1. Introduction

2. Materials and Methods

2.1. Description of the Existing Fish Feed Pelletizing Machines

The existing hand operated fish feed pelletizer is designed for small scale commercial operation. The pelletizing unit consist of a compression chamber, an auger, a die plate and a hand driven pulley. The parts are as shown in Plate 1.

a. 

Compression Chamber

The compression chamber or barrel (Plate 1A) is the housing where the pressure generated is mounted on the material been pelletized. It is made of cast iron and of 10 mm in thickness with an internal diameter of 80 mm and 200 mm in length. It has a hopper opening of 90 mm in diameter at the top which serve as inlet for materials to be pelletized and support for the attachment of the barrel to the frame of the machine.

b. 

Auger

The auger or screw shaft (Plate 1B) conveys material from the bottom of the hopper inside the barrel to the die plate. It is made from cast iron and is 200 mm in length and 25 mm in diameter with nominal diameter of 70 mm, flight of 22.5 mm and pitch of 50 mm. The auger has a clearance of 10 mm between the flight of the auger and the wall of the compression chamber.

c. 

Die Plate

The die plate (Plate 1C) serves as the back wall for retaining the pressure exerted by the auger whilst the perforations on the plate or die holes allow the compressed mash to be forced out of the barrel to form pellets. The die plate is made of mild steel plate, 100 mm in diameter and thickness of 6 mm. The die plate contains 48 cylindrical die holes of 6 mm in diameter each bored onto the plate with a key way cut on its edge which fits to a protruding key on the body of the compression chamber, to prevent the die plate from rotating with the auger and also prevent relative motion between the die plate and compression chamber wall.

Plate 1: Parts of the existing pelletizing machine; (A) compression chamber (B) Auger of the pelleting unit (C) pelleting die.

2.2. Design of Other Components

The other component parts of the modified machine that were designed and fabricated for incorporation with the pelletizing unit are the hopper, power transmission unit (belt drive) and the frame.

2.2.1. Design of Hopper Extension

The hopper extension was designed in the form of cone frustum to allow the hopper extension fit into the opening on the compression chamber of the pelleting unit, with top diameter of 435 mm, base diameter of 90 mm and a height of 280 mm. The base diameter was selected to allow it fit into the chamber opening and slanting at 60 ⁰ angle of repose for fibrous non-free flowing materials [28]. The volume of the hopper was calculated using Equation 1 [29] for volume of a cone frustum thus;

$$V=\raisebox{1ex}{$\pi h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(R^2+Rr+r^2\right)$$

(1)

Where:

R = top radius = 217.5 mm
r = base radius = 45 mm
h = height of frustum = 280 mm

$$V=\frac{3.142\times 280}{3}\left[{217.5}^2+\left(217.5\times 45\right)+{45}^2\right]=0.0173m^3$$

2.2.2. Power Required to Drive the Screw Conveyor

The power required to drive the machine was evaluated using Equation 2 as presented by Singh (2003).

$$P=\frac{QL\left(W_0+\sin \beta \right)}{3600\eta },$$

(2)

Where:

P = power required to drive the machine, kW

Q = capacity of screw conveyor, kg/h.

L = length of screw conveyor = 200 mm (measured).

W_o = material factor = 2.5

β = angle of inclination of screw to the horizontal, 0^o (horizontal screw conveyor) [30]

ᵑ = efficiency of transmission = 0.92.

The capacity of screw conveyor, Q is calculated from Equation 3 as given by Singh (2003)

$$Q=150\pi D^{2}sn\phi \rho C,$$

(3)

Where:

D = nominal diameter of the screw conveyor = 80 mm (measured)

s = pitch of screw conveyor = 0.08 m (for standard screw pitch, S=D) mm

n = speed of rotation of screw = 250 rpm [31]

$\phi$ = loading efficiency = 0.25

ρ = density of material = 1200 kg/m³ [32]

C = inclination factor =1 (for horizontal conveyors) [30]

Hence, from Equation 3,

Capacity,$Q=150\pi \times {0.08}^2\times 0.08\times 250\times 0.25\times 1200\times 1$

$$Q=18.09kg/h$$

Power required, P from Equation 2

$$P=\frac{18095.574\times 0.2\times \left(2.5+\sin 0\right)}{3600\times 0.92}=2.73kW$$

Recall that 1 hp is equal to 746 W, 2.73 kW is equivalent to 3.76 hp, Therefore, a 4 hp electric motor was selected.

2.2.3. Design of Belt Drive System for Power Transmission

(a) 

Speed of driven pulley

A v-belt drive system was designed for the machine and was powered by a 4 hp electric motor with rated speed of 1400 rpm and a driving motor pulley of 100 mm. The pelleting unit has a driven screw conveyor pulley diameter of 300 mm. The speed of the driven pulley is calculated from the velocity ratio of belt drive using Equation 4 [33].

$$\frac{N_1}{N_2}=\frac{d_2}{d_1},$$

(4)

Where:

N₁ = speed of motor pulley, rpm (1400 rpm)

N₂ = speed of screw pulley, rpm

d₁ = diameter of motor pulley, mm (100 mm)

d₂ = diameter of screw pulley, mm (300)

$$\frac{1400}{N_2}=\frac{0.3}{0.1}$$

Speed of the driven pulley, $N_2=466rpm$According to IS:2494-1974 standard, type A belt was selected (Khurmi and Gupta, 2005) for the machine with the following characteristics;

• Power range = 0.7-3.5 kW
• Top width (b) = 0.013 m
• Thickness (t) = 0.008 m
• Cross sectional area= b × t = 0.000104 m²
• Coefficient of friction = 0.25
• Density of rubber belt = 1000 kg/m³
• Permissible stress = 2.8 MPa

(b) 

Belt velocity

The velocity, v of the belt is calculated from Equation 5 [33].

$$V=\frac{\pi d_1{N}_1}{60},$$

(5)

Where:

V= velocity of the belt, ms^-1

d₁ and N₁ are as defined in Equation 4

$$V=\frac{\pi \times 0.1\times 1400}{60}=7.33m/s$$

(c) 

Centrifugal tension of belt

Centrifugal tension of the belt was calculated from Equation 6 [33];

$$T_c={mv}^2,$$

(6)

Where:

T_c = centrifugal tension, N

m = mass per unit length of belt, kg

v = velocity of belt = 7.33 m/s (form Equation 5)

Mass of belt per unit length is calculated from Equation 7 thus;

$$mass=A\times l\times \rho ,$$

(7)

Where:

A= belt cross sectional area = 0.000104m² (IS:2494-1974 standard)

l = unit length of belt (m)= 1.00 m

$\rho$ = density of belt material = 1000 kg/m³ (IS:2494-1974 standard)

$$mass=0.000104\times 1\times 1000=0.104kg$$

Centrifugal tension, T_c from Equation 6.0

$$T_c=0.104\times {7.33}^2=5.59N$$

(d) 

Tension in belt

Tension in the tight side of the belt is calculated from Equation 8;

$$T_1=T-T_c,$$

(8)

Where:

T₁=tension in the tight side of the belt, N

T = maximum tension in belt, N

but maximum tension in the belt is calculated from Equation 9 (Khurmi and Gupta, 2005)

$$T=\sigma \times A,$$

(9)

Where:

σ = permissible stress, 2.8 MPa.

A = cross sectional area, 0.000104 m²

$$T=2.8\times {10}^6\times 0.000104=29.12N$$

Tension in tight side of belt,

$$T_1=29.12-5.59=23.53N$$

Tension in the slack side of the belt is calculated from Equation 10 (Khurmi and Gupta, 2005)

$$2.3\log \left(\frac{T_1}{T_2}\right)=\mu .\theta \csc \beta ,$$

(10)

Where:

T₂ = tension in slack side of belt (N)

µ = coefficient of friction, 0.25

θ = angle of contact, rad.

β = groove angle of belt, 16^o (IS:2494-1974 standard)

Angle of contact, $\theta$ is calculated from Equation 11 [33]

$$\theta =180-2\alpha ,$$

(11)

Where:

$\alpha$ = angle of lap

and α is calculated using Equation 12 thus

$$\sin \alpha =\frac{d_2-d_1}{2C},$$

(12)

Where:

d₁= diameter of driving pulley, 0.1 m

d₂ = diameter of driven pulley, 0.466 m

C = centre distance between the two pulleys, m.

Centre distance between the two pulleys, C is calculated using Equation 13

$${C=\left(2d_2\left(d_1+d_2\right)\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.},$$

(13)

Where:

d₁ = diameter of driving pulley = 0.1 m

d₂ = diameter of driven pulley = 0.466 m

C = Centre distance between pulleys, (mm)

Then, Center distance between the two pulley

$$C=(2\times 0.446\left(0.1+0.446\right){)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$$

$$C=0.698m\approx 698mm\approx 700mm$$

From Equation 12,

$$\sin \alpha =\frac{0.466-0.1}{2\times 0.698}$$

$$\alpha ={\sin }^{-1}0.2479={14.35}^o$$

Angle of contact from Equation 11

$$\theta =180-2\left(14.35\right)={151.30}^o=2.64rad$$

Tension in slack side of belt, from Equation 10.0;

$$2.3\log \left(\frac{23.53}{T_2}\right)=0.25\times 2.64\csc 16$$

$$T_2=1.35N$$

(e) 

Power transmitted by belt

Power transmitted by the belt is calculated using Equation 14;

$$P=\left(T_1-T_2\right)v,$$

(14)

Where:

P = power transmitted by belt, Nm/s

T₁ = tension in the tight side of belt, 23.53 N

T₂ = tension in the slack side of belt, 1.35 N

v = velocity of belt, 7.33 m/s

$$P=\left(23.53-1.35\right)7.33=162.58Nm/s$$

(f) 

Torque transmitted by belt

The torque in the driven pulley shaft is calculated from Equation 15 [33]

$$T=\frac{P\times 60}{2\pi N_2},$$

(15)

Where:

T = torque transmitted by the belt, Nm

N₂= speed of driven pulley, 466 rpm

P = power transmitted by belt, 162.58 W

$$T=\frac{162.58\times 60}{2\pi \times 466}=3.33Nm$$

3. Results and Discusssion

5. Conclusions

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