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 and to make parts proportional. Parts and measurements

should be labeled clearly. Then : 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 = ηCoupling * η Gear * η Belt *(η 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

the input power= Poutηoverall

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 three-phase 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 definite-

purpose 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 4 Kw , Speed =720 rpm

And efficiency =84.1%

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 normal temperatures in dry or

humid environments. The grease’s operating temperature is between -40

and +160°C. 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 FAZ = 2630 N

Accessories

From Electric Motor:

4 kW (5.362 hp)

Rv = (720)/65

= 11.1

The reduction for the gear can be given as following

Belt Gear Theoretical 1-5 1-12Actual 1-3 1-8

When the speed ratio for the belt = 2

The gear speed ratio = 5.55

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, identifies

belts’ drive components in the United States. Other countries use the British,

German, or ISO standards.

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

D 1=(12 xVbelt )

π x n

D 1=(12 x 2500 )

π 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

θ=180±2sin^(-1)〖(D2-D1)/2C〗

the small sheave

θ1=180-2 sin^(-1)〖(D2-D1)/2C〗

θ1=16510o

for large sheave

θ2=180+2 sin^(-1)〖(D2-D1)/2C〗

Θ2=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 (P rated+Padded )

Number of belt= design powerpower 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 1−T 2)v

V=(πD1 n)/60

T 1T 2

=eµθ

for steel µ = 0.25

also

θ1=165.10o = 2.879 rad

T1 = T2eµ θ

T1 = 2.054 T2

power = 4 Kw

40000 =2.054T 2 – T2* ((πD1n)/60

T2 = 298.79 N

T1 = 612.53 N

Gear design

Gear Velocity Ratio RVG = 5.55Input Power Pout = 4kWOutput 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 = 20o

Module = 2

Diametral pitch = 12

Np = 16 teeth

NG = Rvg x Np

NG= 88.8 =89 teeth

The diameter of pinion=NpPd

Dp = 16/12

Dp = 1.34 in

The diameter of gear =N gPd

Dg = 7.41

The circular pitch for gear

P = πDgNg

= 0.2618

The circular pitch for pinion

P= πD pN p

= 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 Dpo= Np+2Pd

= 1.5 in

D G o= N G+2Pd

= 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= π2∗pd

= 0.1308 in

Center distance

C=Ng+Ng2 Pd

= 4.375 in

Base circle diameter

DBp = Dp*cos(20)

=1.27 in

DBg = Dg*cos(20)

= 7.04 in

The pitch line speed=π∗Dp∗n p12

= 126.55 ft/min

Calculate transmitted load

Wt=33000 PVt

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

Sp=(Wt∗Pd )

(F∗Jp )KoKsKmKBKV

Sp = 249.776Ksi

Sg=(Wt∗Pd )( F∗J g )

KoKsKmKBKV

Sg =217.547ksi

The expected contact stress

Sc=CP √ Wt K O kS km 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 xPnpulley

T = 937.04 lb.in

Forces on sheaves

F= T

( Dpulley2 )

2

F= 5.32 lb

Forces on shaft for V-belt

Fb = 1.5 *F

Fb = 8 lb

Wt= TDp2

Wt = 1398.567 lb

Wr = wt *tan(20)

Wr = 454.42 lb

Forces analysis

∑ M = 0

Assume the distance between the force = 6 in

Fn (18) –Rby (12) – Wr(4) = 0.0 wr

FN 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

D=¿¿

DIAMETER VALUE Pulley .78 in Bearing 1.56 in Dnon 3.12 in D pinion 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.