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