Symmetric Six-Phase Stirling System

For a symmetric six-phase system, according to the general formulation in Eq. (5.10), the stiffness matrix K of the linearization is,

TOC o "1-5" h z

—c

подпись: —cMp+(b+c) — b 0 0 0

—c

подпись: —cMp + (b + c) — b 0 0 0

0 —c — + (b + c) —b 0 0

K

подпись: kMp 1 v 1 /

0 0 —c Kp + (b + c) —b 0

0 0 0 —c Kp + (b + c) —b

Mp v /

—b 0 0 0 —c — + (b + c)

Mp

(C.1)

N

X

X)COOOOOC<>C-

<XXXXX>0<XXX

X

X

8

X

……………………………………………………………………… A

XX)0<X>00000>

QOOOOOOOOC*

Ooc*x»oc*x*x

X*o*x

X

X>c<xooc«o<>c-

C<XX>C<>D<XXX

XXX>D<XXXXX

XX>D<X>OODOC-

OOOOC^OOOC’

Oo<*x*xx*c*x

Xc<x

M

X

250

200­

150

100

50­

-50­

-100­

-150­

-200-

C

‘oh

-12

-10

-6 -4 -2

Real part, rad/s

-250

-14

подпись: -250
-14
Figure C.1: Progression of the eigenvalues of six-phase Stirling engine toward the unstable region as the hot side temperature increases. At Th = 79 °C the system becomes self — starting.

And similar to the symmetric three-phase case, a linear transformation x = Tz where

Cos(0)

Sin(0)

Cos(0)

Sin(0)

Cos(0)

Cos(0)

Cos(0)

Sin( 3)

Cos(3)

Sin( f )

Cos( it)

Cos(^)

Cos(0)

Sin( it)

Cos( it)

In(

Si

Cos(f)

Cos(2^)

Cos(0)

Sin( f)

Cos(f)

Cob

In(

Si

Cos( f )

Cos(3^)

Cos(0)

Sin( f)

Cos( f)

Sin( f )

Cos( f )

Cos(4^)

Cos(0)

Sm( f)

Cos(t)

Sin( ^)

Cos()

Cos(5^)

(C.2)

подпись: (c.2)

T

подпись: t

Transforms matrix K into a new state matrix K as below,

Kp_

Mp

0

0

0

0

0

Kp + 1 (6 + c)

Mp 2 ^ ‘

(i

6

1

C)

0

0

0

23

6

1

)c

MP + 1 (b + c)

0

0

0

0

0

Kp + 2 (b + c)

Mp 2

C)

0

0

0

B

1

C)

Mp+1 (b+c)

0

0

0

0

0

0

0

0

0

0

подпись: 0
0
0
0
0

K

подпись: k

Mp+2(6+c)

подпись: mp+2(6+c)

(C.3)

Which simply reveals the following six eigenvalues for the six-phase stiffness matrix K,

P

K

Symmetric Six-Phase Stirling System

(C.4)

(C.5)

(C.6)

(C.7)

 

Fii

 

MP

 

Kp 3 V3„ ,

— + 2(b + c)) ±J^(b — c)

 

1^2,3

 

KP

/14 ————- + 2(6 + c)

MP

Mm^+1(6+c0± ^ %6- c)

 

^5,6

 

Note that a six-phase system includes all the modes of a three-phase system (i. e., fil, fi2, and fi3). In addition, fi4 represents the “pure compression” mode with a frequency that is proportional to the root of the average temperature of compression and expansion spaces. As the average temperature rises, the internal pressure of the engine and, hence, the gas spring stiffness increases, which causes a higher resonant frequency. Remember that /il represents the pure displacement mode which is independent of gas spring stiffness and, therefore, remains unchanged with temperature variations. The remaining two eigenvalues

Symmetric Six-Phase Stirling System

Symmetric Six-Phase Stirling System(a) (b)

Figure C.2: Simulated piston positions of the six-phase system. (a) Startup (b) Steady state.

Correspond to the “forward” and “backward” six-phase operation of the system.

Departing from thermal equilibrium, as Th increases, eigenvalues A5;6 (associated with eigenvalues fi5>6) hit the jw-axis first. Therefore, the relationship between the damping factor and start-up temperature becomes,

11 d /4

Tr = 7F — ~ 3mp(Kp + Kg) (C.8)

Th Tk a V 3

Which indicates that the resonant frequency of the system is set by sum of the stiffnesses Kp and Kg.

The eigenvalue loci of the three-phase Stirling engine system with reverser is shown in Figure C.1 and the simulation result of the same system is depicted in Figure C.2.

Добавить комментарий

Ваш e-mail не будет опубликован. Обязательные поля помечены *