stirling engine - rensselaer polytechnic institute...
TRANSCRIPT
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