compressor design project – preliminary...

Compressor Design Project – Preliminary Report

Project Team 1 Mike Barnes, Karl Dyer, Kelli Martino, Dylan McNally, Scott Storms, Jessica Tryner

Mechanical Engineering Students, Rowan University

Professor Hong Zhang Mechanical Design & Synthesis

Copyright © 2001 Society of Automotive Engineers, Inc.

ABSTRACT Compressors are used daily by most people without their knowledge. Whether the compressor is found in a refrigerator, an air conditioner, or a painter’s spray gun, the fundamentals of the compressor design draw several engineering fields: thermodynamics, fluid mechanics, and mechanical design. Several assumptions from the thermodynamics and fluid mechanics fields are used to shift the focus of this project to the design of the compressors linkage, and piston movement.

INTRODUCTION

Introduction will be written once the full report has been completed.

PROBLEM DEFINITION

Design and construct a multiple cylinder air compressor using materials provided by Rowan University, see list below. The compressor must be powered by a ¼ hp motor attached to a belt driven transmission provided by the University. The compressor shall be designed for optimal performance in two categories, maximum pressure and maximum volumetric flow rate.

AVAILABLE MATERIALS

Raw Materials • Aluminum Tubing (6061-T6; OD-2in; ID-1.5) • Aluminum Rod (6061-T6; OD-2in) • Plastic Rod (Acetal; OD-15/8 & 1 in) • Aluminum stock (1/8”x1”; 1/4” x 1”; 1/4” x 1/2”)

Push Connect Fittings

• Straight (1/8”-NPT; 1/4” tube) • Tee (1/8”-NPT; 2 x 1/4” tube) • Tee (3 x 1/4” tube)

Miscellaneous

• Check Valve (2 x 1/4” tube) • O-ring (Buna-N; assorted sizes) • Dowel Pin (Steel, assorted sizes)

DESIGN METHODOLOGY

A complete MathCAD sheet and supporting MatLab code can be found in Appendix A and B, respectively.

The following sections provide a summary of the theory implemented in the fore mentioned programs.

THERMODYNAMIC ANALYSIS

Ideal gas behavior can be assumed if two criteria are met. First, the operating temperature is greater than twice the critical temperature and second, the pressure is no more than four times the critical pressure of the substance [2]. Compositional analysis shows air is roughly 70% Nitrogen and 20% Oxygen [1]. The critical temperature and pressure of Nitrogen are 126.2 K and 3.39 MPa; of Oxygen 154.6 K and 5.04 MPa. [2]. Air enters the compressor at room temperature, 293.2 K, almost satisfying the first criteria. Although most commercial air compressors can achieve a maximum pressure of 1.2 MPa, the compressor built for this project is significantly less capable and therefore will produce a maximum pressure significantly less than Nitrogen’s or Oxygen’s critical pressure, let alone four times the critical value thus satisfying criteria two. Since both criteria have been met, the Ideal Gas Law was used during the design of the compressor.

The Ideal Gas Law states that the product of the pressure and the volume of an ideal gas must equal the product of the mass, the gas constant, and the temperature of the gas [2], as outlined in Equation 1.

Eq. 1

When air is compressed, the temperature of the air rises [2]. However by adding fins to the cylinders heat transfer from the compressed gas to the surrounding atmosphere is improved. Heat can also be removed from the air between each compression stage by passing the air through an intercooler. When measures such as these are taken to remove heat from the air between each compression stage, the compression process can be considered isothermal [2].

Assuming an isothermal process, meaning T is constant, knowing that the gas constant, R, does not change and the mass of air in the compressor is assumed to be constant the ideal gas laws can be reduced to Equations 2 and 3.

Eq. 2

Eq. 3

This pressure volume ratio is used later in this paper to determine the stroke of each piston.

AVAILABLE WORK FROM MOTOR

The motor has a ¼ hp rating and a rotational speed of 1100 rpm. The output shaft of the motor connects to a belt driven transmission shown in Figure 1.

Figure 1 – Motor and transmission assembly provided for project

Connecting the drive belt to different steps on each sheave changes the angular velocity of the shaft spinning the compressor. Using a diametric ratio, shown in Equation 4, the altered angular velocity can be calculated. Using angular velocity and power the available work from the motor can be calculated, shown in Equation 5. Results of these calculations are summarized in Table 1.

Eq. 4

Eq. 5

Table 1 – Summary of available work and angular velocity obtained at each transmission configurations.

STROKE CALCULATIONS

The team set a project goal of obtaining a max pressure of 793 kPa using three stages. Using atmospheric pressure as the intial pressure the compression ratio was determined by using Equation 6.

Eq. 6

The piston diameter was governed by available materials however the first stage piston’s stroke was chosen to be 2 in, enabling the compressor to still be competitive in the volumetric flow rate of the competition. This choice will be explained later. Using diameter and stroke length, cross-sectional area of the pistons and the volume of the first piston can be found. Knowing the initial pressure and volume, as well as the desired pressure at the end of stage one from the compression ratio, the ideal gas law can be used to produce the final desired volume. Equation 7 shows the rearranged gas law.

Eq. 7

The final volume from stage one becomes the initial volume of stage two. By placing the piston in a circular array 120 degrees apart, each piston will be a third of the way through its stroke when receiving the compressed air from the previous stage, see Figure 2. The multiplying the initial volume of each stage by 3/2 produces the full volume the cylinder will contain when the piston is at bottom dead center. From this volume the stroke can be determined by dividing by the piston’s cross sectional area. Using this methodology the stroke of each successive piston was determined. A detailed analysis for each piston can be viewed in the MathCAD file in Appendix A.

Figure 2 – Volume of each piston with respect to drive shaft’s position

PISTON HOUSING CLEARANCE

To prevent the connecting rod between the piston and crank arm from crashing into the piston housing the geometry of the compressors components was modeled in MathCAD, found in Appendix A. This model was then

Picture Place Holder Filled in for Final Report

adapted into a MatLab code, found in Appendix B, that optimizes the distance from the motor input to the back edge of the piston cylinder housing. This was done by measuring the distance between the connecting rod, modeled as a line, and the corner of the piston housing, modeled as a point, for the entire motion of the connecting rod through one cycle using several iterations of Equation 8. The graphical output of the program displayed in Figure 3.

√

Eq. 8

Where the distance is ; the equation of the line being evaluated must be in the form 0; and the coordinate points of the point in space are and .

Figure 3 – Graphical output of collision checking program

VARIABLE DESIGN PARAMETERS

In order to achieve maximum pressure the compressor must utilize a series of stages where air flow from one piston to the next. To achieve maximum flow rate the tube of the pistons will have to be connected in parallel, allowing each piston to act as a single stage compressor. The compressor was designed to have a crank shaft that can be swapped out easily to allow each piston to have a stroke matching the first piston in the max pressure configuration. By increasing the stroke the maximum amount of air will be displaced. In order to prevent the pistons from bottoming out the piston housing’s on the second and third stage cylinders is also adjustable. These quick modifications are pictured in the design schematics found in Appendix C.

CONCLUSION

Will be completed after Compressor Fest.

REFERENCES

[1] Lide, D. R. and Frederiske, H. P. R., 1997, CRC Handbook of Chemistry and Physics, 86th ed., CRC Press, Boca Raton, http://www.physlink.com/ reference/AirComposition.cfm.

[2] Sonntag, R. E. and Borgnakke, C., 2007, Introduction to Engineering Thermodynamics, 2nd ed., John Wiley & Sons, Inc., Hoboken, pp. 56-273.

APPENDIX A

Stage 2 - Piston F

Start Pressure S3P1 S2P2 399.252 kPa⋅=:=

Start Volume S3V1 S2V2 16.72 cm3⋅=:=

End Pressure S3P2 S3P1 cr⋅ 792.621 kPa⋅=:=

End Volume S3V2S3P1S3P2

S3V1⋅ 8.422 cm3⋅=:=

Determining Geometry of Crank Arms

As long as pistons are out of phase by 120 degrees the piston recieving air from thecompressing piston will be 1/3 of the way through its stroke.

Piston B - Crank Arm OA

Crank Arm LOAPBstroke

225.4 mm⋅=:=

Piston D - Crank Arm OC

Stroke PDstrokeS2V1 1.5⋅

Ac38.383 mm⋅=:=

Crank Arm LOCPDstroke

219.191 mm⋅=:=

Piston F - Crank Arm OE

Stroke PFstrokeS3V1 1.5⋅

Ac19.334 mm⋅=:=

Crank Arm LOEPFstroke

29.667 mm⋅=:=

Determining Cylinder Housing Position

A simulation using matlab was conducted by measuring the distance between a line,representing segment AB, and a point, representing the corner of the piston housing. Thegraph produced is below as well as the length determined. The matlab code is attached.

Connecting Rod Lengths should be 82.4[mm]

LAB 82.4mm:= LCD 82.4mm:= LEF 82.4mm:=

Analysis

Motor Specifications rev 2π rad⋅:=

Power Rating Power 0.25hp 186.425 W=:=

Angular Velocity ω 1100revmin

115.192rads

⋅=:=

Available Workper rotation Wm

Powerω

2⋅ π 10.169 J=:=

Crank Arm Segments

N 240:= Number increments one rotation is divided into

i 1 N..:= Array index

αi 4πiN⋅:= Angle of link OA measured from x-axis

AxiLOA cos αi( )⋅:= Cxi

LOC cos αi( )⋅:= ExiLOE cos αi( )⋅:=

AyiLOA sin αi( )⋅:= Cyi

LOC sin αi( )⋅:= EyiLOE sin αi( )⋅:=

Piston B - Segment AB

δBπ

6:= Coordinate Transformation Angle

βBiasin

LOALAB

sin αi δB−( )⋅⎛⎜⎝

⎞⎟⎠

:= Angle of link AB measured from x-axis

BxiLOA cos αi δB−( )⋅ LAB cos βBi

⎛⎝

⎞⎠

⋅+⎛⎝

⎞⎠

cos δB( )⋅:=

ByiLOA cos αi δB−( )⋅ LAB cos βBi

⎛⎝

⎞⎠

⋅+⎛⎝

⎞⎠

sin δB( )⋅:=

Stroke End Points

TBx max Bx( ):= TBy max By( ):= piston at TDC

BBx min Bx( ):= BBy min By( ):= piston at BDC

LB TBx BBx−( )2 TBy BBy−( )2+ 0.051 m=:= length of stroke

Volume in Cylinder (currently TDC gap = 0, not accounted for)

VBiAc TBx Bxi

−⎛⎝

⎞⎠

2 TBy Byi−⎛

⎝⎞⎠

2+:=

Force Balance

FOAWmLOA

400.34 N=:= Force at motor

FB Ac S1P2⋅ 260.871 N=:= Force at piston

Piston D - Segment CD

δD5π


βDiasin

LOCLCD

sin αi δD−( )⋅⎛⎜⎝

⎞⎟⎠

:= Angle of link CD measured from x-axis

DxiLOC cos αi δD−( )⋅ LCD cos βDi

⎛⎝

⎞⎠

⋅+⎛⎝

⎞⎠

cos δD( )⋅:=

DyiLOC cos αi δD−( )⋅ LCD cos βDi

⎛⎝

⎞⎠

⋅+⎛⎝

⎞⎠

sin δD( )⋅:=

Stroke End Points

TDx min Dx( ):= TDy max Dy( ):= piston at TDC

BDx max Dx( ):= BDy min Dy( ):= piston at BDC

LD TDx BDx−( )2 TDy BDy−( )2+ 0.038 m=:= length of stroke


VDiAc TDx Dxi

−⎛⎝

⎞⎠

2 TDy Dyi−⎛

⎝⎞⎠

2+:=

Force Balance

FOCWmLOC

529.854 N=:= Force at motor

FD Ac S2P2⋅ 517.898 N=:= Force at piston

Piston F - Segment EF

δF3π


βFiasin

LOELEF

sin αi δF−( )⋅⎛⎜⎝

⎞⎟⎠

:= Angle of link AD measured from x-axis

FxiLOE cos αi δF−( )⋅ LEF cos βFi

⎛⎝

⎞⎠

⋅+⎛⎝

⎞⎠

cos δF( )⋅:= Fxi0:=

FyiLOE cos αi δF−( )⋅ LEF cos βFi

⎛⎝

⎞⎠

⋅+⎛⎝

⎞⎠

sin δF( )⋅:=

Stroke End Points

TFx min Fx( ):= TFy max Fy( ):= piston at TDC

BFx max Fx( ):= BFy min Fy( ):= piston at BDC

LF TFx BFx−( )2 TFy BFy−( )2+ 0.019 m=:= length of stroke


VFiAc TFx Fxi

−⎛⎝

⎞⎠

2 TFy Fyi−⎛

⎝⎞⎠

2+:=

Force Balance

FOEWmLOE

1.052 kN⋅=:= Force at motor

FR Ac S3P2⋅ 1.028 kN⋅=:= Force at piston

Graphs

Motion of Pistons - One Cycle

0 200 400 6000.1−

0.05−

0

0.05

0.1Alpha vs X

Bxi

Dxi

Fxi

αi180

π⋅

0 200 400 6000.1−

0.05−

0

0.05

0.1Alpha vs Y

Byi

Dyi

Fyi

αi180

π⋅

0 200 400 600 8000

2 10 5−×

4 10 5−×

6 10 5−×

8 10 5−×

Volume in Pistons vs Alpha

VBi

VDi

VFi

αi180

π⋅

When compressing piston is at TDC the piston it is feeding into is has a volumeequal to 2/3 its total capacity

APPENDIX B

clear format compact Loa=25.4 %Length of crank arm Safety=19.05 %Safety region so piston won't come out during stroke HousingDiam=40.64 %Inside diameter of housing Lab=50; %Inital guess of length AB delta=0.1 %Increment to change segment AB by in loop clearance=5 %gap desired between connector arm and housing bFlag=0 AlphaStart=0; %Starting value of alpha AlphaInc=1; %Angle alpha is incremented by AlphaStop=360; %Ending value of alpha Highlight=10 %Highlighting angle for plots mAlpha=([AlphaStart:AlphaInc:AlphaStop]); while bFlag==0 %Create System Geometry Using a Nominal Length AB %************************************************ %Coordinates of A mAx=Loa*cosd(mAlpha); mAy=Loa*sind(mAlpha); %Angle Beta mBeta=asind(Loa/Lab*sind(mAlpha)); %Coordinates of B mBx=mAx+Lab*cosd(mBeta); By=0; %Place Housing Based on Geometry of the System %*********************************************** %Create Corner Point of Cylinder Housing HxMin=min(mBx)-Safety; Hy=HousingDiam/2; %Create an linear equation representing the location of connecting arm %for each coordinate angle alpha for c0=1:length(mAlpha) m=(mAy(c0)-By)/(mAx(c0)-mBx(c0)); %calculate the slope b=-m*mBx(c0); %calculate y intercept d=abs((m*HxMin-Hy+b)/sqrt(m^2+1)); %calculate the distance if d < clearance Lab=Lab+delta; break; end end if c0==length(mAlpha) bFlag=1; end end %Find other endpoint of housing and construct matrix for plotting wall HxMax=max(mBx); mHx=([HxMin:((HxMax-HxMin)/length(mAlpha)):HxMax]);

%Output final ideal length ['Connecting Rod should be ', num2str(Lab), '[mm] long.'] ['The housing should be positioned ',num2str(HxMin),'[mm] from the origin'] %Plot motion of whole assembly figure axis('equal'); %axis([min(mAx)-10 max(mBx)+10 0 max(mAy)+10]); title('Motion of Entire System','FontSize',16); xlabel('X','FontSize',16); ylabel('Y','FontSize',16); hold on; plot(mHx,Hy,'r-',mHx,-Hy,'r-') for c=1:length(mAlpha) %Create an array to plot the full motion %Point O FullMotion(1,1)=0; FullMotion(1,2)=0; %Point A FullMotion(2,1)=mAx(c); FullMotion(2,2)=mAy(c); %Point B FullMotion(3,1)=mBx(c); FullMotion(3,2)=0; if mod(mAlpha(c),Highlight)==0 plot(FullMotion(:,1),FullMotion(:,2),'bo-') else plot(FullMotion(:,1),FullMotion(:,2),'g.-') end clear FullMotion end

APPENDIX C

1.625

1.000 TYP.

4.842

2.000 TYP.

15.000

9.500

11.563 BOLT CIRCLEEQUALLY SPACED ON3X .219 THRU ALL

120.00°120.00°

R.250 TYP.

.619 SLOT2 PLACES

1/4-20 UNC .5002X .201 .650

9.0001.228

.500

D

C

B

AA

B

C

D

12345678

8 7 6 5 4 3 2 1

DO NOT SCALE DRAWING

MATERIAL

FINISH

DRAWN

CHECKED

MFG APPR.

TEAM 1DATENAME

TITLE:

SCALE: 1:2

J.T. 10-11-09

FRAME

.219 THRU ALL

1.000 1.040

5 4 3 2 1


MATERIAL

FINISH

ALUMINUM

DRAWN

CHECKED

MFG APPR.

TEAM 1DATENAME

TITLE:

SCALE: 2:1

J.T. 10-11-09

CYLINDERSPACER 1

.219 THRU ALL

1.000 .520

5 4 3 2 1


MATERIAL

FINISH

ALUMINUM

DRAWN

CHECKED

MFG APPR.

TEAM 1DATENAME

TITLE:

SCALE: 2:1

J.T. 10-11-09

CYLINDERSPACER 2

1.600 2.000

1.800 BOLT CIRCLEEQUALLY SPACED ON

6-40 UNF .2764X .113 .351

1.800

3.625

.375

.125 TYPICAL 12 PLACES.125 TYPICAL 11 PLACES

CYLINDERDO NOT SCALE DRAWING

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NAME DATE

MFG APPR.

CHECKED

DRAWN

ALUMINUMFINISH

MATERIAL

5 4 3 2 1

TEAM 1

1.600

2.000

1.800 BOLT CIRCLEEQUALLY SPACED

4X .141 THRU ALL .031

GASKETDO NOT SCALE DRAWING

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NAME DATE

MFG APPR.

CHECKED

DRAWN

None

RubberFINISH

MATERIAL

5 4 3 2 1

TEAM 1

2.000

1.800 BOLT CIRCLEEQUALLY SPACED ON4X .141 THRU ALL

1.600

1/8 NPT .332 THRU ALL

45.00°

.375.125

.159 .474#10-32 TAP .380

BOTH SIDES

CYLINDERCAP


10-11-09J.T.

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NAME DATE

MFG APPR.

CHECKED

DRAWN

ALUMINUMFINISH

MATERIAL

5 4 3 2 1

TEAM 1

1.590

.270

.125.125

1.375

.250 THRU ALL

.500

.234

.188.313

.100

1.368

PISTONDO NOT SCALE DRAWING

1011-09J.T.

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TITLE:

NAME DATE

MFG APPR.

CHECKED

DRAWN

NONE

ACETALFINISH

MATERIAL

5 4 3 2 1

TEAM 1

R.250

4.039

2X .250 THRU ALL

.250

5 4 3 2 1


MATERIAL

FINISH

STEEL

DRAWN

CHECKED

MFG APPR.

TEAM 1DATENAME

TITLE:

SCALE: 2:1

J.T. 10-11-09

PISTON ROD

PRESSURE HUB 1

PRESSURE HUB 2

PRESSURE HUB 3

LARGE HUB

MAXIMUMPRESSURE

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2.750 .250 THRU ALL

1.000

.625 THRU ALL

.190

.188

.719

.709

.750

LARGE HUBDO NOT SCALE DRAWING

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NAME DATE

MFG APPR.

CHECKED

DRAWN

ALUMINUMFINISH

MATERIAL

5 4 3 2 1

TEAM 1

2.750

.250 THRU ALL

.756

.750

MAX PRESSUREHUB 1


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NAME DATE

MFG APPR.

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DRAWN

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5 4 3 2 1

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2.7502X .250 THRU ALL

.381.756

.250

MAX PRESSUREHUB 2


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NAME DATE

MFG APPR.

CHECKED

DRAWN

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5 4 3 2 1

TEAM 1

.250 THRU ALL2.750

.381

1.000

.250

MAX PRESSUREHUB 3


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NAME DATE

MFG APPR.

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DRAWN

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LARGE HUBVOLUME HUB

VOLUME HUB

LARGE HUB

MAXVOLUME

CRANKSHAFTDO NOT SCALE DRAWING

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2.750

.250 THRU ALL

1.000

.250

VOLUME HUBDO NOT SCALE DRAWING

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NAME DATE

MFG APPR.

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DRAWN

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MATERIAL

5 4 3 2 1

TEAM 1

.7495.625 - .001

+.000

.190

.188

.517

.502

1.500

.938

R.094

4.875

2.250

.031 , TYP.45.00° X

.6250

DRIVE SHAFTDO NOT SCALE DRAWING

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DRAWN

T.G. & P.

STAINLESS STEELFINISH

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5 4 3 2 1

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.6250

.517

.502

.190

.188

1.345

.938

R.094.031 45.00° X

BOTH ENDS

FRONT SHAFTDO NOT SCALE DRAWING

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5 4 3 2 1

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compressor design project – preliminary...

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