II. Agricultural Machinery Management, Extension and Marketing Flashcards
- Refers to machines collectively.
A. Equipment
B. Tools
C. Machinery
D. Machineries
Answer: C
- The process of dealing with or controlling things.
A. Management
B. Managing
C. Supervising
D. Administering
Answer: A
- The action of promoting and selling products or services, including market research and advertising.
A. Certifying
B. Marketing
C. Endorsing
D. Recommending
Answer: B
III. Agricultural Machinery Design, Fabrication/Manufacturing and Testing
- Which one is a part of a vertical feed mixer?
A. Screw
B. Hopper
C. Discharge chute
D. All of the above
Answer: D
- A feed mixer tested to have mixing coefficient of variation of 10-15% is rated as
A. Poor
B. Fair
C. Good
D. Very good
Answer: C
- As a standard, stripping efficiency of the mechanized stripping machine shall be at least
A. 70%
B. 80%
C. 85%
D. 90%
Answer: B
MODERATE QUESTIONS:
1. A 6-bladed windmill with 8 meters rotor diameter and 45 degrees twist angle is to be modified due to insufficient starting torque. What should be done?
A. Increase blades
B. Decrease blades
C. Increase twist angle
D. Decrease twist angle
Answer: A (since increasing blades increases torque)
- A windmill with 6 meters rotor diameter, 5 cm pump diameter and 2 cm crank arm is to be modified due to low
volumetric efficiency. What should be done?
A. Enlarge rotor
B. Reduce rotor
C. Elongate crank arm
D. Enlarge pump diameter
Answer: C (since elongating crank arm increases pump stroke and volumetric efficiency)
- A 3-bladed 40-meter tall aerogenerator with 20 meters rotor radius designed for 6 m/s wind can’t rotate. Average windspeed at rotor center is 5.5 m/s. Which part needs to be increased?
A. Number of blades
B. Rotor radius
C. Crank arm
D. Tower height
Answer: D (windspeed is higher at higher elevation; increasing rotor radius will severely affect twist angle of entire blade & tip-speed ratio; aerogenerators have no crank arm)
DIFFICULT QUESTIONS:
1. A 2-tonne capacity Jatropha oil expeller was loaded full with dried Jatropha seeds in 3 minutes. It produced 750 kg of crude oil in 3 hours. What is its crude oil production rate?
A. 300 kg/h
B. 250 kg/min
C. 600 kg/h
D. 600 kg/min
Answer: B
Solution:
Poil = Weight of crude oil collected /Total time
= 750 kg /3 h = 250 kg/h
- A castor bean oil expeller was loaded with 4.5 tonnes dried castor bean seeds. It produced 2,250 kg of crude oil
in 4 hours. What is its castor bean cake production rate?
A. 653 kg/h
B. 563 kg/h
C. 365 kg/h
D. 635 kg/h
Answer: B Solution: Pcake = Weight of resulting cake produced / Total time = (4,500 kg seed – 2,250 kg oil) /4 h = 562.5 kg/h
- A 10-tonne capacity coconut oil expeller was loaded full with grated copra. It produced 5,000 kg of crude oil in
10 hours. What is the copra cake production rate of the machine at 95% cake collection efficiency?
A. 745 kg/h
B. 475 kg/h
C. 547 kg/h
D. 574 kg/h
Answer: B Solution: Pcake = [(Wt. of copra input - Wt. of crude oil collected )/ Total time] x Eff = [(10,000 kg - 5,000 kg) /10 h] x 0.95 = 475 kg/h
- What is the weight of agricultural waste that is needed annually to optimally operate a 26-ampere 6-kw vermi
cast production shredder having a capacity of 0.3 tonne of waste per hour? It optimally operates 16 hours daily during the 4-month peak sunny months and 8 h/day at 26 days/month for the rest of the year.
A. 1,057 tonnes
B. 1,075 tonnes
C. 1,750 tonnes
D. 175 tonnes
Answer: C
Solution:
Wt = Input Rate x (Peak Operating Time + Lean Operating Time)
= [0.3 tonne/h][ (16 h/d x 30 d/mo x 4 mo/yr)+ (8 h/d x 26 d/mo x 8 mos/yr)]
= 1,075 tonnes/yr