World Grain - May 2014 - 60

FEED OPERATIONS

Pellet cooler functions
Adjustments can be made to
improve the cooling system
by Fred Fairchild

Pellet cooling is a dynamic process that involves simultaneous heat and mass transfer between the pellets and the cooling air. Pellet cooling occurs as a result of both evaporative
cooling and convective cooling. Evaporative cooling is the
transfer of water from the pellets to the air which results in
both moisture reduction and cooling in the pellets. Convective
cooling also occurs and depends on the temperature difference
between the pellets and the air, the amount of pellet surface
area, and the heat transfer coefficient.
Pellet cooling system performance is affected by many variables. The first are variables found in the air being used and include the air temperature and relative humidity, and the amount
of air passing through the cooler. Pellet variables include the pellet temperature, moisture content, pellet size and density, fines
present, and pellet quantity and quality. Cooler variables include
product bed depth, uniformity of bed depth and system design.
The cooling system must meet several requirements. It must
be designed to effectively cool a range of products by removing
enough heat and moisture for further handling or storage of the
pellets. But, it must also be able to do these tasks while avoiding overdrying of the pellets. Removal of more moisture than
required results in removing extra mass (weight) that results in
losing selling weight. The cooling system itself must be reliable and provide gentle handling of the pellets that minimizes
creation of fines in the process. In some situations, counter-flow
coolers may be equipped with heat exchangers on the entering
air to heat it so greater product drying may be accomplished.
Regardless of the cooler type - horizontal, vertical or
counter-flow - it is very important that the bed depth be
uniform where air passes through the product to keep the air
flowing equally through all the product (see Figure 1, right).
COOLER SYSTEM DESIGN
Assume that it is desired to cool 25 tons per hour of pellets.
If we use 400 cubic feet per minute (CFM) of air per ton to
do the cooling, the amount of air needed is 25 x 400 = 10,000
CFM. The quantity of air used and the cross section area of
the cooler and connecting duct work determine the velocity
of the air. The velocity is determined by the formula Q = VA.
Q = Quantity of Air (CFM), V = Velocity of the air in feet
per minute (FPM), A = Cross-section area of the duct work in
Square Feet.
Assume you are using a Bliss Industries model 10-202-6
60

Figure 1: Bad depth/uniformity
Good

Poor

Op-Flo counter-flow pellet cooler to cool the 25 tons per of
pellets requiring 10,000 cfm of air.
The body of the cooler is round and is 86 inches in diameter.
The cross section area of the body is 40.2 square feet. ( A = π
x r2) Diameter = 2r. Radius r = Diameter/2. Radius (r) for the
cooler body = 86in./2 = r = 43in.; we need to convert this into
feet = 43 in/(12 in. per foot) = r = 3.58 ft. π = 3.14. The cooler
area = π x 3.582 = 40.2 square feet. Using the formula Q = V x
A, we find the velocity (V) of the air passing through the cooler.
V = 10,000 cfm/40.2 square feet = 249 feet per minute.
The exhaust port for the cooler is 10 inches x 36 inches.
Area = Width x Length = 10 x 36 = 360 square inches. You
convert this to square feet by dividing by 144 square inches in
a square foot. A = 360/144 = 2.5 square feet. The velocity of
the air through the exhaust port would be V = Q/A = 10,000
cfm/2.5 square feet = 4,000 feet per minute.
The same information may be used for any shaped cooler
and exhaust outlet to determine the velocities of the air.
The ductwork beyond the cooler is best when of round construction to eliminate any corners in the ductwork. For the
ductwork from the cooler described above, a rectangular to
round transition should be installed to convert the ductwork
to round. Again, you use the formula Q = V x A where Q =
10,000 cfm and V = 4000 fpm. A = Q/A = 10,000/4,000 = 2.5
square feet. Using the formula for a circle,
A = π x r2, 2.5 = 3.14 r2, r2 = 2.5/3.14 = 0.796, r = square root
May 2014 / World Grain / www.World-Grain.com

http://www.World-Grain.com

World Grain - May 2014

Table of Contents for the Digital Edition of World Grain - May 2014