AE BOARD REVIEWERRevised October 8, 2008 by Victorio L. Pasado SUBJECT 3 Rural Electrification, Agricultural Processing, Ag. Structures and Allied Subjects (34%) INSTRUCTIONS: Select the best answer. Shade the corresponding box of your answer in the answer sheet. Give only one answer for each question. Do not make any unnecessary marking in the answer sheet. 1. RURAL ELECTRIFICATION RENEWABLE ENERGY (GENERAL) 1. An international agreement which embodies the Clean Development Mechanism or CDM. A. Geneva Convention B. International Treaty C. Tokyo Protocol D. Kyoto Protocol Answer: D 2. In what year will the Kyoto Protocol expire? A. 2011 B. 2012 Answer: A C. 2015 D. 2020 3. Renewable Energy Act of 2008 is also known as A. RA 9367 B. RA 9513 C. RA 8559 Answer: B D. RA 8435 4. Biofuels Act of 2006 is also known as A. RA 9367 B. RA 9513 Answer: A D. RA 8435 C. RA 8559 5. What energy generating plant does not use turbine for power generation? A. Hydro B. Geothermal C. Ocean-thermal D. Biogas plant Answer: D 6. From the BiGSHOW acronym of renewable energy coverage, what “Bi” stands for? A. Biomass B. Biofuel C. Biomass and biofuel D. Biodiesel Answer: C 7. Aerogenerator, covered lagoon biodigester and minihydro energy projects are evaluated for financial feasibility. Only one feasible project can be funded. Each project needs PHP 30M investment. The financial internal rate of returns are 11%, 8% and 9%, respectively. Annual discount plus inflation rate is 12%. Which of the 3 projects shall be implemented? A. Aerogenerator B. Covered lagoon biodigester C. Minihydro D. None Answer: D (since all their IRRs are less than the discount plus inflation rate) 8. An Ocean Thermal Energy Conversion (OTEC) project was evaluated for financial feasibility. At 12% annual discount rate, it has the following feasibility indicators: net present value of 2.5 billion pesos and internal rate of return of 60%. If annual discount rate drops to 10%, will the project still be feasible? A. No B. Yes C. Maybe D. Insufficient data to determine Answer: B (if it is feasible at 12% discount rate, the more it will be feasible at 10%) What diode is needed as protection against lightning in solar installations? A. 5 Vdc C. what switch is useful in the instrumentation? A. LM 333 Answer: D 14. Shading diode D. INSTRUMENTATION AND CONTROL 9. A. +12 Vdc input B. 7 Vdc D. Fiber optics B. Shading diode D. CQ 556 D. A kind of IC included as an important part of electronic instruments for gathering speed. Power switch D. Transmission of data to a remote line-of-sight point is best done using A. Automatic transfer switch B. Radio frequency Answer: D 16. Pressure switch C. What part of solar module should be included such that solar charging is continuous at slower rate even when partially shaded by trees? A. A switch for automatic starting of pumps with sealed water tank in drip irrigation systems. Ground D. Which DC voltage is best for remote instrumentation? A.ELECTRONICS. TD 8062 B. Binary output Answer: C 12. Voltage reducer B. 9 Vdc Answer: A 15. Time switch . Pressure switch C. +5 Vdc output C. Which one is an integrated circuit (IC) used in instrumentation and control? A. 7 or 9 Vdc B. Sensor D. Variable resistor Answer: B 13. Bypass diode C. Time switch Answer: A 17. E-mail D. Automatic transfer switch B. To automate control of irrigation dam gates. Timer C. Lightning diode Answer: B 10. Blocking diode B. A. velocity and discharge real-time data. Lightning diode Answer: A 11. Power switch Answer: B 2 D. Fax C. A. 5. ES 8813 C. Bypass diode C. Blocking diode B. What pin is common in integrated circuits (ICs). Ecliptic B. Rotor D. Reciprocating Answer: D 22. Reciprocating pump C. Drag force B. Crank arm Answer: C 25. Side force D. What type of pump is appropriate for multi-bladed windmills? A. Hinged side vane Answer: B 3 C. Rotor D. Lift force C. Hydraulic B. What efficiency can exceed 100% while the windmill is in operation? A. Volumetric C. Diaphragm C. The component of force used to rotate the rotor in multi-bladed windmills. A. A. Pneumatic D. Side force D. Drop force Answer: B 20. Transmission B. High wind D. Drag force B. Part of multi-bladed wind pump which is responsible in converting rotational movement to reciprocating action. Turbine C. No tail . Drop force Answer: A 21. Terrestrial wind B. Which type of multi-bladed wind pump tail is more effective for windy areas? A. Jet stream Answer: D 24. A wind energy resource located about 10-16 km from Earth surface with velocity over 150 kph A. Connecting rod D. The component of force used to rotate the rotor in savonius windmills. Governor Answer: A 19. A. Rotor B. Manual folding D. Lift force C. Extra-terrestrial wind C. What part of an aerogenerator exceeding 100 kw capacity is responsible in converting rotational movement to electricity? A. Crank Answer: B 23. Alternator B.WIND POWER 18. Transmission C. 15-17% D.26. Rotor diameter C. Increase blades C. Jig D. 75 B. Connecting rod length B. what part should be increased? A. An 4-bladed windpump with 7 meters rotor diameter and 45 degrees twist angle is to be modified due to insufficient starting torque. 1 ft/s C. 3. To have a tip speed ratio of 1:1 in windmills. Energy demand survey Answer: C 27. Rotor B. In big wind farm projects.3 m/s? A. 1 m/s B. 60 C. If typical overall efficiency of multi-bladed windmills is 8-16%. Rotor blade B. what should be the average rotor blade twist angle in degrees? A. which should be done first? A.3 ft/s Answer: C Solution: Windspeed = tip speed. Tower Answer: C 28. 90-95% B.3 m/s D. 3. what is the typical overall efficiency of manufactured aerogenerators or electricity generating windmills? A. 7-10% Answer: A 31. 45 D. What should be done? A. Crank arm Answer: D 29. since tip-speed ratio is 1:1 = 3. 18-21% C. Well diameter D. If a wind mill rotor has a tip-speed ratio of 1:1. what should be done first? A. In multi-bladed wind mill fabrication. Decrease twist angle Answer: B (since increasing blades increases torque) 4 .3 m/s 32. To increase water displacement volume in wind mills. Decrease blades D. Rotor frame C. what windspeed is required for its rotor tip to move at 3. Wind velocity measurement D. Increase twist angle B. 30 Answer: C 30. 015 kwh/kg 5 D. 0. Reduce rotor C. What is the electrical energy consumption of a 30-kw vermicast production shredder plugged in a 240-volt source if it requires 1 hour to shred 2 tonnes of farm waste? A. 6 cm pump diameter and 2. Average windspeed at rotor center is 6 m/s.2 cm crank arm is to be modified due to low volumetric efficiency. 0. aerogenerators have no crank arm) CONVENTIONAL ENERGY 35. 0. A windpump having 8 meters rotor diameter.015 kwh/kg C. 0.06 kwh/kg Answer: B Solution: Ec = PcTo / Wi = 30 kw (1 h) / 2.000 kg = 0.33. Which part needs to be increased? A. What should be done? A.2 m/s wind can’t rotate. A 3-bladed 50-meter tall aerogenerator with 25 meters rotor radius designed for 6. Tower height D. increasing rotor radius will severely affect twist angle of entire blade & tip-speed ratio. Enlarge rotor B. Crank arm Answer: C (Note: Windspeed is higher at higher elevation. Number of blades B. Rotor radius C.08 kwh/kg .02 kwh/kg B. Enlarge pump diameter D. Elongate crank arm Answer: D (Note: Elongating crank arm increases pump stroke and volumetric efficiency) 34. what is the possible system voltage reading of a solar home system when all electrical loads are switched on? A. Photovoltaic conductor material used in manufacturing solar modules. 20-22 V Answer: D 41.SOLAR POWER 36. Load D. During 8:00 P. 11-13 V B. what is the possible voltage reading of solar pump without battery when electrical load is in switched on? A. which positive polarity wire should be connected last to the battery control unit (BCU)? A. Boron C. Battery Answer: B C. Battery appropriate for slow charging and slow discharging energy systems? A. what is the possible voltage reading of solar battery charger when no battery is being charged? A. A. When a bamboo shades half of solar module. 16-19 V D. Silicon with little boron D. 11-13 V B. Zero amperage . During a clear sky noontime in the Philippines. Automotive battery C.M. Equal silicon and boron 39. Zero voltage Answer: C B. Sealed lead-acid battery B. During a clear sky noontime in the Philippines. During solar panel installation. in the Philippines. NiCad Battery D. 16-19 V D. which condition is true? A. Low amperage 6 D. 14-15 V C. Maintenance-free battery B. Low voltage C. Silicon Answer: C B. Monitor 43. Kind of battery useful in solar home systems. 14-15 V C. 20-22 V Answer: A 42. 16-19 V D. Solar battery Answer: D 37. A. 14-15 V C. Solar battery Answer: C 38. All lead-acid batteries C. 11-13 V B. 20-22 V Answer: C 40. Solar module B. Automotive battery C. A. How do you classify a 260 kw hydro power plant? A. Hydro power plant D. Geophysical survey C. 100 kw B. Rotating part of a geothermal power plant which moves the generator. In an Ocean Thermal Energy Conversion (OTEC) system. Tidal C. Jet propulsion D. Drilling B. Fluid pressure C. Water tank Answer: A 50. The Ocean Thermal Energy Conversion (OTEC) operating principle resembles a A. Small hydro C. Transtidal Answer: C 7 D. Wave B. Microhydro B. Gasoline engine C. Minihydro D. Energy demand survey D. what causes OTEC fluid to flow and rotate the turbine? A. Rotor B. Chemical test Answer: B 48. Turbine Answer: D OCEAN ENERGY 49. Hot air balloon B. The geothermal power system resembles in principle to a A.HYDRO POWER 44. Bouyancy B. 50 kw D. Temperature difference Answer: D 51. Thermal . Big hydro Answer: C 45. Which stage comes first in a geothermal power plant development? A. Underwater current D. Using Philippine classification. Gas pump C. Rotator C. Which one is not a form of ocean energy. A. 75 kw C. 25 kw Answer: A GEOTHERMAL ENERGY 46. specify the maximum power rating for microhydro? A. Refrigerator B. Motor D. Kettle Answer: D 47. 600 kg/h D. 450 kg/h C.600 kg oil) /4 h = 400 kg/h 8 D. Hopper C.2. Which one is not a part of a vertical feed mixer? A. Serial number Answer: C B. 300 kg/h B. A Jatropha oil expeller was loaded with 2.200 kg seed . A 2.8-tonne capacity Jatropha oil expeller was loaded full with dried Jatropha seeds in 3 minutes. Fair D. bottling of virgin coconut oil prints batch numbers. Expiry indicator 53. Product recall D. 300 kg/h . 85% C. 90% B. It produced 600 kg of crude oil in 4 hours. Discharge chute 54. Conveyor D. AGRICULTURAL PROCESSING 52. In the factory. 700 kg/h B. the minimum stripping efficiency of the mechanized stripping machine shall be A. 80% D. Good C. Poor 55. Feed mixers evaluated to have 10-15% mixing coefficient of variation is rated as A. It produced 900 kg of crude oil in 2 hours. 70% Answer: C 56. 400 kg/h Answer: C Solution: Pcake = Weight of resulting Jatropha cake produced / Total time = (2. Screw Answer: C B. Very good Answer: B B. 500 kg/h C. What is the main purpose of batch numbers? A. What is its Jatropha cake production rate? A. Bottle identification C.2 tonnes dried Jatropha seeds. What is its crude oil production rate? A. Based on standards. 300 kg/min Answer: B Solution: Poil = Weight of crude oil collected / Total time = 900 kg /2 h = 450 kg/h 57. 000 kg of crude oil in 20 hours. 3312 B. 588 kg/h D.Wt.000 kg) /20 h] x 0. 700 kg/h B.95 = 665 kg/h 59. A.2 tonne/h][ (16 h/d x 30 d/mo x 3 mo/yr)+ (8 h/d x 26 d/mo x 9 mos/yr)] = 662 tonnes /yr 9 .000 kg . How many tonnes of agricultural waste is needed annually to optimally operate a 25-ampere 5-kw vermicast production shredder having a capacity of 0. 665 kg/h C. It produced 7. of crude oil collected )/ Total time] x Eff = [(21. What is the copra cake production rate of the machine at 95% cake collection efficiency? A.7. 885 kg/h Answer: B Solution: Pcake = [(Wt. 1440 D. 662 C.58.2 tonne of waste per hour? It optimally operates 16 hours daily during the 3-month peak sunny months and 8 h/day at 26 days/month for the rest of the year. of copra input . 1872 Answer: B Solution: Wt = Input Rate x (Peak Operating Time + Lean Operating Time) = [0. A 21-tonne capacity coconut oil expeller was loaded full with grated copra. How many 40-kg bags of cement are needed per cubic meter concrete in standard (Class A) open channel construction? A. Class AA B. how many cubic meters of sand should be mixed to produce one cubic meter concrete? A.3. 5 cm C. 12 D. Class C Answer: B 62. What is the standard diameter of both vertical and horizontal reinforcing bars in open channel construction? A. A. 12 cm B. 16 mm C. 12 . 16 B. 6 B. 12 mm D. 0. 18 Answer: C 65. B or C concrete mix. Class A C. 30 cm D. 0. Class B D.50 Answer: D 66. 25 cm C. What is the standard concrete wall thickness of lateral trapezoidal open channel in open channel design? A. 8 mm Answer: C 61. 15 cm Answer: D 64.25 B.5 C. 50 cm Answer: A 63. What is the standard spacing between vertical bars in open channel construction? A. 0. 9 Answer: C 10 D. 10 cm D. When using class A. What class of concrete mix should be used in open channel construction? A.0 C. 10 mm B.75 D. The appropriate gage number of galvanized iron (GI) tie wire for open channel construction. 10 C. 1. AGRICULTURAL BUILDINGS AND STRUCTURES IRRIGATION STRUCTURES 60. 20 cm B. 7. 000 C. 14.5 cubic meters concrete per linear meter? 1 truck load is 2. 53.750 B. concrete x Trucks/cu. 36.000 B. Estimate the truck loads of gravel needed for an 8-km main irrigation canal requiring 4 cubic meters concrete per linear meter. In open channel construction.6 cu. How much should be budgeted for cement for 20-km irrigation canal if it requires half cubic meter concrete per linear meter and cement costs Php210/bag? Use 5% safety factor.080 truck loads 70.6 cubic meter concrete per linear meter? Use 5% safety factor.m.m. 35. A.1 x 1 cu./cu.000 B. 18.m/m x 20. Php 18.750 truck loads 69. 14.5 cubic meters. x 1 truck load/2.400 bags 68.05 x 9 bags/cu.m.5 cu. x (0.000 m) = 14.5 cu.200 C.000 m = 113.000 m) = 36.080 D.m. A. Use 5% safety factor.05 x 0. 180.m.945.m. How many truck loads of sand are needed for 50-km main irrigation canal requiring 3.000 Answer: A Solution: Truck loads = Safety factor x Cu. Use 10% safety factor. concrete x Concrete volume x Unit Cost = 1. x Concrete volume = 1. 114.000 D. x (4 cu.m. Php 18.000 D. A.5 cu.800 Answer: C Solution: Truck loads = Safety factor x Cu.5 cubic meters.m/m x 8.m.800 B. x Concrete volume = 1.000 11 .67.000 Answer: B Solution: Budget = Safety factor x Bags cement/cu.m. Php 19.m. 1 truck load is 2.000 m) x Php 210 = Php 19. concrete x Trucks/cu./cu. sand/cu. Php 19.m.m.05 x 9 bags/cu.m.400 Answer: D Solution: Bags = Safety factor x Bags cement/cu. gravel/cu.845.m.900. 108.5 cu. x 0.m. x (3. 36.000 C. how many 40-kg bags of cement are needed for a 20-km lateral channel if it requires 0.845. A. 113. concrete x Concrete volume/m x m length = 1.570 D. 12.m/m x 50.m.300 C.m/m x 20. x 1 truck load/2.5 cu.800. 88 74.615 B. C. x 60. The water exerts 4. concrete x Concrete volume x Unit Cost = 1.615.8. 81. Offensive stress = Weight/Area = 2 x (75.000 m) x (Php 200 + Php 10) = Php 13. stress = Sinking FS x Offensive Stress.71. A concrete dam with 60. Php 12. Answer: A Solution: Sinking Factor of safety = Min. 8.88 D.606 D. Coefficient of friction at dam base is 0. Use 5% safety factor.615 72.m. 5. 7. A gravity dam weighing 75. 2.1 tonnes/sq. defensive horizontal force/Offensive horizontal force Min. Cement costs Php200/bag while transport costs Php 10/bag.5 tonnes/sq.m.236. 81.m.m/m x 10.000 sq.m.m. Php 12. submerged vertical upstream wall is to be constructed against a 546 tonnes/sq. Give cement budget estimate for 10-km canal requiring 2/3 cubic meter concrete per linear meter. defensive bearing stress/Offensive bearing stress Min.m.236 Answer: A Solution: Budget = Safety factor x Bags cement/cu.000 /(4050 x 30) = 2. A. 2.000 tonne-meters. Php 13.m.5 tonnes/square meter 73. B.000 tonnes B.8 Weight = 81. 1.300 C.000 tonnes 12 .000 tonnes C.900 tonnes Answer: B Solution: SlidingFactor of Safety = Min. x (0.) / 0. A. A.000 tonnes is to be constructed with length along dam axis of 1 km and base of 100 meters.606. 0. defensive force = Coef.520.75 tonnes/sq.m. 65. A 2-meter thick trapezoidal section of a concrete dam has resisting moment about the dam toe of 350.08 Answer: C Solution: Overturing Factor of Safety = Resisting moment/Overturning moment = 350.520 tonnes D. def.667 cu.m.900.5 tonnes/sq.300. of friction x Weight = FS x Offensive force Weight = FS x Offensive force / Coef.000 tonnes / (1. What should be the minimum soil bearing stress? Use factor of safety against sinking of 2. of friction Weight = 2 x (546 tonnes/sq.05 x 9 bags/cu.050 tonnes horizontally at 30 meters above the dam toe. What is the factor of safety against overturning? A.000 sq. average water pressure. 1. Php 13.236.80 C.900.m. 65. D. What should be the minimum dam weight? Use factor of safety against sliding of 2.22 B.000 m x 100 m) = 1. Fixed dome B. Linear low density polyethylene (LLDPE) Answer: D 79. Carbon trading value D. 16 times Answer: B 80. Value of fee to project owners from Clean Development Mechanism (CDM) fund for carbon emission reduction projects. 70% D. 50% Answer: B 83. The toxic gaseous by-product of geothermal power plants that is trace in biogas. Polyvinyl Chloride (PVC) D. What is the acronym for the Inter-Governmental Panel for Climate Change? A. 40% B. Covered lagoon Answer: D 78. Tube-bag D. If water-manure ratio is 1:1. 50% B. The minimum methane content of biogas such that biogas-fueled generators will run. & CLEAN DEVELOPMENT MECHANISM (CDM) 75. 21 times C. IPCC D. A. In covered lagoon biodigester where solids is less than 3%. Floating type C. A. 10 times D. Emission value B. Kind of plastic sheet for constructing balloon and bed lining in covered lagoon biogas plants. 65% Answer: C 13 D. IPFCC Answer: C 77. what is approximate methane content in biogas? A. 50% C. Acetate C. Hydrogen sulfate C. B. 70% Answer: D 82. Hydrogen sulfide D. Best type of biogas plant for large-scale animal farms having low solid content of effluents. WASTE MGT.4. In the environment. by how many times is methane destructive than carbon dioxide? A. A. what is the approximate methane content of biogas? A. Sulfuric acid Answer: C 81. Carbon sequestration Answer: B 76. 2 times B. High density polyethylene (HDPE) A. 60% C. Carbon credit value C. IGPFCC B. 85% C. IGPCC C. A. 40% B. 60% D. Carbon monoxide B. 75% . The consumption of each gas burner is 1.063 m3/kg./hour. B = biogas consumption of device. M = manure production.of heads.000 heads of porkers of mixed ages. 40 D.2 m3 14 . Estimate the biogas production in cubic meters per day from a large balloon type digester using 1:1 water-manure ratio. G = Specific gas production of the manure for a specific retention period P = (2000 heads)(2. N = No. N = No. 84 C. 162 D.84.2 cu. Determine the daily biogas consumption in cubic meters of an industrial boiler using 5 units of 50-cm gas burner 8 hours a day. A.2 kg/head while the specific gas production for 30-day manure retention period is 0.m.2 kg/head)(0. The source farm has 2.063 m3/kg) = 277. Use a retention period of 30 days. 48 B. T = Time C = (5 units)( 1. 277 Answer: D Solution: P = NMG where P = biogas production potential. 80 Answer: A Solution: C = NBT where C= Consumption.400 B. A. The mean daily manure production of porkers is 2. of units. 126 C. 4.2 m3/hr)(8 hrs) = 48 m3 85. 800. C = biogas production.5 x G for chicken dung at 25-day retention period = 0. C. XXX (2) Renewable Energy Journal. Vols. M = manure production. DOE and JICA. of heads. 7 C.200. 2008 Philippine Agricultural Engineering Journal. The mean daily manure production of breeding cattle is 13 kg/head while its specific gas production for 25day manure retention period is half of that of chicken dung. 1992 Fundamentals of Environmental Science by G.000 tonnes of methane plus 3.06 m3/kg = 0. to avoid gas shortage round up to 193 heads 87. US$ 204. how much will the project owner of a new covered lagoon biodigester receive annually from CDM fund for the first 10 years if the digester produces 7. Tambong. 4.03 m3/kg C = (85 m3)/[(13 kg/head)(0.000 tonnes of carbon dioxide annually.000 + 21(7. A. G = Specific gas production of the manure for a specific retention period G = 0.000 C. 34 Answer: C Solution: N = P/(MG) where N = No. various issues & authors Manual for Micro-Hydropower Development.86.03 m3/kg)] = 217. US$ 120. US$ 1. Philippine Agricultural Engineering Standards (PAES).000 B. Assume current carbon credit costs US$12 per tonne of carbon equivalent. How many heads of breeding cattle are needed as source of manure to generate 85 m3 of biogas a day if the retention period is 25 days and water-manure ratio is 1:1. Catchillar. Amount = Carbon cost per tonne x [(tonnes carbon/year + 21 (tonnes of methane/year)] = US$ 12 [3.000)] = US$ 1. When approved as Clean Development Mechanism (CDM) project. 5.800. 2004 15 . 6.000 Answer: A Solution: Considering that methane is 21 times more destructive than carbon. 2. A. 3.9 heads. US$ 1.5 x 0.000 D. Vol.000/year REFERENCES: 1. 218 D. 33 B. 1-8 Biogas Plant Design by Arthur It.