Volume 3-2016-Agricultural, Food, And Feed Process Engineering

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Agricultural, Food

and Feed Process

Engineering

A G R I C U L T U R A L E N G I N E E R I N G R E V I E W E R

VOL

UME

A.T.Belonio D.A.H. Belonio 2016

Draft: May 2016

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Agricultural, Food and Feed

Process Engineering

AGRICULTURAL ENGINEERING REVIEWER VOLUME 3

by

Alexis T. Belonio, MS, PAE, ASEAN Engineer

Daniel Alexis H. Belonio, PAE

Copyright  2016

No part of this book is allowed to be photocopied or reproduced in any form without any written permission from the author.

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Revised Edition

Acknowledgment/Prayer:

We are very much thankful to you God the Father for inspiring us to prepare this review material to help those who graduated the agricultural engineering degree to pass the Professional Agricultural Engineering Licensure Examination. May you grant them the desires of their hearts to pass the exam and be used by You in the future years to come! All glory and honor belongs to You! In Jesus name, Amen!

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PREFACE

This book, Agricultural Engineering Comprehensive Board Exam Reviewer Volume 3, was prepared primarily for the agricultural engineering graduates who are preparing for the professional agricultural engineering licensure examination. With this material, they can be refreshed and updated on the new principles and developments in agricultural engineering. Also, they can develop their analytical ability in analyzing the problems related to current practices in agricultural engineering.

The contents of the book are series of questions and problems that compel the students to review the fundamentals, theories, and concepts in

agricultural engineering. The questions and problems are grouped into subtopics. Most questions are practical in nature but tricky to test whether students have thorough understanding of the principles in each of the different subject matters in agricultural and food process engineering. The answer for each question and problem is provided at the end of every topic.

This book is the third of the six volumes of the Reviewer. It focuses on the different topics on agricultural, food, and feed process engineering. It also includes specific topics on refrigeration, cold storage, heat transfer, and thermal insulation.

This book is still in draft form and is produced in limited copies. Additional items will be included in the future to make this material more

comprehensive. Comments and suggestions are welcome for the improvement of this book.

May this book become useful to the aspiring Agricultural Engineers as they prepare for the Agricultural Engineering Board Examination. God bless!

ALEXIS T. BELONIO

DANIEL ALEXIS H. BELONIO

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No. of

Pages

General Information

Basic Engineering

Properties of Agricultural and Food Products

Moisture Content

Cleaning, Grading, and Sorting

Size Reduction and Sieving

Mixing and Blending

Air Moving Device

Material Handling

Crop Drying

Crop Storage

Rice Milling

Corn Milling

Coconut Flour Processing

Sugar Processing

Food Process Engineering

Feed Milling

Refrigeration and Cold Storage

Heat Transfer and Thermal Insulation

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General Information 1. A study that deals with the principles

and practices of processing agricultural products suitable for food and feeds. a. Agricultural processing

b. Food processing c. Feed processing d. All of the above

2. A study that deals with the

application and practices in converting agricultural products into different kinds and forms of food.

a. Agricultural processing b. Food processing c. Feed processing d. All of the above

3. A study that deals with the

application and practices in converting agricultural products into different kinds and forms of food suitable for animal consumption.

a. Agricultural processing b. Food processing c. Feed processing d. All of the above

4. These are crops that are produced and harvested with normally low moisture content of about 20 to 30% and do not easily deteriorate or spoil. a. Perishable crops

b. Durable crops c. Flexible crops d. None of the above

5. These are crops that have high moisture content of 30% and more such as fruits and vegetables including dairy, meat, poultry, and fish and easily deteriorate or spoil.

a. Perishable crops b. Durable crops c. Flexible crops d. None of the above

6. Processing operation which does not significantly change the physical

characteristics of the product such as drying and dehydration of fruits and vegetables.

a. Primary processing b. Secondary processing c. Tertiary processing d. None of the above

7. Processing operation which changes the physical properties of the product such as processing banana into catsup. a. Primary processing

b. Secondary processing c. Tertiary processing d. None of the above

8. Which country do not belong to the ASEAN Free Trade Agreement?

a. China b. Japan c. Korea

d. All of the above e. None of the above

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Basic Engineering

1. The study that deals with fluids at rest such as those fluids like coconut oil, milk, etc.

a. Hydrodynamic b. Hydrostatic c. Hydro pneumatic d. None of the above

2. The study that deals with the various factors affecting the relationship

between the rate of flow of fluid and the various pressure tending it to cause or inhibit its flow.

a. Hydrodynamic b. Hydrostatic c. Hydro pneumatic d. None of the above 3. Fluids include ___. a. solid, gas, and liquid b. gas and liquid c. liquid

d. all of the above

4. Which of the following resources are considered fluid?

a. Biogas b. Producer gas c. Bioethanol d. Coconut oil e. Sugar cane juice f. All of the above g. None of the above

5. Which of the following resources are not fluid?

a. Biomass

b. Grains and Cereals c. Fruits and vegetables d. All of the above

6. Newton’s law of motion states that: a. Everybody continues in a state of rest or of uniform motion in a straight line unless compelled by force to change that state.

b. The rate of change of momentum is proportional to the force applied and took place in the direction of the force application.

c. To every action there is always an equal and opposite reaction.

d. All of the above

7. Mass flow rate states that:

a. The rate of flow of fluid is constant at any point in the system and there is no accumulation or depletion of fluid within the system.

b. The rate of flow of fluid is not constant at any point in the system and there is accumulation or depletion of fluid within the system.

c. The rate of flow of fluid increases at any point of the system and there is accumulation and no depletion within the system.

d. None of the above

8. What is the mass flow rate of coconut oil in a 2-in. pipe if the velocity of the fluid is measured at 0.02 m/s? The specific weight of oil is 1500 kg/m3_.

a. 3.54 kg/sec b. 3.13 kg/sec c. 2.35 kg/sec

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9. Which of the following factors affect the flow of fluid in pipes?

a. Characteristics of fluid b. Size of pipe

c. Shape of pipe

d. Condition inside the surface of the pipe

e. Fluid velocity f. All of the above g. None of the above

10. When fluids flow in parallel elements, it is classified as ___. a. turbulent flow

b. laminar flow

c. mixture of laminar and turbulent flow

11. When fluids flow in elemental swirl or eddies, it is classified as ___.

a. turbulent flow b. laminar flow

c. mixture of laminar and turbulent flow

12. This refers to the internal resistance of fluid to shear.

a. Viscosity

b. Coefficient of friction of fluid to fluid c. All of the above

13. Reynolds number is a function of the ___.

a. inside diameter of the pipe b. velocity of fluid inside the pipe c. specific weight of the fluid d. fluid viscosity

e. All of the above f. None of the above

Basic Engineering 14. Reynolds number equation. a. Re = DVδ/μ

b. Re = DV/μ c. Re = DV/δμ

15. This refers to the characteristics of fluid that is linearly related to shear force.

a. Newtonian fluid b. Non-Newtonian fluid c. All of the above d. None of the above

16. This refers to the characteristics of fluid not linear with the shear force. a. Newtonian fluid

b. Non-Newtonian fluid c. All of the above d. None of the above

17. An example of Newtonian fluid. a. Coconut oil

b. Butter c. Slurries

18. An example of Non-Newtonian fluid. a. Purees

b. Jathropha oil c. Coconut water d. None of the above

19. The characteristic of agricultural products that determines the angle by which it will move freely by gravity with respect to the material in which it is held, such as hopper bottom of a storage bin structure.

a. Angle of repose b. Angle of friction c. Angle of slide d. None of the above

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20. The angle formed by agricultural product itself with respect to the horizontal axis which varies with the moisture content and the amount of foreign matter.

a. Angle of repose b. Angle of friction c. Angle of slide d. None of the above 21. The unit of pressure. a. Psi

b. Inches of water c. Inches of mercury d. All of the above e. None of the above

22. High pressure is usually expressed in ___.

a. psi

b. inches of water c. inches of mercury d. All of the above e. None of the above

23. Low pressure is usually expressed in ___.

a. psi

24. Very low pressure is usually expressed in ___.

a. psi

Basic Engineering 25. The plenum chamber of a flatbed dryer registers 2 inches of water. How much is the equivalent pressure of the chamber in psig?

a. 0.072 psig b. 1.201 psig c. 1.537 psig

26. The pressure taken perpendicular from the direction of fluid where sample application of which is taking the velocity of flow of fluid in pipes using pitot- tube manometer.

a. Static pressure b. Dynamic pressure c. Isothermic pressure d. None of the above

27. The pressure taken from the direction of fluid which is also the pressure resulting from force due to change in velocity of the fluid. a. Static pressure

b. Dynamic pressure c. Isothermic pressure d. None of the above

28. A simple device used to measure pressure by taking the difference in the height of fluid inside a tube.

a. Pyschrometer b. Hydrometer c. Manometer d. None of the above

29. An open tube pointing to the direction of flow of the fluid that is usually used with manometer to measure static and dynamic pressures. a. Pitot-Tube

b. Orifice c. Weir

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30. Which of the following instruments can measure the flow of fluid?

a. Pitot tube b. Venturi meter

c. Hot wire anemometer d. All of the above e. None of the above

31. Which of the following fluids is considered as highly incompressible? a. Flue gases

b. Biogas c. Vegetable oil d. All of the above

32. Which of the following fluids is considered as compressible? a. Producer gas

b. Vegetable oil c. Molasses

33. The mass flow rate equation: a. Q = AVδ

b. Q = AV c. Q = Aδ

34. The available energy due to the elevation above a reference plane. a. Pressure Energy

b. Kinetic Energy c. Potential Energy d. None of the above

35. The available energy due to the internal pressure.

a. Pressure Energy b. Kinetic Energy c. Potential Energy d. None of the above

Basic Engineering 36. The energy available from moving fluids.

a. Pressure Energy b. Kinetic Energy c. Potential Energy d. None of the above

37. In fluid mechanics, the total hydraulic energy is the sum of ___ energy plus the work supplied by the pump or blower less friction (conduit and fittings) in the system.

a. potential and kinetic b. pressure and kinetic

c. potential, kinetic, and pressure d. None of the above

38. Which of the following statements is true?

a. The velocity of fluid flowing in a pipe is highest at the center and decreases towards its surfaces.

b. The velocity gradient for streamlined flow in a long circular conduit is

parabolic in shape.

c. The velocity gradient for turbulent flow flattens and the relationship between the maximum and the average velocity changes.

39. The process which results in the diffusion of substances from the region of high concentration to the region of low concentration.

a. Energy transfer b. Mass transfer c. Heat transfer d. None of the above

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40. The amount of heat transfer from within the product.

a. Thermal diffusivity b. Thermal conductivity c. Thermal expansion d. None of the above

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Conversion of Units

Inches of water Psi x 27.648 Inches of water In. mercury x 13.6 psi In. of water x 0.0361 psi In. mercury x 0.491 Inches mercury Psi x 2.036

Inches of mercury In. of water x 0.0736

Basic Engineering

Static Pressure P = W H where:

P - intensity of pressure, kg/m2

W - unit weight of liquid, 1000 kg/m3

H - depth of water, m

Continuity Equation Q = A V where:

Q - discharge, m3_/sec

A - cross sectional area of pipe, m2

V - average velocity of water, m/s Velocity of Flow V = [2 g H] 1/2 where: V - velocity of flow, m/s g - gravitational acceleration, m/s2 H - height of water, m

Friction Loss in Pipe

Hf = [f L V2 ] / [2 g D]

where:

H_f - pressure loss in pipe, m f - friction factor

L - length of pipe, m

V - average velocity of water in pipe, m/s g - gravitational acceleration, 9.8 m/s2

D - pipe diameter, m Flow from Vertical Pipe (50- to 200-mm Pipe

Diameter with H = 0.075 to 0.1m ) 0.87 D2_H1/2 Q = --- 287 where: Q - pipe discharge, lps D - pipe diameter, mm

H - vertical rise of water jet, m

Flow from Vertical Pipe (50- to 200-mm Pipe Diameter with H = 0.3 to 0.6m ) 0.97 D2_H1/2 Q = --- 287 where: Q - pipe discharge, lps D - pipe diameter, mm

H - vertical rise of water jet, m Flow from Horizontal Pipe

Q = 3.6 A X/ y ½

where:

Q - discharge, gpm

A - cross sectional area of water at the end of the pipe, in2

X - coordinate of the point on the surface measured parallel to the pipe, in y - vertical coordinate, in

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Problem 1

What is the rate of flow of coconut oil in a 2- in. pipe if the velocity of the fluid is measured at 0.02 m/s? The specific weight of oil is 1500 kg/m3_.

Given:

Inside diameter of pipe - 2 in. Fluid velocity - 0.02 m/s Specific weight of fluid - 1500 kg/m3

Required: Mass flow rate Solution: Q = A V d = 3.14 (2 in.)2/4 x 0.025 m/in. x 0.02 m/s x 1500 kg/m3 = 2.35 kg/sec Basic Engineering

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Problem 2

A coconut oil is to be pumped into a 6meter-high tank from the pump at 100 kg per minute rate. The container where the oil is contained is located 1 m below the pump. The estimated dynamic head loss in the system is 0.2 m of water. What is the horsepower required to pump the oil? Assume a 0.93 kg/l specific weight of oil. Also ,compute the brake horsepower and motor horsepower, if the efficiencies for pump, transmission, and motor are 60%, 80%, and 90%, respectively. Given: Q - 100 kg/min Hsd - 6 m Hss - 1 m Hd - 0.2 m Y - 0.93 kg/l Required: Fluid Horsepower Solution: FHP = Q H Y / 33,000 ft-lb/min Q = 100 kg/min = 220 lb/min H = (6 m + 1 m + 0.2 m) x 3.28 ft/m = 23.6 ft FHP = 220 lb/min x 23.616 ft / 33,000 ft- lb/min-hp = 0.158 hp BHP = FHP / Ep = 0.158 hp / 0.6 = 0.263 hp MHP = BHP / (Em x Et) = 0.263 hp / (0.9 x 0.8) = 0.36 hp

Therefore, use ½-hp pump.

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Problem 3

A rotary pump is used in pumping vegetable oil to a tank at a rate of 1 gpm. The pump is driven directly by an electric motor with an input power of 1.5 hp running at 1700-rpm speed. The total head loss of the system is 10 m. If the motor drives the pump at 2200 rpm, what is the rate of flow of the pump? What is the change in the total head delivered by the pump? What would be the increase in the input power of the pump? Given: Q₁- 1 gpm HP1 - 1.5 hp N₁- 1700 rpm H₁- 10 m N₂- 2200 rpm Required: Q2, HP2, and H2 Basic Engineering Solution: N₁/N₂= Q₁/Q₂ Q₂ = N₂Q₁/N1 = 2200 rpm ( 1gpm)/1700 rpm = 1.294 gpm N₁2_/N 22 = H1/H2 H₂= H₁ (N₂2_{) / N} 12 = 10 m (2200 rpm)2_{/ (1700 rpm)}2 = 16.75 m N₁3_/N 23 = HP1/HP2 HP₂ = HP₁ (N₂3_{) / N} 13 = 1.5 hp (2200 rpm)3_{/ (1700 rpm)}3 = 3.25 hp

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1. The unhulled grain of Oryza sativa. a. Palay

b. Paddy c. Rough rice d. All of the above

2. The fibrous layer of paddy when hull is removed and is commonly known as the silver skin of brown rice.

a. Lemma and Palea b. Pericarp

c. Bran layer

3. An angle with tangent equals the coefficient of friction between surfaces of the stored materials.

a. Angle of contact b. Angle of friction c. Angle of repose d. All for the above

4. The coefficient of friction of rice to steel material is 0.41. What would be the minimum angle of the hopper required for the system?

a. 22 deg. b. 30 deg. c. 50 deg.

5. The drying temperature of copra is ___.

a. less than 50C b. between 50° to 95C c. above 95C

6. The density of pig and poultry feeds is approximately ___.

a. 250 kg/m3

b. 450 kg/m3

c. 650 kg/m3

Properties of Agricultural Products 7. The slender type paddy has ___ length to width ratio.

a. less than 2.0 b. between 2 to 3 c. more than 3.0 d. None of the above

8. The free space between the husk of rough rice and the brown rice kernel greatly affects milling in the sense that ___.

a. the larger the free space the greater is the amount of broken grains

b. the smaller the free space the lesser its abrasive effect to the rubber-roll huller

c. the larger the free space, the lesser the breakage and losses d. All of the above

9. The uneven expansion and

contraction of the inner and the outer layers of the grains that cause fissuring is a factor of ___.

a. mechanical stress b. thermal stress c. physical stress d. All of the above e. None of the above

10. The specific gravity of rice. a. 1.11 to 1.12

b. 1.19 to 1.30 c. 1.22 to 1.26 d. None of the above

11. The specific gravity of corn. a. 1.11 to 1.12

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12. The specific gravity of sorghum a. 1.11 to 1.12

13. The bulk density of rough rice. a. 25 lbs/ft3

b. 36 lbs/ft3

c. 45 lbs/ft3

14. The bulk density of shelled corn. a. 25 lbs/ft3

b. 36 lbs/ft3

c. 45 lbs/ft3

15. The physical properties of important agricultural processing. a. Surface area b. Porosity c. Bulk density d. Angle of repose e. Angle of friction f. All of the above

16. The angle of repose of rough rice. a. 27 degrees

b. 33 degree c. 36 degrees

17. The angle of repose of shelled corn. a. 27 degrees

18. The angle of repose of sorghum. a. 27 degrees

Properties of Agricultural Products 19. The angle of friction of rough rice in steel.

a. 0.41 b. 0.37 c. 0.36

20. The angle of friction of shelled corn in steel.

a. 0.41 b. 0.37 c. 0.36

21. The angle of friction of soybean in steel.

a. 0.41 b. 0.37 c. 0.36

22. Which of the following thermal properties are important in agricultural processing?

a. Thermal conductivity b. Specific heat

c. Coefficient of thermal expansion d. Heat of vaporization

e. Thermal diffusivity f. All of the above

23. The thermal conductivity of paddy grain ___ linearly with temperature, bulk density, and moisture content. a. increases

b. decreases

c. do not increase or decrease d. None of the above

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24. The amount of heat required to raise the temperature of the product one degree higher when heat is added. a. Heat of vaporization

b. Specific heat c. Thermal diffusivity d. None of the above

25. The amount of heat required to remove a certain amount of moisture from the product.

a. Heat of vaporization b. Specific heat

c. Thermal diffusivity d. None of the above

26. Hygroscopic properties important in agricultural processing.

a. Equilibrium moisture content b. Mass transfer coefficient c. Dry matter loss

d. Tempering period e. All of the above

27. The property of a product that holds moisture when subjected to a prolong condition of known temperature and relative humidity.

a. Dry matter loss

b. Mass transfer coefficient c. Equilibrium moisture content d. None of the above

29. The mass transfer among

agricultural products is a function of ___.

a. air temperature b. air relative humidity c. airflow rate

d. thickness e. All of the above

Properties of Agricultural Products 30. The equivalent loss in the dry matter of a product due to respiration during storage.

a. Moisture loss b. Dry matter loss c. Respiration loss d. None of the above

31. The pressure drop in the product during drying is a function of ___. a. airflow

b. amount of foreign matter present c. moisture content

d. depth of drying e. All of the above

32. The terminal velocity of rough rice. a. 1.3 to 5.2 m/s

b. 6.0 to 7.1 m/s c. 7.5 to 9.0 m/s d. None of the above

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Paddy Porosity

P_m = 69.05 - 0.885 M Pl = 65.55 - 0.475 M

where:

P_m – porosity for medium paddy, % P_l– porosity for long paddy, %t M – moisture content wet basis, %

Thermal Conductivity of Paddy Grains K = 0.0500135 + 0.000767 M where:

K – thermal conductivity, BTU/hr-ft-F M – moisture content, % wet basis Specific Heat of Paddy

C = 0.22008 + 0.01301 M where:

C – specific heat, BTU/lb-F M – moisture content, % wet basis

Length of Paddy (Short Grain) 11.21%M21.89%

L = 0.7318 + 0.00122 M where:

L - length of paddy, cm

M – moisture content of paddy, % Width of Paddy (Short Grain)

11.21%M21.89%

W = 0.3358 + 0.00089 M where:

W - width of paddy, cm

M – moisture content of paddy, %

Thickness of Paddy (Short Grain) 10.40%M22.59%

T = 0.2187 + 0.000089 M where:

T - thickness of paddy, cm

M – moisture content of paddy, % Coefficient of Thermal Expansion of Milled

Rice (For Temperature Below 53C) Ck = 0.0002403 perC

where:

C_k – coefficient of thermal expansion at storage moisture over a temperature of 30-70C

Coefficient of Thermal Expansion of Milled Rice (For Temp. Equal and Above 53C) Ck = 0.0003364 perC

where:

C_k – coefficient of thermal expansion at storage moisture over a temperature of 30-70C

Latent Heat of Vaporization of Paddy HV = 2.32 [1094-1.026 x

(T+17.78)] x [1 + 2.4962 Exp (-21.73M)] where:

HV – latent heat of vaporization, KJ/kg T – air temperature, C

M – moisture content, decimal dry basis

Equilibrium Moisture Content Md = E – F ln [ -R ( T + C) ln RH ]

where:

Md – moisture content, decimal dry basis E – constant, 0.0183212 to 0.480920 F – constant, 0.026383 to 0.066826 R – universal gas constant, 1.987 T – temperature, C

C – constant, 12.354 to 120.098 RH – relative humidity, decimal Mass Transfer Coefficient of Paddy

K_g = 0.008489-0.000225T +0.000236 RH – 0.00042 Q where:

K_g – mass transfer coefficient, moisture decimal drybasi-cm2_/h-m2_-kg

T – temperature of drying air, C RH – relative humidity, %

Q – airflow rate of drying air, m3_/min

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Problem 2

A receiving pit for rough rice is to be designed for a rice mill. If concrete material is to be used, what is the minimum angle of inclination of the pit to make it self-emptying? Consider a 0.52 coefficient of friction for rough rice to smooth concrete. If steel will be used instead of concrete with 0.46 coefficient of friction, what will be the inclination needed for the pit?

Given:

Material - rough rice Coef. of friction to concrete - 0.52 Coef. of friction to steel - 0.46 Required: Minimum angle of friction to concrete and steel materials Solution:

Angle of friction for concrete = tan -1 0.52 for concrete

= 27.47 degrees

Angle of friction for steel = tan -1 0.46 for steel = 24.70 degrees

Problem 1

What is the porosity of a medium-sized paddy having 21% moisture content wet basis? If the paddy is to be dried to 14% wet basis, what is the percentage increase in the porosity of the sample?

Given:

MC wb 1 - 21% MCwb 2 - 14%

Required: Porosity and percentage increase if sample is dried to 14% Solution: At 21% MC, PM = 69.05 - 0.885 MC = 69.05 – 0.885 (0.21) = 68.86% At 14% MC, PM = 69.05 – 0.885 MC = 69.05 – 0.885 (0.14) = 68.92% % Increased = (68.92% - 68.86%) x 100 / 68.86% = 0.1%

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Problem 3

Ten tons of shelled corn is to be stored in a cylindrical silo. Loading is from the bottom to the top of the cylindrical container, without considering the additional volume of the cone due to the effect of the angle of repose of the sample. If the silo requires a ratio of 2D = H, what would be the dimension of the silo? Assume a 45 lb/ft3 _{bulk density for shelled corn.}

Given:

Sample - shelled corn Weight of sample - 10 tons Ratio - 2D = H Density of sample - 45 lb/ft3

Required: Dimension of silo Solution:

Volume of paddy = Weight/Density

= [10 tons x 1000 kg/ton]/[45 lb/ft3_{x kg/2.2lb x (3.28 ft/m)}3_] = 13.85 m3 D2_{/4 x H = 13.85 m}3 D2_{/4 x 2 D = 13.85 m}3 D3 _{= (8/}_{) 13.85 m}3 D = 2.07 m H = 4.14 m

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Problem 4

What is the amount of heat

needed to raise the temperature of a 1.5-tons medium-sized paddy from 27°C to 45°C? The moisture content of paddy is

14% wet basis.

Properties of Agricultural Products

Given:

Sample - paddy Weight of sample - 1.5 tons Initial temperature of sample - 27°C (106.2°F) Final temperature of sample - 45°C (138.6°F) Moisture content - 14% wet basis Required: Heat energy required to raise the temperature of the sample Solution: C = 0.22008 + 0.1301 M = 0.22008 + 0.1301 (0.14) = 0.22190 BTU/lb-F Qs = 1.5 tons x 0.2219 BTU/lb-°F x (138.6°F – 106.2°F) = 23,725.548 BTU

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Problem 5

What is the latent heat of vaporization of paddy at 28% moisture content wet basis and at 40°C temperature. If 200-kg moisture will be removed from the bulk of paddy, what is the amount of heat required to remove it?

Given:

Moisture content - 28% wet basis Temperature - 40°C

Moisture to be removed - 200 kg Required: Latent heat of vaporization

Amount of heat required to remove the moisture Solution:

HV = 2.32 [1094-1.026(T+17.78)][1+2.4962 – 21.73 M_]

= 2.32 [1094 – 1.026 (40°C + 17.78][1+2.496 – 21.73 (0.28)] = 2,409.731 kJ/kg of water

Heat to be removed = HV x Moisture to be removed = 2,409.731 kJ/kg x 200 kg = 481,946.2 kJ

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Problem 6

Five hundred tons of paddy, at 16% wet basis, is to be stored in a silo for 6 months. If the storage temperature in the warehouse will be maintained at 95°F, what would be the expected weight of the paddy after the storage period? Neglect losses due to insect and pests.

Given:

Product - paddy Weight of product - 500 tons Storage time - 6 months Moisture content - 16% Storage temperature - 95°F Required: Weight of paddy after storage Solution:

DML = 1- exp [(A x TC_{) exp [D (T-60)] exp (E (W-0.14))]}

= 1-exp [(-0.000914 x 4.32 0.6540) exp [0.03756 (95-60)] Exp [33.61 (0.16 – 0.14)]]

= 1 – exp [-(0.00237)(3.715)(1.958)] = 1 – 0.983

= 0.01

Therefore, the weight of paddy after 6 months is 491.45 tons.

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Problem 7

If the coefficient of friction of rice to steel material is 0.41, what is the minimum angle of hopper required for the system?

Given:

Coefficient of friction - 0.41

Required: Angle of friction of the hopper Solution: tan θ = 0.41 θ = tan -1_0.41 = 22.3° Problem 8

A grain dryer was used to reduce the moisture content of four tons of paddy rice from 25% to 15% wet basis. What is the amount of moisture removed?

Given:

MC initial - 25% MC final - 14% Initial weight - 4 tons

Required: Weight of moisture removed Solution: Wt. of Moisture removed = W_i - W_f Wt. final = Wi (1 – MCi) / (1 – MCf) = 4 tons (1- 0. 25) / (1 -0.14) = 4(0.75) / (0.86) = 3.53 tons

Wt. of moisture removed = 4 tons – 3.53 tons = 0.47 ton

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1. The amount of moisture in the grain expressed as a percentage of the total weight of the samples.

a. Moisture content dry basis b. Moisture content wet basis c. Moisture content

d. All of the above

2. The amount of moisture in the grain expressed as a percentage of the weight of dry matter.

a. Moisture content dry basis b. Moisture content wet basis c. Moisture content

3. A moisture meter that determines the moisture content of the product based on the ability of the current to pass through the material.

a. Capacitance-type moisture meter b. Resistance-type meter

c. Infrared moisture meter d. All of the above

4. The moisture found on the surface of the material.

a. Unbound water b. Bound water c. Free moisture d. All of the above

5. When the amount of water in a product is determined based on its dry matter content, the moisture content is expressed in ___.

a. wet basis b. dry basis c. semi wet basis d. None of the above

6. The method of measuring the moisture content of the product by direct extraction of water.

a. Primary method b. Secondary method c. Tertiary method d. None of the above

7. If a product has 12% moisture content expressed in wet basis, the equivalent percentage moisture content in dry basis is ___.

a. 13.6% b. 16.3% c. 13.4%

8. A product has 25% moisture content expressed in dry basis. What is the percentage moisture content of the product in wet basis?

a. 20% b. 22% c. 26%

9. An expression of moisture content commonly used in commercial scale. a. Moisture content wet basis b. Moisture content dry basis c. Moisture content

d. All of the above

10. A plot or a graph representing the equilibrium moisture content of grains at different relative humidity but with constant temperature.

a. Isothermal graph b. Sorption isotherms c. Isothermal humidity d. None of the above

(27)

11. An important index to determine whether the paddy is ready for harvesting, storage or milling. a. Color

b. Moisture content c. Hardness

12. The moisture inside the tissue of a material that can only be removed by heating or reducing the vapor pressure within the material.

a. Chemically bound moisture b. Bound moisture

c. Free moisture d. All of the above

13. The difficult-to-remove moisture from the product that requires other method than heating in order to remove it.

a. Chemically-bound moisture b. Bound moisture

c. Free moisture d. All of the above

14. Which of the following is considered primary method in determining

moisture content? a. Oven method b. Distillation method c. Infrared method d. All of the above e. None of the above

15. Which of the following is considered secondary method in determining moisture content?

a. Electrical resistance method b. Electrical capacitance method c. Chemical method

d. Hygrometric method e. All of the above f. None of the above

16. The appliance usually used in calibrating resistance- and capacitor-type moisture meters.

a. Oven b. Distillation c. Infrared

d. All of the above

17. The device used by the industry in determining moisture content of grains. a. Oven

b. Electrical resistance and capacitance-type meter

c. Infrared moisture meter d. All of the above

18. If 20 kg of water is to be removed from 110-kg paddy, what is the

percentage moisture content of the paddy in wet basis?

a. 14.5% b. 16.9% c. 18.18%

19. What is the percentage moisture content of the paddy in Item 18 above when expressed in dry basis?

a. 16.4% b. 19.9% c. 22.2%

20. Determine the amount of moisture that can be removed from 9 tons of paddy with 24% initial moisture content dried to 14%.

a. 1.05 tons b. 1.26 tons c. 1.57 tons

(28)

Moisture Content (Wet Basis) MCw = (Ws – Wdm)100 / Ws where:

MCw - moisture content wet basis, % MCd - moisture content dry basis, % Ws - weight of sample, g

Wdm - weight of dry mater, g

Moisture Content (Dry Basis)

MCd = (Ws – Wdm) 100 / Wdm where:

MCw - moisture content wet basis, % MCd - moisture content dry basis, % Ws - weight of sample, g

Wdm - weight of dry mater, g Wet Basis to Dry Basis

MCd = MCw 100 / (100 – MCw) where:

MCd - moisture content dry basis, % MCw - moisture content wet basis, %

Dry Basis to Wet Basis

MCw = MCd 100 / (100 + MCd) where:

MCd - moisture content dry basis, % MCw - moisture content wet basis, % Moisture Loss

Wi ( 1 – MCi ) = Wf ( 1- MCf ) ML = Wi - Wf where:

Wi - initial weight of sample, kg Wf - final weight of sample, kg Mci - initial moisture content, % wb MCf - final moisture content, % wb ML - moisture loss, kg

(29)

Problem 2

What is the percentage moisture content of paddy in Problem 1 above when expressed in dry basis? Given: MCw - 14.5% Required: MCd Solution: MCd = (MCw x 100 )/ (100 - MCw) = (14.5 x 100) / (100-14.5) = 16.95% Problem 1

What is the amount of moisture to be removed from a material with 14.5% moisture content and 76 kg dry matter weight?

Given:

MCi - 14.5% Wdm - 76 kg

Required: Amount of moisture to be removed Solution: MC = Wmr x 100 / (Ww + Wdm) Wmr = 0.145 (Wm + 76 kg) Wmr = 0.145 Wmr + 11.02 kg 0.855 Wmr = 11.02 kg Wmr = 11.02 kg / 0.855 = 12.89 kg Moisture Content

(30)

Problem 3

What is the equilibrium moisture content of paddy stored in bags at 27°C storage temperature with 90% humidity. If the moisture content of paddy stored is 14% wet basis, is there a danger for the grains to deteriorate when the condition prevails for a longer period?

Given:

Storage temperature - 27°C Storage humidity - 90% Moisture content of sample - 14% Required: Equilibrium moisture content Solution:

Md = E – F ln [-R (T+C) ln RH)

= 0.325 – 0.046 ln [-1.987 (27 +35.703) ln (0.90) = 0.206 or 20.66% moisture dry basis

Mw = Md 100 / (100 – Md)

= 20.6% x 100 / (100 – 20.6%) = 17.08%

Problem 4

If 20 kg of water is removed from a 110-kg paddy, what is the percentage moisture content of paddy in wet basis?

Given:

Wm - 20 kg Wi - 110 kg Required: % MC wet basis Solution:

% MCw = ( Wi – Wf ) x 100 / Wi = ( 20 kg / 110 kg ) x 100 = 18.18 %

(31)

Problem 5

If a product has 12% moisture content wet basis, what is the equivalent percentage moisture content in dry basis? Given: MCw - 12% Required: MCd Solution: MCd = MCw x 100 / (100 – MCw) = 12 x 100 / ( 100 – 12 ) = 13.64% Problem 6

A product has 25% moisture content, expressed in dry basis . What is the percentage moisture content of the product in wet basis?

Given: MCd - 25% Required: MCw Solution: MCw = MCd x 100 / (MCd + 100) = ( 25 x 100 ) / (25 + 100 ) = 20% Moisture Content

(32)

Problem 7

If 9 tons of a product is to be dried from 24% to 14%, what would be the final weight of the product?

Given:

Wi - 9 tons MCi - 24% MCf - 14% Required: Weight final Solution: Wi ( 1 – Mci ) = Wf (1- MCf) Wf = Wi (1-MCi) / (1-MCf) = 9 tons (1-0.24) / (1.014) = 7.9 tons Problem 8

What is the moisture loss of the product in Problem 7 above?

Given:

Wf - 7.9 tons Required: Moisture loss Solution: ML = Wi – Wf ML = 9 tons – 7.9 tons = 1.05 tons Moisture Content

(33)

Problem 9

Shelled corn initially at 24% is to be dried to 14%. If the initial weight of the shelled corn is 1000 kilos, what is its final weight? Given: Grain - corn MC initial - 24% MC final - 14% Initial weight - 1000 kg Required: Final wt. of the corn Solution: W_i (1 – MC_i) Wt. = (1 - MC_f) 1000 kg (1 – 0.24) = (1 – 0.14) = 883.72 kg Problem 10

Referring to Problem 9, what is the moisture loss of corn grains?

Given:

Wt. - 883.72 kg Required: Moisture loss Solution: Moisture loss = Wt_i - Wt_f = 1000 kg – 883.72 kg = 116.28 kg Moisture Content

(34)

1. The percentage of grains that are free from foreign matter.

a. Percentage foreign matter b. Percentage Cleanliness c. Purity

2. A type of screen cleaners that uses air blast to assist in the cleaning. a. Screen separator

b. Fanning mill c. Blower

3. The process of separating lower density materials from grains. a. Aspiration

b. Cleaning c. Fanning

d. All of the above

4. A primary cleaner which separates larger particles from rice, also known as “rough cleaning” of rough rice.

a. Rotary sieve b. Scalper

c. Vibrating screen d. All of the above

5. The recommended size of scalper for paddy is ___.

a. 0.2-0.25sq.m. per ton b. 0.3-0.35sq.m. per ton c. 0.4-0.45sq.m. per ton d. None of the above

6. The amount of foreign materials in a sample of grains.

a. Impurities b. Chalky grain c. Dockage

Cleaning, Grading, and Sorting 7. The rough cleaning of paddy

removing most foreign materials prior to drying and storage.

a. Aspiration b. Scalping c. Sieving

d. All of the above

8. A kind of cleaner that uses air in separating lesser-density materials from corn kernel/corn grits such as floured corn, germ and bran.

a. Aspirator

b. Oscillating screen c. Cleaner

d. Cyclone separator e. All of the above

9. The process of removing contaminants from the product. a. Sorting

b. Grading c. Cleaning

10. The process of separating cleaned products in various quality fractions that maybe defined based on the size, shape, texture, colors, and other factors.

a. Sorting b. Grading c. Cleaning

11. Which of the following parameters are taken into consideration during cleaning and grading products? a. Size b. Weight c. Surface Texture d. Affinity to liquid e. Color f. Shape

(35)

12. The material commonly used in separating grains from chaffs according to size.

a. Indented plate b. Perforated sheet c. Steel screen d. All of the above

13. When removing light particles from threshed grains that goes with the chaffs during operation, what would be the best course of action that you can recommend as an Agricultural Engineer in order to achieve better cleaning and separation?

a. Install additional perforated sheets in the cleaning system.

b. Provide an air blast system using a blower so it can carry light particles. c. Reduce the feeding of threshed grains.

14. When using a cylindrical slotted rotating screen or sheet, the material can be separated in terms of ___. a. weight

b. shape c. thickness

15. Indented rotating cylinders are usually used for separating grains in terms of ___.

a. thickness

b. width and length c. size

16. Air blast separation is effective for separating the product based on ___. a. size

b. weight c. length

Cleaning, Grading, and Sorting 17. A separator used for particles with low and high terminal velocity.

a. Air blast separator b. Aspirator

c. Fanning mill d. None of the above

18. Vibrator separator is recommended for separating grains based on ___. a. weight

b. surface texture c. size

d. none of the above

19. A grain separator for seed based on the affinity for liquid or on the rate at which the surface of the seed absorbs liquid.

a. Vibrator separator b. Aspirator

c. Magnetic separator d. None of the above

20. The machine used to separate grains according to color or reflectivity.

a. Color meter

b. Electronic color sorter c. Electric color separator d. None of the above

21. Precleaners are provided for the rice milling plant primarily to ___ and to improve the quality of the final product. a. add for the machine requirement b. protect other processing equipment c. reduce the load

22. In a tray-type paddy separator, brown rice moves at the ___ of the tray. a. bottom

b. middle c. top

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23. Self -cleaning sieves used for rice mills basically consist of ___.

a. a brush to remove particles that clogs the screen of the cleaner

b. one or two balls that continuously hammer the top sieve of the cleaner c. assigned operator to do the cleaning while the machine is operating

24. A machine used to remove hard materials of almost the same size as milled rice through gravity separation using air stream to cause the rice grains to be carried away by the air stream leaving heavy materials behind. a. Aspirator

b. De-stoner c. Sifter

25. Magnetic separator basically is installed into the rice milling plant before the ___.

a. precleaner b. husker c. whitener d. polisher

e. None of the above

26. A separation equipment coupled to the end of a pneumatic conveyor to separate solid particles from the air stream.

a. Magnetic separator b. Cyclone Separator c. Vibratory separator d. None of the above

Cleaning, Grading, and Sorting 27. Efficiency of ordinary cyclone ranges from ___.

a. 80 to 90% b. 95 to 98% c. 98 to 99%

(37)

1. The process of making too-large-to- be-used solid materials usable. a. Size reduction

b. Grinding c. Cutting

d. All of the above

2. Size-reduction processes include ___. a. cutting

b. crushing c. grinding

d. All of the above

3. Raw materials for feeds are processed primarily using a size- reduction equipment to ___.

a. increase the surface area of contact of the material to facilitate taking place of reaction

b. make it palatable when used as feed for animals

c. facilitate conveyance operation at the feed mill

f. all of the above

4. A physical mechanism used in size- reduction process.

a. Compression b. Impact c. Attrition d. Cutting

f. All of the above

5. Which of the following statements is true in relation to size-reduction

operation?

a. Size reduction is the most inefficient unit operation in terms of energy use. b. Size reduction is the most efficient unit operation in terms of energy use. c. Size reduction is neither efficient nor inefficient unit operation in terms of energy use

6. Which of the following statements is true in relation to reducing particle size of a material?

a. Reduction of materials to very fine sizes is more costly in terms of energy as compared with reduction to relatively coarse particles.

b. Reduction of materials to very fine sizes is much cheaper in terms of energy as compared with reduction to relatively coarse particles.

c. Reduction of materials to very fine sizes has the same cost as compared with reduction to relatively coarse particles.

7. Which of the following is an example of size-reduction processes?

a. Chopping corn fodder b. Slicing of sweet potato c. Milling of cassava into flour d. All of the above

8. A trade term used relative to the reduction of grains into meal or flour. a. Shredding

b. Crushing c. Milling

9. Size-reduction machines are characterized based on ___. a. capacity

b. power required unit material reduce c. size and shape of the product after reduction

d. range of size and shape of resulted product.

e. All of the above 10. A burr or a plate mill. a. Micro mill

b. Attrition mill c. Roller mill d. All of the above

(38)

11. The process of reducing the size of materials by applying force to the product in excess of its strength. a. Shearing

b. Crushing c. Milling

12. The process of reducing the size of materials by pushing or forcing a thin sharp knife into it.

a. Shearing b. Crushing c. Cutting

d. All of the above

13. An indicator of the uniformity of ground materials in the resultant product.

a. Uniformity index b. Particle size c. Fineness modulus d. All of the above

14. The process of reducing the size of materials by cutting and crushing actions.

a. Milling b. Grinding c. Shearing

15. Hammer mills are designed purposely to ___.

a. chop forage materials b. reduce the sizes of granular products

c. compact powdered product d. None of the above

Size Reduction 16. Vertical-axis hammer mill is more advantageous than horizontal-axis for the reason that:

a. It is more effective in grinding grains with higher moisture content of up to 25%.

b. The power requirement of the vertical-axis hammer mill is much lower than that of the horizontal-axis

hammer mill.

c. Less broken grains can be derived from vertical-axis.

17. A grain milling machine consisting of two cast-iron disk plates with teeth or serration on one or on both faces. a. Hammer mill

b. Attrition mill c. Roller mill d. All of the above

18. A grain milling machine suitable for grinding grains with moisture content of up to 30%.

a. Hammer mill b. Attrition or bar mill c. Roller mill

d. All of the above

19. A type of mill used for crushing grains by allowing it to pass through two rotating-cylinder steels, one of which is smaller than the other. a. Hammer mill

b. Roller mill c. Plate mill d. All of the above

(39)

20. When reduced material size is between 1/8 in. or more, it is classified as ___.

a. microscopic range b. sieve range c. dimension range d. None of the above

21. When reduced material size is between 0.125 to 0.0029 in., it is classified as ___.

a. microscopic range b. sieve range c. dimension range d. None of the above

22. When reduced material size is less than 0.0029 in., it is classified as ___. a. microscopic range

b. sieve range c. dimension range d. none of the above

23. A device used for classifying granular materials by passing through series of screens.

a. Tyler Sieves b. Tyler screen c. Tyler Separator d. None of the above

24. Sieves used in determining the fineness modulus of a material includes 3/8-, 4-, 8-, 14-, 28-, 48-. and ___ -in. mesh.

a. 60 b. 80 c. 100

Size Reduction 25. In hammer milling, the fineness of a material is determined by the ___. a. number of hammers

b. speed of hammer

c. size of hole of the screen d. None of the above

26. The speed of hammer mills. a. 200 to 1000 rpm

b. 1500 to 4000 rpm c. 4500 to 6000 rpm d. None of the above

27. The operating speed of burr mills. a. Less than 1200 rpm

b. 1200 to 2400 rpm c. 2400 to 3200 rpm d. None of the above

28. Performance characteristics of size- reduction equipment are based on ___. a. size uniformity

b. temperature rise c. power requirement d. trouble-free operation e. All of the above

29. The power requirement in reducing the size of grains is ___ for moist grains than for dried grains.

a. lesser b. the same c. higher

30. The power requirement of size reduction equipment is ___ for fibrous than crystalline materials.

a. lesser b. the same c. higher

(40)

31. The energy requirement of a hammer mill in grinding shelled corn. a. 7.4 kw-hr/ton

b. 5.8 kw-hr/ton c. 2.3 kw-hr/ton d. None of the above

32. A cassava processing plant is required to grind fresh cassava into flour, what size-reduction machine would you recommend as an Agricultural Engineer? a. Hammer mill b. Roller mill c. Burr mill

33. The size-reduction machine usually used in feed milling to break dried grains into suitable size for feeds.

a. Hammer mill b. Roller mill c. Burr mill

34. In burr milling, the size of grind can be varied by ___.

a. varying the size of the screen opening

b. varying the pressure on the plates c. varying the opening between rolls d. None of the above

35. During grinding, the temperature of the material ___ when grinding fibrous than granular material.

a. decreases b. increases

c. does not change d. None of the above

Size Reduction 36. The factor affecting the power requirement of grinding mill. a. Type of material

b. Moisture content of material c. Fineness of grinding of material d. Rate of feeding the material e. Type and condition of mill f. All of the above

37. The most suitable machine for reducing size of materials like forages, straw, stalks, and weeds.

a. Hammer mill b. Roller mill c. Burr mill d. Cutter mill e. All of the above

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1. The blending of ingredients or of materials in agricultural processing operation.

a. Sorting b. Molding c. Mixing

d. All of the above

2. The auger that is centrally located in a vertical feed mixer and usually rotates at ___ speed.

a. 100-250 rpm b. 250-400 rpm c. 400-650 rpm d. None of the above

3. Horizontal feed mixer is not advantageous for feed mixing as compared with vertical feed mixer for the reason that ___.

a. it has lower capacity

b. it has high power requirement c. it has shorter life span

d. none of the above

4. If mixing is needed during conveying, the suitable conveyor to use is ___ type. a. bucket

b. screw c. belt

d. All of the above

5. A feed mixer characterized by high capacity, short mixing time, and high power requirement.

a. Horizontal mixer b. Inclined mixer c. Vertical mixer d. All of the above

Mixing and Blending 6. A feed mixer characterized by low capacity, longer mixing time and low power requirement.

a. Horizontal mixer b. Inclined mixer c. Vertical mixer d. All of the above

7. A feed mixer with an auger that elevates the feed to the top of the mixing bin and spreads them evenly throughout by gravity for another mixing cycle.

a. Horizontal mixer b. Vertical mixer c. Inclined mixer d. All of the above

8. A mixer with U-shaped bin containing a central mixing blade or a ribbon mounted on a rotating shaft. a. Horizontal mixer

b. Inclined mixer c. Vertical mixer d. All of the above

9. The assembling and measuring of the needed amount of every solid raw feed material in the formulation of the desired composition of a mixture. a. Mixing

b. Blending c. Pelleting

d. All of the above

10. The process of combining different materials until a certain degree of homogeneity is achieved.

a. Milling b. Mixing c. Feeding

(42)

11. Feed materials are purposely mixed to ___.

a. obtain a homogenous feed to ensure that animals are given the same proportions of nutrient b. reduce the density of the feed material so that it can be easily transported

c. improve the digestability of feeds

d. None of the above

12. Basically, the purpose of mixing is ___.

a. to promote the transfer of heat between hot and cold products b. to obtain good contact between materials being mixed

c. to promote reactions between reactants

d. all of the above e. any two of the above

13. A type of mixer suitable for either free flowing or non-free flowing materials.

a. Rotating mixer without stirrers b. Rotating mixer with stirrers c. All of the above

14. An indicator of satisfactory mixing. a. Produce uniform mixture

b. Less time in mixing materials c. Less cost

15. During mixing, heavier particles tend to remain ___ of the container.

a. near the bottom b. away from the bottom c. at the middle

Mixing and Blending 16. During mixing, round or small particles tend to move towards the ___ of the container.

a. top b. middle c. bottom

d. All of the above

17. Difficulty in mixing may result if the solid particles has ___.

a. the same size but of different specific gravity

b. different sizes c. different shapes d. All of the above

18. The type of mixer satisfactory for extensively large operations.

a. Batch-type mixer

b. Semi-continuous type mixer c. Continuous-type mixer d. None of the above

19. The type of mixer suitable for moderate to small operation in which overhead cost is low and labor cost is not critical.

a. Batch-type mixer

b. Semi-continuous-type mixer c. Continuous-type mixer d. None of the above

(43)

1. The measure of the power output of a fan in relation to its power input. a. Fan thermal efficiency

b. Fan performance index c. Fan efficiency

d. All of the above

2. An air-moving device that produces high airflow but low head.

a. Fan b. Blower c. Compressor d. All of the above

3. An air-moving device suitable for grain cleaning.

a. Axial fan b. Propeller fan c. Cross-flow fan d. All of the above

4. If the diameter of the blower is increased, the pressure will consequently ___.

a. increase b. decrease c. be the same d. None of the above

5. An air-moving device commonly used for drying which produces high pressure but low airflow.

a. Blower b. Fan

c. Compressor d. None of the above

6. Which of the following operations employs air-moving devices?

a. Drying

b. Material handling c. Refrigerating d. All of the above

Air-Moving Device 7. A Class 1 air-moving device operating at pressure equal to or more than 1 psi. a. Blower

b. Fan c. Exhauster d. All of the above

8. A Class 2 air-moving device operating at pressure less than 1 psi.

a. Blower b. Fan

c. Compressor d. All of the above

9. An air-moving device in which the air flow is at the same axis with the shaft of the device.

a. Propeller fan b. Axial flow fan c. Tube axial fan d. Vane axial fan e. All of the above f. None of the above

10. An axial-type fan that delivers large volume of air at low head and has few narrow blades.

a. Propeller fan b. Axial-flow fan c. Tube-axial fan d. Vane-axial fan e. None of the above

11. An axial-type fan with enlarged hubs and warped blades for better efficiency in order to operate against higher pressure than that of the propeller fan. a. Propeller fan

b. Axial-flow fan c. Tube-axial fan d. Vane-axial fan e. None of the above

(44)

12. An type fan with wider axial-flow wheel capable of delivering higher pressure than that of the axial-flow fan and is driven either by belt or by direct drive system.

a. Propeller fan b. Axial-flow fan c. Tube-axial fan d. Vane-axial fan e. None of the above

13. An axial-flow-type fan with set of guide vanes located before or after the wheel and is capable to deliver higher pressure than that of the tube-axial fan. a. Propeller fan

b. Axial-flow fan c. Tube-axial fan d. Vane-axial fan e. None of the above

14. An air-moving device in which airflow enters the axis of the shaft of the rotating blades and leaves in perpendicular manner.

a. Propeller fan b. Axial-flow fan c. Tube-axial fan d. Centrifugal blower e. None of the above

15. A centrifugal-type blower operating at low speed at several inches of pressure than that of the axial-type fan and has a squirrel-cage rotor and large number of blades.

a. Forward curved-blade centrifugal blower

b. Radial or straight-blade centrifugal blower

c. Backward curved-blade centrifugal blower

d. All of the above

Air-Moving Device 16. A centrifugal blower with blades positioned at the same plane with the rotating shaft and has larger housing than other types of centrifugal blower and is capable of handle dirty air and conveys materials that go through the fan.

a. Forward curved-blade centrifugal blower

b. Radial or straight-blade centrifugal blower

d. All of the above

17. A type of centrifugal blower with few blades tilted backward from the

direction of wheel rotation and is inherently high speed with self-limiting horsepower characteristics, and delivers higher pressure than that of the forward and radial-type centrifugal blowers. a. Forward curved-blade centrifugal blower

b. Radial or straight- blade centrifugal blower

d. All of the above

18. A low-pressure high-airflow centrifugal-type blower that delivers uniform airflow along its length. a. Axial-type fan

b. Centrifugal blower

c. Cross-flow centrifugal blower d. All of the above

(45)

19. If the width of the blade of an axial-type fan is increased, the airflow will relatively ___.

a. increase b. decrease

c. remain the same d. None of the above

20. If the width of the impeller of a centrifugal-type blower is increased, the airflow will relatively ___.

21. If the diameter of the impeller of the centrifugal-type blower is increased, the pressure will relatively ___.

c. remain the same d. none of the above

22. If the diameter of the blade of an axial-type fan is increased, the airflow will relatively ___.

23. Noise level is highest for ___ centrifugal blower.

a. forward-curved b. radial

c. backward-curved d. None of the above

Air-Moving Device 24. For the same shaft rpm, which of the following centrifugals delivers higher pressure?

a. Forward-curved b. Radial

c. Backward-curved d. None of the above

25. The air-moving device commonly used for refrigeration and ventilation systems.

a. Forward-curved b. Radial

c. Backward-curved d. None of the above

26. The air-moving device commonly used for 6-ton flatbed paddy dryers. a. Forward-curved centrifugal blower b. Propeller fan

c. Vane-axial fan d. None of the above

27. The air-moving device commonly used for pneumatic conveyor systems. a. Backward-curved centrifugal blower b. Propeller fan

c. Vane-axial fan d. None of the above

28. The air-moving device commonly used for tunnel-ventilated poultry farms. a. Radial centrifugal blower

b. Propeller fan c. Vane-axial fan d. None of the above

(46)

29. The factor to consider in selecting an air-moving device.

a. Quantity of air to be moved per unit time

b. Estimated system resistance or head c. Amount of noise permitted

d. Space available e. Economic implication f. All of the above g. None f the above

(47)

Specific Speed

N_s = [ N Q 0.5_{] / [Ps}0.75_]

where:

N_s – specific speed, dmls

N - speed of air-moving unit, rpm Q - airflow, cfm

P_s – pressure requirement, in. H₂O

Impeller Width (Centrifugal and Mixed Flow Blower)

175 Q W = ---

 N D2

where:

W – width of impeller, in. Q - airflow, cfm

N - speed of impeller, rpm D - diameter of impeller, in.

 - flow coefficient, 0.01 to 0.80 Impeller Diameter (2.35) 108 P_s D = [---] 0.5  N2 where:

D - diameter of impeller, in. P_s – pressure requirement, in. H₂O

 - pressure coefficient, 0.05 to 2.0 N - speed of impeller, rpm

Impeller Width (Traverse Flow) 550 Q

W = ---  N D2

for 0.5  W/D  10 where:

W – width of impeller, in. Q - airflow, cfm

 - flow coefficient, 0.01 to 0.80 Pitch Angle for Axial Fan

350 Q

 = Sin –1_---

 N D3

where:

 - pitch angle, deg Q - airflow, cfm

 - flow coefficient, 0.01 to 0.80

Casing Dimension (Forward Curved Centrifugal)

H_c = 1.7 D B_c = 1.5 D

W_c = 1.25 W + 0.1 D where:

H_c – height of casing, in. B_c - breadth of casing, in W_c– width of casing, in. D – diameter of impeller, in W - width of impeller, in

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Casing Dimension (Narrow Backward Curved Centrifugal) Hc = 1.4 D B_c = 1.35 D W_c = W + 0.1 D where:

H_c – height of casing, in. Bc - breadth of casing, in

W_c– width of casing, in. D – diameter of impeller, in W - width of impeller, in

Casing Dimension (Wide Backward Curved Centrifugal)

Hc = 2.0 D

B_c = 1.6 D W_c = W + 0.16 D where:

H_c – height of casing, in. Bc - breadth of casing, in

W_c– width of casing, in. D – diameter of impeller, in W - width of impeller, in Casing Dimension (Mixed Flow)

H_c = 2.0 D B_c = 2.0 D

Wc = 0.46 D

where:

H_c – height of casing, in. B_c - breadth of casing, in W_c– width of casing, in. D – diameter of impeller, in

Casing Dimension (Traverse Flow) H_c = 2.2 D

B_c = 2.2 D Wc = W + [D/4]

where:

H_c – height of casing, in. B_c - breadth of casing, in W_c– width of casing, in. D – diameter of impeller, in Casing Dimension (Vane-Axial Flow)

W_c = 1.2 D where:

Wc – width of casing, in.

D – diameter of impeller, in

Casing Dimension (Tube-Axial Flow) W_c = 1.0 D

where:

D – diameter of impeller, in Casing Dimension (Partially-Cased Fan)

W_c = 0.5 D where:

D – diameter of impeller, in

Mechanical Efficiency f = AHP / BHP

where:

_f - fan efficiency, decimal AHP - air horsepower, hp BHP - brake horsepower, hp Air Horsepower Q V H AHP = --- 33,000 where:

AHP - air horsepower, hp Q - airflow rate, cfm

V - specific weight of air, lb/ft3

H - total head, ft Brake Horsepower Q P_a BHP = --- 6360 f where: BHP - brake horsepower, hp Q - airflow rate, cfm

P_a - static pressure, in. water

_f - fan efficiency, decimal

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Fan Laws D₂5 _N 2 3 HP2 = HP1 --- --- D₁5 _N 13 where: HP – fan horsepower, hp D - fan diameter, in. N - speed of impeller, rpm Fan Laws Q₁1/2 _H 2 3/4 N2 = N1 --- --- H₁3/4 _Q 2½ where: N – impeller speed, rpm H - fan head, in. H₂0 Q - air flow rate, cfm Propeller Fan Pitch

P = 2  r tan  where:

P - pitch in. r - fan radius, in.

 - angle of fan blade twist, deg

Fan Laws H₁1/4 _Q 2 1/2 D2 = D1 --- --- Q₁1/2 _H 2¼ where:

D – impeller diameter, in. H - fan head, in. H₂0 Q - air flow rate, cfm