Introduction
The engineering design process involves a series of steps that
lead to the development of a new product or system. In this
design challenge, students are to complete each step and
document their work as they develop their lunar plant growth
chamber. The students should be able to do the following:
Firstly, Identify Criteria and Constraints -- Students should specify
the design requirements (criteria). Example: Our growth chamber
must have a growing surface of 10 square feet and have a
delivery volume of 3 cubic feet or less. Students should list the
limits on the design due to available resources and the
environment (constraints). Example: Our growth chamber must
be accessible to astronauts without the need for leaving the
spacecraft. Then, Brainstorm Possible Solutions -- Each student in
the group should sketch his or her own ideas as the group
discusses ways to solve the problem. Labels and arrows should be
included to identify parts and how they might move. These
drawings should be quick and brief. Also Generate Ideas -- In this
step, each student should develop two or three ideas more
thoroughly. Students should create new drawings that are
orthographic projections (multiple views showing the top, front
and
one
side)
and
isometric
drawings
(three-dimensional
depiction). These are to be drawn neatly, using rulers to draw
straight
lines
measurements
and
to
should
make
be
parts
labeled
proportional.
clearly.
Then
Parts
:
and
Explore
Possibilities -- The developed ideas should be shared and
discussed among the team members. Students should record pros
and cons of each design idea directly on the paper next to the
drawings.
Finally : Select an Approach -- Students should work in teams and
identify the design that appears to solve the problem the best.
Students should write a statement that describes why they chose
the solution. This should include some reference to the criteria
and constraints identified above.
Project description
Design motor speed reducers consist of two stages the first stage
(High speed belt transmission) the second (Low Stage gear
transmission) knowing that
Output Power: 3.2 kW
Output R.P.M: 65 r.p.m
Vertical Position.
Project requirements
This project includes design the following
1) Motor selection
2) Belt design
3) Gears design
4) Shafts Design
5) Bearing Selection
6) Keys Design
7) Case Design
8) Accessories i.e (rings, seals, etc)
Efficiency calculations
Overall efficiency = Coupling * Gear * Belt *( Bearing) 2
So the these efficiencies was assumed to be as following
output Power
3.2 Kw
Speed output 65 rpm
Coupling efficiency 0.98
Gear efficiency
0.99
Belt efficiency 0.90
Bearing efficiency
0.98
So the overall effeciency = 85.7 %
So
theinput power=
Pout
overall
The input power = 3.734 Kw
The input power = 5hp
Low voltage motor selection
The AC electric motor used in a VFD system is usually a threephase induction motor. Some types of single-phase motors can be
used, but three-phase motors are usually preferred. Various types
of synchronous motors offer advantages in some situations, but
three phase induction motors are suitable for most purposes and
are generally the most economical choice. Motors that are
designed for fixed-speed operation are often used. Elevated
voltage stresses imposed on induction motors that are supplied
by VFDs require that such motors be designed for definitepurpose inverter-fed. ABB is one of only a handful of leading
motor manufacturers in Europe to have a motor range to meet or
exceed the minimum efficiencies stated in the highest level of the
EU agreement of LV motors.
The selection motor is M3AA 160M with Power
=720 rpm
And efficiency =84.1%
4 Kw , Speed
The terminal box is made of aluminum and is located on top of
the stator as standard. It is provided with two knockout openings
(one Pg and one metric) and can be turned 4x90. Cable glands
are not included. The size of the box is
the same in size 56 and 63.also The motors are supplied with
bearings that are
lubricated for life with a bearing grease for use at
temperatures in dry or humid environments.
operating temperature is between -40
normal
The greases
and +160C. The life time of the grease L10 is defined as the
number of operating hours after which 90% of the bearings are
sufficently well lubricated. 50% of the
bearings can achieve a grease life time that is twice as long. The
maximum life time of the grease should, however, be considered
to be 40,000 hours, equivalent to around 5 years.
The Position will be set For Electric Motor
Permissible axial forces
The forces were founded using the table which is below at Model
number as shown
FAD= 4730 N
2630 N
FAZ =
Accessories
From Electric Motor:
4 kW (5.362 hp)
Rv = (720)/65
= 11.1
The reduction for the gear can be given as
following
Theoretical
Actual
Belt
1-5
1-3
When the speed ratio for the belt = 2
The gear speed ratio = 5.55
Gear
1-12
1-8
Belt drivers design
Belt drives are recommended for power trains that operate at
high speeds and drives that have shock loads. Power
is
transmitted by a flexible belt traveling around two or more
pulleys. Standard belts are flat, V, or cog (tooth). Flat belts have
limited use in modern machines because the lower
amount of
friction between the belt and pulley reduces the amount of power
that can be transmitted. The use of V-belts increases the power
that can be transmitted because of the wedging action between
the belt and the pulley, and they can operate at a higher speed.
A cogged belt is used when a belt drive is desired and
the
components must stay in time. The Rubber Manufacturing
Association, or RMA,
United States.
standards.
identifies
belts drive components in the
Other countries use the British, German, or ISO
V- Belt Service Factor Selection:
The service factor was selected as following
Pin = 5.36 hp
So the design power can be calculated using service factor
Design Power = 5.362x1.6 = 8.6 hp
The speed of the proposed belt in range (2000-5000) ft/min
Vbelt = 2500 ft /min
D1=
( 12 xVbelt )
xn
D1=
( 12 x 2500 )
x 720
D1 = 13.26 in
D2 = Rvb * D1
D2 = 2 * 13.26
D2 = 26.53 in
D2 <C<3(D2+D1)
26.53>C>119.37
we select C as 30 in To find the length of the belt L =
2C+1.57(D1+D2)+ (D2-D1)^2/4C
So
L= 121.2 in
To find the correction factor B
B = 4L-6.28(D1+D2)
B =234.91 in
the actual center distance can be calculated using the correction
B,
C=(B+(B^2-32(D2-D1)^2 ))/16
C = 28.59 in
The contact angle
=1802sin^(-1)(D2-D1)/2C
the small sheave
1=180-2 sin^(-1)(D2-D1)/2C
1=16510o
for large sheave
2=180+2 sin^(-1)(D2-D1)/2C
2=194.88 o
The length of the span:
S=(c^2-((D2-D1)/2)^2 )
S = 27.81 in
Now the
Pcorrected =C e C l ( Prated + Padded )
Number of belt=
design power
power corrected
To find Ce at the first contact angle
Ce = 0.96
To find Cl at length 125
CL = 1 at 5V
Now to find the rated power based on D1 or small sheave
diameter
P rated = 56 hp , now to find the power added at speed ration ,so P
added =0.6 hp
P design = 54.336 hp
Now
The number of belt = 0.098
So one belt is suitable to be used , for the type of belt we select
type narrow industrial
Tension in the belt can be calculated
P(T 1T 2)v
V =(D 1 n)/60
T 1
=e
T2
for steel = 0.25
also
1=165.10o = 2.879 rad
T1 = T2e
T1 = 2.054 T2
power = 4 Kw
40000 =2.054T2 T2) *(D1n)/60
T2 = 298.79 N
T1 =
612.53 N
Gear design
Gear Velocity Ratio RVG = 5.55
Input Power Pout = 4kW
Output Speed nG = 65 rpm
Rvg = np /ng
np = 5.55*65
np = 360.75 rpm
to find the module the standards
Where the pressure angle was assumed to be = 20 o
Module = 2
Diametral pitch = 12
Np = 16 teeth
NG = Rvg x Np
NG= 88.8 =89 teeth
The diameter of pinion=
Np
Pd
Dp = 16/12
Dp = 1.34 in
The diameter of gear
Ng
Pd
Dg = 7.41
The circular pitch for gear
P=
Dg
Ng
= 0.2618
The circular pitch for pinion
P=
D p
Np
= 0.2618
In order to find the dimensions of the gear the following
procedure will be used
Firstly
Addendum = a = 1/Pd
Addendum = 0.084 in
Dedendum = b= 1.25/pd
Dedendum = .104 in
Clearance = 00.25/Pd
Clearance = .02808 in
the outside diamter
for pinion
= 1.5 in
Dpo=
Np+2
Pd
DG o=
N G+2
Pd
= 7.583 in
The root diameter
Dpr = DP-2b
=1.34-2*.104
=1.132 in
Dgr = Dg-2b
= 7.41-2*.104
=7.202 in
Whole diameter
Ht =a+b
= .188 in
Working depth
Hk = 2a
= 0.168 in
Tooth thickness
t=
2pd
= 0.1308 in
Center distance
C=
Ng+ Ng
2 Pd
= 4.375 in
Base circle diameter
DBp = Dp*cos(20)
=1.27 in
DBg = Dg*cos(20)
= 7.04 in
The pitchline speed=
Dpn p
12
= 126.55 ft/min
Calculate transmitted load
Wt=
33000 P
Vt
Wt = 1398.212 lb
According to application Qv = 8
The Vt = 126.55 ft/min
Kv = 1.16
Choose load factor Ko
So
Ko = 1.75
the size factor was selected to be 1 as shown in table
The face width
F = 12/Pd
= 12/12 = 1 in
Which is in range 8/Pd < F<16/Pd
8/12= 0.667
16/12 = 1.334
Km was assumed to be 1
The rim thickness factor Kb
Kb = 1.98
Now to find the geometry factor
At Np and Ng
Jp = .27
And
Jg =0.31
Bending stress
S p=
( WtPd )
KoKsKmKBKV
( FJp )
Sp = 249.776Ksi
Sg=
( WtPd )
KoKsKmKBKV
( FJ g )
Sg =217.547ksi
The expected contact stress
Sc=C P
Wt K O k S k m K v
F DP I
Cp = 2300
I = .12 ,
Sc = 305.576 ksi
Shaft design
Input Power = 4 Kw
Output speed nG = 65 rpm
Input Speed nP = = 360.75 rpm
for pulley
T=
63000 xP
npulley
T = 937.04 lb.in
Forces on sheaves
F=
T
Dpulley
2
F= 5.32 lb
Forces on shaft for V-belt
Fb = 1.5 *F
Fb = 8 lb
Wt=
T
Dp
2
Wt = 1398.567 lb
Wr = wt *tan(20)
Wr = 454.42 lb
Forces analysis
=0
Assume the distance between the force = 6 in
Fn (18) Rby (12) Wr(4) = 0.0
FN
wr
RBy
RDy
RBy = 143.49 lb
RDy = 305.61 lb
RBx = Wr / 2
= 227.21lb
RDx = Wt RBx
= 227.2 lb
The shaft material was selected to be 1040 QQT1300
Sn = SnCsCr
Where
Cr = 0.81
Cs = 0.93
Sn = 38 Ksi
Sn = 28.6254 ksi
For pully the bending moment
Mx = wt/2 (4)
Mx = 2797.128 lb
For pinion
Mx = Wt *4
Mx = 5594.256 lb
Now by using the followin formulla the diamters the properties of
the shaft was estemated based on each moment
KtM
'
(Sn )
3 T
(())+
4 SY
( )
32 N
1
3
D=
DIAMETER
Pulley
Bearing
Dnon
D pinion
VALUE
.78 in
1.56 in
3.12 in
1.8
Bearings selection
A bearing is a mechanical device which provides relative motion
between two or more parts. Bearings are widely used in various
industrial as well as in our day-to-day applications. Bearings
permit four common types of motions like linear motion (for eg.
drawer), spherical motion (for e.g. ball and socket joint), axial
motion (for eg. shaft rotation) and hinge motion (for eg. door,
elbow, knee). There are different types of bearings used in
mechanical devices. Using a suitable bearing in many applications
help in improving accuracy, efficiency, reliability, operation speed,
purchasing costs of operating machinery.
Types of Bearings
There are many types of bearings each used for different
aplications and different purposes. Bearings can be used either
singularly or in combinations. Bearings are classified broadly
according to their motions, operating principle, load capacity and
speed and size they can handle. Accordingly, we find bearings of
different shapes, speed, lubrication, materials and so on. The
most popular bearing types are given below.
Input Power = 4 Kw
Input Speed nG = 65 rpm
Input Speed nP = 360.75 rpm
the life was selected to be 3000 0 hr
Based on these data the capacity of the bearing was calculated
For gear shaft
C = 50134.23 lb
For pinion shaft
C= 40533.652 lb
Mechanical seals
Mechanical seals can be classified into several tvpes and
arrangements:
PUSHER:
Incorporate secondary seals that move axially along a shaft or
sleeve to maintain contact at the seal faces. This feature
compensates for seal face wear and wobble due to misalignment.
The pusher seals' advantage is that it's inexpensive and
commercially available in a wide range of sizes and
configurations. Its disadvantage is that ft's prone to secondary
seal hang-up and fretting of the shaft or sleeve. Examples are
Dura RO and Crane Type 9T.
UNBALANCED:
They are inexpensive, leak less, and are more stable when
subjected to vibration, misalignment, and cavitation. The
disadvantage is their relative low pressure limit. If the closing
force exerted on the seal faces exceeds the pressure limit, the
lubricating film between the faces is squeezed out and the highly
loaded dry running seal fails. Examples are the Dura RO and
Crane 9T.
CONVENTIONAL:
Examples are the Dura RO and Crane Type 1 which require setting
and alignment of the seal (single, double, tandem) on the shaft or
sleeve of the pump. Although setting a mechanical seal is
relatively simple, today's emphasis on reducing maintenance
costs has increased preference for cartridge seals.
NON-PUSHER:
The non-pusher or bellows seal does not have to move along the
shaft or sleeve to maintain seal face contact, The main
advantages are its ability to handle high and low temperature
applications, and does not require a secondary seal (not prone to
secondary seal hang-up). A disadvantage of this style seal is that
its thin bellows cross sections must be upgraded for use in
corrosive environments Examples are Dura CBR and Crane 215,
and Sealol 680.
BALANCED:
Balancing a mechanical seal involves a simple design change,
which reduces the hydraulic forces acting to close the seal faces.
Balanced seals have higher-pressure limits, lower seal face
loading, and generate less heat. This makes them well suited to
handle liquids with poor lubricity and high vapor pressures such
as light hydrocarbons. Examples are Dura CBR and PBR and Crane
98T and 215.
CARTRIDGE:
Examples are Dura P-SO and Crane 1100 which have the
mechanical seal premounted on a sleeve including the gland and
fit directly over the Model 3196 shaft or shaft sleeve (available
single, double, tandem). The major benefit, of course is no
requirement for the usual seal setting measurements for their
installation. Cartridge seals lower maintenance costs and reduce
seal setting errors
Speed reducers
speed reducers are a component of many mechanical, electrical,
hydraulic and biological motors. It is easiest to think of a speed
reducer as a gear or series of gears combined in such a manner
to increase the torque of an engine. Basically, the torque of an
engine increases in direct proportion to the reduction of the
engines rotations per minute. If you decrease the rotation without
slowing down the engine, you increase the force generated. The
concept of using gears goes back thousands of years, and the
idea of using gears to control torque can be traced at least as far
back as the Renaissance period with inventors such as Leonardo
De Vinci.
Function
Imagine building an old-fashioned grain mill by alongside a river.
You build the mill, install the wheel, and now you have a very
rudimentary motor--the spinning wheel. However, there is a
problem. The wheel is spinning so fast that you cannot use it. If
you try, you risk endangering yourself or ruining the product you
built the mill to process. You cannot slow down your engine
because your motor is driven by a natural force, the river, so you
must instead figure out a means to convert or reduce the speed
of the motor to a working speed. The solution is a speed reducer
Effect
the purpose of a speed reducer is not just to increase torque, but
to reach the ideal torque for the machine in use. This is done by
reducing the speed input rotation to a ratio of "1/X." In this case,
"X" represents the reduction ratio. The variable "X" is then
multiplied against the torque of the unreduced engine. This gives
you the torque of the engine after a speed reducer has been
applied to it. That torque can now be applied to drive whatever
machine it is intended for
types
There are as many types of speed reducers as there are types of
gears. Examples of the types gears used to make speed reducers
are:
Spur: a gear wheel having radial teeth parallel to the axle.
Worm: a rotating screw that meshes with the teeth of another
gear on an inclined plane.
Bevel and spiral bevel gears: a gear wheel that meshes with
another at an angle between 90 and 180.
Uses
Speed Reducers are used in any industry that uses machinery,
whether it is hydraulic or electric. Some examples of uses of
speed reducers are in running conveyor belts, medical machines,
food processors, printing devices, computers, automotive engines
and construction-related machinery.
Considerations
The type used is dependent on the type of engine, and when
replacing a worn-out gear, it is important to replace the reducer
with the same type used by the original equipment manufacturer,
or seek expert advice.
