CHAPTER 6

SYSTEM DYNAMIC ANALYSES

The device was in the form of an inverted beam engine, and incorporated the characteristic phase shift between the displacer and piston that we see all Stirling engine today. The engine also featured the cyclic heating and cooling of the internal gas by means of an external heat source, but the device was not yet known as Stirling engine.

In Beta type Stirling engine, it consists single power piston and displacer, whose ideal purpose is to displace the working gas at constant volume and shuttle it between the expansion and the compression spaces than the series arrangement cooler, regenerator and heater.

For Rhombic linkage, the drive utilizes a jointed rhomboid to convert linear work from a reciprocating piston to rotational work.

The connecting rod of the piston is rigid as opposed to a common reciprocating engine which directly connects the piston to the crankshaft with a flexible joint in the piston. Instead, the rod is connected to the one of the corner of a rhombus. When force is applied to the piston, it pushes down, at the same time, the outer corners of the rhomboid push out. They push on two cranks or flywheels which causes them to rotate, each in opposite directions. As the wheels rotate the rhombus progresses its change of shape from being flattened in the direction to the piston axis at top dead centre to being flattened in the perpendicular direction to the piston axis at the bottom dead centre. Dead centre is when the value of clearance is nearly same as the power maximum then the engine is assume as stable condition where there is no vibration.

Figure XX

The schematic diagram of Stirling engine with rhombic linkage

30

6.1 TORQUE ANALYSIS

The torque applied on the crankshaft,

T total=T gas−T inertia

SinceT=Iα, therefore,

T total=Iα, whereby

α=¿Angular acceleration of crankshaft

The angular velocity, ω have to be maintained in order to make the movement of the flywheel constant, therefore Iα=0

Thus, T gas=T inertia

Since the gas inside the cylinder is the source of the movement of the flywheel, the torque applied at the flywheel is the same as the torque of the gas. T inertiaincluded torque of the mass flywheel and the movement occur in the system.

T flywheel=r× Fp=Iα

Where,

F p=¿Force of the piston and exerted to the lower shaft of rhombic system

r=¿Radius of the flywheel

6.1.1 DETERMINING VALUE F p

Using the Schmidt’s analysis, by evaluating the area enclosed in the pressure/volume indicator diagrams. The starting point of the analysis is that the total mass of working gas in the machine is constant, hence:

31

M=me+mc+mk+mh+mr

Using the ideal gas law given by

m= pVRT

From the ideal gas equation itself, it can be written as

M=

p(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r

T r)

R

Where effective regenerator T rgiven by T r=¿¿

Substituting the T requation into the ideal gas equation, the equation becomes

M=p¿¿

Solving for working gas pressure p therefore obtaining

p=MR (V e

Te

+V c

Tc

+V k

T k

+V h

T h

+V r ln(T h

T k)

(T h−T k ) )−1

The volume for compression and expansion, V cand V e can be determined from calculation in thermo analyses part meanwhile

V k=¿Volume of cooler

V h=¿Volume of heater

32

V r=¿Volume of regenerator

From the figure,

Force generated from the piston, F p=PA p

Whereby Apis the area of the displacer

pis the pressure of the displacer

In rhombic system, the displacer pressure exerted is the same as the pressure on the displacer and the area of the displacer is the same with the area of the power piston since diameter of piston is equal to the diameter of displacer ( D p=Dd )

From Schmidt’s analysis, the pressure

p=MR (V e

Te

+V c

Tc

+V k

T k

+V h

T h

+V r ln(T h

T k)

(T h−T k ) )−1

Substituting the pressure equation into the force generated from the piston equation

F p=MR

A p(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln(Th

T k)

( Th−Tk ) )33

At the crankcase where the force exerted on the rhomboid linkage, the force is divided into two since the rhomboid is symmetrical in shape. Therefore force acting at one side of the crank is

F p=MR

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln (T h

T k)

(T h−T k ) )Comparing with Newton’s 2nd law F=ma, thus, the piston acceleration,

a p=R

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln (T h

T k)

(T h−T k ) )

6.2 MASS BALANCE

34

From the vector diagram,

V f +V p=V fp

From the given equation, since the piston is moving downwards, therefore,

V p j=V fp−V f

From previous Schmidt’s analysis, the piston acceleration are obtained

a p=R

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln (T h

T k)

(T h−T k ) )From the acceleration of the piston itself, the velocity of the piston can be determined since the acceleration is the derivative of the velocity with respect to time

a=dvdt

35

To get the velocity, the acceleration should be integrated with respect to time

v=∫ adt

The same way applies to get the velocity of the piston,

v p=∫ R

2 Ap(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln(T h

T k)

(T h−T k ) )dt

The equation for piston velocity, v p can be determined

v p=R

2 Ap(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln(T h

T k)

(T h−T k ) ) t

Given that v fp=ωfpk × rfp

From trigonometry, the r fpcan be determined,

r fp=(e−r sin θ) i+¿

Using the cross product

v fp=ωfpk × ¿

v fp=ωfp (e−rsinθ ) j−ωfp¿

Since the rotation of the flywheel is anticlockwise during the piston moves, thus,

v fp=−¿

v fp=−ωfp (e−rsinθ ) j+ωfp¿

36

Given the velocity of the flywheel

v f=ωf x rf

v f=−ωf k ×(rsinθi+rcosθ j)

v f=−ωf (rsinθ ) j+ωf (rcosθ ) i

From the equation of velocity of the piston

v p j=v fp−v f

Substituting the value,

v p j=¿

v p j=¿

v p j=¿

Thus, from the vector, pairing can be made

ωfp¿

−ωfp ( e−rsinθ )+ωf (rsinθ)=v p

From equation 1,

ωfp=ωf (rcosθ)

√l2−¿¿¿

Then, using substitution method, substitute the equation for ωfpinto equation 2

−¿

ωf ¿

From the equation derived, then, the angular velocity of the flywheel can be obtained

ωf =v p

−¿¿

6.3 DYNAMIC BALANCING

37

1

2

From the free body diagram on the figure of one side crank, the vector position of the piston force is

F p=F pcos θi+F p sinθ j

Substituting the value

F p=MR

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln (T h

T k)

(T h−T k ) ) into the equation, thus,

F p=MR

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln (T h

T k)

(T h−T k ) )cosθ i+ MR

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

T h

+V r ln (T h

T k)

(T h−T k ) )sinθ j

From region 0−90 °

r=rsin∅ i+rcos∅ j

Therefore torque produced is the torque from gas with equation

T=r × F p

38

T=(rsin∅ i+rcos∅ j)×[MR

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

Th

+V r ln(T h

T k)

(T h−T k ) )cosθ i+ MR

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

Th

+V r ln(T h

T k)

(T h−T k ) )sinθ j ]

Using the cross product rule,

T=[rsin∅MR

2 A p(V e

Te

+V c

T c

+V k

Tk

+V h

T h

+V r ln(T h

T k)

(Th−T k) )sinθ ]k−[rcos∅ MR

2 A p(V e

T e

+V c

T c

+V k

T k

+V h

Th

+V r ln(T h

T k)

(T h−T k ) )cosθ ]k

Meanwhile that, T inertia is equal to the T gas since Iα=0

39

6.4 BETA TYPE FREE PISTON MODELLING

One of the novel applications of the Stirling cycle is in free piston configurations and indeed the configuration is one which holds immediate promise.

Free piston engines operate without physical linkage instead they rely only on the gas pressure and in some cases mechanical springs to impart the correct motions to the reciprocating elements. Such machines have the advantage of simplicity, low cost, ultra reliability, and freedom or working gas leakage over the conventional Stirling engines. Depending on particular configurations, these engines may also be designed to operate at constant frequency.

md xd=Ad ( Pc−Pe)−F spring1

m p xp=A p ( Pb−Pc)−F spring2+c p x p

md xd=F pressure ,d−F spring 1

40

m p xp=Fpressure , p−F spring2+F load x p

md xd+F spring 1=F pressure, d

m p xp+F spring 2−c p x p=Fpressure , p

md xd+kd xd=F pressure ,d

m p xp+k p x p−c p x p=F pressure , p

PART II

6.4.2 STATE SPACE REPRESENTATION

Piston,

x p=up

x p=v p

x p= v p

up−v p=0

F pressure, p=m p v p−cd v p+k p up

1m p

Fpressure , p

=v p−cd

m p

v p+k p

mp

u p

State vector, z=(up

v p)

z=(up

v p)

(1 00 1)(u p

v p)+( 0 −1

−cd

mp

k p

mp)(u p

v p)=( 0

1m p

)F pressure , p

41

z+Az=B F pressure, p

Where, A=( 0 −1−cd

m p

k p

m p)

B=( 01mp

)Displacer,

xd=ud

xd=vd

xd=vd

ud−vd=0

md xd+kd xd=F pressure ,d

1md

Fpressure, d

= vd+kd

md

ud

State vector, z=(ud

vd)

z=(ud

vd)

(1 00 1)(ud

vd)+(0 −1

0kd

md)(ud

vd)=( 0

1md

)F pressure, d

z+Az=B F pressure, d

Where, A=(0 −1

0kd

md)

B=( 01md

)ud=xd=Xe=−QEsin [γ (ϕ )]

vd= xd=ve=−QEcos γ (ϕ )

ud

−QE=sin [γ ( ϕ)]

42

(ud

−QE)

2

=sin2[γ (ϕ )]

vd

−QE=cosγ (ϕ )

(vd

−QE)

2

=cos2[γ (ϕ )]

(ud

−QE)

2

+(vd

−QE)

2

=1

Phase portrait,

Assume that the model is as below:

43

θθR

k

Assume as a spring

configuration with a constant k.

Applying Newton’s second law,

m xp=m p g−F

m p g=kδ

x=Rθ

J θ=FR−k (δ+Rθ ) R

¿ FR−k (δ+Rθ ) R

¿ FR−kδR−kθ R2

¿ FR−mp gR−kxR

¿(F−mp g)R−kxR

¿m p x p R−kxR

J θ=−mp θ R2−kθ R2

J θ+m p θ R2+kθ R2=0

(J+m p R2) θ+kθR2=0

θ+ k R2

(J+mp R2)θ=0

θ+wn2θ=0

Natural frequency,

wn=√ k R2

(J+m p R2)

44

Mp

Xp

mg

F

F Mg

However applying the law of conservation of energy,

The kinetic energy of the system:

T=12

m p ẋ2+ 12

M ẏ2+ 12

J θ2

¿ 12

m p ẋ2+ 18

M ẋ2+ 14

M R2( ẏR

)2

¿ 12

m p ẋ2+ 216

M ẋ2

The potential energy of the system,

U 0=12

k yδ2+Mgy+mgx

U=12

k ( y¿¿δ + y )2+Mg ( y− y0 )∓mg ( x−x0 )¿

k yδ=Mg+2mg

U=U 0−12

k y2=U 0−18

k x2

Applying the law of conservation of energy,

T+U=12

m p ẋ2+ 216

M ẋ2+U 0−18

k x2=constant

m p ẋ p x p+38

M ẋ p x p+14

k x p ẋ p=0

(m p+38

M ) x p+14

k x p=0

x p+2k

8mp+3 Mx p=0

x p+wn2 x p=0

Natural frequency,

wn=√ 2k8m p+3 M

6.5 SYSTEM DYNAMIC ANALYSIS BASED ON ASSUMPTION

45

From the analysis done with numerical iteration with the software Excel, thus, the data is presented in the form of line graph.

1 18 35 52 69 86 1031201371541711882052222392562732903073243413580

0.020.040.060.08

0.10.120.140.160.18

0.2

yc vs crank angle

crank angle

yc

1 18 35 52 69 86 103120137154171188205222239256273290307324341358

-0.2

-0.18

-0.16

-0.14

-0.12

-0.1

-0.08

-0.06

-0.04

-0.02

0

ye vs crank angle

crank angle

ye

46

0.5 1 1.5 2 2.5 3 3.5 4 4.50

0.005

0.01

0.015

0.02

0.025

Tmax vs cycle

cycle

Tgas

1 18 35 52 69 86 103120137154171188205222239256273290307324341358

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Tgas(comp) vs crank angle

crank angle

Tgas

(com

p)

47

1 18 35 52 69 86 103120137154171188205222239256273290307324341358-5

0

5

10

15

20

25

Tmass vs crank angle

Tmass(comp) Tmass(exp)

crank angle

Tmas

s

1 18 35 52 69 86 103120137154171188205222239256273290307324341358-5

0

5

10

15

20

25

30

Ttotal vs crank angle

crank angle

Ttot

al

48

0.5 1 1.5 2 2.5 3 3.5 4 4.50.00E+00

5.00E+02

1.00E+03

1.50E+03

2.00E+03

2.50E+03

3.00E+03

3.50E+03

rpm vs cycle

cycle

rpm

0.5 1 1.5 2 2.5 3 3.5 4 4.50

0.001

0.002

0.003

0.004

0.005

0.006

0.007mass gas vs cycle

cycle

mas

s gas

49

0.001 0.002 0.003 0.004 0.005 0.006 0.0070.00E+00

5.00E+02

1.00E+03

1.50E+03

2.00E+03

2.50E+03

3.00E+03

3.50E+03

rpm vs mass gas

mass gas

rpm

50