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This is project presentation for Physics course. Instructor and project supervisor was Prof. Alpana Vishvajit at Aliah University. It includes: Bicycle, Pumps, Air, Transfer, Stroke, Cylinder, Design, Spring, Seal, Piston, Compressed
Typology: Slides
2011/2012
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1
Presentation Layout
1 •^ Introduction
2 •^ Types of bicycle pump
3 •^ Mass of air inside the tube at 14.7 & 100 psi
4 •^ Mass of air transfer per stroke and total strokes
5 •^ Cylinder design
6 •^ Rod design
7^ •^ Spring design
8^ •^ Seal selection
2
Introduction
The pressurized air then transferred from
the cylinder to a receiving tube or tire
The volume of air compressed per stroke
of the piston is generally a function of the
diameters of the piston and the cylinder,
as well as the length of the cylinder.
4
Type of Bicycle Pump
Manually operated bicycle pumps
generally fall into two classes
5
Mass of Air inside Tube at 14.7psi
Mean dia of cycle tire= 69 cm
Perimeter of tube= 216.77 cm
Diameter of the tube = cm
Volume of tube = 2724 cm^3
Pv = RT
Sp. Volume
of air at 14.
psi
Mass of air in the tube =3.206 g
7
Mass of Air inside Tube at 100 psi
8
Clearance & Swept Volume of Pump
length of clearance = 5 cm
Clearance volume = 79.5 cm^3
Swept volume ΔV = V 1 – V 2
ΔV = 636 - 79.5 = 556.5 cm^3
10
Mass of Air in Pump
Mass of air after Sucking Air
let the valve is close to 130 psi
m 1 = m 2
PV = m RT
1
m = 0.7485 g
11
Mass Transfer per Stroke
Let the valve close at P 3 = 100 psi
Let change of temperature
Using ideal gas law
Mass left in pump
Now mass transfer per stroke
P V = m RT 3 3 3 3
0 0 ΔT = 10 C = 10 K T = T - 3 2 ΔT
0 T = 322 K 3
m = 0.5931 3 g
Δm = m - m 2 3 Δm = 0.1554 g
13
Total Number of Strokes
total mass in the tube at 100 psi
total number of stroke =
mass per stroke
total number of stroke =
total number of stroke = 140 strokes
14
Cylinder Design
Generally in the thin wall cylinder we
assume there is no radial stress and
We consider hoop or tangential and
longitudinal stresses.
2 t
P × d σ = 2 × t
σ = 12.215 MPat
l
P× d σ = 4 × t l
σ = 6.108 MPa
16
Material Selection
17
Rod Design
Force applied on the rod ends = F
let FS = 2
Length of rod = 38 cm
π 2 F = P× × d 4
F = 1.426 kN
Fcr = 2.852 kN
eff
L = 0.65L L^ eff = 24.7 cm
19
Material Selection for Rod
Sy =
220 MPa
E = 172
GPa
Select
ASTM
A
20