Thermodynamic and Dynamic Analysis of Free Piston Stirling ...courses.washington.edu/mengr535/Sample Presentations/Stirling_Engine.pdf · Numerical Techniques ME 535 PROJECT REPORT

UNIVERSITY OF WASHINGTON

Thermodynamic and Dynamic Analysis of

Free Piston Stirling Engine Using

Numerical Techniques

ME 535 PROJECT REPORT

Spring 2014

Amrit Om Nayak & Nitish Sanjay Hirve

Master of Science in Mechanical Engineering

University of Washington

ME 535 Computational Techniques in Mechanical Engineering – Project Report

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Table of Contents

1. Introduction ................................................................................................................. 2

Problem Description .................................................................................................... 2

2. Mathematical Modeling of the System ......................................................................... 3

Thermodynamic analysis ............................................................................................. 3

Dynamic analysis ......................................................................................................... 8

Equations of motion ................................................................................................... 10

Displacer 1 ............................................................................................................. 10

Equation of motion for the working piston .............................................................. 12

Solving for roots ......................................................................................................... 13

Deriving the oscillation criterion.............................................................................. 14

Operating frequency equation ............................................................................... 15

Equation for stroke ratio and power ........................................................................... 16

3. Numerical analysis or iterative solution methodology ................................................ 17

4. Results ...................................................................................................................... 18

Thermodynamic results .......................................................................................... 18

MATLAB OUTPUT ................................................................................................. 18

Graphical results .................................................................................................... 19

Dynamic results ............................................................................................................. 20

MATLAB OUTPUT for dynamic analysis ....................................................................... 20

Case 1 ........................................................................................................................... 20

Case 2 ........................................................................................................................... 20

Case 3 ........................................................................................................................... 20

Graphical results ........................................................................................................... 21

5. Inferences & Conclusions .......................................................................................... 24

6. References ................................................................................................................ 25

7. Appendix ................................................................................................................... 26


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Thermodynamic and Dynamic Analysis of Free Piston Stirling Engine

Using Numerical Techniques

Amrit Om Nayak*, Nitish Sanjay Hirve*

*University of Washington, Master of Science in Mechanical Engineering

Abstract:

This report deals with studying, understanding and building a mathematical model to

describe and predict the performance of a real time free piston Stirling system. A free

piston Stirling engine with inherent problems and flawed performance was chosen as part

of the case study. Various mathematical models, methodologies and approaches were

studied in order to devise an analytical framework for the effective study of the

thermodynamics and dynamics surrounding this engine so as to understand the causes

for the failure of the system and troubleshoot/identify the exact problems plaguing this

system. Numerical analysis served as the basis for carrying out this entire study. Various

controlling parameters were iterated to understand the system. Taylor series expansion

of varying orders was used to understand the accuracy of the model.

Keywords – Oscillation, Critical frequency, Damping, Displacer, Working medium, power

piston/working piston, adiabatic analysis, semi-adiabatic analysis, energy balance,

viscous forces, Isothermal analysis, thermal coupling, efficiency, power.

1. Introduction

Problem Description

A free piston Stirling engine is analyzed here. This particular engine is a real time machine

which did not perform well. A detailed thermodynamic analysis of this engine enables us

to estimate the maximum possible work output of this engine in theory. A dynamic

analysis of the same, complementing the thermodynamic analysis, sheds light on the

actual real time losses which result in reduced engine performance. In this particular case,

the engine was not able to sustain itself and would die out. This was indicative of inherent

damping. Analysis of the dynamic engine data, will help locate the exact points of losses


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and understand the magnitude of losses due to damping. The entire process is a

numerical analysis wherein various parameters are iterated to understand and predict

engine performance.

2. Mathematical Modeling of the System

Thermodynamic analysis:

There are multiple thermodynamic models for the analysis of a Stirling engine which is an

external combustion engine. The basic analysis involves dividing the engine into 5 control

volumes. The hot end expansion volume, the hot end heat exchanger volume, the

regenerator volume, the cold end heat exchanger volume and the cold end compression

volume. The aim of thermodynamic analysis is to establish the thermodynamics

properties in each of these volumes. Finding analytical solutions is extremely difficult.

Hence using numerical techniques instantaneous pressure, volume, mass and

temperature are be found in all the control volumes. There are 2 analysis that are

performed numerically.

1) Isothermal analysis: Here the temperature of the expansion and compression

spaces is assumed to remain constant (which is not the case in reality). Constant

pressure is a simplifying assumption to get an approximate idea about the working

of the engine.

2) Adiabatic analysis: Here the temperature of expansion and compression spaces is

not assumed to be constant but instead adiabatic expansion and compression

processes are assumed to occur in these spaces. This is closer to the real

operation of a Stirling engine.

The volume variation is assumed to be sinusoidal, giving instantaneous volume as a

function of crank angle, and then instantaneous pressure, temperature and mass using

ideal gas law (working fluid is He for which ideal gas law holds) are determined. The

Pressure-Volume work and heat input and heat rejection is found and ultimately

integrated by iteration over one complete cycle, giving the engine thermodynamic

performance in terms of work per cycle and thermal efficiency.


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Primarily, the Stirling engine volume is divided into 5 control volumes and energy balance

equation is applied to these control volumes. The variables are pressure, temperature

and volume. Using the sinusoidal variation of volume we eliminate the volume variable,

& temperature by using adiabatic relation. Before going with 3 variables, that is pressure,

volume and temperature, we implement a simpler analysis under isothermal conditions,

constant temperature. If we consider variable temperature and assume expansion and

compression processes in the engine to be adiabatic, then the analysis method is called

adiabatic analysis. Energy balance diagram is shown below:

A detailed adiabatic analysis yields more accurate results as compared to the isothermal

analysis. The adiabatic analysis predicts a work output after subtracting the inherent

irreversibility of the thermodynamic system. This puts a theoretical limit on the maximum


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possible work output from the engine. The pressure-volume plot after analysis is shown

below-

Figure: PV diagram for engine

The initial guess for the change in pressure is given as below-

Dp = −

γp ((DVcTck

) + (DVeThe

))

(VcTck

) + γ(VceTce

+VrTr

+VheTh

) +VeThe

It is iterated for a few times over the entire cycle to get stable results.

Nomenclature and symbols:

2 x R2= stroke of power piston, in cm

3 x RC= stroke of displacer, in cm

AL=phase lag

DB= displacer diameter, in cm


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DC= diameter inside engine cylinder, in cm

Dp= change in pressure, in MPa

f=crank angle

gAck=mass flow from compression space to cold end heat exchanger space

gAhe= mass flow from hot end heat exchanger space to expansion space

gAkr= mass flow from cold end heat exchanger to regenerator space

gArh= mass flow from regenerator space to hot end heat exchanger space

h= clearance between cylinder top/bottom and displacer top/bottom when displacer is at

topmost/bottommost position respectively.

LC= cold end heat exchanger length

LH= hot end heat exchanger length

M= total mass of working gas present in the engine, in gram mole

mc=compression space mass

me=expansion space mass

mec=cold end heat exchanger mass

meh=hot end heaat exchanger mass

mr=regenerator mass

p=instantaneous pressure

R= universal gas constant= 8.314 J/gmol-K

r=Cp/Cv for the working gas

Tc=compression space temperature

Tce=cold end heat exchanger temperature


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Te=expansion space temperature

The=hot end heat exchanger temperature

Tr=regenerator temperature

Vc=Cold end volume

Vcd=cold end dead volume

Vce=cold end heat exchanger volume

Vd=cold end maximum displacer volume

Ve=Maximum Hot end volume

Vhd=hot end dead volume

Vhe=Hot end heat exchanger volume

Vp=cold end maximum piston volume

Vr=regenerator volume

DVc= change in volume in compression space

DVe= change in volume in expansion space

Dmc= change in mass in compression space

Dme= change in mass in expansion space

Dmce= change in mass in cold end heat exchanger space

Dmhe= change in mass in hot end heat exchanger space

Dmr= change in mass in regenerator space

Tck, The= conditional temperatures varying as per instantaneous mass flow direction


8

Dynamic analysis:

Schematic of the engine

*NOTE: We assume the displacer 2 to be non-existent to simplify our present analysis.

We also assume that the same type of springs are used throughout. The orientations of

the springs are symmetric or parallel subject to their placement in the design.

LEGEND

SYMBOL DESCRIPTION

A Cylinder Area

Ap Working Piston Area

D Damping Constant


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Fp Pressure Forces

Fsl Sliding Friction Force

i √−1

K Spring Constant

m Mass of Component

mtg Total mass of gas

p Pressure

ps Pressure due to spring

Po Mean cycle pressure

P Power

Q Stored energy/energy dissipated per cycle

R Gas constant

s Laplace transform variable

t Time

Tc Cold end temperature

Th Hot end temperature

Tr Regenerator effective temperature

V Phasor, volume

Vc Compression space volume

Ve Expansion space volume

Vh Heater volume

Vk Cooler volume

Vr Regenerator void volume

Vcs Compression space swept volume

Ves Expansion space swept volume

W Work

x Displacement

�̇� Velocity


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�̈� Acceleration

𝑥 Transformed displacement (laplacian)

X Amplitude, phasor

Φ Phase angle (piston to displacer)

θ Phasor angle

Ω Frequency (rad/s)

Subscripts

c Cylinder, Operating conditions

d Displacer

p Working piston

s Spring

vd Viscous damping

sd Spring damping

Equations of motion

Displacer 1:

Our force balance is given by:

Ftotal = -Fvd + Fnet(gas & spring) - Fsd

We now get our initial equation of motion as:

md1�̈�d1 = -Dd1�̇�d1 + Ad1(ps1 – p)

Here, Dd1 is damping on displacer due to viscous forces on the moving gas. For ‘n’ number

of springs attached to the displacer, we get:

md1�̈�d1 = -Dd1�̇�d1 + Ad1(nps1 – p)

But we know that here ‘ps1’ acts over the entire area Ad1 and its ‘Sin ϒ’ component acts

over the displacer.


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Actual spring pressure acting on displacer 1 = nps1Sin ϒ = 𝐧𝐊𝒔𝟏𝒙𝐝𝟏𝐒𝐢𝐧𝟐ϒ

𝐀𝐝𝟏

On generalizing the above equation for any orientation of the springs and replacing ‘𝐒𝐢𝐧𝟐ϒ’

by ‘δ1’ to represent the angular component and get:

md1�̈�d1 = -Dd1�̇�d1 + Ad1(𝐧𝐊𝒔𝟏𝒙𝐝𝟏𝛅𝟏

𝐀𝐝𝟏 – 𝐩)

Again, considering spring damping and other factors, we rewrite as:

md1�̈�d1 = -Dd1�̇�d1 – Ds1�̇�d1𝛅𝟏 + Ad1(𝐧𝐩𝒔𝟏𝛅𝟏 +𝐦𝐝𝟏

𝐦𝐜𝐩 – 𝐩)

[Where Ds1 = 2𝛏√𝐦𝒅𝟏𝐊𝒔𝟏 and 𝛏 = damping ratio]

Hence, the equation of motion for displacer 1 on linearizing is:

Taking Laplace transform and rewriting, we get:

{Where, (as + b)d1 = initial conditions and a = f(xd1,o) and b = f(xp,o);

ωd1 = undamped resonance frequency = √Kd1

md1;

And Kd1 = Kext,d1(1 + md1

mc) + Ad1(

∂pc

∂xd1−

md1

mc

∂pc

∂xc− nδ1

∂ps1

∂xd1)

*Note: mc is mass of cylinder and xc is displacement of cylinder.

But practically, by order of magnitude analysis:

Kd1 ≈ Ad1(∂pc

∂xd1−

md1

mc

∂pc

∂xc− nδ1

∂ps1

∂xd1)

Qd1 = (stored displacer 1 energy/energy loss per cycle at ωd1) = ωd1

2π

md1

Dd1;

Δp = drop in pressure across the heat exchanger loop;

Dd1,p = displacer/piston viscous coupling

Dext = incidental damping (Ns/m) and we assume Dd1 = Ddd1

Now, αp ≈ Ad1 (∂p

∂xp−

mp

mc

∂p

∂xc) }

md1[𝐬𝟐 + 𝛚𝐝𝟏

𝟐𝛑(

𝟏

𝐐𝐝𝟏+

𝛅𝟏

𝐐𝐒𝟏) 𝐬 + ωd1

2 ] �̂�𝐝𝟏 + 𝛂𝐩�̂�𝐩 + Dd1,ps�̂�𝐝𝟏 = (as + b)d1

md1�̈�d1 = (-Dd1 – Ds1δ1)�̇�d1 + Ad1[(𝐧𝛛𝐩𝐬𝟏

𝛛𝐱𝐝𝟏𝛅𝟏 +

𝐦𝐝𝟏

𝐦𝐜

𝛛𝐩

𝛛𝐱𝐜 –

𝛛𝐩

𝛛𝐱𝐝𝟏) 𝐱𝐝𝟏 –

𝛛𝐩

𝛛𝐱𝐩 𝐱𝐩 ]


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Now let, Td1(s) = md1[𝒔𝟐 + 𝝎𝒅𝟏

𝟐𝛑(

𝟏

𝑸𝒅𝟏+

𝜹𝟏

𝑸𝑺𝟏) 𝒔 + 𝝎𝒅𝟏

𝟐 ] �̂�𝒅𝟏 + Dd1,ps�̂�𝒅𝟏

Hence, we now get,

Equation of motion for the working piston:

The linearized unbalanced pressure force on the working piston, Fp, is given by:

Fp = -Fd1,p – Fnet = -Fd1,p – (Fg – Fs2)

Where Fg = force due to gaseous working medium

On linearizing, we get,

(This is the working piston equation of motion)

Here, 𝛅𝟐 = Angular component based on spring orientation in the working piston chamber.

The first term in the above equation represents a thermodynamic coupling between piston

force and displacer motion. Considering damping forces on the piston:

-Dp�̇�p = damping force on piston from electric generator (or other useful load)

-Ds2�̇�p𝛿2 = damping force on piston due to attached springs

And Kp ≈ Ap(𝜕𝑝𝑐

𝜕𝑥𝑝−

𝑚𝑝

𝑚𝑒

𝜕𝑝𝑐

𝜕𝑥𝑒− 𝑛𝛿2

𝜕𝑝𝑠2

𝜕𝑥𝑝) ;

On taking Laplace transform of the working piston equation of motion we get:

mp[𝒔𝟐 + 𝝎𝒑

𝟐𝛑(

𝟏

𝑸𝒑+

𝜹𝟐

𝑸𝑺𝟐) 𝒔 + 𝜔𝑝

2 ] �̂�𝒑 + 𝜶𝒑�̂�𝒅𝟏 + Dd1,ps�̂�𝒑 = (ys + z)p

{Where, (ys + z)p = initial conditions and y = f(𝑥𝑝,𝑜) and z = f(𝑥𝑑1,𝑜);

𝜔𝑝 = undamped resonance frequency = √𝐾𝑝

𝑚𝑝; And Kp = Kext,p(1 +

𝑚𝑝

𝑚𝑒) +

Ap(𝜕𝑝𝑐

𝜕𝑥𝑝−

𝑚𝑝

𝑚𝑐

𝜕𝑝𝑐

𝜕𝑥𝑐− 𝑛𝛿2

𝜕𝑝𝑠2

𝜕𝑥𝑝)

Td1(s)�̂�d1 + 𝜶𝒑�̂�𝒑 = (as + b)d1

Fp = -Ap[𝝏𝒑

𝝏𝒙𝒅𝟏 𝒙𝒅𝟏 + (

𝝏𝒑

𝝏𝒙𝒅𝟏− 𝐧

𝝏𝒑𝒔𝟐

𝝏𝒙𝒑𝛅𝟐 ) 𝒙𝒑]


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But practically, by order of magnitude analysis:

Kp ≈ Ap(𝜕𝑝𝑐

𝜕𝑥𝑝−

𝑚𝑝

𝑚𝑐

𝜕𝑝𝑐

𝜕𝑥𝑐− 𝑛𝛿2

𝜕𝑝𝑠2

𝜕𝑥𝑝) And Qp = (stored working piston energy/energy loss

per cycle at 𝜔𝑝) = 𝜔𝑑1

2π

𝑚𝑑1

𝐷𝑑1; Δp = drop in pressure across the heat exchanger loop; Dd1,p

= displacer/piston viscous coupling; Now, 𝛼𝑇 ≈ 𝐴𝑝 (𝜕𝑝

𝜕𝑥𝑑1−

𝑚𝑑1

𝑚𝑐

𝜕𝑝

𝜕𝑥𝑐) }

Now let, Tp(s) = mp[𝑠2 + 𝜔𝑝

2π(

1

𝑄𝑝+

𝛿2

𝑄𝑆2) 𝑠 + 𝜔𝑝

2 ] �̂�𝑝 + Dd1,ps�̂�𝑝

Hence, we now get, the following two equations:

1) Equation of motion of displacer 1: 𝑻𝒅𝟏(𝐬)�̂�𝒅𝟏 + 𝜶𝒑�̂�𝒑 = (𝐚𝐬 + 𝐛)𝒅𝟏

2) Equation of motion of working piston : 𝑻𝒑(𝐬)�̂�𝒑 + 𝜶𝑻�̂�𝒅𝟏 = (𝐲𝐬 + 𝐳)𝒑

Solving for roots

Now solutions of 𝑥𝑑1(𝑡) and 𝑥𝑝(𝑡) are given by (locating in the complex plane) the 4 roots

of:

𝑻𝒅𝟏(𝐬)𝑻𝒑(𝐬) − 𝜶𝒑𝜶𝑻 = 𝟎

We take Dd1,p = 0 here for simplicity.

*Note:

For oscillation, at least 2 roots must be in the Right Half Plane (RHP). We check this by

examining the complex plane of LHS in above equation where, ‘s’ moves along the

‘Bromwich contour’ in a clockwise sense. Since the LHS has no RHP poles, number of

clock wise rotations about the origin is equal to the number of RHP roots. We know that

in reality, 𝜶𝑻 < 𝟎 (as pressure decreases when displacer moves up). We infer that number

of RHP roots is equal to the number of rotations of the point -|𝛼𝑝𝛼𝑇| + 𝑖(0) in a graph of

𝑇𝑑1(s)𝑇𝑝(s) where ‘s’ moves along the ‘Bromwich contour’. Hence we now get:

𝑻𝒅𝟏(𝐬)𝑻𝒑(𝐬) = (𝐬 – 𝛌+)(𝐬 – 𝛌−)(𝐬 – 𝛃+)(𝐬 – 𝛃−)


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Where λ and β are the zeroes or roots of Td(s) = 0 and Tp(s) = 0 respectively. By using

quadratic formula we now get after making consideration for practical machines:

𝛌+, 𝛌− = 𝜔𝑑1 {−1

4𝜋(

1

𝑄𝑑1+

𝛿1

𝑄𝑠1) ± i }

𝛃+, 𝛃− = 𝜔𝑝 {−1

4𝜋(

1

𝑄𝑝+

𝛿2

𝑄𝑠2) ± i}

Locations of roots in complex plane (V1, V2, V3 and V4 are volume phasors)

Deriving the oscillation criterion:

Now, we start mapping the locus of ‘s’ at s = 0 and get 𝜃1 + 𝜃2 + 𝜃3 + 𝜃4 = 0

And [Td1(0)Tp(0)] is real and positive.

*Note: As s → i(∞), then (𝜃1 + 𝜃2 + 𝜃3 + 𝜃4) → 2𝜋; Clearly, this graph circles the origin

counterclockwise and returns to the real axis at +infinity. Now, the contour traverses an

infinite RHP semi-circle. For ‘s’ moving from +∞ to –∞ in clockwise direction, (𝜃1 + 𝜃2 +

𝜃3 + 𝜃4) decreases from 2𝜋 to -2𝜋 and the locus of Td1(s)Tp(s) is given by 2 circles of

infinite radius in clockwise direction.


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For position 1, 2 RHP roots exist → 2 clockwise encirclements of −|𝛼𝑝𝛼𝑇|

For position 2, no RHP roots exist → 2 clockwise and 2 counterclockwise encirclements

and hence no net rotation and hence they are not RHP roots of the equation.

Clearly locus of the final part of hte contour is negative imaginary axis from -𝑖∞ to origin.

In this part the graph is a complex conjugate of the graph in the positive imaginary axis

part. Hence we get, 𝜃1(𝜔𝑐) + 𝜃2(𝜔𝑐) + 𝜃3(𝜔2) + 𝜃4(𝜔𝑐) = 𝜋

Hence a necessity for oscillation in our system is :

𝑇d1(i𝜔𝑐)𝑇𝑝(i𝜔𝑐) < |𝛼𝑝𝛼𝑇| OSCILLATION CRITERION

Operating frequency equation (𝜔𝑐):

Also, from our diagram, we get, 𝜔𝑝< 𝜔𝑐 < 𝜔𝑑

Hence, 𝜃1(𝜔𝑐) = −𝜃3(𝜔𝑐) and for practical cases, @ ω = ωc, 𝜃2 + 𝜃4 ≈ 𝜋; Now, on solving,

𝝎𝒄 ≈ 𝝎𝒑𝝎𝒅𝟏{𝑸𝒅𝟏𝑸𝒔𝟏(𝑸𝒔𝟐 + 𝜹𝟐𝑸𝒑) + 𝑸𝒑𝑸𝒔𝟐(𝑸𝒔𝟏 + 𝜹𝟏𝑸𝒅𝟏)}

𝝎𝒑𝝎𝒅𝟏𝑸𝒔𝟏(𝑸𝒔𝟐 + 𝜹𝟐𝑸𝒑) + 𝝎𝒅𝟏𝑸𝒑𝑸𝒔𝟐(𝑸𝒔𝟏 + 𝜹𝟏𝑸𝒅𝟏)

If Qd1≫Qp → 𝜔𝑐 ≈ 𝜔𝑑1 And if Qp≫Qd1→ 𝜔𝑐 ≈ 𝜔𝑝

Also we get the ‘General Criterion for Oscillation of the System’:

𝒎𝒅𝟏𝒎𝒑(𝝎𝒄 + 𝝎𝒅𝟏)(𝝎𝒄 + 𝝎𝒑) [(𝝎𝒄 − 𝝎𝒅𝟏)𝟐 + {

𝝎𝒅𝟏

𝟒𝝅(

𝟏

𝑸𝒅𝟏+

𝜹𝟏

𝑸𝒔𝟏)}

𝟐

]

𝟏𝟐

[(𝝎𝒄 − 𝝎𝒑)𝟐+ {

𝝎𝒑

𝟒𝝅(

𝟏

𝑸𝒑+

𝜹𝟐

𝑸𝒔𝟐)}

𝟐

]

𝟏𝟐

< |𝜶𝒑𝜶𝑻|


16

If ωc = ωp = ωd1, then we get,

𝒎𝒅𝟏𝒎𝒑𝝎𝟒

𝟒𝝅𝟐(

𝟏

𝑸𝒅𝟏+

𝜹𝟏

𝑸𝒔𝟏)(

𝟏

𝑸𝒑+

𝜹𝟐

𝑸𝒔𝟐) < |𝜶𝒑𝜶𝑻|

Equation for stroke ratio and power:

Similarly on solving we obtain,

𝑥𝑑1(𝑡) = 𝑋𝑑1eⅈ𝑤𝑐𝑡 and 𝑥𝑝(𝑡) = 𝑋𝑝eⅈ𝑤𝑐𝑡 And on simplifying we get:

𝑋𝑑1

𝑋𝑝=

−𝛼𝑝

𝑇𝑑1(i𝜔𝑐)

Where 𝑋𝑑1and 𝑋𝑝 are complex amplitudes by considering only the real parts. Hence, we

now get stroke ratio and piston/displacer phase angle by:

|𝑿𝒅𝟏

𝑿𝒑

| =(𝜶𝒑)

𝑲𝒅𝟏

{[𝟏 − (𝝎𝒄

𝝎𝒅𝟏)

𝟐

]

𝟐

+ (𝝎𝒄

𝝎𝒅𝟏

𝟏

𝟐𝝅(

𝟏

𝑸𝒅𝟏+

𝜹𝟏

𝑸𝐬𝟏))

𝟐

}

−𝟏𝟐

We now get,

𝛟 = 𝐭𝐚𝐧−𝟏

[ (

𝝎𝒄

𝝎𝒅𝟏

𝟏𝟐𝝅(

𝟏𝑸𝒅𝟏

+𝜹𝟏

𝑸𝒔𝟏))

(𝝎𝒄

𝟐

𝝎𝒅𝟏𝟐 − 𝟏)

]

For practical engines, we generally have 40o<ϕ<90o

And also for ωc ≈ ωp = ωd1, we have ϕ = 𝜋

2

And

|𝑿𝒅𝟏

𝑿𝒑

| =(𝜶𝒑)

𝛚√

𝟏

𝟐𝝅(

𝟐𝝅𝑸𝒔𝟏

𝟐𝝅𝑸𝒔𝟏𝑫𝒅𝟏 + 𝜹𝟏𝛚𝒎𝒅𝟏)

And also we get,

𝑷 = 𝛚|𝜶𝑻|

𝟐|𝑿𝒅𝟏||𝑿𝒑| 𝐬𝐢𝐧𝛟

And we observe that P = Pmax @ 𝛟 = 90o such that ωp = ωd1.


17

3. Numerical analysis or iterative solution methodology

A heuristic iterative approach is used to solve this unique problem. Heuristic refers to

experience-based techniques for problem solving, learning, and discovery that give a

solution which is not guaranteed to be optimal. Where the exhaustive search is

impractical, heuristic methods are used to speed up the process of finding a satisfactory

solution via mental shortcuts to ease the cognitive load of making a decision. Examples

of this method include using a rule of thumb, an educated guess, an intuitive judgment,

stereotyping, or common sense. In more precise terms, heuristics are strategies using

readily accessible, though loosely applicable, information to control problem solving in

human beings and machines.

This unique problem is solved by using an explicit approach driven by heuristic

inferences. Since there is no analytical solution to the complex energy equation or the

dynamic mathematical model of the system which is developed, one has to go with

numerical methods to solve these equations and predict engine performance. The idea

is to express the desired variables in the form of an explicit equation, and then eliminate

the extra variables using thermodynamics and dynamic relations.

For example in the numerical analysis, we assume the volume inside the engine to vary

sinusoidally and then calculate instantaneous thermodynamic properties, namely

pressure, temperature, gas mass from ideal gas law. Using instantaneous properties, we

calculate instantaneous work done by/on the system. Then we integrate over the entire

cycle, that is, from crank angle of 0o to 360o, to get the work output per cycle from the

engine along with heat given to and heat rejected from the engine. The initial guess for

the change in pressure is given below (also mentioned previously)-

Dp = −

γp ((DVcTck) + (

DVeThe))

(VcTck) + γ(

VceTce +

VrTr +

VheTh ) +

VeThe

We need to iterate for a few times over the entire cycle to get stable results.

http://en.wikipedia.org/wiki/Rule_of_thumb

http://en.wikipedia.org/wiki/Hypothesis

http://en.wikipedia.org/wiki/Common_sense

http://en.wikipedia.org/wiki/Problem_solving


18

In the dynamic analysis, Taylor or McLaurin series expansion of 𝑒𝑖𝜔𝑡 is used to varying

orders to improve solution accuracy.

We however note that the results obtained with higher orders of Taylor’s series expansion

also differ by a very small value from the results obtained directly by iteration without their

use. This indicates inherent accuracy in the numerical analysis techniques and intuitive

approach utilized in our heuristic approach.

4. Results

Thermodynamic results

MATLAB OUTPUT:

Matlab output for Isothermal Analysis-

--------------------------------------

Work per cycle in J by numerical integration :

17.2093

Work per cycle in J by Mayers relation : 13.5294

Work per cycle in J as per Senft equation :

18.1755 Work per cycle in J by Cooke-Yarborough relation :

17.5579

Matlab Output for Adiabatic Analysis –

--------------------------------------

Enter the crank angle increment in degrees : 0.01 Enter number of iterations : 9

Compression ratio: 1.3859

Work output in J/cycle: 16.6654 Heat input in J/cycle: 29.6967

Heat rejected in J/cycle: -15.5192

Regenerator heat in J/cycle: -2.1517e-04 Efficiency of engine: 56.1189


19

Graphical results:

Variation of power output and efficiency with respect Variation of power and efficiency with respect

to regenerator effectiveness to phase angle

Power and efficiency for variable L/D and comp. ratio Power and efficiency Vs clearance height

Work and heat in the cycle vs volume phase angle Mass variations in the control volumes vs

Crank/Volume phase angle


20

Dynamic results:

MATLAB OUTPUT for dynamic analysis

Case 1:

Enter the crank angle increment in degrees : 0.1

Enter number of iterations : 6

Enter the required taylor series order : 8

Average Power from engine in Watts = 57.320961565215846

Average operating frequency of engine = 13.148354491987975

Stroke-ratio of engine = 1.093457321442433

Optimal Phase difference between displacer and working piston in

degrees = 82.212576286029801

Case 2:




Average Power from engine in Watts = 57.320961565082868




degrees = 82.212576286029801

Case 3:




Average Power from engine in Watts = 2.224851581656202e+02




degrees = 3.941512051486042


21

Graphical results:

The following graphical results are for a crank angle increment of 0.1 degree and Taylor

series order of 8.


22


23

-100%

0%

100%

200%

300%

400%

500%

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Erro

r %

Step size (crank angle increment in degrees)

Error % Vs. Step Size

Error in Power (%) Error (%) in operating frequency

-3.5E-06

-0.000003

-2.5E-06

-0.000002

-1.5E-06

-0.000001

-5E-07

0

0.0000005

0 2 4 6 8 10 12 14

Erro

r %

Taylor Series Order

Error % Vs. Taylor series order

Error (%) in operating frequency Error(%) in power


24

5. Inferences & Conclusions

1) Maximum theoretical average power output of engine predicted by adiabatic

method of thermodynamic analysis is 249.98 Watts. However, dynamic analysis

shows an average power output of just 57.321 Watts. This means that remaining

power is lost in damping, vibrations and viscous forces within the system. This is

consistent with observed poor performance of the free piston Stirling engine in

question.

2) The operating frequency after iteration is predicted by the dynamic analysis to be

about 13.15 Hz which is close to the design operating frequency of 15 Hz.

However, for higher number of iterations and smaller step size (smaller crank angle

increments), we obtain an operating frequency of about 8 Hz which is consistent

with the loss in power due to damping, vibrations and viscous forces.

3) The general oscillation criterion graphs helps us clearly pin point the problematic

regions where the criterion is violated. These are 130 degrees to 200 degrees and

270 degrees to 325 degrees of displacer positions. These are also reflected in the

power output graphs where sudden drop in power is observed in these regions.

We infer that, no net power is generated in these regions and damping forces

dominate. The system continues its motion just by momentum and energy stored

in the springs. This is consistent with the flattened regions in the power graph.

4) We infer that, viscous damping, spring damping need to be reduced while spring

stiffness needs to be increased in the system. Further, it is also necessary to

reduce friction along sliding surfaces.

5) Numerical analysis using Taylor/McLaurin series expansions has been done and

error % of power as well as operating frequency is plotted against step size as well

as Taylor series order. Increase in step size leads to changes in error % of power

produced which has a nature similar to a damped sinusoidal curve. However,

increase in step size leads to a steady increase in error % in operating frequency.

This is because increased step size leads to reduced number of iterations.


25

6) We also observe that changing the order of Taylor series has no impact on the

error % of operating frequency and has very miniscule impact on the error % of

power generated. We can therefore, conclude that use of Taylor expansion has a

minimal impact on the accuracy of results obtained here. We infer that the inherent

heuristic iterative approach is efficient without use of Taylor expansion.

7) At very small step size (say 0.01 crank angle increment or lower), the power (118

Watts) and operating frequency (8 Hz) values undergo appreciable changes and

are more realistic. This suggests that the dynamic model requires a very large

number iterations (>50000) to obtain stable solutions due to the complexity of the

mathematical model. However, such low step sizes require more than 2000

seconds to process and the time required increases drastically with reduction in

step size. Therefore, the process is expensive and more powerful computing

resources are required to evaluate the model more accurately.

8) Phase difference value of about 82.215 degrees obtained here is almost similar

the adiabatic analysis prediction of about 81 degrees.

6. References

[1] G. Walker., Stirling Engines, (1980),17, Oxford Univ. Press.

[2] Schmidt theory for stirling engines, Koichi Hirata,National Maritime Research Institute

[3] Martini, W. R. Stirling Engine Design Manual. Richland : Martini Engineering, 1983.

[4] Berchowitz et al., ‘Linear dynamics of free piston stirling engines’, Proceeding of Institution of

Mechanical Engineers, Vol 199, No A3, pp 203 – 213, 1985.

[5] Stirling Engine Design Manual, Second Edition, William Martini, January 1983.

[6] Kraitong, Kwanchai (2012) Numerical Modelling and Design Optimisation of Stirling Engines

for Power Production. Doctoral thesis, Northumbria University.

[7] Sastry, S., 1999, Nonlinear Systems: Analysis, Stability, and Control, 1st edition, Springer,

New York.

[8] Der Minassians, A., and Sanders, S. R., 2007, “A Magnetically-Actuated Resonant-Displacer

Free-Piston Stirling Machine,” Fifth International Energy Conversion Engineering Conference,

IECEC.

[9] Der Minassians, A., 2007, “Stirling Engines for Low-Temperature Solar-Thermal-Electric


26

Power Generation,” Ph.D. thesis, EECS Department, University of California, Berkeley,

Berkeley, http://www.eecs.berkeley.edu/Pubs/TechRpts/2007/EECS-2007-172.html.

7. Appendix

The following is an illustration of the Free Piston Stirling Engine analyzed in this

project report.

ENGINE PARAMETERS USED:

Operating temperature = 550oC = 823 K

Operating frequency (design) = 15 Hz

Engine starting pressure = 3 bar

Stationary mass (not reciprocating) in engine = 3.602 kg

Viscous damping coefficient = 6.8 N s/m

Spring damping coefficient = 3.8 N s/m

Displacer stroke = 35 mm

Working Piston stroke = 30 mm

Working Piston diameter = 60 mm

Displacer diameter = 70 mm

Spring Stiffness = 109360 N/m

Documents

Thermodynamic and Dynamic Analysis of Free Piston Stirling ...courses.washington.edu/mengr535/Sample Presentations/Stirling_Engine.pdf · Numerical Techniques ME 535 PROJECT REPORT