36
Thermo-elasticy in a cylindrical jet engine nozzle MT 09.16 D.V. Wilbrink July 7, 2009
Thermo-elasticy in acylindrical jet engine nozzle
MT 09.16
D.V. Wilbrink
July 7, 2009
Abstract
The companies AMT Netherlands and Draline have developed a compact jet engine, that is de-signed to be fitted to hang gliders. This so-called PSR Jet System can provide propulsion to thehang glider on demand of the pilot.
In the initial design, the blades of a nozzle in the engine that directs hot exhaust gases fromthe combustion chamber to the turbine blades turned out to crack under prolonged operation.This cracking is caused by the high temperature gradients that occur in the nozzle. The goal ofthis research is to gain insight in the thermo-mechanical response of the nozzle. This is done bymaking an analytical description of the stresses resulting from the temperature gradients and thegiven geometry.
For this analytical stress model, general expressions are first derived for cylindrical objects,containing both solid cylinders and cylinders consisting of a set of blades. Subsequently, these ex-pressions are applied to a few imaginary situations, allowing the reader to get more familiar withthermo-mechanical behavior in cylindrical objects. This is followed by an analysis of the stressdistributions in the PSR Jet System nozzle. The stresses for two different temperature conditionsare computed and subsequently the influence of the stiffness of the solid parts of the nozzle on thestresses is analyzed.
The nozzle analysis showed that stresses involved with temperature gradients due to operationare high, with levels near the yield stress level, and therefore most likely cause cracking. Thestresses involved with shutting down the turbine and thereby suddenly cooling the nozzle bladeswere even higher than stresses from operating conditions. Furthermore, reducing the complianceof the nozzle parts connecting to the blades proved to be an effective way to reduce stress levelsand therefore the risk of cracking in the blades.
This analytical modeling does not provide stress profiles of high accuracy, as it based on a set ofassumptions. However, it can be used to get a clear general impression of the thermo-mechanicalbehavior of PSR Jet System engine nozzle, useful as a guidance for future numerical analysis.
Contents
1 Introduction 3
2 Thermo-elastic analysis of cylindrical bodies 52.1 Individual cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.1.1 Solid cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1.2 Bladed cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Composite cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2.1 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2.2 Resulting system of equations for piecewise constant temperature . . . . . . 9
3 Examples 123.1 Open cylinder with a linear temperature distribution . . . . . . . . . . . . . . . . . 123.2 Double concentric cylinder with piecewise constant temperatures . . . . . . . . . . 15
3.2.1 Cylinders of equal thickness . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2.2 Cylinders of different thickness . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.3 Piecewise constant approximation of a linear temperature profile . . . . . . . . . . 19
4 Analysis of the Turbine Nozzle 214.1 Operating conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.2 Cooldown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4.2.1 Reduced flange stiffness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2.2 Reduced nozzle wall stiffness . . . . . . . . . . . . . . . . . . . . . . . . . . 264.2.3 Reduced flange and engine wall stiffness . . . . . . . . . . . . . . . . . . . . 27
5 Conclusions 29
A Matlab-script to determine the stress profile for various solid composite cylin-ders 30
B Matlab-script to determine the stress profile for the 3-cylinder PSR Jet Systemnozzle model 33
Chapter 1
Introduction
Sometimes hang gliders can be short of lift for a period of time. Having a lightweight boosteraboard to provide power for some time when needed greatly increases the range of flight andlanding safety of such aircraft. An increasing number of new gliders is therefore equipped withsmall two-stroke engines.
However, these engines are relatively large and heavy, requiring an especially designed gliderfuselage. A compact gas turbine engine, similar to those used in small rockets or model airplanes,is a good alternative as it has a higher power-to-weight ratio. The companies of ASR and Dralinehave developed such an engine as a hang glider booster, called PSR Jet System. For both newand existing hang gliders this engine can be fitted inside the fuselage, to be deployed and providepropulsion on demand of the pilot as shown in Figure 1.1.
Figure 1.1: The PSR Jet System mounted on a hang glider
Like any modern jet engine, the PSR Jet System engine (see Figure 1.2) contains a compressorwhich sucks air into the combustion chamber. Hot gas exits this chamber at the rear, thrustingthe plane forward. These hot gases are guided through a nozzle to deflect their path towards theblades of a turbine, causing the turbine wheel to spin rapidly. The turbine powers the compressorat the front by a shaft leading through the center of the jet engine.
Figure 1.2: The PSR Jet System engine
4
During testing of the engine, the blades of the nozzle (shown in Figure 1.3) turned out to crackunder prolonged operation. Apparently the geometry of the component does not allow the usedmaterial, a nickel-based steel, to cope with the stresses due to the major temperature differences inthe nozzle. The goal of this research is to gain insight in the thermo-mechanical stress distributionunderlying this cracking behavior in order to clearly identify the problem.
Figure 1.3: Nozzle of the PSR Jet System engine, viewed from the rear side
The key to the thermo-mechanical stress analysis consist in a model that enables us to de-termine the stress distributions in a cylindrical object with prescribed geometry and temperaturedistributions, which is done in Chapter 2. This analytical model is applied to both single andcomposite concentric cylinders of various geometries and temperature distributions in Chapter 3,to provide more insight on thermo-elastic behavior of cylindrical objects. By applying the analyt-ical model to the PSR Jet System engine nozzle under different conditions in Chapter 4, we cananswer two key questions of the research:
• Are temperature gradients in the material the cause of cracking near the nozzle blades?
• Will this cracking most likely be caused at operating or cooldown conditions?
• How does reducing the stiffness of solid parts in the nozzle affect the stresses?
Finally, in Chapter 5 answers are given to the above questions and conclusions are drawn on theapplicability of the model.
Chapter 2
Thermo-elastic analysis ofcylindrical bodies
In this chapter, a general expression is derived for the stresses occurring in cylindrical objects,being subject to imposed temperature gradients. First general stress expressions are derivedfor individual cylinders. Subsequently, these expressions are linked to one another in objectsconsidered to consist of multiple cylinders, which are referred to as composite cylinders. Anexample of a composite cylinder is depicted in Figure 2.1.
Figure 2.1: A composite of (three) concentric cylinders
2.1 Individual cylinders
In this section, general stress functions for individual cylinders are derived analytically. This isfirst done for solid cylinders, then for cylinders representing a set of blades.
2.1.1 Solid cylinders
The analysis will be done in cylindrical coordinates along the vector basis {~er, ~eθ, ~ez}, where ~ez isalong the axis of the cylinder. The general displacement vector is:
~u = u(r, θ, z)~er + v(r, θ, z)~eθ + w(r, θ, z)~ez (2.1)
Because we only consider axisymmetrical situations we can drop the tangential displacement(v = 0) and therefore reduce the problem from 3D to 2D, considering an (imaginary) infinitelysmall slice of material taken from the disk at an arbitrary angle θ. The axial strain componentcan be eliminated assuming a situation of plane stress: axial stresses are neglected compared tothe radial and tangential stresses σr and σθ, so σz = 0. The remaining deformation is now entirelycharacterized by the radial displacement vector which is a function of the radius r at an arbitraryangle θ:
~u = u(r)~er(θ) (2.2)
2.1 Individual cylinders 6
This displacement gives rise to a two-dimensional strain tensor ε:
ε =12
(~∇~u+ ~∇~uT ) (2.3)
~∇ =∂
∂r~er +
1r
∂
∂θ~eθ (2.4)
ε = εr ~er ~er + εθ ~eθ ~eθ =∂u
∂r~er ~er +
u
r~eθ ~eθ (2.5)
The assumptions imply that bending effects (deforming the disk-shape) are negligible and thecylinder is free to contract in z-direction, giving rise to an axial strain εz, to meet the plane stresscondition.
Each of the three principal strains εr, εθ and εz consist of a mechanical part due to mechanicalstress and a thermal part caused by thermal expansion. Assuming that only elastic deformationoccurs, the mechanical strains are linked to the corresponding stresses by Hooke’s Law. Further-more, thermal strain is the product of the thermal expansion coefficient of the material α and thetemperature difference T with respect to a reference temperature. The three principal strains arethus given by:
εr =1E
(σr − νσθ) + αT (2.6)
εθ =1E
(σθ − νσr) + αT (2.7)
εz = − νE
(σr − σθ) + αT (2.8)
In these relations E is Young’s modulus of elasticy and ν Poisson’s ratio of lateral contraction.Both material parameters are assumed to be constant, as is α. Another used assumption is thatσz = 0. Equations (2.6) and (2.7) can be re-written to:
σr =E
1− ν2(εr + νεθ − (1 + ν)αT ) (2.9)
σθ =E
1− ν2(εθ + νεr − (1 + ν)αT ) (2.10)
Substituting (2.5) now allows one to express the stresses in terms of the radial displacement u:
σr =E
1− ν2
(du
dr+ ν
u
r− (1 + ν)αT
)(2.11)
σθ =E
1− ν2
(u
r+ ν
du
dr− (1 + ν)αT
)(2.12)
These two stress components must satisfy the equation ~∇ · σ = ~0. In the case of axisymmetry asconsidered here, this equation reduces to:
dσrdr
+σr − σθ
r= 0 (2.13)
Substituting the two stress components according to (2.11) and (2.12) finally yields the followingdifferential equation in terms of u(r):
d
dr
(1r
d(ru(r))dr
− (1 + ν)αT (r))
= 0 (2.14)
2.1 Individual cylinders 7
To obtain the deformation and stress state due to an arbitrary (axisymmetric) temperature dis-tribution T (r), we need to derive the general solution u(r) of this differential equation.As a first step towards the solution, (2.14) can be integrated (once) to give:
d(ru)dr
− (1 + ν)αrT (r) = 2(1− ν)c1r (2.15)
where c1 is an integration constant and the factor 2(1 − ν) has been introduced to simplify thefurther analysis. To obtain u(r), we need to integrate this equation once more. As the primitivefunction of rT (r) can only be determined if T (r) is known, we introduce the function Q(r) as aprimitive, so that dQ
dr = rT (r). This allows us to write:
ru− (1 + ν)αQ(r) = (1− ν)c1r2 + c2(1 + ν) (2.16)
where c2 has been introduced as an integration constant. The solution u(r) can now be writtenas:
u(r) = (1− ν)c1r + (1 + ν) (c2 + αQ(r))1r
(2.17)
The general displacement solution thus obtained can be used to determine the two components ofthe strain tensor (2.5):
εr =du
dr= (1− ν)c1 − (1 + ν)
c2r2− (1 + ν)α
Q(r)r2
+ (1 + ν)αT (r) (2.18)
εθ =u
r= (1− ν)c1 + (1 + ν)
c2r2
+ (1 + ν)αQ(r)r2
(2.19)
Substituting these results in equations (2.9) and (2.10) yields the following general expressions forthe radial and tangential stress:
σr(r) = E
(c1 − (c2 + αQ(r))
1r2
)(2.20)
σθ(r) = E
(c1 + (c2 + αQ(r))
1r2− αT (r)
)(2.21)
The constants of integration c1 and c2 must be determined from the boundary conditions resultingfrom the geometry of objects that these stress functions apply to, as we will see in Section 2.2.
2.1.2 Bladed cylinders
For some cylindrical objects, one of the modeled composite cylinders is actually a ring or set ofmultiple blades (provided that there are sufficient blades to assume axisymmetry), as illustratedin Figure 2.2. For this situation, the stress profile is different as we will see in this Section. Asindividual blades are not in contact with their neighbors, no tangential stresses can be transmittedso σθ = 0. Note that blades may have some internal tangential stress, but for the collection ofblades as a whole we can neglect it. With this assumption relation (2.6) reduces to:
εr =σrE
+ αT (r) (2.22)
which can be re-written asσr = E(
du
dr− αT (r)) (2.23)
The equilibrium equation ~∇ · σ = ~0 reduces to:
dσrdr
= 0 (2.24)
2.2 Composite cylinders 8
r1
r2
r1
r2r3
r0
Figure 2.2: Illustration of a composite of two solid cylinders and one bladed cylinder
Substitution of the radial stress function (2.23) in this equation yields:
d
dr
(E(
du
dr− αT (r))
)= 0 (2.25)
This differential equation will be solved to obtain an expression for u(r). As a first step towardsthe solution, we now have:
du
dr− αT (r) = c1 (2.26)
The temperature profile T (r) is yet unknown, so to make the next step in formulating an expressionfor u(r) we introduce an alternative primitive function Q(r) for the blades, so that dQ
dr = T (r).We now have the general displacement function:
u(r) = c1r + c2 + αQ(r) (2.27)
Substitution of this result in the radial stress distribution (2.23) yields:
σr(r) = Ec1 (2.28)
The constants c1 and c2 are again to be determined by the boundary conditions resulting fromthe geometry. The above solution will be used in Chapter 4 for the stress analysis of the PSR JetSystem nozzle.
2.2 Composite cylinders
In the previous chapter we formulated general stress distributions for cylindrical objects experienc-ing temperature gradients. The constants featuring in these expressions can be determined by theboundary conditions for the given geometry of the object, that will be formulated in Section 2.2.1.These conditions result in a set of equations that determines the constants of integration ci1 andci2, where i = 1, 2, ..., n indicates the individual concentric cylinders, as shown in Section 2.2.2.The obtained expressions for the constants can be substituted in the general stress distributionsto determine their precise form for the given geometry and loading.
2.2.1 Boundary conditions
The boundaries for which conditions are prescribed can be subdivided in two types: free boundariesand interfaces.
2.2 Composite cylinders 9
Free boundaries
Every (composite) cylinder has has an inner surface r = r0 and an outer surface r = rn, wheren is the total number of individual cylinders in the composite cylinder (see for example r0 andr3 in Figure 2.2). At these free boundaries, the material is free to move in radial direction. Theassumption of static equilibrium implies that for these surfaces, we have:
σr(r0) = σr(rn) = 0 (2.29)
This boundary condition enables us to eliminate one integration constant in each of the inner andouter cylinder’s stress distributions.
Interfaces
Composite cylinders have interfaces, i.e. surfaces r = ri with i = 1, 2, 3, ..., n − 1 where twoconcentric cylinders make contact with each other. Examples are r1 and r2 in Figure 2.2. Ateach of these interfaces, two conditions can be formulated. These 2(n− 1) conditions allow us toeliminate the remaining 2(n− 1) integration constants.One condition follows from the requirement that the (radial) forces acting on each interface fromboth sides must be in equilibrium. For interfaces between two solid concentric cylinders, this canbe formulated as:
tiσr,i(ri) = ti+1σr,i+1(ri) (2.30)
If we formulate the total cross-sectional area of bladed cylinders at the interface as mA, where mis the number of blades in the cylinder and A is the cross-sectional area of a single blade, the forceequilibrium can be formulated as:
ritiσr,i(ri) = mAσr,i+1(ri) (2.31)
A second condition at the interface follows from the continuity of material, as we assume norupture occurs. This means the displacements at both sides of the interface must be equal:
ui(ri) = ui+1(ri) (2.32)
2.2.2 Resulting system of equations for piecewise constant temperature
For a composite cylinder consisting of n individual cylinders, we have to specify 2n constants ofintegration to determine the stress distribution in every concentric cylinder it contains. This isdone by solving the system of 2n equations consisting of 2 free boundary conditions and 2(n− 1)interface conditions. In this section, this system is formulated for composite cylinders that consistsof concentric cylinders which each have uniform temperature Ti. This is first done for systemsconsisting of solid cylinders and subsequently for boundaries where bladed cylinders are involved.
Solid cylinders
Using the temperature distribution T (r) = Ti for ri−1 < r ≤ ri, the function Qi(r) for the i-thcylinder can be determined to be:
Qi(r) =12Tir
2 (2.33)
Substituting this expression in the displacement solution (2.17) and stress distributions (2.20) and(2.21) yields:
ui(r) = (1− ν)ci1r + (1 + ν)ci21r
+12
(1 + ν)αTir (2.34)
σr,i(r) = E
(ci1 −
ci2r2− 1
2αTi
)(2.35)
2.2 Composite cylinders 10
σθ,i(r) = E
(ci1 +
ci2r2− 1
2αTi
)(2.36)
As a first step in determining the constants, the free boundary conditions σr(r0) = σr(rn) = 0from Section 2.2.1 are applied to the radial stress expression (2.35), resulting in the followingequations:
c11 −c12r20
=12αT1 (2.37)
cn1 −cn2
r2n=
12αTn (2.38)
Substitution of the radial stress expression (2.35) in (2.30) yields for the interface at ri:
ti(ci1 −ci2r2i
)− tj(cj1 −cj2r2i
) =12α(tiTi − tjTj) (2.39)
where j = i+ 1.Finally, equation (2.32), after substitution of the displacement expression (2.34) reads:
1− ν1 + ν
(ci1 − cj1)r2i + (ci2 − cj2) =12α(Tj − Ti)r2i (2.40)
We now have 2n equations to solve 2n constants of integration, enabling us to determine the stressdistributions for composite cylinders these equations apply to.Calculation by hand easily becomes tedious for composite cylinders with n ≥ 3, so we let computerpower take over at this point. The Matlab-script added in appendix A takes material parametersand vectors containing radius-, thickness- and temperature values, solves the above expressionsfor the integration constants and draws the graphs of the radial and tangential stress in red andgreen respectively.
Bladed cylinders
Around cylinders that consist of a (by approximation) axisymmetric set of blades as for exampleshown in Figure 2.2, the equations resulting from the interface boundary conditions are differentthan those for solid cylinders.We define the bladed cylinder at uniform temperature Ti and thickness ti at ri−1 < r ≤ ri. Withthe temperature known, we can determine Q(r):
Q(r) = rTj (2.41)
Substituting this in the displacement expression (2.27) yields:
ui(r) = (ci1 + αTi)r + ci2 (2.42)
The radial stress distribution is provided by (2.28) and becomes:
σr(r) = Eci1 (2.43)
We can now use the boundary conditions to determine the constants, starting with the forceequilibrium at the inner boundary of the bladed ring r = ri. Using j = i+1, the force equilibriumequation (2.31) can be re-written to get:
ci1 −ci2r2− mA
riticj1 =
12αTi (2.44)
The displacement continuity equation (2.32) can be specified by substitution of the displacementfunctions (2.34) and (2.42):
(1− ν)ci1ri + (1 + ν)ci2ri− cj1ri − cj2 = αri
(Tj −
12
(1 + ν)Ti
)(2.45)
2.2 Composite cylinders 11
Analogically, the displacement continuity at r = rj can be written as:
(1− ν)ck1rj + (1 + ν)ck2rj− cj1rj − cj2 = αrj
(Tj −
12
(1 + ν)Tk
)(2.46)
This set of equations provides us a way of determining the constants of integration and thereforethe exact stress distributions. This will be done in Chapter 3 for some imaginary situations togain insight on thermo-elastic stress behavior of cylindrical objects. In Chapter 4, this system isapplied to obtain stress distributions for a model of the PSR Jet System engine nozzle.
Chapter 3
Examples
To provide a better understanding of thermo-elastic stresses predicted by the solutions that werederived in Chapter 2, the stress distributions in a number of typical cylindrical objects are eval-uated in this chapter. First a single cylinder with a hole and a linear temperature distribution isanalyzed. In the second example, stresses in two concentric cylinders with piecewise constant tem-perature distribution are computed. Finally, the assumption of piecewise constant temperatureis used to numerically simulate the linear temperature profile of the first example cylinder. Forall situations the material parameters listed in Table 3.1 are assumed, unless indicated otherwise.The reference temperature, at which objects are assumed to be stress-free, is taken as 20◦C, i.e.T = 0 implies a temperature of 20◦C. The given yield stress can be used relate the calculatedstresses to.
Parameter ValueE 190GPaν 0.3α 16 ∗ 10−6K−1
σy 760MPa
Table 3.1: Material properties of nickel based alloy Inconel 713 LC around 540◦C
3.1 Open cylinder with a linear temperature distribution
Let us take for example a single open cylinder as illustrated in Figure 3.1 with a linear temperaturedistribution between T = T0 at the inner surface (r = r0) and T = T1 at the outer surface (r = r1),as shown in the temperature profile of Figure 3.2.
r1r0
z
r
Figure 3.1: A single cylinder
The temperature distribution in the material, i.e. for r0 < r < r1, is thus:
T (r) =∆T∆r
r + T0 (3.1)
3.1 Open cylinder with a linear temperature distribution 13
T(r)
r1r0
T0
T1
r
T
Figure 3.2: Linear temperature profile T (r) within the cylinder
where ∆T = T1 − T0 and ∆r = r1 − r0. The primitive Q(r) of rT (r) can be determined as:
Q(r) =13
∆T∆r
r3 +12T0r
2 (3.2)
The radial stress as given by equation (2.20) becomes in this case:
σr(r) = E
(c1 −
12αT0 −
c2r2− 1
3α
∆T∆r
r
)(3.3)
Using the two free boundary conditions we can determine the constants c1 and c2. Starting withσr(r0) = 0 we obtain:
c1 −12αT0 =
c2r20
+13α
∆T∆r
r0 (3.4)
Substituting this result in (3.3) to eliminate c1 yields:
σr(r) = E
(c2(r−2
0 − r−2)− 13α
∆T∆r
(r − r0))
(3.5)
The second boundary condition, σr(r1) = 0, allows us to also eliminate c2 as:
c2 =13α∆T
1− ( r0r1 )2r20 (3.6)
so that we finally obtain:
σr(r) =13αE∆T
(1− ( r0r )2
1− ( r0r1 )2−
rr1− r0
r1
1− r0r1
)(3.7)
This means that the radial stress in proportionally dependent on α, E and ∆T . This thirdparameter reveals the cylinder is stress-free at a homogenous reference temperature (not strain-free). Furthermore, this stress distribution also implies that the radial stress has a hyperbolicshape, as shown in Figure 3.3 using the geometry- and temperature parameters shown in Table3.2. This distribution can be understood as follows. As warmer material at the outside of thering tends to expand more than that at the inside, a tensile stress exists in the material in theradial direction. This radial stress vanishes at the inner and outer surface to satisfy the stress-freeboundary condition.
The tangential stress can be obtained by substituting (3.1) and (3.2) in the general expression(2.21) to get:
σθ(r) = E
(c1 +
c2r2− 1
2αT0 −
23α
∆T∆r
r
)(3.8)
3.1 Open cylinder with a linear temperature distribution 14
r0 r1 T0 T1
0.04 m 0.10 m 400◦C 700◦C
Table 3.2: Properties of the single cylinder with linear temperature distribution. The given tem-peratures do not include the reference temperature
0.04 0.05 0.06 0.07 0.08 0.09 0.10
10
20
30
40
50
60
70
80
90
100
110
r [m]
radi
al s
tres
s [M
Pa]
Figure 3.3: Radial stress profile σr(r) in the cylinder with linear temperature distribution
0.04 0.05 0.06 0.07 0.08 0.09 0.1−400
−300
−200
−100
0
100
200
300
400
500
600
r [m]
tang
entia
l str
ess
[MP
a]
Figure 3.4: Tangential stress profile σθ(r) in the cylinder with linear temperature distribution
Then use expressions (3.4) and (3.6) to obtain:
σθ(r) =13αE∆T
(1 + r20
r2
1− ( r0r1 )2−
2 rr1− r0
r1
1− r0r1
)(3.9)
As we can see in Figure 3.4, in which this distribution has been plotted, the tangential stressgradually changes from being positive at the inner surface to negative at the outer surface. Thisis also a result of increasing thermal expansion as we move outwards: the inner material partiallyfollows the outer layers of material as they tend to expand more. As a result, inner layers arestretched tangentially. On the other hand, the outer part of the cylinder is restricted in itsthermal expansion by the cooler inner part and therefore adopts a smaller radius than it would ifunconstrained; a compressive tangential stress results.Figure 3.5 demonstrates that, for the cylinder to be in equilibrium, the net tangential force acting
3.2 Double concentric cylinder with piecewise constant temperatures 15
dr
dz
Figure 3.5: Tangential forces acting on a cross-section in a r-z-plane
on an arbitrary cross-section in the r-z-plane must vanish. This tangential force balance meansthat the tangential stresses integrated over the area of this cross-section have to be zero. As westated before that stresses are constant in z-direction and only depend on the radius r, this meansthe tangential stresses integrated over the radius have to be zero:∫ r1
r0
σθ(r) dr = 0 (3.10)
We can see that this relation indeed holds in Figure 3.4.
3.2 Double concentric cylinder with piecewise constant tem-peratures
In this example we consider a composite cylinder which consists of two concentric uniform cylin-ders, as depicted in Figure 3.6. Both cylinders have their own constant uniform temperatureT (r) = Ti for ri−1 < r ≤ ri with i = 1, 2, as illustrated in Figure 3.7.
z
r
r1r0 r2
t 1 t 2
Figure 3.6: A composite of two concentric cylinders
T(r)
r1r0
T1
T2
r2r
T
Figure 3.7: Temperature profile T (r) within the double cylinder for T1 < T2
T0 EN T1 PLAATJE AANPASSEN General stress expressions for this temperature profilewere derived in Section 2.2.2. Also the equations resulting from the boundary conditions were
3.2 Double concentric cylinder with piecewise constant temperatures 16
formulated. From the free boundary conditions (2.37) and (2.38) we obtain for this n = 2 situation:
c11 −12αT1 =
c12r20
(3.11)
c21 −12αT2 =
c22r22
(3.12)
Substitution of these expressions in the radial stress distributions yields:
σr,1(r) = c12E(r−20 − r−2) (3.13)
σr,2(r) = c22E(r−22 − r−2) (3.14)
To determine c12 and c22 we use the two interface conditions at r = r1. The interface forceequilibrium (2.30) can be written as:
c12 =t2t1
( r0r2 )2 − ( r0r1 )2
1− ( r0r1 )2c22 (3.15)
Furthermore, the interface displacement continuity condition (2.40) can be expressed as:
1− ν1 + ν
(c11 − c21)r21 + (c12 − c22) =12α(T2 − T1)r21 (3.16)
With these four equations all the four integration constants can be expressed in terms of thegiven parameters. Substituting the results in the stress distributions (2.20) and (2.21) gives thestress expressions for this situation as:
σr,1(r) = Eα∆T( r0r1 )2 − ( r0r2 )2
t1t2A+B
(1−
(r0r
)2)
(3.17)
σr,2(r) = Eα∆T1− ( r0r1 )2
A+ t2t1B
((r0r
)2
− (r0r2
)2)
(3.18)
σθ,1(r) = Eα∆T( r0r1 )2 − ( r0r2 )2
t1t2A+B
(1 +
(r0r
)2)
(3.19)
σθ,2(r) = Eα∆T1− ( r0r1 )2
A+ t2t1B
((r0r
)2
+ (r0r2
)2)
(3.20)
whereA = (1− (
r0r1
)2)(1− ν)(r0r2
)2 + (1 + ν)(r0r1
)2 (3.21)
B = ((r0r1
)2 − (r0r2
)2)((1− ν) + (1 + ν)(r0r1
)2) (3.22)
Note that ∆T = T2 − T1 is now the temperature difference between the two cylinders. Thesequadratic functions are rather elaborate, containing many dimensionless products of geometry.In the following two subsections we will visualize these stress profiles and compare them withthose generated by the Matlab script of Appendix A. This is first done for two cylinders ofequal thicknesses and then for different thicknesses. The parameter values used (unless indicatedotherwise) are given in Table 3.3.
3.2 Double concentric cylinder with piecewise constant temperatures 17
r0 r1 r2 t1 t2 T0 T1
0.04 m 0.10 m 0.15 m 0.004 m 0.01 m 400◦C 700◦C
Table 3.3: Properties of a double cylinder with piecewise constant temperature distribution
3.2.1 Cylinders of equal thickness
For t1 = t2 the dimensionless geometry terms in the stress distributions significantly simplify to:
σr,1(r) =12Eα∆T
r22 − r21r22 − r20
(1− r20
r2
)(3.23)
σr,2(r) =12Eα∆T
r21 − r20r22 − r20
(r22r2− 1)
(3.24)
σθ,1(r) =12Eα∆T
r22 − r21r22 − r20
(1 +
r20r2
)(3.25)
σθ,2(r) = −12Eα∆T
r21 − r20r22 − r20
(r22r2
+ 1)
(3.26)
By filling in the parameter values from Table 3.3, except for t1, which is taken equal to t2 = 0.10m,we can determine the stress distributions. The figures representing these distributions which aredrawn here were provided by running the Matlab-file given in Appendix A with the used param-eter values as entry, indicating the file works properly.
z
r
r1r0 r2
0.04 0.06 0.08 0.1 0.12 0.14 0.16−50
0
50
100
150
200
250
r [m]
radi
al s
tres
s [M
Pa]
Figure 3.8: Radial stress profile σr(r) in the double cylinder of uniform thickness
3.2 Double concentric cylinder with piecewise constant temperatures 18
0.04 0.06 0.08 0.1 0.12 0.14 0.16−600
−400
−200
0
200
400
600
r [m]
tang
entia
l str
ess
[MP
a]
Figure 3.9: Tangential stress profile σθ(r) in the double cylinder of uniform thickness
Figure 3.8 shows how the profile of the radial stress starts at zero at the inner and outersurfaces and reaches a maximum value at the interface r = r1. In contrast to the radial stress,the tangential stress is not a continuous function as Figure 3.9 demonstrates. For this system of acold inner cylinder and a hot outer cylinder, the relatively cold inner cylinder resists the expansionof the hot outer cylinder, giving it a positive tangential stress. This is similar to the situation oflinear temperature profile from Section 3.1, but now the temperature suddenly changes, resultingin a discontinuous change from positive to negative tangential stress at the interface. As can beconcluded from the graph, the tangential force balance over the total cross-section neverthelessstill applies: ∫ r2
r0
σθ(r) dr =∫ r1
r0
σθ,1(r) dr +∫ r2
r1
σθ,2(r) dr = 0 (3.27)
3.2.2 Cylinders of different thickness
For the more general case of two concentric cylinders with different thicknesses we take the pa-rameter values given in Table 3.3. The radial and tangential stresses are as depicted in Figure3.10 and Figure 3.11 respectively.
Compared to Figure 3.8 for uniform thickness, the radial stress profile is no longer continuousat the interface. As the force balance (2.30) prescribes, it shows a jump by a factor which is theinverse of the thickness ratio t2
t1between the two cylinders. Compared to the uniform case,the
radial stresses in the inner cylinder have increased while those in the outer cylinder have decreased.The tangential stress profile (Figure 3.11) also remains of the same form as for uniform thickness.The tangential force equilibrium (3.27) still applies, provided that we now take the thicknessdifference into account, again confirming the situation to be in static equilibrium:
t1
∫ r1
r0
σθ,1(r) dr = −t2∫ r2
r1
σθ,2(r) dr (3.28)
3.3 Piecewise constant approximation of a linear temperature profile 19
0.04 0.06 0.08 0.1 0.12 0.14 0.16−50
0
50
100
150
200
250
300
350
400
450
r [m]
radi
al s
tres
s [M
Pa]
Figure 3.10: Radial stress profile σr(r) in the double cylinder of different thickness
0.04 0.06 0.08 0.1 0.12 0.14 0.16−500
0
500
1000
r [m]
tang
entia
l str
ess
[MP
a]
Figure 3.11: Tangential stress profile σθ(r) in the double cylinder of different thickness
3.3 Piecewise constant approximation of a linear tempera-ture profile
To demonstrate the use of the piecewise constant temperature profile, the cylinder of Section 3.1with 0.04m ≤ r ≤ 0.10m and 400◦C ≤ T ≤ 700◦C is subdivided in 12 parts. This means thatevery ∆r = 5mm the temperature steps up by ∆T = 25◦C.
The stress profiles shown in Figure 3.13 and Figure 3.14 now clearly resemble Figure 3.3 andFigure 3.4 respectively. As becomes especially clear from the tangential stress distribution, ahigher accuracy can be obtained by taking more, but smaller temperature steps as input.
3.3 Piecewise constant approximation of a linear temperature profile 20
0.4 0.5 0.6 0.7 0.8 0.9 1350
400
450
500
550
600
650
700
750
r [−]
Tem
pera
ture
[K]
Figure 3.12: Temperature profile T (r) used in the piecewise constant approximation of a linearprofile
0.4 0.5 0.6 0.7 0.8 0.9 10
10
20
30
40
50
60
70
80
90
100
r [m]
radi
al s
tres
s [M
Pa]
Figure 3.13: Radial stress profile σr(r) obtained from the piecewise constant temperature
0.4 0.5 0.6 0.7 0.8 0.9 1−400
−300
−200
−100
0
100
200
300
400
500
r [m]
tang
entia
l str
ess
[MP
a]
Figure 3.14: Tangential stress profile σθ(r) obtained from the piecewise constant temperature
Chapter 4
Analysis of the Turbine Nozzle
Now that it is possible to predict the stress distributions for many cylindrical geometries and tem-perature profiles, we can analyse the actual nozzle of the PSR Jet System as shown in Figure 4.1.We first describe the nozzle model geometry and material properties, to use those in the followingparagraphs to determine the stress distributions for various conditions.
Figure 4.1: Nozzle of the PSR Jet System engine, viewed from the front side
To construct the model, the nozzle geometry is simplified to a composite cylinder consistingof three concentric cylinders as shown in Figure 2.2. From the inside to the outside the cylindersrepresent:
1. The inner flange, modeled as a solid cylinder
2. The blades, modeled as a bladed cylinder
3. The nozzle wall; modeled as a solid cylinder
The second cylinder is modeled by imaginarily tilting each of the blades to align them parallel tothe planes of the other two cylinders. Together they then approximately form a cylinder of thesame thickness of the cylinder’s individual blades, see Figure 2.2. This cylinder is furthermoretangentially compliant, as individual blades are not able to transfer stresses in tangential directiononto one another. For the composite cylinder, the geometry given in Table 4.1 is used, more orless representing the actual sizes of the nozzle.
4.1 Operating conditions 22
The material that the nozzle consists of is Inconel 713, a nickel-based steel. The material prop-erties are assumed to be constant at the values given in Table 3.1. The reference temperature, atwhich objects are assumed to be stress-free, is taken to be 20◦C.
0 1 2 3ri[mm] 30.0 60.0 84.8 88.0ti[mm] - 3 1.5 55.4
Table 4.1: Geometry properties of the PSR Jet System nozzle model
To determine the stress distributions for various conditions in the nozzle, we can use the setof equations that determine the constants of integration for both solid and bladed cylinders fromSection 2.2.2. Solving the resulting system of 6 equations for 6 integration constants and substi-tuting the results in the general stress distributions yields the stress profiles for every situation.The added Matlab-script in Appendix B is used to do these calculations, determine the stressprofiles and draw the plots for these turbine nozzle models.First, the stresses at operating conditions are evaluated in Section 4.1. As cracking is suspected tobe initiated under cooldown conditions, the stresses in this situation are examined subsequentlyin Section 4.2. Finally, the geometry of the inner and outer cylinder is adapted in the followingthree sections to examine the influence of their stiffness on the stress profile during cooldown.
4.1 Operating conditions
When the engine is running at full throttle, hot exhaust gases are assumed to heat the blades toapproximately 700◦C, the outer nozzle wall to 650◦C while the inner flange remains at 250◦C asshown in Table 4.2.
1 2 3Ti[◦C] 250 700 650
Table 4.2: Estimated temperature distribution in the PSR nozzle model under operating conditions
Figure 4.2 shows the radial stress distribution predicted for these temperatures. A positiveradial stress exists in every part of the nozzle, so the whole composite is under tension. Theminimum values are, as they should be, at the radial-stress-free inner and outer surface of thenozzle. The stress profile clearly shows a constant and maximum radial stress in the blades,which at 628 MPa is at least a factor two higher than anywhere in the flange or the wall and closeto the yield stress, which is 670 MPa. This confirms the blades to be the most likely parts to crack.
The tangential stress profile of Figure 4.3 shows how the bladed cylinder has no tangentialstress. The relatively cold flange has a positive tangential stress, as it resist the expansion of theother two (hot) cylinders. In contrast, the outer cylinder has a negative tangential stress, as it isrestricted in its expansion by the flange inside.
Note that the maximum value of tangential stress, located near the inner surface, is higherthan the maximum radial stress. From a physical perspective this is unexpected, as the cold insideof the flange is the least likely part of the nozzle to deform plastically. This result is probably dueto the inaccurate geometrical modeling of the inner flange, but a more accurate explanation is yetto be found.
4.2 Cooldown 23
0.03 0.04 0.05 0.06 0.07 0.08 0.090
100
200
300
400
500
600
700
r [m]
radi
al s
tres
s [M
Pa]
Figure 4.2: Radial stress profile σr(r) of the PSR nozzle model under operating conditions
0.03 0.04 0.05 0.06 0.07 0.08 0.09−400
−200
0
200
400
600
800
1000
r [m]
tang
entia
l str
ess
[MP
a]
Figure 4.3: Tangential stress profile σθ(r) of the PSR nozzle model under operating conditions
4.2 Cooldown
When the fuel supply of the PSR Jet is shut down to end a boost, the engine is still running ataround 150.000 rpm, suddenly sucking cold outside air through the turbine. As a consequence,the nozzle blades cool rapidly while the engine wall is still at the high operating temperature,not being capable of releasing its heat as rapidly as the blades. We assume for this situation thetemperature of the blades changes to 450◦C, yielding the temperature distribution shown in Table4.3. This situation is expected to be the most critical for cracking, so the stress profile is examined
4.2 Cooldown 24
in this section.
1 2 3Ti[◦C] 250 450 650
Table 4.3: Temperature profile of the PSR nozzle model under cooldown conditions
0.03 0.04 0.05 0.06 0.07 0.08 0.09−100
0
100
200
300
400
500
600
700
800
900
r [m]
radi
al s
tres
s [M
Pa]
Figure 4.4: Radial stress profile σr(r) of the PSR nozzle model under cooldown conditions
0.03 0.04 0.05 0.06 0.07 0.08 0.09−600
−400
−200
0
200
400
600
800
1000
1200
r [m]
tang
entia
l str
ess
[MP
a]
Figure 4.5: Tangential stress profile σθ(r) of the PSR nozzle model under cooldown conditions
4.2 Cooldown 25
The radial stress depicted in Figure 4.4 shows a similar profile to that of the full operatingconditions, but now with higher values. The maximum radial stress, located in the blades hasincreased from 628 MPa to 800 MPa. This means cracking is indeed more likely to occur in theblades in this cooldown situation than under operating conditions. The tangential stress profileof Figure 4.5 shows again how the hot nozzle wall’s expansion is resisted by the flange and thetangentially weak bladed cylinder, this time with increased absolute values compared to the op-erating conditions.
Based on the above observations, a logical way to avoid cracking in the nozzle’s blades is tomake the parts connected to the blades more compliant. This can be achieved by either makingthe flange more compliant, or the wall, or both. This will be done in the following subsections forthe present cooldown situation. More compliance is introduced by reducing the radial dimensionsof the solid cylinders.
4.2.1 Reduced flange stiffness
To represent a more flexible flange, in this Section the center hole radius is imaginarily increasedto r0 = 45mm, thus reducing the radial thickness of the flange by 50%. All other parameters ofthe cooldown situation are still assumed to apply. The radial geometry for this situation is givenin Table 4.4.
i 0 1 2 3ri[mm] 45.0 60.0 84.8 88.0
Table 4.4: Radial geometry of the cooldown PSR nozzle model with enlarged r0
0.03 0.04 0.05 0.06 0.07 0.08 0.09−100
0
100
200
300
400
500
600
700
800
900
r [m]
radi
al s
tres
s [M
Pa]
Figure 4.6: Radial stress profile σr(r) of the cooldown PSR nozzle model with enlarged r0
The shape of the radial stress profile shown in Figure 4.6 is similar to that of the reference(cooldown) situation from Figure 4.4, which is depicted in dashed lines here. However, values have
4.2 Cooldown 26
0.03 0.04 0.05 0.06 0.07 0.08 0.09−600
−400
−200
0
200
400
600
800
1000
1200
1400
r [m]
tang
entia
l str
ess
[MP
a]
Figure 4.7: Tangential stress profile σθ(r) of the cooldown PSR nozzle model with enlarged r0
decreased significantly; the maximum radial stress in the blades has dropped by 34%. The flangestiffness clearly has a significant influence on the radial stresses in the blades.
The tangential stress profile of Figure 4.7 increases in the flange as is gets smaller. Simultane-ously, leaving out a part of the flange material causes the absolute tangential stress in the nozzlewall to decrease as the tangential force balance prescribes as given in the equation:
t1
∫ r1
r0
σθ,1(r) dr = −t3∫ r3
r2
σθ,3(r) dr (4.1)
where an increase in r0 causes σθ,3 to decrease.
4.2.2 Reduced nozzle wall stiffness
Instead of reducing the flange stiffness, we can also weaken the nozzle wall. Again the radialthickness of the solid cylinder representing it is reduced by 50%, so that r3 = 86.4mm, as shownin Table 4.5.
0 1 2 3ri[mm] 30.0 60.0 84.8 86.4
Table 4.5: Radial geometry of the cooldown PSR nozzle model with reduced r3
Compared to the (dashed) reference stress distribution, the radial stress with reduced wallstiffness as given in Figure 4.8 has again decreased, this time by 28% in the blades. Removingthe outer half of the wall therefore has a similar effect on the radial stresses as removing the innerhalf of the flange, as both changes reduce the radial stress.
The change in tangential stress is the opposite of the change depicted in Figure 4.7, as Figure4.9 shows.
4.2 Cooldown 27
0.03 0.04 0.05 0.06 0.07 0.08 0.09−100
0
100
200
300
400
500
600
700
800
900
r [m]
radi
al s
tres
s [M
Pa]
Figure 4.8: Radial stress profile σr(r) of the cooldown PSR nozzle model with reduced r3
0.03 0.04 0.05 0.06 0.07 0.08 0.09−600
−400
−200
0
200
400
600
800
1000
1200
r [m]
tang
entia
l str
ess
[MP
a]
Figure 4.9: Tangential stress profile σθ(r) of the cooldown PSR nozzle model with reduced r3
4.2.3 Reduced flange and engine wall stiffness
Knowing that splitting the flange or engine wall radially in half reduces the radial stresses inthe blades by about one third, we will now see what happens if the two stiffness reductions arecombined.
Figure 4.10 shows how the radial stress values now have been reduced to 52% of their originalvalue at cooldown. This 48% reduction is the same as reducing the stress by 34% and 28% suc-
4.2 Cooldown 28
0 1 2 3ri[mm] 30.0 60.0 84.8 86.4
Table 4.6: Radial geometry of the cooldown PSR nozzle model with enlarged r0 and reduced r3
0.03 0.04 0.05 0.06 0.07 0.08 0.09−100
0
100
200
300
400
500
600
700
800
900
r [m]
radi
al s
tres
s [M
Pa]
Figure 4.10: Radial stress profile σr(r) of the cooldown PSR nozzle model with enlarged r0 andreduced r3
cessively, so the stiffness reductions from both sides add up.
Chapter 5
Conclusions
The analytical model that is constructed int his paper showed that stressess in the material directlyproportionally depend on temperature differences with a reference temperature. Other parame-ters on which the stresses depend directly proportional are the Youngs modulus and the thermalexpansion coefficient, both properties that indicate the ability of the material to cope with strainsand temperature gradients. For the turbine nozzle of the PSR Jet System, this meant that tem-peratures involved with turbine operation reached levels near the yield stress level, confirming thedesigners suspicions that cracking of the nozzle blades under prolonged operation is most likelycaused by these temperatures. Compared to operating conditions, stress levels increased when theturbine is shut down so the nozzle blades cool rapidly while the adjacent material is still at hightemperature.
One way to reduce stress levels in the nozzle blades, and therefore reduce the risk of cracking,is to make the adjacent parts of the nozzle more compliant. This solution, modeled by reducingthe radial thickness, proved to reduce stresses by significant amounts.
The model composed and applied to analyse the thermo-mechanical behavior of the PSR JetSystem nozzle has a number of limitations. Stress levels are not to be taken as very accurate, asthe model is based on a set of assumptions that simplify stress results, for example the assump-tions that the nozzle parts do not bend and are in a situation of plane stress. The single-cylindermodeling of the inner flange is another considerable simplification.
What these results do provide is a global impression of the thermo-mechanical behavior of theturbine nozzle. It tells under which conditions cracking is probably caused and how it can be pre-vented. To make a more detailed analysis with accurate stress numbers, a numerical model of theturbine nozzle could be made. The analytical results from this paper can be used to locate areasof interest for numerical stress analysis of the nozzle, for example where the blades are attachedto the flange and nozzle wall. Knowing that cooldown of the turbine is most likely to make theblades crack, a more detailed analysis of thermal-mechanical behavior during this (repeated) pro-cess can be done. Additionally, the flange can be modeled in more detail to examine the influenceof compliance more thoroughly.
Appendix A
Matlab-script to determine thestress profile for various solidcomposite cylinders
clear allclose all
E = 1.9e5; % [MPa]v = 0.3; % [-]a = 16e-6; % [K^-1]Tref = 20; % [deg. C]
% 3-cylinder PSR Jet Engine nozzler0=0.03;rvec=[0.060 0.0848 0.088]’; % vector r1;r2;..;rntvec=[0.003 0.0015 0.0554]’; % vector t1;t2;..;tnTvec=[400 700 400]’-Tref; % vector T1;T2;..;Tn
N=length(rvec);
% ENTER FREE BC’S SYSTEMCSF=zeros(2,2*N+1);% BC sigmar @ r0:CSF(1,1)=1;CSF(1,2)=-1/(r0^2);CSF(1,2*N+1)=a*Tvec(1)/2;% BC sigmar @ rN:CSF(2,2*N-1)=1;CSF(2,2*N)=-1/(rvec(N)^2);CSF(2,2*N+1)=a*Tvec(N)/2;
% ENTER INTERFACE STRESS BC’S SYSTEMCS=zeros(N-1,2*N+1);% BC sigmar @ riRivec=[];for i=1:(N-1)
CS(i,2*i-1)=tvec(i);
31
CS(i,2*i)=-tvec(i)/(rvec(i)^2);CS(i,2*i+1)=-tvec(i+1);CS(i,2*i+2)=tvec(i+1)/(rvec(i)^2);CS(i,2*N+1)=1/2*a*(tvec(i)*Tvec(i)-tvec(i+1)*Tvec(i+1));
end
% ENTER INTERFACE DISPLACEMENT BC’S SYSTEMCU=zeros(N-1,2*N+1);% BC u @ rifor i=1:(N-1)
CU(i,2*i)=1;CU(i,2*i+2)=-1;CU(i,2*i-1)=(1-v)/(1+v)*rvec(i)^2;CU(i,2*i+1)=-(1-v)/(1+v)*rvec(i)^2;CU(i,2*N+1)=1/2*a*rvec(i)^2*(Tvec(i+1)-Tvec(i));
end
% CREATE AND SOLVE TOTAL SYSTEM OF CONSTANTSC=[CSF;CS;CU];Cans=rref(C);
% PLOT RADIAL STRESS GRAPHSfigurerstep=0.0001;plot(r0:rstep:rvec(length(rvec)),0,’k’) % plot black r-axishold on% PLOT RADIAL STRESS GRAPH CYLINDER 1r=r0:rstep:rvec(1);sr1=E*(Cans(1,2*N+1)-Cans(2,2*N+1)./(r.^2)-1/2*a*Tvec(1));plot(r,sr1,’Color’,[.8 0 0],’LineWidth’,2)% PLOT OTHER RADIAL STRESS GRAPHSfor i=2:N
r=rvec(i-1):rstep:rvec(i);sr=E*(Cans(2*i-1,2*N+1)-Cans(2*i,2*N+1)./(r.^2)-1/2*a*Tvec(i));plot(r,sr,’Color’,[.8 0 0],’LineWidth’,2)
endgrid onxlabel(’r [m]’)ylabel(’radial stress [MPa]’)hold offprint -depsc sr.eps
% PLOT TANGENTIAL STRESS GRAPHSfigureplot(r0:rstep:rvec(length(rvec)),0,’k’) % plot black r-axishold on% PLOT TANGENTIAL STRESS GRAPH CYLINDER 1r=r0:rstep:rvec(1);st1=E*(Cans(1,2*N+1)+Cans(2,2*N+1)./(r.^2)-1/2*a*Tvec(1));plot(r,st1,’Color’,[0 .5 0],’LineWidth’,2)% PLOT OTHER TANGENTIAL STRESS GRAPHSfor i=2:N
32
r=rvec(i-1):rstep:rvec(i);st=E*(Cans(2*i-1,2*N+1)+Cans(2*i,2*N+1)./(r.^2)-1/2*a*Tvec(i));plot(r,st,’Color’,[0 .5 0],’LineWidth’,2)
endgrid onxlabel(’r [m]’)ylabel(’tangential stress [MPa]’)hold offprint -depsc st.eps
Appendix B
Matlab-script to determine thestress profile for the 3-cylinderPSR Jet System nozzle model
clear allclose all
E = 1.9e5; % [MPa]v = 0.3; % [-]a = 16e-6; % [K^-1]Tref = 20; % [deg. C]
% 3-ring PSR-nozzler0=0.03;rvec=[0.060 0.0848 0.088]’;tvec=[0.003 0.0015 0.0554]’;Tvec=[250 700 650]’-Tref;
% Enter free BC’s systemCSF=zeros(2,7);% BC sigmar @ r0:CSF(1,1)=1; % c11CSF(1,2)=-1/(r0^2); % c12CSF(1,7)=1/2*a*Tvec(1);% BC sigmar @ r3:CSF(2,5)=1;CSF(2,6)=-1/(rvec(3)^2);CSF(2,7)=1/2*a*Tvec(3);
% Enter interface stress BC’s systemCS=zeros(2,7);% BC sigmar @ r1:CS(1,1)=1; % c11CS(1,2)=-1/(rvec(1)^2); % c12CS(1,3)=-tvec(2)/tvec(1); % c21CS(1,7)=1/2*a*Tvec(1);% BC sigmar @ r2:
34
CS(2,3)=-rvec(1)*tvec(2)/(rvec(2)*tvec(3)); % c21CS(2,5)=1; % c31CS(2,6)=-1/(rvec(2)^2); % c32CS(2,7)=1/2*a*Tvec(3);
% Enter interface displacements BC’s systemCU=zeros(2,7);% BC u @ r1CU(1,1)=(1-v)*rvec(1)^2; % c11CU(1,2)=(1+v); % c12CU(1,3)=-rvec(1)^2; % c21CU(1,4)=-rvec(1); % c22CU(1,7)=a*(rvec(1)^2)*(Tvec(2)-1/2*(1+v)*Tvec(1));% BC u @ r2CU(2,3)=-rvec(2)^2; % c21CU(2,4)=-rvec(2); % c22CU(2,5)=(1-v)*(rvec(2)^2); % c31CU(2,6)=(1+v); % c32CU(2,7)=a*(rvec(2)^2)*(Tvec(2)-1/2*(1+v)*Tvec(3));
% CREATE AND SOLVE TOTAL SYSTEM OF CONSTANTSC=[CSF;CS;CU];Cans=rref(C);c=Cans(:,7);
rstep=0.0001;r1=r0:rstep:rvec(1);r2=rvec(1):rstep:rvec(2);r3=rvec(2):rstep:rvec(3);
% PLOT RADIAL STRESS GRAPHSfigureplot(r0:rstep:rvec(3),0,’k’) % plot black r-axishold on
sr1=E*(c(1)-c(2)./(r1.^2)-1/2*a*Tvec(1));sr2=E*c(3);sr3=E*(c(5)-c(6)./(r3.^2)-1/2*a*Tvec(3));
plot(r1,sr1,’Color’,[.8 0 0],’LineWidth’,2)plot(r2,sr2,’Color’,[.8 0 0],’LineWidth’,2)plot(r3,sr3,’Color’,[.8 0 0],’LineWidth’,2)
grid onxlabel(’r [m]’)ylabel(’radial stress [MPa]’)hold off% AXIS([0.03 0.09 -100 900])print -depsc sr.eps
35
% PLOT TANGENTIAL STRESS GRAPHSfigureplot(r0:rstep:rvec(3),0,’k’) % plot black r-axishold on
st1=E*(c(1)+c(2)./(r1.^2)-1/2*a*Tvec(1));st2=0;st3=E*(c(5)+c(6)./(r3.^2)-1/2*a*Tvec(3));
plot(r1,st1,’Color’,[0 .5 0],’LineWidth’,2)plot(r2,st2,’Color’,[0 .5 0],’LineWidth’,2)plot(r3,st3,’Color’,[0 .5 0],’LineWidth’,2)
grid onxlabel(’r [m]’)ylabel(’tangential stress [MPa]’)hold offprint -depsc st.eps
