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1 Introduction
A pump body forming a substantially cylindrical chamber having an inner surface and having at least one inlet to enable one-way gaseous communication between internal and external the pump body, the inlet comprising a flex valve (bend repeatedly) seat and a flex valve comprising a sealing face and a plurality of flex arms defining a plurality of apertures there between, the flex valve seat and flex valve cooperating to substantially seal the chamber by gas pressure pressing the sealing face against the flex valve seat when gas pressure proximate the inlet is higher internal to the pump body than gas pressure proximate the inlet external to the pump body, the flex valve seat and flex valve further cooperating to enable the introduction of gas into the chamber by gas pressure pressing the sealing face away from the flex valve seat causing the flex arms to extend thereby opening the apertures there between and allowing gas to pass there through when gas pressure proximate the inlet is higher external to the pump body than gas pressure proximate the inlet internal to the pump body, a substantially piston rod forming an internal passage, the piston rod having a first end positioned in the chamber of the pump body and the piston rod having at least one opening proximal the first end to enable gaseous communication between the internal passage and the chamber of the pump body, the piston rod further having a second end connectable to a transmission hose to enable gaseous communication between the internal passage and the transmission hose, and a piston head assembly attached to the piston rod proximal the first end and positioned in the chamber of the pump body, the piston head assembly comprising a pair of flexible u- cup seals each comprising a concave face and each attached to the piston rod proximal the first end with the at least one opening of the piston rod there between and with the concave faces facing together, the u-cups seals being sized such that their perimeters slid ably fit within the chamber of the pump body and being flexible to enable gas pressure within the chamber to deform the seals to allow gas to pass between at least one of the seals and the inner surface and to substantially prevent gas from passing between at least one of the seals and the inner surface. Comprising the steps of
(a) Connecting the transmission hose to an item to be inflated,
(b) Pivoting the pump body in a first direction comprising a first stroke which causes the linkage and piston rod to pivot on the bracket and causes the piston head assembly of the piston rod to slide within the chamber of the pump body thereby forcing gas from the chamber through the internal passage of the piston rod and the transmission hose to the item to be inflated, (c) Pivoting the pump body in a second direction comprising a second stroke which causes the linkage and piston rod to pivot on the bracket and causes the piston head assembly of the piston rod to slide within the chamber of the pump body thereby forcing gas from the chamber through the internal passage of the piston rod and the transmission hose to the item to be inflated, and (d) Repeating steps (b) and (c) until the desired gas pressure is attained within the item to be inflated [1].
Figure1. 1: hand operated bicycle pump
2.2 Mass inside the Tube at 14.7 Psi
From ideal gas equation Pv RT
Or v 1 ^ ^^ RTP
1
v (^) 101325
^
v 1 0.8497 m^3 /kg …………… (2.3)
As we know that
m 1 Vv 1 ^ ^ ^ 6 1
m 0.
^
m 1 ′= 3.206 × 10-6^ kg m 1 ′= 3.206 g ………………(2.4)
2.3 Mass at 100 Psi inside the Tube
Specific volume inside the tube at 100 psi
2 v RT P
^ ^
2
v (^) 689285.
^
v 2 0.1249 m^3 /kg …………………(2.5)
Hence mass is
(^22) m V v
^ ^
^
6 2
m 0.
^
m 2 0.02181 Kg
m 2 21.81 g ………………….(2.6)
3.1.3 Clearance Volume of the Pump
Let length of the clearance = L 2 = 5 cm Clearance volume V 2 = A × L 2 V 2 = 15.9 × 5 = 79.5 cm^3 …………….. (3.3)
3.1.4 Swept Volume of the Pump
Swept volume ΔV = V 1 – V 2 ΔV = 636 - 79.5 = 556.5 cm^3 …….. (3.4)
3.2 Mass of Air in Pump
3.2.1 Mass after Sucking Air by the Pump
At room temperature using ideal gas equation
PV 1 (^) m RT 1
Or m 1 ^ PVRT^1
6 1
m 287 300
^
m 1 (^) 0.7485 g ……………………..(3.5)
3.2.2 Specific Volume after Compression of the Pump
Let the valve is close still after full compression. So the mass after compression is the same as after sucking i.e.
m 1 = m 2
Specific volume after compression is v 2
2 2 2
v V m
6 (^2 )
v 0.7485 10
v 2 (^) 0.1062 m^3 /kg
Or v 2 (^) 106.2 cm^3 /g ……………….. (3.6)
3.2.3 Temperature after Compression
Using ideal gas equation
P v 2 2 (^) RT 2
T 2 P v^2 R
After compression the pressure increase to P 2 = 130 psi = 896318 pa
2
T 287
T 2 332 K …………………. (3.7)
3.2.4 Mass Transfer per Stroke
Let the valve close at P 3 = 100 psi Let the change of temperature
T 100 C 100 K
Hence T 3 (^) T 2 T
d 0 (^) 48.26 mm
t 1.651 mm d 48.26 2 1. d 45 mm
Check for thin and thick wall cylinder.
Applying condition for thin wall cylinder
t ^ d
Hence the condition is satisfied and the cylinder is thin wall. Now we calculate the stress in the cylinder. Generally in the thin wall cylinder we assume there are no radial stress and only hoop or tangential stress and longitudinal stress.
tangential stress = t
2 t 2
P d
t
2 3
t 2 1.651 10
^
t 12.215 MPa ……………..(3.12)
longitudinal stress = l
l 4 P d (^) t
2 3
l 4 1.651 10
^
l 6.108 MPa ………….(3.13)
Material selection
Select ASTM A47 which is called “malleable cast iron”
Sut 345 MPa
S y 220 MPa
In thin wall pressure vessel the condition of plane stress is satisfied i.e. 3 0
Hence 1 12.215 MPa
2 6.108 MPa
3 0 MPa
Failure theory
Applying maximum shear stress theory (Tresca). Taking factor of safety is unity.
1 ^ 3 FSS^ y
…………………[3]
12.215 ^0 ^ ^2201
Hence the condition is satisfied. So the cylinder in thin wall and is safe under the applied pressure.
Figure3. 2: Mohr’s Circle
3.6 Rod Design
Force applied on the rod ends = F
F P 2 A
2 F P (^) 4 d
^
F 4
^ ^
F 1.426 kN ……………………….(3.14) Let factor of safety is two i.e. FS = 2 Critical force = Fcr
Fcr 2 F
Fcr 2 1.
Fcr 2.852 kN ………………(3.15)
Rod acts as a column and it’s both ends are fixed.
Length of the rod (column) = L
L = ∆ L + L 3 ; where ∆L = swept length of the pump
L 3 = length for spring rest on the pump Let L 3 = 3 cm L = 35 + 3 = 38 cm
Effective length of the rod = Leff
Leff 0.65 L
Leff 0.65 38
Leff 24.7 cm ………………..(3.16)
Material for the rod
Select ASTM A47 which is called “malleable cast iron”
E 172 GPa S (^) y (^220) MPa
Moment of inertia for the rod = I
4 I (^) 64 dr
^
I ^2 A r
eff^2 E y
L E
^ S
^
^
9 6
eff E
L
^
eff 124. E
L
For our calculation
24.7 102 4
eff r
L
d
^
2 3
L eff
L eff 146.154 > eff
E
L
Hence our guess is right and it is an Euler column. As shown in the following figure.
Figure3. 3: Euler curve and Johnson’s parabolic approximation
3.7 Threads For The Rod And Nut
We have diameter of the rod = dr = 6.76 mm
Taking coarse series from table 11.1 of Machine Design by Khurmi [5]
Select M Pitch = 1 mm Core diameter = 5.773 mm Core diameter = 5.918 mm for Nut
Table3. 1:pipe thread sizes[6]
3.9 Spring Design
Force applied on the spring = Fs
Fs P 3 A
3 2 F s P (^) 4 d
^
F s 4
^ ^
Fs 1.097 kN ……………….(3.19)
Assume solid length of the spring Ls = 3 cm Let spring wire diameter = dw
Mean diameter of the spring = D
D dr dw
D 6.76 dw mm …………….(3.20)
Using torque formula
T F ^ D
T ^ d^ w T 3707.86 548.5 dw ………………….(3.21)
Spring material
Steel selection from Machine Design by Khurmi 651 MPa G 80 GPa Using torsion formula
3 T (^) 16 dw
^
dw (^) 16 dw
3707.86 548.5 dw 127.8 dw^3
By hit and trial we get
dw 3.535 mm
Standard value of the spring wire diameter is
