quad Energy consumed (2.769 10 kWh) 15 3600 kJ 1 kWh 1 GJ 1 toe 9 13 2.381 10 toe Energy consumed (2.769 10 kWh) 6 41.868 GJ 1 10 kJ EES SOLUTION "Given" Electricity_world=19028E9 [kWh] "1 tWh=1E12 Wh = 1E9 kWh" Efficiency=0.38 f_OECD=0.553 "Solution" Electricity_OECD=f_OECD*Electricity_world EnergyConsumed=Electricity_OECD/Efficiency EnergyConsumed_tWh=EnergyConsumed/1E9 "tWh" EnergyConsumed_Quad=EnergyConsumed*convert(kWh, Quad) EnergyConsumed_toe=EnergyConsumed*Convert(kWh, GJ)/41.868 "1 toe = 41.868 GJ"

  8. 1-8 smtb98@gmail.com 1-29 The annual amount of electricity saved for 10 million households is Electricity saved (Number of homes)(Old electricity consumption (1 10 )(1800 1.35 10 kWh The annual operating hours of power plant is Annual hours = (0.9)(36524 h) = 7884 h The additional power that would be needed to meet the extra demand is smtb98@gmail.com New electricity consumption) 6 450)kWh 10 10 Electricity saved Annual hours 1.35 10 kWh (0.8)(7884 h) 6 2.14 10 kW Additional power f load which is equivalent to 2140 MW. EES SOLUTION "Given" Electricity_old=1800 [kWh] Electricity_new=450 [kWh] NumberOfHomes=10E6 f_load=0.8 f_time=0.9 "Solution" ElectricitySaved=Electricity_old-Electricity_new TotalElectricitySaved=ElectricitySaved*NumberOfHomes AnnualHours=f_time*365*24 [h] AdditionalPower=(TotalElectricitySaved/f_load)/AnnualHours 1-30 The natural gas price in $/kWh is       $ therm 1 Btu 1 3600 kJ = =     0.0512 $/kWh         Gas price (1.5 ) therm 100,000 Btu 1.055 kJ kWh 1 EES SOLUTION Price_gas=1.5 [$/herm]*convert($/therm,$/kWh)

  9. 1-9 1-31 The resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. Then the total amount of electric energy used in 3 hours becomes Total energy = (Energy per unit time)(Time interval) = (4 kW)(3 h) = 12 kWh Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ, Total energy = (12 kWh)(3600 kJ/kWh) = 43,200 kJ Note that kW is a unit for power whereas kWh is a unit for energy. EES SOLUTION W_dot_e=4 [kW] time=3 [h] Energy=W_dot_e*time Energy_kJ=Energy*Convert(kWh,kJ) 1-32 (a) We note that 1 toe = 41.868 GJ. Using these unity conversion ratios and others available at the end of the book, we express the wind power capacity in the world in various units as follows:   1  6 10 kW     8 = =  7.43 10 kW Wind power capacity (743 GW) GW 1       GJ/s 1 MJ 1 3600 s 9 = =  2.675 10 MJ/h             Wind power capacity (743 GW) − 3 GW 1 h 1  1 10 GJ       GJ/s 1 3600 s toe 1 = = 63,900 toe/h             Wind power capacity (743 GW) GW 1 h 1 41.868 GJ The quantity in large parentheses in these equations is a unity conversion ratio. (b) The percentage of offshore wind capacity with respect to total power capacity in the world 31.9 GW = = = 0.258% Percent offshore capacity . 0 ( 06 ) 002576 . 0 743 GW (c) What is the approximate total power capacity in the world is 743 GW = = 12,383 GW Global power capacity 0.06 EES SOLUTION WindCapacity=743 [GW] f_wind=0.06 OffshoreWindCapacity=31.9 [GW] WindCapacity_kW=WindCapacity*convert(GW, kW) WindCapacity_MJperh=WindCapacity*convert(GW, MJ/h) OffshoreFraction=(OffshoreWindCapacity/WindCapacity)*f_wind WorldPowerCapacity=WindCapacity/f_wind

  10. 1-10 But the actual generation is 4370 TWh or 4.37106 GWh. Then, the annual capacity factor of solar power installations become smtb98@gmail.com 1-33 (a)If the hydropower installations produced power 365 days a year and 24 hours a day at the rated power of 1.34 TW (1340 GW), the amount of power generation would be smtb98@gmail.com Annual generation = (1340 GW)(36524 h) = 1.173107 GWh 6  4.37 10 GWh 7 = = = 37.2% Capacity factor . 0 372  1.173 10 GWh That is, the solar power installations operate 37.2% of the time at the rated power. (b) Noting that 1 exajoules = 11018 J, the global amount of hydropower generation in exajoules (XJ) is         TJ/s 1 J 1 1 XJ 3600 s = =         15.73 XJ         Hydropower generation (4370 TWh) − 12 18 TW 1 h 1   1 10 TJ 1 10 J EES SOLUTION HydroCapacity=1.34 [TW]*1000 "GW" HydroGeneration=4370 [TWh]*1000 "GWh" AnnualHours=365*24 FullHydroGeneration=HydroCapacity*AnnualHours CapacityFactor=HydroGeneration/FullHydroGeneration 1-34 If the nuclear power plants produced power 365 days a year and 24 hours a day at the rated power of 393 GW, the amount of power generation would be Annual generation = (393 GW)(36524 h) = 3.443106 GWh But the actual generation is 2553 TWh or 2.553106 GWh. Then, the annual capacity factor of nuclear power installations become 6  2.553 10 GWh = = = 74.2% Capacity factor . 0 742 6  3.443 10 GWh That is, the nuclear power plants operate 74.2% of the time at the rated power. EES SOLUTION NuclearCapacity=393 [GW] NuclearGeneration=2553 [TWh]*1000 "GWh" AnnualHours=365*24 FullNuclearGeneration=NuclearCapacity*AnnualHours CapacityFactor=NuclearGeneration/FullNuclearGeneration

  11. 1-11 1-35 If the nuclear power plants produced power 365 days a year and 24 hours a day at the rated power of 393 GW, the amount of power generation would be Annual generation = (15.6 GW)(36524 h) = 136,700 GWh But the actual generation is 89,300 GWh. Then, the annual capacity factor of geothermal power plants become 89,300 GWh = = = 65.4% Capacity factor . 0 654 136,700 GWh That is, the geothermal power plants in the world operate 65.4% of the time at the rated power. EES SOLUTION GeothermalCapacity=15.6 [GW] GeothermalGeneration=89300 [GWh] AnnualHours=365*24 FullGeothermalGeneration=GeothermalCapacity*AnnualHours CapacityFactor=GeothermalGeneration/FullGeothermalGeneration 1-36 First we calculate maximum power generation of the solar power plant assuming nonstop operation 365 days a year and 24 hours a day at the rated power of 393 GW: Maximum power generation = (200 MW)(36524 h) = 1,752,000 MWh Using the capacity factor, the actual power generation becomes Actual power generation = (Capacity factor)(Maximum power generation) = (0.25)(1,752,000 MWh) = 438,000 MWh Repeating calculations for wind farm: Maximum power generation = (110 MW)(36524 h) = 963,600 MWh Actual power generation = (Capacity factor)(Maximum power generation) = (0.40)(963,600 MWh) = 385,440 MWh EES SOLUTION SolarCapacity=200 [MW] SolarCapacityFactor=0.25 WindCapacity=110 [MW] WindCapacityFactor=0.40 AnnualHours=365*24 MaxSolarGeneration=SolarCapacity*AnnualHours ActualSolarGeneration=SolarCapacityFactor*MaxSolarGeneration MaxWindGeneration=WindCapacity*AnnualHours ActualWindGeneration=WindCapacityFactor*MaxWindGeneration

  12. 1-12 1-38 Select the correct unit conversions. smtb98@gmail.com 1-37 Which of the following is a unit for the amount of energy? smtb98@gmail.com (a) kW (b) kJ/h (c) J/min (d) kWh (e) Btu/h Answer: (d) kWh I. 1 kJ = 1 kPam3 II. 1 kWh = 3600 kJ III. 1 kW = 1 kJ/min (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (b) I and II 1-39 Which of the following unit conversion is incorrect? (d) 1 J = 1 Nm (e) 1 GJ = 106 J (a) 1 J/kg = 1 m2/s2 (b) 1 kPa = 1 kN/m2 (c) 1 W = 1 J/s Answer: (e) 1 GJ = 106 J 1-40 A solar panel operates for a period of 10 hours at an average power rating of 1000 W. How much power is generated during this period? (a) 10 kWh (b) 10 kW (c) 10 kJ (d) 10,000 W (e) 100 W/h Answer: (a) 10 kWh Power generation = (1 kW)(10 h) = 10 kWh 1-41 A wind turbine with an average power rating of 50 kW generates 600 kWh power in a given 24-hour period. What is the capacity factor of this turbine during this period? (a) 0.1 (b) 0.25 (c) 0.5 (d) 0.6 Answer: (c) 0.5 Maximum power generation = (50 kW)(24 h) = 1200 kWh Capacity factor = Actual power generation/Maximum power generation = 600 kW/1200 kW = 0.5 1-42 Express 100 toe energy in kWh unit. 1 toe = 41.868 GJ. (a) 1,163,000 kWh (b) 1163 kWh (c) 323 kWh Answer: (a) 1,163,000 kWh (e) 1 (d) 323,000 kWh (e) 100,000 kWh      6     41.868 GJ 1 10 kJ kWh 1     = = 1,163,000 kWh 100 toe       100 ( toe) toe 1 1 GJ 3600 kJ

  13. 1-13 1-43 Express 999 GWh energy in toe unit. 1 toe = 41.868 GJ. (a) 23.9 toe (b) 23,860 toe Answer: (d) 85,899 toe (c) 85.9 toe (d) 85,899 toe (e) 999 toe   1  6       10 kWh 3600 kJ 1 GJ toe 1     = = 999 GWh 85,899 toe             999 ( GWh) 6 GWh 1 kWh 1 41.868 GJ  1 10 kJ Fossil Fuels and Nuclear Energy 1-44 Main energy sources include coal, oil, natural gas, nuclear energy, and renewable energy. Among these coal, oil and natural gas are fossil fuels. 1-45 Common coal types are bituminuous (soft coal), subbituminuous, antracite (hard coal), and lignite (brown coal). 1-46 Carbon burns according to C + ½ O2→CO and CO + ½ O2→ CO2 reactions. If some of carbon monoxide (CO) cannot find sufficient oxygen to burn with by the time combustion is completed, some CO is found in the combustion products. This can happen even in the presence of stoichiometric or excess oxygen due to incomplete mixing and short time of combustion process. 1-47 Coal is mostly used for electricity generation in power plants and petroleum products are mostly used as fuel in transportation vehicles. 1-48 The categories of oil used in power plants and industrial heating applications and their characteristics are: Distillate oils: These are higher quality oils which are highly refined. They contain much less sulfur compared residual oils. Residual oils: These oils undergo less refining process. They are thicker with higher molecular mass, higher level of impurities, and higher sulfur content. 1-49 Natural gas is mostly transported in gas phase by pipelines between the cities and countries. Sometimes, natural gas is first liquefied into about −160C before being carried in large insulated tanks in marine ships. 1-50 Natural gas is used in boilers for space heating, hot water and steam generation, industrial furnaces, power plants for electricity production, and internal combustion engines.

  14. 1-14 smtb98@gmail.com smtb98@gmail.com 1-51 The tremendous amount of energy associated with the strong bonds within the nucleus of the atom is called nuclear energy. The best-known fission reaction involves the split of the uranium atom (the U-235 isotope) into other elements. When a uranium-235 atom absorbs a neutron and splits during a fission process, it produces a cesium-140 atom, a rubidium-93 atom, 3 neutrons, and 3.2  10−11 J of energy. Nuclear energy by fusion is released when two small nuclei combine into a larger one. When two heavy hydrogen (deuterium) nuclei combine during a fusion process, they produce a helium-3 atom, a free neutron, and 5.1  10−13 J of energy. 1-52 Vast majority of devices, equipment, and appliances used by people operate on electricity. From a thermodynamic point of view, electricity can be converted to other forms of energy (such as mechanical energy and heat) in a 100% conversion efficiency. 1-53 The incorporation of wind power and solar power into the grid involves some irregularities and uncertainties due to changing wind and solar conditions on hourly, daily, and seasonal basis. This requires a more flexible electrical grid system than the existing conventional system in order to accommodate inconsistent supply of renewable electricity. This new grid system is called smart grid, which is an important area of research and development for electrical engineers. 1-54 The combustible constituents in the coal are carbon C, hydrogen H2, and sulfur S. The higher and lower heating values of hydrogen are 141,800 kJ/kg and 120,000 kJ/kg, respectively and the heating value of carbon is 32,800 kJ/kg (Table A-7). Using their mass fractions, the higher heating value of this particular coal is determined as HHV mf HHV mf HHV mf HHV (0.7961)(32,800 kJ/kg) (0.0466)(141,800 kJ/kg) 32,770 kJ/kg C C H2 H2 S S (0.0052)(9160 kJ/kg) Similarly, the lower heating value of the coal is LHV mf LHV (0.7961(32,800 kJ/kg) 31,750 kJ/kg mf LHV mf LHV C C H2 H2 S S (0.0466)(120,000 kJ/kg) (0.0052)(9160 kJ/kg) The difference between the higher and lower heating values is about 3 percent.

  15. 1-15 1-55 The higher and lower heating values of hydrogen are 141,800 kJ/kg and 120,000 kJ/kg, respectively, and the heating value of carbon is 32,800 kJ/kg (Table A-7). The molar mass of carbon (C) is 12 kg/kmol and that of hydrogen (H2) is 2 kg/kmol, respectively (Table A-1). Using the chemical formula, the higher heating value of gasoline is determined as HHV HHV HHV m m m m C C H2 H2 C8H15 C H2 (8 12 kg)(32,800 kJ/kg) (15 1 kg)(141,800 kJ/kg) (15 1 kg) (8 12 kg) 47,530 kJ/kg The lower heating value is LHV LHV m m m C C H2 H2 LHV C8H15 m C H2 (8 12 kg)(32,800 kJ/kg) (15 1 kg)(120,000 kJ/kg) (15 1 kg) (8 12 kg) 44,580 kJ/kg 1-56 The higher and lower heating values of hydrogen are 141,800 kJ/kg and 120,000 kJ/kg, respectively, and the heating value of carbon is 32,800 kJ/kg (Table A-7). The molar mass of carbon (C) is 12 kg/kmol and that of hydrogen (H2) is 2 kg/kmol, respectively (Table A-1). Using the chemical formula, the higher heating value of light diesel fuel is determined as HHV HHV HHV m m m m C C H2 H2 C12H22 C H2 (12 12 kg)(32,800 kJ/kg) (22 1 kg)(141,800 kJ/kg) (22 1 kg) (12 12 kg) 47,245 kJ/kg The lower heating value is LHV LHV m m m C C H2 H2 LHV C12H22 m C H2 (12 12 kg)(32,800 kJ/kg) (22 1 kg)(120,000 kJ/kg) (22 1 kg) (12 12 kg) 44,355 kJ/kg

  16. 1-16 The amount of heat released as 1.2162 kg water is condensed is latent H2O (1.2162 kg H O/kg CH )(2442 kJ/kg H O) fg Q m h smtb98@gmail.com 1-57 The molar masses of C8H15 and H2O are 111 and 18 kg/kmol, respectively. When 1 kmol of C8H15 is burned with theoretical air, 7.5 kmol of water (H2O) is formed. Then the mass of water formed when 1 kg of C8H15 is burned is determined from (7.5 kmol)(18 kg/kmol) (1 kmol)(111 kg/kmol) N M smtb98@gmail.com N M H2O H2O 1.2162 kg H O/kg C H m H2O 2 8 18 C8H18 C8H15 2970 kJ/kg C H 2 4 2 8 15 Then the lower heating value of octane becomes LHV HHV 44,330 kJ/kg 47,300 kJ/kg 2970 kJ/kg Q latent 1-58 The total amount of heat input to coal plants is determined from the plant efficiency as 12 electric W Q electric W 1.51 10 kWh 3600 kJ 0.34 16 1.599 10 kJ Q plant in 1 kWh in plant The amount of coal consumed per year is 16 Q 1.599 10 kJ 25,000 kJ/kg 11 6.395 10 kg in HV Q m m in coal coal coal HV coal 1-59 Which one cannot be considered as a fossil fuel? (a) Coal (b) Natural gas Answer: (d) Hydrogen 1-60 Which is not a fuel? (a) Oil (b) Natural gas (e) Carbon dioxide (CO2) Answer: (e) Carbon dioxide (CO2) 1-61 Which is not a coal type? (a) Bituminuous coal (b) Subbituminuous coal (e) Green coal Answer: (e) Green coal 1-62 Which coal type is of the lowest quality? (a) Bituminuous coal (b) Subbituminuous coal (e) Hard coal Answer: (d) Lignite (c) Oil (d) Hydrogen (e) None of these (c) Coal (d) Carbon monoxide (CO) (c) Antracite coal (d) Lignite (c) Antracite coal (d) Lignite

  17. 1-17 1-63 Electricity is mostly produced from _________ burning power plants in the world. (a) Coal (b) Natural gas Answer: (c) Coal 1-64 The most common use of petroleum products is (a) Motor vehicles (b) Electricity generation (e) Industrial furnaces Answer: (a) Motor vehicles 1-65 Which fuel is the most polluting fuel and the largest contributor to global carbon dioxide emissions? (a) Coal (b) Natural gas (c) Oil Answer: (c) Coal 1-66 Which fuel has the highest heating value? (a) Coal (b) Natural gas (c) Oil Answer: (d) Hydrogen 1-67 Desirable characteristics of coal are (a) High sulfur content, high heating value (c) Low sulfur content, high heating value Answer: (c) Low sulfur content, high heating value 1-68 Order the following fuels from higher values of heating value to lower values. I. Coal II. Gasoline III. Natural gas (a) I, II, III (b) I, III, II (c) II, I, III Answer: (e) III, II, I 1-69 In order to accommodate the inconsistent supply of renewable electricity, a more flexible electrical grid system than the existing conventional system is required. This is called (a) Smart grid (b) Flexible electricity (c) Consistent grid supply (e) None of these Answer: (a) Smart grid (c) Oil (d) Nuclear (e) Solar (c) Space heating (d) Steam generation (d) Nuclear (e) Solar (d) Hydrogen (e) Sulfur (b) High sulfur content, low heating value (d) Low sulfur content, low heating value (d) II III, I (e) III, II, I (d) Intelligent transmission

  18. 1-18 smtb98@gmail.com 1-70 The total amount of heat input to a coal-burning power plant is 5000 GJ. The heating value of coal is 25,000 kJ/kg. The amount of coal consumed per year is (a) 5000 ton (b) 500 ton (c) 200 ton (d) 200,000 ton (e) 25,000 ton Answer: (c) 200 ton smtb98@gmail.com 9  Q 5 10 kJ in = = = = 200 ton 200 000 , kg m coal HV 25,000 kJ/kg coal

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