animals
b. improve its digestability c. prolong its storage period d. None of the above
42. A machine that separates paddy from brown rice.
a. Screen sifter b. Paddy separator c. Paddy husker d. Rice whitener e. None of the above
43. The capacity of a compartment- type separator is ___.
a. 20-30 kg. brown rice per hour b. 40-60 kg. brown rice per hour c. 70-80 kg. brown rice per hour d. None of the above
44. A one-ton-per-hour rice mill is to be designed. The milling recovery is 68% while the hulling efficiency is 90%. What size of paddy separator will you
recommend?
a. 26 compartments b. 32 compartments c. 40 compartments d. None of the above
Hulling Coefficient Ch = Wbr / Wp where:
Ch – hulling coefficient, decimal Wbr – weight of brown rice, grams Wp – weight of paddy, grams
Wholeness Coefficient Cw = Wwbr / Wbr where:
Cw – wholeness coefficient, decimal Wwbr – weight of whole brown rice, grams Wbr – weight of brown rice, grams Hulling Efficiency
h = Ch Cw where:
h – hulling efficiency, decimal Ch – hulling coefficient, decimal Cw – wholeness coefficient, decimal
Percentage Brown Rice Recovery % BRR = (Wbrr / Wp ) x 100 where:
%BRR – percentage brown rice recovery, % Wbrr – weight of brown rice, kg
Wp – weight of paddy, kg Percentage Broken Milled Rice
%BR = (Wbr / Wmr) 100 where:
%BR – percentage broken rice, % Wbr – weight of broken rice, kg Wmr – weight of milled rice, kg Throughput Capacity Ct = 0.2 Wp / To : brown rice Ct = [Wp MR]/To: milled rice where: Ct - throughput capacity, kg/hr Wp - weigh t paddy input, kg To - operating time, hr
MR – milling recovery, decimal 0.60 to 0.69 Percentage Brewer’s Rice
%BrR = (Wbrr / Wmr ) 100 where:
%BrR – percentage brewer’s rice, % Wbrr – weight of brewer’s rice, kg Wmr – weight of milled rice, kg
Head Rice Recovery %HR = (Whr / Wmr ) 100 where:
%HR – head rice recovery, % Whr – weight of head rice, kg Wmr – weight of milled rice
Rice Milling
Milling Recovery
% MR = ( Wmr / Wp ) 100 where:
% MR – milling recovery, % Wmr – weight of milled rice, % Wp – weight of paddy, kg
Speed of Low Speed Rubber Roller Ns = Nh - [0.25 / Nh]
where:
Ns - speed of slower rubber roller, rpm Nh - speed of faster rubber roller, rpm
Number of Compartments for Paddy Separator NC = Cb / 40 : long grain NC = Cb / 60 : short grain where: NC - number of compartments
Cb - throughput capacity, kg brown rice per hour
Number of Brake for Vertical Abrasive Whitener
NB = [D / 100] : Germany
NB = [D / 100] : Italy where:
NB – number of brakes, units D - cone diameter, mm
Problem 1
A single-pass rice mill was tested for 30 min. and gave the following results: input paddy, 1000 kg; milled rice produced, 650 kg; broken rice produced, 150 kg; and weight of head rice, 200 kg. Compute the following: (a) Throughput capacity, (b) % Milling recovery, (c) % Broken grains, and (d) % Head rice recovery.
Given: Wp - 1000 kg Wmr - 650 kg Whr - 200 kg Wbr - 150 kg Time - 30 min.
Required: Throughput capacity % Milling recovery % Broken grains
% Head rice recovery
Solution:
Throughput Capacity = Wp x MR / To
= 1000 kg x 0.65 / 30 min/60 min/hr = 1300 kg milled rice per hr
% Milling recovery = Wmr 100 /Wp = 650 kg x 100 / 1000 kg = 65% % Broken rice = Wbr 100 / Wmr = 150 kg x 100 / 650 kg = 23%
% Head rice recovery = Whr 100/Wmr = 200 kg x 100 / 650 kg = 31%
Problem 2
Five tons of paddy milled in 6 hours produces 3950 kg of brown rice and 3250 kg of milled rice. What is the milling recovery?
Given:
Wt of paddy - 5 tons Milling time - 6 hors Wt of brown rice - 3950 kg
Wt of milled rice - 3250 kg Required: Milling Recovery Solution:
MR = Wt milled rice x 100 / wt of paddy = 3250 kg x 100 / 5000 kg
= 0.65 x 100 = 65% Problem 3
What is the hulling coefficient of the huller in Problem 2?
Given:
Weight of brown rice - 3950 kg Weight of paddy - 5000 kg Required: Hulling Coefficient
Solution:
Coef. of hulling = Wt of brown rice/Wt of paddy = 3950 kg / 5000 kg
= 0.79
Problem 4
If the head rice recovery of the paddy in Problem 2 is equal to 85%, what is the amount of broken grains?
Given:
Head rice recovery - 85% Required: Weight of milled rice Solution: Wt. of broken grains = Wt mr (100 – HRR) = 3250 (100 – 0.85) = 487.5 kg Problem 5
Referring to Problem 2, what is the amount of rice hull produced during milling?
Given:
Wt. of paddy - 5000 kg Required: Weight of rice hull Solution:
Wt. of RH = Wt. of paddy x 20% = 5000 kg x 0.20 = 1000 kg
Problem 6
A disk huller with 600-mm diameter will be installed as return huller for a rice milling plant. As an Agricultural Engineer, determine the diameter of the pulley required for the huller. The motor for the huller will have 4-inches pulley diameter and is expected to run at 1740 rpm.
Given:
Diameter of huller - 600 mm Motor pulley - 4 in. Motor speed - 1740 rpm Required: Huller pulley diameter Solution:
Huller rpm = (14 m/s ) (60 s/min) /[ 3.14 (0.6 m)] = 445.8 rpm
Huller Pulley D = 4 in (1740 rpm) / 445.8 rpm = 15.6 in. use 16 in.
Problem 7
A rice milling plant is to be designed to run at 5tons-per-hour throughput rate. The design milling recovery is 69% while the hulling efficiency is 95%. How many compartments are needed for the paddy separator of the rice mill? Assume a 50-kg br/hr-comp.
Given:
Throughput capacity - 5 tph Milling recovery - 69% Hulling efficiency - 95% Required: No. of compartments Solution:
C p = 5 tph / 0.69 = 7.25 tph of paddy C br = 7.25 tph x 0.95 = 6.9 tph brown rice
No. of compartments = 6.9 tph x 1000 kg/ton x 1 comp/50 kg-hr = 138 compartments
Problem 8
A rubber roll husker was tested to determine its performance. The husker ran for 2 hours at a rate of 1000 kg of input paddy. The weight of brown rice obtained after passing the husker was 805 kg. The weight of whole brown rice obtain per kg of sample was 900 grams. What is the hulling coefficient, wholeness coefficient, and hulling efficiency of the huller? What is the throughput capacity of the husker, in kilogram per hour of brown rice?
Given:
Weight of paddy - 1000 kg Operating time - 2 hours Weight of brown rice - 805 kg Weight of whole brown rice - 900 g per kg of brown rice
Required:
Hulling coefficient, wholeness coefficient, husking efficiency, and throughput rate Solution:
Hulling coefficient = 805 kg/1000 kg = 0.805
Wholeness coefficient = (900 g brown rice x 1kg/1000 g) x 805 kg brown rice / 805 kg brown rice
= 0.9
Hulling efficiency = (0.805 x 0.90) x 100 = 72.45%
Throughput capacity = 805 kg / 2 hours = 402.5 kg/hr
1. The process of reducing corn kernel into grits, germ, and pericarp with or without conditioning.
a. Dry milling b. Hammer milling c. Attrition milling d. None of the above
2. The major component of a corn mill that reduces corn kernels into grits. a. Burr mill
b. Hammer mill c. Steel roller mill d. All of the above
3. The milled corn kernel where the outer covering and germs have been removed and with particle size of not less than 0.86 mm.
a. Broken corn kernel b. Cracked corn kernel c. Corn grits
d. None of the above
4. The main product in milling corn.