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Solenoid Operated Piston Pump Engineering Essay

Disclaimer: This work has been submitted by a student. This is not an example of the work written by our professional academic writers. You can view samples of our professional work here.

Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of UK Essays.

Published: Mon, 5 Dec 2016

This project is aimed at analysing and designing a solenoid operated piston pump which is capable of delivering solution (this report assumes water) at a flow rate of 1 litre/min. However, the customer usage requires the flow rate to remain between 0.9 and 1.1 litre/min at an ambient pressure of about 1 bar.

The operation mode of the piston pump is described below using the diagram: OscillPistonPump

Fig 1.1 Solenoid Operated Piston Pump1

The solenoid coil (4) generates an electromagnetic field by the single wave diode rectified current flowing through the coil. Each current pulse moves the piston (5) against the pressure spring (3). This movement reduces the volume in the suction chamber causing an increase in pressure (P a 1/V), which opens the valve (6) in the piston, thereby allowing the liquid to run into the pressure side. When the current acting on the solenoid pulse is off, the pressure spring pushes back the piston toward the pressure side. The increase of pressure caused by the piston movement closes the piston valve (6) and the liquid flows through the valve (7) set in the pressure connection (8) and into the pressure pipe. The piston movement also simultaneously increases the volume in the suction chamber, thereby reducing the pressure below the chamber. The low pressure in the suction chamber opens the valve (2) set in the suction connection (1), and the liquid is sucked into the pump and the cycle starts again. The piston size and the length of its displacement define the flow rate. The pump will run without damage when the liquid flow is stopped momentarily1.

This design concentrated on the piston, suction chamber and pressure springs’ design. Although references were made to the valves and solenoid force, engineering analysis were not carried out on them.

CHAPTER 2

INITIAL ENGINEERING DESIGN ANALYSIS

This section considered the engineering analysis of the operation of the piston pump to achieve the require specifications. The given specifications are;

Flow rate Q = 1 Lit/min

Frequency F = 60 cycles/sec

Ambient Pressure = 1 bar

Using the above specifications, the length of stroke of the piston, which is also termed as the “Swept Volume”, can be calculated using the relation below:

Q = Volumetime=Volume ×frequency

= π d2 L4 ×f

∴L= 4Qπd2f

Where: Q = Flow Rate =1 lit/min= 1.667 ×104 mm3/sec

f = Frequency (cycles/sec)

L = Length of stroke/Swept volume (mm)

d = Diameter of piston/suction chamber (mm)

The diameter was varied from 1 to 20 mm and the corresponding lengths of stroke were obtained at different frequencies of 40, 45, 50, 55 and 60 cycles/sec. The results obtained were plotted (See appendix 1). After careful look, the frequency at 40 cycle/sec, so subsequent calculations would be based on this. It was also noticed that reasonable pair of dimensions of the diameter and length occurred around the diameters 5-10mm, therefore subsequent calculations were based on this range.

2.1 LOAD ANALYSIS

The load analysis was carried out on each component designed as indicated below:

A. Piston: The load analysis on the piston was done by isolating the piston and analysing the forces acting it. The different forces acting on the piston are as shown below:

  • Force on piston causing acceleration
  • Magnetic force from solenoid coil
  • Resultant spring force
  • Kinematic frictional force
  • Gravitational force
  • Resultant hydraulic force (including assumed viscous effect)
  • This is assuming that atomic, initial static frictional force and temperature effects are negligible.

The force analyses were carried out considering three different cases under which the pump operation can undergo. The intake and ejection strokes were also analysed separately to reduce complications. The difference between the intake and ejection stroke is that, the magnetic force from the solenoid is zero during ejection, because the solenoid is off:

  • Case I: This is when the piston pump is used horizontally, that is, it is used to pump fluid on the same datum. This means that the gravitational effect and the height difference in the hydraulic force will be zero. The relationship between the forces will therefore be:

Intake stroke:

Force causing motion = Force from solenoid – Resultant spring force – Resultant hydraulic force – Frictional force

Ejection stroke:

Force causing motion = Resultant spring force – Resultant hydraulic force – Frictional force

  • Case II: This considered the case when the pump is used to transfer fluid from a higher level to a lower level. This means that the gravitational effect will favour the direction of flow therefore reducing the force needed to drive the piston. The relationship between the forces will therefore be:

Intake stroke:

Force causing motion = Force from solenoid – Resultant spring force – Resultant hydraulic force – Frictional force – Gravitational force

Ejection stroke:

Force causing motion = Resultant spring force – Resultant hydraulic force – Frictional force + Gravitational force

  • Case III: This considered the case when the piston pump is used to deliver fluid from a lower level to a higher level. The difference between this case and case II is in the gravitational effect and the datum difference in the hydraulic effect. The design load analysis was done under this circumstance because pumps are usually used for this particular purpose. Even with this design concept, the pump can still be used for other cases, but it might deliver fluid at a higher flow rate, which could still be in the boundaries of the given tolerance of the flow rate. The relationship between the forces will therefore be:

Intake stroke:

Force causing motion = Force from solenoid – Resultant spring force – Resultant hydraulic force – Frictional force + Gravitational force

Ejection stroke:

Force causing motion = Resultant spring force – Resultant hydraulic force – Frictional force – Gravitational force.

The different forces were calculated as follows using the free body diagram of the piston shown below:

Figure 2.1 Boundary conditions of intake and ejection strokes

Force from solenoid coil= Fs

Force on piston causing motion = Mpa

Where Mp = mass of piston kg and a = acceleration of piston (mm/s2)

Mp= ρ ×V

ρ = Density of material (Stainless steel) =8×10-6 (kg/mm3)

V=Volume of fluid displced in one stroke mm3= Q ×t= Qf

where f=45 cycles/sec=90 strokes/sec (2 strokes=1 cycle)

Mp= ρ × Qf=8×10-6 × 1.667 ×10490=1.482×10-3

From law of motion; v2= u2+ 2aS

u = 0 ∴a=v22S

Also v= St= S ×f

v=Velocity (mm/s) and S= L=Length of stroke (mm)

∴a=L ×f22L= L × f22= L × 9022

The length was varied from 5 to 10 mm, and different accelerations were obtained (See appendix 2).

Resultant spring force = K2∆x- K1∆x= ∆xK2- K1= ∆x∆K

Where K1 and K2=Stiffness of springs 1 and 2 respectively (N/mm)

∆x=L=Stoke length (mm)

Kinematic frictional force = μk×N= μk×Mpg

Where μk=Coefficient of kinematic friction

N=Normal force= Mpg

g=acceleration due to gravity=9810 mm/s2

Gravitational force = Mpg

Hydraulic force = Total Change in Pressure ∆P (N/mm2)Surface Area of Piston A (mm2)

From Bernoullli’s equation  P1ρg+ V122g+ Z1= P2ρg+ V222g+ Z2

∆P= P1-P2=ρV22-V122+ ∆Zρg

Q= A1V1= A2V2 ,  then V2= QA2= A1V1A2 and V1= QA1

∆P= ρA1V1A22-V122+ ∆Zρg= V12ρ2 A1A22- 1+ ∆Zρg

∆P= ρ Q22A12A1A22- 1+ ∆Zρg

Where Q= Flow rate (mm3/s) , ρ =density of water =1×10-6 (kg/mm3)

A1and A2=Area mm2  and V1 and V2=Velocity (m/s)

∆Z=L=Length of Stroke mm

Including the discharge coefficient C = 0.98 to account for viscous effect, ∆P therefore becomes:

∆P= ρ Q22C2A12A1A22- 1+ Lρg

∴ Hydraulic force = ρ Q22C2A12A1A22- 1+ LρgSurface Area of Piston A mm2

= ρ Q22C2A12A1A22- 1+ LρgA2- A1

The forces were algebraically added according the ejection stroke equation developed above (case III) to obtain ?K at different diameter of pistons, fixing inner diameter of Piston D2 (corresponding to A2) = 0.5, 1, 1.5, 2 and 2.5mm (See appendix 3).

Force causing motion = Resultant spring force – Resultant hydraulic force – Frictional force – Gravitational force.

Mpa= L ∆K- ρ Q22C2A12A1A22- 1+ LρgA2- A1- μkMpg- Mpg

∆K= 1LMpa+ μkg+g+ ρ Q22C2A12A1A22- 1+ LρgA2- A1

The hydraulic effect is due to the fluid forced out from the suction chamber into the outlet. Therefore the A1 and A2 will be the area of the piston and the outlet, corresponding to diameters D1 and D2 respectively. Also the outlet diameter was assumed to be equal to the inner diameter of the piston.

The results obtained for difference in stiffness ?K above, were used to obtain the force from solenoid coil Fs using the injection stroke equation above. Also different diameter of piston were used while varying the inner diameter of piston D2 (corresponding to A2) = 0.5, 1, 1.5, 2 and 2.5mm (See appendix 4).

Considering the intake stroke equation for case III:

Force causing motion = Force from solenoid – Resultant spring force – Resultant hydraulic force – Frictional force + Gravitational force

Mpa= Fs-L∆K- ρ Q22C2A12A1A22- 1+ LρgA1- μkMpg+ Mpg

Fs= Mpa+ μkg-g+L∆K+ ρ Q22C2A12A1A22- 1+ Lρg A1

The hydraulic effect is due to the change in pressure as the fluid passes through the piston, because of the reduction in area. Therefore the A1 and A2 will be the area of the piston outer and inner diameter, corresponding to diameters D1 and D2 respectively.

B. Pressure Springs: The load analysis of the spring was also done by isolating the spring and analysing the forces acting it. Considering the ejection stroke of upper spring (spring 1), the different forces acting on the spring are as shown below:

  • Force on piston causing acceleration
  • Spring force
  • Resultant hydraulic force (including assumed viscous effect)
  • This is assuming that the frictional force on spring is negligible because the surface area contacting the wall is small.

Force causing motion = Spring force + Resultant hydraulic force

Mpa= L×K1+ ρ Q22C2A12A1A22- 1+ Lρg A1

K1=1LMpa- ρ Q22C2A12A1A22- 1+ Lρg A1

∴K2=K1+∆K

Where Force on springs Fsk=K×Length of stroke

The values of stiffness of springs 1 and 2 were calculated using the relationships above at different outer and inner diameters of the piston. The graphs were plotted to see the variations (See appendix 5 and 6).

C. Inlet Valve and Spring: Considering also the inlet valves and analysing the forces acting it, the injection stroke is caused by an increase in volume of the suction chamber, causing a corresponding decrease in pressure. Therefore the different forces acting on the inlet valve are given below:

  • Inlet spring force at compression
  • Resultant hydraulic force (including assumed viscous effect)
  • This is assuming that the frictional force and gravitational force on the valve is negligible because the valve is light.

Resultant Pressure Change= ?P

From Gas Law: P1V1= P2V2

P1 and P2 are the initial and final pressures of both the inlet and suction chamber respectively (N/mm2). The initial pressure P1 is assumed to be equal to the external pressure which is given to be equal to the atmospheric pressure Pa = 1 bar = 0.1 N/mm2. That is why fluid is not flowing because there is no pressure difference, or P1 was higher than Pa

P2= P1V1V2= PaV1V2 where V2=V1+Vs and Vs=Swept Volume per stoke

Vs=Flow rateFrequency in stroke/sec=1.667×10490 =185.22 mm2/stroke

P2= P1V1V1+Vs

∆P1=Change in pressure due to swept volume= Pa-P2

∆P1=Pa-PaV1V1+Vs=Pa V1+Vs-PaV1 V1+Vs=PaV1-PaV1+PaVsV1+Vs=PaVsV1+Vs

Where V1 = VT and it is the total volume of the inlet spring area, suction chamber and the inner space of the piston.

∆P2=Pressure Change due to area changes

∆P2=ρ Q22C2A12A1A22- 1+ Lρg

The above pressure change is the sum of the pressure changes from the inlet through suction chamber and into piston’s inner diameter. This is negligible because the pressure drops as it enters the suction chamber and increases as it enters the inner diameter of piston, thereby almost cancelling out.

∆P=∆P1=PaVsVT+Vs

Hydraulic force=spring force at compression

∆P1A3=PaVsA3VT+Vs= K3x3

PaVs=K3x3A3VT+ K3x3A3Vs

VT=PaVs- K3x3A3VsK3x3A3

Where A3=Inlet area mm2, K3=Inlet Spring Stiffness (N/mm)

and x3=Spring movement=Valve lifting mm

The values the total internal volume VT was obtained at different values of the diameter of the inlet D3 (corresponding to A3). The value of the spring force K3x3 was varied from 0.01 to 0.05 N and the variations were plotted to see an appropriate one (See appendix 7).

2.2 Component Design and Selection

The component design has been carried out along with the load analysis shown above. The desired dimensions for different components were then selected after a careful study and analysis of the graphs plotted. The dimensions were selected based on those that satisfy the required specifications, reasonably able to be manufactured and can be selected from the manufacturer’s catalogue as in the case of the springs2. Below are the component dimensions:

Solenoid:

Solenoid Frequency: 45cycles/sec = 90 strokes/sec

Force from solenoid coil: 108.8N

Length of stroke: 7.367 mm

Piston:

Piston outer diameter: 8 mm

Piston inner diameter: 2 mm

Springs:

Pressure spring 1 rate = 5.771 N/mm

Force on spring 1 = Rate * length of stroke = 5.771 * 7.367 = 42.515 N

Pressure spring 2 rate = 14.683 N/mm

Force on spring 1 = Rate * length of stroke = 14.683 * 7.367 = 108.17 N

From the above calculations and estimated values of the spring rates, the most accurate spring chosen from the compression spring catalogue are (see appendix 8 and 9):

Spring 1: C6609150

Wire diameter: 1.02 mm

Outer Diameter: 7.62 mm

Free length: 15.88 mm

Rate: 5.81 N/mm

Spring 2: D22110

Wire diameter: 1.25 mm

Outer Diameter: 7.55mm

Free length: 17mm

Rate: 15.03 N/mm

Inlet:

Inlet spring stiffness = 0.02 N/mm

Inlet spring length = 9.804 mm

Inlet diameter = 1.78 mm

2.3 Stress Analysis

The stress analysis was carried out on just two components as shown below. This was because these are the two components whose failure affects the pump operation most.

A. Piston: The two stresses acting on the piston are normal and shear stresses which is given as.

Stress (N/mm2) sij= Force (N)Area (mm2)

The notation is to differentiate between the direction and plane of action, where the first digit represents the plane of action and the second digit represents the direction of force. When the notations are different, it signifies shear stress and when the notations are the same it means normal stress.

The force on the piston varies as the piston goes through the cycle, therefore the different forces and principal stresses were calculated as the spring compresses and stretches. This was shown in appendix 10 and 11, but the calculations of the maximum and minimum principal stresses at the spring’s peak are shown below. The principal stresses were calculated because they are the cause of fracture in a component3.

Considering the piston and spring 1:

Fig 2.2: Stresses acting on piston from spring 1 and wall3

s11= 0 because there is no horizontal force in that direction

s12= Force from SolenoidSurface area of piston= Fsp Do Lp= 108.8p×8×15=0.2886 N/mm2

Where D0=Outer diameter of piston mm, Lp=Length of Piston (mm)

s22= Force from spring 1Outer Area-Inner Area= K1Lp4 Do2- Di2

s22=5.771 ×7.367p4 82- 22= 42.51547.1239=0.9022 N/mm2

s21= 0 because there is no horizontal force in that direction

Considering the piston and spring 2:

s11= 0 because there is no horizontal force in that direction

s12= Force from SolenoidSurface area of piston= Fsp Do Lp= 108.8p×8×15=0.2886 N/mm2

Where D0=Outer diameter of piston mm, Lp=Length of Piston (mm)

s22= Force from spring 2Outer Area-Inner Area= K2Lp4 Do2- Di2

s22=14.638 ×7.367p4 82- 22= 107.838147.1239=2.2884 N/mm2

s21= 0 because there is no horizontal force in that direction

The total principal stress which is the usual cause of fracture was calculated using the total normal stresses from the springs and the shear stress from solenoid.

Total shear stresses:

Ts12=s12 from Spring 1+ s12 from Sprig 2=0.2886+0.2886= 0.5772

Total normal stresses:

Ts22=s22 from Spring 1+ s22 from Sprig 2=0.9022+2.2954= 3.1976

Therefore the principal stresses:

s11s22- s(s11+s22)+s2-s122=0

0×3.1976- s(0+3.1976)+s2-0.57722=0

s2-3.1976s-0.3331=0

Principal stresses; smin=-0.101 N/mm2, smax=3.2986 N/mm2

B. Pressure Springs: The major stress acting on the spring is shear stress acting on the coils. The force and consequentially the shear stress on the springs vary as the piston deflection (i.e. length of stroke) increases and decreases. The various forces and shear stresses were calculated and the graph plotted (see appendix 12). But the calculation of the maximum shear stress, which occurs at the full deflection is shown below4:

Fig 2.4: Force acting on spring4

Shear stress tmax= 8FDWpd3

Where F=Force on spring N

D=Mean outer diameter of spring mm

d=diameter of spring coil mm

W = Wahl Correction Factor which accounts for shear stress resulting from the spring’s curvature

W=4C-14C-4+0.615C

C=Dd

Considering Spring 1

Fmax= K1×Length of stroke=5.771×7.367=42.515 N/mm2

D=7.62 mm and d=1.02 mm ?C=Dd= 7.621.02=7.4705

W=4C-14C-4+0.615C= 4×7.4705-14×7.4705-4+0.6157.4705=1.1982

tmax= 8FmaxDWpd3= 8×42.515 ×7.62×1.1982p×1.023=931.113 N/mm2

Considering Spring 2

Fmax= K1×Length of stroke=14.638×7.367=108.17 N/mm2

D=7.55 mm and d=1.25 mm ?C=Dd= 7.551.25=6.04

W=4C-14C-4+0.615C= 4×6.04-14×6.04-4+0.6156.04=1.2506

tmax= 8FmaxDWpd3= 8×108.17 ×7.55×1.2506p×1.253=1331.119 N/mm2

CHAPTER 3

INITIAL MANUFACTURING DESIGN ANALYSIS

3.1 Dimensions

The dimensions of all the main components; piston, springs, cylinder and valves had been obtained from the calculations and graphical analysis made above. However, the detailed dimensions of all components namely; pump body (left and right side), cylinder and liners, piston, springs and valves are shown in the CAD drawing in appendix 13.

3.2 Tolerances

Tolerance for Stroke Length

The statistical tolerance of the stoke length was calculated using integral method, which is much more effective than an additional tolerance. Given the tolerance of the flow rate as ± 0.1litres/min, the tolerance of the frequency was assumed to be ± 5 cycles/sec under normal distribution condition. The tolerance of the stroke length was calculated as follows:

Standard deviation s=Tolerance3 ×Cp where Cp=process capability index

In general manufacturing industry, a process capability index (Cp) of 1.33is considered acceptable. Therefore Cp< 1.33, the process is not capable. A Cp of 1.33 represents a situations where about 32 parts per million made will be outside the required tolerance band.

Flow rateQ=1 ±0.1 lit/min= 1.667 ×104 ±1.667 ×103mm3/sec

                            σQ=3.33 ×1033 ×1.33=8.356 ×102

Frequency F= 45 ±5 cycles/sec (Assuming a Normal distributed variable)

                            σf=103 ×1.33=2.506

Therefore the flow rate and frequency could be written as;

Q ~ N 1.667 ×104 , 8.356 ×102 mm3/sec

f ~ N 45 , 2.506 cycles/sec

Q = Volumetime=Volume ×frequency

= π d2 L4 ×f

∴L= 4Qπd2f

Using differential tolerance:

σ∅2= i=1n∂∅∂xi2 σxi2

σL2= ∂L∂Q2σQ2+ ∂L∂f2σf2+ ∂L∂d22σd2

σL2= 4π 1μd2 ×μf2σQ2+ μQμd2 ×μf22σf2+ μQμd3 ×μf2σd2 ×2

∴Tolerance=σ3 ×Cp

The standard deviations and tolerances of the stoke length were calculated using the above equations, while varying the diameter from 1 to 20 mm, and the results were plotted out (see appendix 14).

Tolerance for Piston Principal Stress

Assuming a normally distributed around the maximum principal stress acting on the piston, the standard deviation and the tolerance of the maximum principal stress was calculated using the load distribution obtained in appendix 11.

∴3σ=3.2918-0.5772=2.7146

Tolerance=Cp×3σ=1.33×2.7146=3.6104 N/mm2

Upper and lower limit=3.61042=± 1.8052 N/mm2

Tolerance for Springs Shear Stress

Also assuming a normally distributed around the maximum shear stress acting on the springs, the standard deviation and the tolerance of the maximum shear stress was calculated using the load distribution obtained in appendix 12.

For spring 1:

∴3σ=931.113-0=931.113

Tolerance=Cp×3σ=1.33×931.113=1238.38 N/mm2

Upper and lower limit=1238.382=± 619.19 N/mm2

For spring 2:

∴3σ=1331.119-0=1331.119

Tolerance=Cp×3σ=1.33×1331.119=1770.39 N/mm2

Upper and lower limit=1770.392=± 885.195 N/mm2

3.3 Fits

The components that are fitted into the cylinder, namely; cylinder liner, piston springs 1 and 2 are almost of equal diameter. But because of the consideration of the fits and limits to give some allowance a transition fit was chosen from “Data Sheet 4500A – British Standard selected ISO Fits-Hole Basis”. Since it fell in between the nominal size of 0 – 6 mm, the transition fit selected was H700.015 for the hole and k60-0.009 for the shaft5.

3.4 Material Selection

Piston and Cylinder

The piston and the cylinder are to be made of stainless steel grade 431. This is due to the prevention of fracture which could be caused by principal stress. From the maximum principal stress obtained for the piston above (3.2986 N/mm2 = 3.2986 MPa), it is sure that the material which has a yield strength of 655 MPa will be able to prevent failure. Also the other reason for choosing this material is because of its high resistance to corrosion6. Since the piston and cylinder interacts with the fluid, which increases the tendency for corrosion to occur, it is quite safe to use a highly corrosion resistance material like this. It is also very easily machined in annealed condition. The properties of the stainless steel grade 431are shown in appendix 156.

Springs

The springs are to be made of stainless steel grade 316. This is also due to the strength of the grade in preventing fracture, breakage and buckling of the spring due to the shear stress acting on it. From the maximum shear stress calculated above (1331.119 N/mm2 = 1.331 GPa), it is sure that this grade of stainless steel with an elastic modulus of 193 GPa will be able to withstand the compression. The material is also highly corrosion resistant and relatively easy to machine. The other properties of the stainless steel grade 316 are shown in appendix 156.

Valves

The valves are to be made of polytetrafluoroethylene – PTFE, which is a thermoplastic. This was chosen because the material has to be light and can easily float. Also, it has very low coefficient of friction, which reduces the fluid drag force and wears on the piston and spring.

3.5 Surface Finish

The surface finishing chosen for the manufacturing of the parts was to be 0.8 µm Ra. This is to reduce friction and rate of wear, because there are lots of parts moving against each other. The grinding process requires a very great accuracy because it is a relatively delicate manufacturing process.

3.6 Geometric Tolerance

In obtaining the tolerance of the components, since algebraic addition of tolerances is very unrealistic and will not be efficient, the tolerances of components that fit into each other were taken from the “Data Sheet 4500A – British Standard selected ISO Fits-Hole Basis”5. These are show below

S/No

Parts

Dimensions

(mm)

Tolerances (mm)

Drawings

1

Cylinder

11.00

+ 0.015

 

2

Cylinder liner

8.00

– 0.009

 

3

Piston

2.00

– 0.006

 

4

Spring 1

17.00

± 0.0015

3.7 Process Selection

The manufacturing processes of the various parts of the pump will be very important aspects of the design.The parts to be manufactured are pump body, cylinder liners and piston. It will take a great deal of accuracy in carrying out the process, because the geometry of the parts is very delicate. Any wrong dimension will affect the output or operation of the pump.

There are three steps in manufacturing the components mentioned above. Firstly, all the components would be manufactured by casting, which would probably not give the accurate dimensions. Then a turning/boring process can then be carried out, using a CNC or lathe machines, to achieve better dimension. The last process is the surface finish, which gives a smoother and precise dimension.

It is relatively easier to manufacture the components by this method because of the intricacies of the geometry and dimensions, and also the materials chosen are easily machined. The manufacturing process of the springs would not be considered in this report because they are provided by suppliers.

CHAPTER 4

DESIGN OPTIMISATION

4.1 Component Manufacturing Risk Assessment

Component Name

Pump Body

(Left & Right Side)

Calculation of qm

Drawing number

001

mp = 1 × 1.6 = 1.6

gp = 1.7 × 1 × 1 × 1 × 1.1 × 1.1 = 2.057

Ajustable tol= Design tolmp×gp

= + 0.0151.6 ×2.057=+0.00455

 

tp = 1.7×1 = 1.7

sp = 1 × 1.3 = 1.3

qm = 1.7 × 1.3 = 2.21

Manufacturing variability risk, qm = 2.21

Material

431 Stainless Steel

Manufacturing Process

Turning/Boring

Characteristic Description

Holes at centre to edge

Characteristic Dimension

8 mm

Design Tolerance

+ 0.015

Surface Roughness

0.8µm Ra

 

Component Name

Piston

Calculation of qm

Drawing number

005

mp = 1 × 1.6 = 1.6

gp = 1 × 1 × 1 × 1 × 1 × 1.1 = 1.1

Ajustable tol= Design tolmp×gp

= 0.0061.6 ×1.1=0.0034

 

tp = 1.7×1 = 1.7

sp = 1 × 1 = 1

qm = 1.7 × 1 = 1.7

Manufacturing variability risk, qm =1.7

Material

431 Stainless Steel

Manufacturing Process

Turning/Boring

Characteristic Description

Holes at centre to edge

Characteristic Dimension

2 mm

Design Tolerance

– 0.002, -0.008

Surface Roughness

0.8µm Ra             

 

The values of the component manufacturing risk analysis obtained above are considerably with a low risk. This shows that the processes chosen for the manufacturing of the components are acceptable.

4.2 Failure Mode and Effects Analysis (FMEA)

The failure mode and effects analysis (FMEA) is an analytical technique performed to ensure that all possible failure modes of the piston pump have being identified and address. Below are the predicted failure modes of each components of the piston pump, the caused, effects and the suggested solutions:

It can be seen from the FMEA above that the spring breakage has the greatest severity, but the wear on all the components has the greatest risk priority number. This is because wear would be experience by the customer over time of use which made the risk priority number very high. Therefore, while desig


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