Stirling Engine

C. Sean Bohun (Mentor) Kevin Del Bene, Mike Caiola, Tyson DiLorenzo, Zhenyu He, Longfei Li, Arthur Mitrano, Ivana Seric, Liyan Yu

What is this?

Problem Statement

!   Handheld Stirling Engine !   Runs off the heat between hand

and room temperature

!   Heat difference creates pressure differences in reaction chamber which drives working piston !   Displacer “moves” gas to cold

side or hot side to create pressure changes

Handheld Stirling Engine

Problem Approach

!   Model Pressure Changes in Reaction Chamber

!   Model Mechanical Linkage !   Couples motion of working piston with displacer

!   Non-dimensionalize !   Validate magnitude of dimensional constants and

choose characteristic quantities

!   Solve Numerically

Assumptions

!   Ideal gas law as opposed to adiabatic gas law

!   Hot and cold gas is well mixed !   Gas heats and cools instantaneously

!   Linear friction function

!   Massless linkages

Peclet Number

!   Diffusion in reaction chamber is negligible compared to advection in reaction chamber

@T

@t=

@2T

@y2+ Pe

@T

@y

Pe =hv

� 1

h = 2.2⇥ 10�2m

= 2.2⇥ 10�5m2

s

v = 2.2⇥ 10�2m

s

Height of chamber

Thermal Diffusivity

Characteristic Speed

Reaction Chamber

PV = n ¯RT Ideal Gas Law

P =

¯RT

Mgas⇢ Pressure and Density Relation

mp∙g

F

A∙P0

Stages in Engine Cycle Stage 1 Stage 2 Stage 3 Stage 4

Fnet

= P1

Ap �mpg � P0

Ap Force on Piston

P1

=

1

Mgas

⇢0

¯RThot

Pressure on Piston when � = 0

P1

=

1

Mgas

⇢0

¯RThot

+ Tcold

2

Pressure on Piston when � =

1

2

P1

=

1

Mgas

⇢0

¯RTcold

Pressure on Piston when � = 1

!   This can be parameterized by , the position of the displacer, as:

�

P1

=1

Mgas

⇢0

R̄ [(1� �)Thot

+ �Tcold

]

Fnet

= P1

Ap �mpg � P0

Ap

Linkage

!   Need a way to determine the position of the displacer by knowing the position of the working piston

!   Reduces down to knowing the position of the flywheel

Equation of Motion for Flywheel

!   Position on the flywheel is given by:

!   Need to find the net force on the flywheel from the piston in the direction tangential to the flywheel

I ✓̈ = �k✓̇ +G(t)b

✓(0) =⇡

2

✓̇(0) = !0

Geometry of Flywheel and Linkage

!

"L

b

FT

F

Decomposition of Force

!   By using the geometry of the flywheel and linkage, the force from the working piston can be decomposed to the force along the linkage

!   The force along the linkage is decomposed again to find the force on the flywheel in the tangential direction !   This force drives the flywheel

!

φL

b

FT

F

~FL = F cos� h� sin�, cos�i

FT =

~FL · ˆt = F cos� (sin ✓ sin�+ cos ✓ cos�)

!   Using the geometry we can also find expressions that relate position on the flywheel to the angle between the working piston and displacer linkages.

sin� =

b

L| cos ✓|

cos� =

q1� sin

2 �

!

φL

b

FT

F

Position of Displacer

!   Using the geometry we can find an expression for

�

!   Alpha is the angle from the working piston to the displacer

� =

2

4s

1�✓b

L

◆2

cos

2(✓ � ↵)�

✓b

L

◆sin(✓ � ↵)� 1

3

5 L

2b+

1

2

=

1

2

(� sin(✓ � ↵) + 1) +O

✓b

L

◆

Non-Dimensionalization

!   We non-dimensionalize the equation of motion for the flywheel and use the fact that at rest, the piston should not move:

¨✓ =(1� �) cos� (sin ✓ sin�+ cos ✓ cos�)

� mdgb cos�

�

� kpbI�

˙✓

� = (Thot

� Tcold

)

A ¯R⇢0

Mgas

t = ⌧ t?

⌧ =

rI

b�

⇢0

=

(ApP0

+mpg)Mgas

¯RApTcold

Simplification

!   Using the fact that at rest, the piston should not move and the temperature difference should thus be zero:

¨✓ =(1� �) cos� (sin ✓ sin�+ cos ✓ cos�)

� mdgb cos�

�

� kpbI�

˙✓

⇢0

=

(ApP0

+mpg)Mgas

¯RApTcold

Size of Parameters

!

φL

b

FT

F

a = 5⇥ 10�2m

b =1

10a

Lp = 2a

Ld = 2a+1

5a

h =2

5a

hd =1

5a

hp = 2⇥ 10�3m

Tcold

= 293K

Thot

= Tcold

+�T = Tcold

+ 5K

¯R = 8.314kg m s

�2

K

�1

mol

�1

Mair

⇡ 28⇥ 10

�3

kg mol

�1

⇢0

= 1kg m

�3

P0

= 10

5

Pa

Ap = 3⇥ 10

�4

m

2

mp = 2⇥ 10

�3

kg

md = 10

�2

kg

mf = 15⇥ 10

�3

kg

I =

1

2

mfa2

= 1.8⇥ 10

�5

kg m

2

Results

0 50 100 150 200 250 300 350 400 4500

0.5

1

1.5

2

2.5

3x 104 k = 1e ! 6, !(0) = !2 , "(0) = 1

t (s)

Rev

olut

ions

Effect of Friction

0 20 40 60 80 100 120 140 160 1800

500

1000

1500

2000

2500

3000

3500

4000

t (s)

rpm

k=1e−6k=8e−6

Phase Portrait

−1 −0.5 0 0.5 135.6

35.7

35.8

35.9

36

36.1k = 1e ! 6, !(0) = !

2 , "(0) = 1

cos(e)

rpm

Stalling

0 2 4 6 8 10 12 14 16 180

1

2

3

4

5

6

7k = 1e ! 5, !(0) = !

2 , "(0) = 1

t (s)

Rev

olut

ions

0 2 4 6 8 10 12 14 16 18−600

−400

−200

0

200

400

600

800k = 1e ! 5, !(0) = !

2 , "(0) = 1

t (s)

rpm

−0.04 −0.03 −0.02 −0.01 0 0.01−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08k = 1e ! 5, !(0) = !

2 , "(0) = 1

cos(e)

rpm

No Initial Velocity

0 2 4 6 8 10 12 14 16 180

100

200

300

400

500k = 1e ! 6, !(0) = !

2 , "(0) = 0

t (s)

Rev

olut

ions

Reversing Motor

0 2 4 6 8 10 12 14 16 18−500

−400

−300

−200

−100

0

100k = 1e ! 6, !(0) = !

2 , "(0) = 0, # = !4

t (s)

Rev

olut

ions

P-V Diagram

Future Work

!   Gas Dynamics !   High speeds, equations of state

!   Choice of gas in chamber

!   Optimization of various geometric parameters !   Change timing of working piston and displacer

!   Configuration of engine

Thank You

Andrew Ross

This is a Skunk