Design manually operated pedal powered pump to demonstrate a functional reciprocating pump in rural people of Ethiopia. To facilitate the local people by providing water for various purposes and to optimize the use of natural resources.

2. Pedal Reciprocating Piston Powered Pump
Instructor: Ms.Mastewal
Group Member: ID
Wondie Chanie………………….…………..0502760
Tsegaye Chemere ………................................0502706
Wako Abduba………………………………..0502737
Tibebu Dagnew ………………………………0502647
Wondimagegn Wassie……………………..…0502764
Contact:
Mobile: +251932270893
E-mail: romeochanie@gmail.com

3. INTRODUCTION
In our country Ethiopia there is a lot of natural
resources like, lakes but we didn't develop our
agricultural process in an appropriate way. So we are
going to design and develop Manually Operated
Pedal Reciprocating Piston Powered Pump.

4. OBJECTIVE
General Objective:
Design manually operated pedal powered pump to demonstrate a
functional reciprocating pump in rural people of Ethiopia. To
facilitate the local people by providing water for various purposes
and to optimize the use of natural resources.
Specific Objective:
 To design chain and sprocket
 To design cylinder
 To design piston and rubber cup
 To design structure
 To design shaft
 Bearing selection

5. PROBLEM STATEMENT
The introduction of pedal powered reciprocating
piston pump must be essential for all rural
communities in which electric power is not available.
In most rural communities there is no an electric
power, so the uses of motor powered pumps are not
applicable and fuel powered pumps also not
economical for lower level farmers.

6. APPLICATION AREA
Our application area is in south gonder, gumara river
near woreta city. And it uses for the irrigation purpose
for their agricultural process. And in addition it can
operate in all irrigation purpose for any place.

7.  The main components of pedal powered reciprocating
piston pump
1. Cylinder
2. Piston
3.Connecting rode
4.Valves
5.Pedal
6. Chain
7. Seal
8. Crank
9. Manifold

8. WORKING PRINCIPLE

9. DESIGN ANALYSIS
Design of chain and sprocket:
 we assume:
The driver sprocket rotates at N1 =60rev/min and the
driven sprocket rotates at N2 =200rev/min

10.  Then the speed ratio =N2/ N1 =200/60=3 we find
that for the roller chain, the number of teeth on the
smaller sprocket or pinion (T1) for a velocity ratio of
3 are 25.Number of teeth on the larger sprocket or
gear, T2=T1*N2/ N1 =25*(200/60) =83
Design of chain and sprocket con…

11. From table we find that corresponding to a pinion speed of
200rpm the power transmitted for chain No. 08 is kW per
strand. Therefore, a chain No.08 with two strands can be
used to transmit the required power.
 Pitch, p=12.7mm
 Roller diameter, d=8.51mm
 Minimum width of roller, w=7.75mm
 Breaking load =17.8KN

12. D1=p cosec (180/T)
D1 =12.7cosec (180/25)
D1 =100mm
D2 =p cosec(180/T1)
D2 =12.7 cosec (180/83)
D2 =335mm

13. L=1514.5mm=1.5145m
K=119
L=K*P=119*12.7
L=1514.5mm=1.5145m
 Centre distance between the
sprockets,=30p=30*12.7=381mm
 In order to accommodate initial sag in the chain, the
value of centre distance is reduced by 2 to 5mm
 Therefore Correct centre distance x=381-3=378mm
 We know that the number of chain links

14. DESIGN OF THE CYLINDER
Material selection: cold rolled stainless steel
 Design Specification:
 Outside cylinder diameter=115mm
 Cylinder length=200mm
 Design verification
 Pressure developed (p)=force on the piston/cross
sectional area of the cylinder
 Discharge(v)=
(Volume of the cylinder/stroke)*(number of
strokes/second)

15.  Output specifications:
 Pressure developed=108.24kpa
 Discharge(Q)=2.07lit/sec
Design of cylinder con…

16. DESIGN OF PISTON AND RUBBER CUP
Objective: continuous forth and back movement of
pistons throughout the cylinders with much better
sealing.
Sealing material=rubber cup
The length of the piston must same extent larger than
length of the cylinder.
Overall height=240mm and the diameter of the piston
rod is 20mm
taking the clearance as the seal thickness 3mm
Piston diameter=112mm

17. DESIGN OF SHAFT
The diameter of the shaft is determined from the
equation:
 The material for shaft is selected as mild steel
 its allowable shear stress is: 42Mpa

18. Design of shaft con…
We get: T=18.9Nm
Then: T=Fr we get F=189N
T=
𝜋
16
*𝜏*d3 for solid shaft
Calculated value : d=12.8mm
 So it is preferable to use diameters greater
than this value, let us take standard shaft
diameter 15mm
Using this formula: P=
2𝜋𝑁𝑇
60

19. DISIGN OF STRUCTURE
 To manufacture the structure first the overall external
force and dimensions of all supporting structures
must be specified.
 We can assume that the operator weight as
approximately 60kg and the force applied by the
operator by his/her legs 1082 N

20. α
β
Design of structure con…
The free body diagram is shown below
D
C
B
A

21.  For ergonomic comfortability the angle of the
seating is α=750 from the horizontal.
Design of structure con…
 Then finally calculating all the force in each
members of the structure, these all dimensions
which calculating by this method is further
important for calculating the dimensions such as
diameters.
 ΣM=0
 ΣF=0
 To calculate the force of each members of the
structure.

22.  According to American Society of Mechanical
Engineers (ASME) code for the design of a fixed
load supporting and shaft ,the maximum
permissible working stress in tension or
compression a hallow steel rod may be taken
84Mpa with allowance.
Design of structure con…
 To calculate the diameters of each members of
the structure.
max=
𝐹𝑚𝑎𝑥
𝐴

23. SELECTION OF BEARING
A bearing is a machine element that constrains relative
motion between moving parts to only the desired
motion.
The design of the bearing may, for instance provide for
free linear movement of the moving part or for free
rotation around a fixed axis or it may prevent a motion
by controlling the vectors of load to be supported.

24. The thrust ball bearings are used for carrying thrust
loads exclusively and at speeds below 2000rpm, the
load on the bearing is only axial or thrust load, there is
no radial load is applied.
fig.0.0 Thrust ball bearing
Selection of bearings con…

25.  Then the total load is given by
W=YWA
W=1.3*1313.2N
W=1707.16N
In order to select a most suitable ball bearing, first of all,
the basic dynamic axial load is calculated. Then it is
multiplied by the service factor (KS) to get the design
basic dynamic axial load capacity. The service factor is
1.5. Therefore the design dynamic equivalent load should
W=1707.16*1.5
W=2560.74N

26. We find that for a single thrust ball bearing number
202, the basic dynamic capacity,
C=6.30KN
=6300N
We know that rating life of the bearing in revolutions,
L=(c/w)^k*10^6
=(6300/2570.7)^3
=14.7 ∗ 10^6 Revolution …(k=3, for ball
bearings)
We select life of bearing, in hours, LH=4000
The relationship between the life in revolutions (L) and
the life in working hours (LH) is given by
N=L/(70*LH)=(14.7*10^6)/(70*4000)
N=61.25rpm

27. CONCLUSION
This project focused on the construction and
operation of the reciprocating piston pump. Our
achievement is something we term a moderate
success. Our project is easy to operate and cost
effective (2763.68 birr).
By the use of this manually pedal powered pump we
can save money and we supply water in irrigation and
other agricultural uses.

28. It is clearly seen that the overall cost for the pump can
be using energy efficiency. Therefore we recommend
that the over all design should be manufactured and
distribute for our agricultural process.
RECOMMENDATION