NOVATEUR PUBLICATIONS

INTERNATIONAL JOURNAL OF INNOVATIONS IN ENGINEERING RESEARCH AND TECHNOLOGY

[IJIERT]
ISSN: 2394-3696
VOLUME 2, ISSUE 4APR.-2015

BIDIRECTIONAL MIXER

Mr. Raghunath Rajaput
Mr. Tamboli Najirkhan
Prof. S.T Waghmode
Department of Mechanical Engineering,
Shivaji University / A.I.T.R.C. Vita, India

ABSTRACT

In conventional method of mixing the metal oxide powder and vehicle mixing is
carried out on Unidirectional Stirring Machine The stirrer of conventional machine
rotates in one direction only which creates a particular flow pattern in the fluids hence
the particles tend to stick to the walls of container owing to the centrifugal force rather
than mixing thoroughly in mixture of paint, ultimately results into poor quality mixture
of paints there by poor quality output of paint .In order to have a homogeneous mixing
would be appropriate to have a directions of rotation of stirrer shaft which will rotate
stirrer blades in opposite directions in one cycle this will form turbulent flow pattern
there by leading to creation of irregular flow pattern and resulting into thoroughly
mixed paint mixture preparation which will create the good quality paint.

KEY WORDS: Unidirectional, Homogeneous mixture, Uniform, Periphery of blades
Shape, oscillating motion

INTRODUCTION
Process industries like chemical plants, food processing plants, paint industry
etc.Largely employ mechanical mixers to carry out mixing of powders, semisolid jelly
fluids etc.Mixing is a process where powder or jellies are mixed together through in the
form of uniform mixture where stirring is the process to mix the fluid and powder to
dissolve the powder thoroughly in given mixture and form a uniform product or output.
In either of above cases thorough mixing of material is desirable to give and good and
uniform quality output. Mixing of powders of different material in order to form a
uniform product or a powder mix is quiet easy but when it is desirable to mix powder in
a fluid matter specially when the density of powder is high the problem occurs due to
heavy weight of particles of powder has a tendency to settle down, so we make
bidirectional mixer which move opposite direction in one cycle. For that motion we
using the crank and fork mechanism. Which form the turbulence in mixer and make
homogeneous mixture .Mixing is one of the qualities of the product,

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NOVATEUR PUBLICATIONS
INTERNATIONAL JOURNAL OF INNOVATIONS IN ENGINEERING RESEARCH AND TECHNOLOGY
[IJIERT]
ISSN: 2394-3696
VOLUME 2, ISSUE 4APR.-2015

EXPERIMENTAL SRTUP
The mechanism as shown in the figure is developed to produce an oscillating motion in
the vertically suspended output shaft trough the continuously rotating horizontal input
shaft. The input shaft carries an input crank that engages with inputshaft at one end and
the fork at other. The fork is coupled to output shaft by means of fork pin. During 0 to
180 degree rotation of the input shaft the crank and the fork together make output shaft
to rotate in clockwise direction by 60 degrees, whereas during 180 to 360 degrees of
input the output changes direction and returns to mean position

Fig 1.Auto CAD of Assembly Set-up

Fig 2.Actual set-up

DESIGN PROCEDURE OF BIDIRECTIONAL MIXER

Careful design approach has to be adopted. The total design work has been split
up into two parts;
System design
Mechanical Design.
SYSTEM DESIGN:
System design mainly concerns the various physical constraints and ergonomics,
space requirements, arrangement of various components on main frame at
system, man and machine interactions.

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ISSN: 2394-3696
VOLUME 2, ISSUE 4APR.-2015

MECHANICAL DESIGN:
For design parts a detailed design is done & designation thus obtain are compared to
the next highest dimension which is ready available in market. The processes charts are
prepared & passed on to the work are specified. The parts to be purchased directly are
selected from various catalogues & specification so that anybody can purchase the
same from the retail shop with the given specifications
SELECTION OF MOTOR
Input data (Ref. www.engineering toolbox.com)
1. Kinematic viscosity of Paint = 2.4 poise
=2.4/0.01 centipoise
= 240 centipoise
2. Specific gravity of paint = 1.59 kg/lit
In design of spatial agitator the approach to design would be to calculate the torque
required at the output shaft for stirring, and based on this torque selecting an appropriate
motor after incorporating a suitable factor of safety. The torque calculation will be
based on two analogies namely; torque required to overcome the viscous force by virtue
of the fluid viscosity and secondly the torque required to overcome the static total
pressure on each blade owing to the stationary fluid i.e., paint.
Output shaft

Total torque on
= Torque owing
+ Torque owing
to viscous force
Static pressure

Calculation of Torque Owing To Viscous Force At Periphery Of Blades

The blade tip traces a loci of points which is a circle; hence the motion of the bracket
due to oscillation of the output shaft can be considered to be a cylinder (assuming blade
angle =00), which is moving against another cylinder i.e., the container both separated
by a fluid film of thickness of 30 mm.
We can put up the above problem as follows;
Problem: Find the force and power required to move a shaft of diameter 10 cm against
a journal of internal diameter 14 cm, separated by a fluid of kinematic viscosity 2.4
poise. Shaft rotates at 80 rpm.
Solution:

Given: = 2.4 poise = 1/10*2.4 = 0.24 Ns/m2

Speed of shaft = 80 rpm
Tangential speed of shaft = u = DN/80
= x 0.10 x 80/60

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=.0418 m/sec
Now,
= du/dy
Where;
= Shear stress (N/m2)
du = Change in speed = u-0 = 0.418 m/sec
dy = Distance between shaft and journal = 0.01m
= 0.24 x 0.418/0.01 = 10.032 N/m2
Area of the cylinder that is exposed to this shear intensity will be the circumferential
area ;( assuming width of blade = 40 mm)
A= c x D x w
= x 0.1 x 0.04
=0.012 m2
Shear force (F) = Shear stress x Shear area
= 10.032 x 0.012
= 0.126 N
Power = F x u
=0.126 x 0.418
=0.0526 Watt
Calculation of Torque owing to viscous force at top and bottom ends of blades
The blades along the length when rotated along with bracket will trace an annular ring
at the either ends of the blade.
We can put up the above problem as follows;
Problem: Find the force and power required to move a plate of width 5 cm and length
100 cm against a stationary plate extending infinitely, separated by a fluid of kinematic
viscosity 2.4
Poise at a distance of 1 cm. Plate moves at 0.418 m/sec.

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ISSN: 2394-3696
VOLUME 2, ISSUE 4APR.-2015

Solution :

Given : = 2.4 poise

= 1/10*2.4
= 0.24 Ns/m2
Speed of plate = 0.418 m/sec
Now,
= du/dy
Where;
= Shear stress (N/m2)
du = Change in speed = u-0 = 0.418 m/sec
dy = Distance between shaft and journal = 0.01m
= 0.24 x 0.418/0.01
= 10.032 N/m2
Area of the cylinder that is exposed to this shear intensity will be
A=LxB
= 1 x 0.05
=0.05 m2
Shear force (F) = Shear stress x Shear area
= 10.032 x 0.05
= 0.528 N
Total shear force = 3 x F
= 1.584 N
Power = F x u
=1.584 x 0.418 = 0.662 Watt

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INTERNATIONAL JOURNAL OF INNOVATIONS IN ENGINEERING RESEARCH AND TECHNOLOGY
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ISSN: 2394-3696
VOLUME 2, ISSUE 4APR.-2015

Calculation of Torque owing to Static total pressure acting on the blades by virtue
of the stationary fluid:
In calculation of torque due to static force exerted by the fluid we use the following
analogy;
Problem:Determine the total pressure on a flat plate of length 50and width 40mm
which is placed vertically in such a way that the centroid of plate is at a distance of
100mm below the free surface of fluid of specific gravity 1.59kg/lit
Solution :

Given : sp gr. =1.59 kg / lit=1.59*1000kg/m3

Total pressure is given by;
F = g A h
=1590 x 9.81 x0.05 x0.04 x 0.1
=3.115N
There are three such blades,
Thus the total force = 3 x 3.115
= 9.35 N

The torque that each pinion has to overcome to rotate about its own axis is given by;
T = 9.35 x 0.04
=0.374 N-m
Power required at the output shaft to overcome the static resistance of fluid is ,
Ps = 2NTs/60
=2 x 3.142 x80 x0.374/60
=3.13 Watt
Thus the net power required at the output shaft is the summation of the above three
powers;
Pnet= 0.0526+0.662 +3.13
=3.8446 Watt approximately 4 watt
The mechanism used for converting rotary power into oscillatory energy is tested for its
efficiency so assuming only 50% efficiency and assuming that further power is lost in
friction at the bush bearings we shall assume overall efficiency to be 30 % only thus
power required by machine will be 28 watt approximately. As we intend to run
machine at various speeds hence we shall employ a commentator motor speed of which
can be varied by placing an rheostat in series.

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DESIGN OF OPEN BELT DRIVE

Motor pulley diameter (D2) = 10 mm
IP _ shaft pulley diameter (D1) = 36 mm
Reduction ratio = 5
Coefficient of friction = 0.23
Maximum allowable tension in belt = 200 N
Center distance (X) = 182
V = DN/(60 x 1000)
= 4.188m/sec
Sin

R1 - R2
X

Sin =(18 5) / 182

So,
=9.8069
= (180 2) /180
= (180 29.8069) /180
= 2.7964rad
Now,
2.3log [T1/T2]=
=0.232.7964
=0.64319
log[T1/T2]=0.64312.3
=0.2796
T1/T2=1.9039
=200/1.9039
T2=136N

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Result Table
Tension in tight side of belt (T1) = 200 N
Tension in slack side of belt (T2) = 136 N

DESIGN OF INPUT SHAFT.

MATERIAL SELECTION:Ref: - PSG (1.10 & 1.12) + (1.17)
DESIGNATION

EN 24

ULTIMATE TENSILE
STRENGTH N/mm2

YEILD STRENGTH
N/mm2

900

700

Table1: Material Selection

ASME CODE FOR DESIGN OF SHAFT.
According to ASME code permissible values of shear stress may be calculated form
various relations.
Fsmax = 0.18 fult
= 0.18 x 900
= 162 N/mm2
OR
Fsmax = 0.3 fyt
=0.3 x 700
=210 N/mm2
Considering minimum of the above values;
Fsmax = 162 N/mm2
Shaft is provided with key way; this will reduce its strength. Hence reducing above
value of allowable stress by 25%
Fsmax = 121.5 N/mm2
This is the allowable valve of shear stress that can be induced in the shaft material for
safe operation.

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CHECK FOR TORSIONAL SHEAR FAILURE OF SHAFT.

Assuming minimum section diameter on input shaft = 16 mm
d = 16 mm
Td =/16 x Fsact x d3
Fsact = (16 x Td) / x d 3
= (16 x 1.19 x 10 3) / x (16) 3
Fs act = 1.47 N/mm2
As

Fs act < FSs all

I/P shaft is safe under torsional load.

SELECTION OF BEARING
Shaft bearing will be subjected to purely medium radial hence we shall use ball
bearings for our application.
Selecting; Single Row deep groove ball bearing as follows Series 62

No

Bearing of basic D
design no (SKF)

D1

D2

Basic capacity

17BC02

6203

21

40

36

12

4440

17

7500

Table2: Selection of Bearing

P = X Fr+ Y F a
For our application F R Belt Tension =196+49 = 245 N
P = X Fr+ Y F a
As; F a/ Fr< e X =1, Y=1
P = Fr
Max radial load = Fr =245 N.

P=245N
Calculation dynamic load capacity of brg

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( C ) p, where p= 3 for ball bearings

L=

P
When P for ball brg
For m/c used for eight hr of service per day;
LH = 12000- 20000hr
But;
L10= (60 n L10H) / 106
L10 = (6080012000) /106
L10=576 mrev
Now;
L10=(C/P) ^1/p
576=(c/245) ^1/3
C = 2038.48 N
Here,
C = 2038.48 N < 4440 N
As the required dynamic capacity of brg is less than the rated dynamic capacity of
brg;

DESIGN OF OUTPUT SHAFT.

MATERIAL SELECTION: Ref:- PSG (1.10 & 1.12) + (1.17)
DESIGNATION

EN 24

ULTIMATE TENSILE YEILD STRENGTH

STRENGTH N/mm2

N/mm2

900

700

Table3: Material Selection

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NOVATEUR PUBLICATIONS
INTERNATIONAL JOURNAL OF INNOVATIONS IN ENGINEERING RESEARCH AND TECHNOLOGY
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ISSN: 2394-3696
VOLUME 2, ISSUE 4APR.-2015

ASME CODE FOR DESIGN OF SHAFT.

According to ASME code permissible values of shear stress may be calculated from
various relation.
Fs max = 0.18 fult
= 0.18 x 900 = 162 N/mm2
OR
Fs max = 0.3 fyt
=0.3 x 700
= 210 N/mm2
Considering minimum of the above values

Fs max = 162 N/mm2

Shaft is provided with key way; this will reduce its strength. Hence reducing above
value of allowable stress by 25%

Fs max = 121.5 N/mm2

This is the allowable valve of shear stress that can be induced in the shaft material for
safe operation.
CHECK FOR TORSIONAL SHEAR FAILURE OF SHAFT.
Assuming minimum section diameter on input shaft = 16 mm
d = 16 mm
Td = /16 x Fs act x d3
Fsact = (16 x Td) / x d 3
= (16 x 1.19 x 10 3)/ X (16) 3
Fs act = 1.47 N/mm2
As
Fs act <Fs all
I/P shaft is safe under tensional load

ADVANTAGES
1. Stirrer has bi-directional ie,it rotates in both directions; this gives uniform mixing.
2. Quality of mixing is very high
3. Low cost of production because it does not require an gear box.
4. Fast production rate
5. Compact size so minimal space requirements

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APPLICATIONS
1. Mixing of multiple colour paint in paint industry.
2. Mixing of metallic powders in pigment in preparation of ionic paints.
3. Can be used as skimming machine.
4. Dairy applications with suitable change in stirrer material.
5. Mixing applications in pharmaceutical industry.

FUTURE SCOPE
1) In this project if you connect the two mutually perpendicular stirrer which moving in
perpendicular direction then it increase the efficiency of Bi-directional mixer.
2) Increasing the angle of opposite side rotation increasing the efficiency

MIXING SYATEM
The mixing system consists of the container and impeller spun by a motor drive .The dc
motor is mounted on the base plate. With help of the pulley and belt transfer the it
motion to the stirrer. Container is glass beaker. When undergoes mixing the upper side
of the tank closed with plate the mixing is undergoes for paint and lassi as fallow.
1. First tank filled with 500grm oil bond then in it added the seven drop of the sterner to
identify proper mixing with help of the colour. Then output shaft take place inside the
tank centrally. Then start the switch varies speed of the motor.
2. For another mixing system we take curd, water, sugar. Firstly we take 350ml water
in container then add the 500grm crud and add 100grm sugar. Then outer shaft place in
the container centrally and on the switch, and varies speed of the motor.

MIXING RESULT
The mixing result for paint and lassi as fallow.
1. By mixing observation we get that oil bound properly mix with colour and we get
the homogeneous mixture of the oil bound and sterner.
2. The curd, sugar water properly mix and form the homogeneous mixture. Sugar will
not seetal down at the bottom due to turbulence created in the container.

CONCLUSION
Mixing process has been performed which conform that the proposed mixing prevent
the formation of segregated region hence shorten the mixing time than other mixing
method (constant speed, manual mixing, sinusoidal bidirectional). Also by using the bidirectional mixer in container create turbulent flow of mixture and we get the
homogeneous mixture.

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NOVATEUR PUBLICATIONS
INTERNATIONAL JOURNAL OF INNOVATIONS IN ENGINEERING RESEARCH AND TECHNOLOGY
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ISSN: 2394-3696
VOLUME 2, ISSUE 4APR.-2015

REFERENCES
[1] .S.Ye. And T. Chau (2007),Chaoization of DC Motors for Industrial MixingIEEE
Transaction on Industrial Electronic, Vol-54, No 4, August 2007
[2]. Konstaintain Anatolyevich Yudin (2013):-Modelling Mixers Gyroscopic
Type Middle-East Journal of Scientific Research 17 (8):1125, 2013
[3].Narayanan, S., Bhatia, V. K., Guha, D. K. and Rao, M. N (1969), "Suspension of
Solids by Mechanical Agitation" diem Eng Sci, Vol 24, pp 223-230, 1969
[4]. Shamlou, P. A. and Zolfagharian, A (1987)., "Incipient Solid Motion in Liquids in
Mechanically Agitated Vessels" I Chem E Symposium Series No. 108, pp 195-208, 1987
[5].Voit, H. and Mersmann, A., "General Statement for the Minimum Stirrer Speed
During Suspension" German Chem. Eng, Vol 9, pp 101-106, 1986
[6]. Design of Machine Elements:V. B.Bhandari
[7] P S G Design Data

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