Thermo-elasticy in a cylindrical jet engine nozzle MT 09.16

Transcription

1 Thermo-elasticy in a cylindrical jet engine nozzle MT D.V. Wilbrink July 7, 2009

2 Abstract The companies AMT Netherlands and Draline have developed a compact jet engine, that is designed to be fitted to hang gliders. This so-called PSR Jet System can provide propulsion to the hang glider on demand of the pilot. In the initial design, the blades of a nozzle in the engine that directs hot exhaust gases from the 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 of this research is to gain insight in the thermo-mechanical response of the nozzle. This is done by making an analytical description of the stresses resulting from the temperature gradients and the given 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 expressions are applied to a few imaginary situations, allowing the reader to get more familiar with thermo-mechanical behavior in cylindrical objects. This is followed by an analysis of the stress distributions in the PSR Jet System nozzle. The stresses for two different temperature conditions are computed and subsequently the influence of the stiffness of the solid parts of the nozzle on the stresses is analyzed. The nozzle analysis showed that stresses involved with temperature gradients due to operation are high, with levels near the yield stress level, and therefore most likely cause cracking. The stresses involved with shutting down the turbine and thereby suddenly cooling the nozzle blades were even higher than stresses from operating conditions. Furthermore, reducing the compliance of the nozzle parts connecting to the blades proved to be an effective way to reduce stress levels and 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 of assumptions. However, it can be used to get a clear general impression of the thermo-mechanical behavior of PSR Jet System engine nozzle, useful as a guidance for future numerical analysis.

3 Contents 1 Introduction 3 2 Thermo-elastic analysis of cylindrical bodies Individual cylinders Solid cylinders Bladed cylinders Composite cylinders Boundary conditions Resulting system of equations for piecewise constant temperature Examples Open cylinder with a linear temperature distribution Double concentric cylinder with piecewise constant temperatures Cylinders of equal thickness Cylinders of different thickness Piecewise constant approximation of a linear temperature profile Analysis of the Turbine Nozzle Operating conditions Cooldown Reduced flange stiffness Reduced nozzle wall stiffness Reduced flange and engine wall stiffness Conclusions 29 A Matlab-script to determine the stress profile for various solid composite cylinders 30 B Matlab-script to determine the stress profile for the 3-cylinder PSR Jet System nozzle model 33

4 Chapte Introduction Sometimes hang gliders can be short of lift for a period of time. Having a lightweight booster aboard to provide power for some time when needed greatly increases the range of flight and landing safety of such aircraft. An increasing number of new gliders is therefore equipped with small two-stroke engines. However, these engines are relatively large and heavy, requiring an especially designed glider fuselage. 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 Draline have developed such an engine as a hang glider booster, called PSR Jet System. For both new and existing hang gliders this engine can be fitted inside the fuselage, to be deployed and provide propulsion 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 compressor which sucks air into the combustion chamber. Hot gas exits this chamber at the rear, thrusting the plane forward. These hot gases are guided through a nozzle to deflect their path towards the blades of a turbine, causing the turbine wheel to spin rapidly. The turbine powers the compressor at the front by a shaft leading through the center of the jet engine. Figure 1.2: The PSR Jet System engine

5 4 During testing of the engine, the blades of the nozzle (shown in Figure 1.3) turned out to crack under prolonged operation. Apparently the geometry of the component does not allow the used material, a nickel-based steel, to cope with the stresses due to the major temperature differences in the nozzle. The goal of this research is to gain insight in the thermo-mechanical stress distribution underlying 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 determine the stress distributions in a cylindrical object with prescribed geometry and temperature distributions, which is done in Chapter 2. This analytical model is applied to both single and composite 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 analytical model to the PSR Jet System engine nozzle under different conditions in Chapter 4, we can answer 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 the applicability of the model.

6 Chapter 2 Thermo-elastic analysis of cylindrical 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 derived for individual cylinders. Subsequently, these expressions are linked to one another in objects considered to consist of multiple cylinders, which are referred to as composite cylinders. An example 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 is first done for solid cylinders, then for cylinders representing a set of blades Solid cylinders The analysis will be done in cylindrical coordinates along the vector basis { e r, e θ, e z }, where e z is along the axis of the cylinder. The general displacement vector is: u = u(r, θ, z) e r + v(r, θ, z) e θ + w(r, θ, z) e z (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) infinitely small slice of material taken from the disk at an arbitrary angle θ. The axial strain component can be eliminated assuming a situation of plane stress: axial stresses are neglected compared to the radial and tangential stresses σ r and σ θ, so σ z = 0. The remaining deformation is now entirely characterized by the radial displacement vector which is a function of the radius r at an arbitrary angle θ: u = u(r) e r (θ) (2.2)

7 2.1 Individual cylinders 6 This displacement gives rise to a two-dimensional strain tensor ɛ: ɛ = 1 2 ( u + u T ) (2.3) = r e r + 1 r θ e θ (2.4) ɛ = ɛ r e r e r + ɛ θ e θ e θ = u r e r e r + u r e θ e θ (2.5) The assumptions imply that bending effects (deforming the disk-shape) are negligible and the cylinder is free to contract in z-direction, giving rise to an axial strain ɛ z, to meet the plane stress condition. Each of the three principal strains ɛ r, ɛ θ and ɛ z consist of a mechanical part due to mechanical stress and a thermal part caused by thermal expansion. Assuming that only elastic deformation occurs, the mechanical strains are linked to the corresponding stresses by Hooke s Law. Furthermore, thermal strain is the product of the thermal expansion coefficient of the material α and the temperature difference T with respect to a reference temperature. The three principal strains are thus given by: ɛ r = 1 E (σ r νσ θ ) + αt (2.6) ɛ θ = 1 E (σ θ νσ r ) + αt (2.7) ɛ z = ν E (σ r σ θ ) + 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 that σ z = 0. Equations (2.6) and (2.7) can be re-written to: σ r = σ θ = E 1 ν 2 (ɛ r + νɛ θ (1 + ν)αt ) (2.9) E 1 ν 2 (ɛ θ + νɛ r (1 + ν)αt ) (2.10) Substituting (2.5) now allows one to express the stresses in terms of the radial displacement u: σ r = σ θ = E ( du 1 ν 2 dr + ν u ) (1 + ν)αt r E ( u 1 ν 2 r + ν du ) (1 + ν)αt dr (2.11) (2.12) These two stress components must satisfy the equation σ = 0. In the case of axisymmetry as considered here, this equation reduces to: dσ r dr + σ r σ θ r = 0 (2.13) Substituting the two stress components according to (2.11) and (2.12) finally yields the following differential equation in terms of u(r): d dr ( ) 1 d(ru(r)) (1 + ν)αt (r) = 0 (2.14) r dr

8 2.1 Individual cylinders 7 To obtain the deformation and stress state due to an arbitrary (axisymmetric) temperature distribution 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)c 1 r (2.15) where c 1 is an integration constant and the factor 2(1 ν) has been introduced to simplify the further analysis. To obtain u(r), we need to integrate this equation once more. As the primitive function of rt (r) can only be determined if T (r) is known, we introduce the function Q(r) as a primitive, so that dq dr = rt (r). This allows us to write: ru (1 + ν)αq(r) = (1 ν)c 1 r 2 + c 2 (1 + ν) (2.16) where c 2 has been introduced as an integration constant. The solution u(r) can now be written as: u(r) = (1 ν)c 1 r + (1 + ν) (c 2 + αq(r)) 1 (2.17) r The general displacement solution thus obtained can be used to determine the two components of the strain tensor (2.5): ɛ r = du dr = (1 ν)c 1 (1 + ν) c 2 (1 + ν)αq(r) r2 r 2 + (1 + ν)αt (r) (2.18) ɛ θ = u r = (1 ν)c 1 + (1 + ν) c 2 + (1 + ν)αq(r) r2 r 2 (2.19) Substituting these results in equations (2.9) and (2.10) yields the following general expressions for the radial and tangential stress: σ r (r) = E (c 1 (c 2 + αq(r)) 1r ) 2 (2.20) σ θ (r) = E (c 1 + (c 2 + αq(r)) 1r ) 2 αt (r) (2.21) The constants of integration c 1 and c 2 must be determined from the boundary conditions resulting from the geometry of objects that these stress functions apply to, as we will see in Section Bladed cylinders For some cylindrical objects, one of the modeled composite cylinders is actually a ring or set of multiple blades (provided that there are sufficient blades to assume axisymmetry), as illustrated in Figure 2.2. For this situation, the stress profile is different as we will see in this Section. As individual blades are not in contact with their neighbors, no tangential stresses can be transmitted so σ θ = 0. Note that blades may have some internal tangential stress, but for the collection of blades as a whole we can neglect it. With this assumption relation (2.6) reduces to: which can be re-written as ɛ r = σ r E The equilibrium equation σ = 0 reduces to: + αt (r) (2.22) σ r = E( du αt (r)) (2.23) dr dσ r dr = 0 (2.24)

9 2.2 Composite cylinders 8 r2 r2 r3 r1 r1 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 E( du ) αt (r)) = 0 (2.25) dr dr This differential equation will be solved to obtain an expression for u(r). As a first step towards the solution, we now have: du dr αt (r) = c 1 (2.26) The temperature profile T (r) is yet unknown, so to make the next step in formulating an expression for 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) = c 1 r + c 2 + αq(r) (2.27) Substitution of this result in the radial stress distribution (2.23) yields: σ r (r) = Ec 1 (2.28) The constants c 1 and c 2 are again to be determined by the boundary conditions resulting from the geometry. The above solution will be used in Chapter 4 for the stress analysis of the PSR Jet System nozzle. 2.2 Composite cylinders In the previous chapter we formulated general stress distributions for cylindrical objects experiencing temperature gradients. The constants featuring in these expressions can be determined by the boundary conditions for the given geometry of the object, that will be formulated in Section These conditions result in a set of equations that determines the constants of integration c i1 and c i2, where i = 1, 2,..., n indicates the individual concentric cylinders, as shown in Section The obtained expressions for the constants can be substituted in the general stress distributions to determine their precise form for the given geometry and loading Boundary conditions The boundaries for which conditions are prescribed can be subdivided in two types: free boundaries and interfaces.

10 2.2 Composite cylinders 9 Free boundaries Every (composite) cylinder has has an inner surface r = r 0 and an outer surface r = r n, where n is the total number of individual cylinders in the composite cylinder (see for example r 0 and r 3 in Figure 2.2). At these free boundaries, the material is free to move in radial direction. The assumption of static equilibrium implies that for these surfaces, we have: σ r (r 0 ) = σ r (r n ) = 0 (2.29) This boundary condition enables us to eliminate one integration constant in each of the inner and outer cylinder s stress distributions. Interfaces Composite cylinders have interfaces, i.e. surfaces r = r i with i = 1, 2, 3,..., n 1 where two concentric cylinders make contact with each other. Examples are and r 2 in Figure 2.2. At each of these interfaces, two conditions can be formulated. These 2(n 1) conditions allow us to eliminate the remaining 2(n 1) integration constants. One condition follows from the requirement that the (radial) forces acting on each interface from both sides must be in equilibrium. For interfaces between two solid concentric cylinders, this can be formulated as: t i σ r,i (r i ) = t i+1 σ r,i+1 (r i ) (2.30) If we formulate the total cross-sectional area of bladed cylinders at the interface as ma, where m is the number of blades in the cylinder and A is the cross-sectional area of a single blade, the force equilibrium can be formulated as: r i t i σ r,i (r i ) = maσ r,i+1 (r i ) (2.31) A second condition at the interface follows from the continuity of material, as we assume no rupture occurs. This means the displacements at both sides of the interface must be equal: u i (r i ) = u i+1 (r i ) (2.32) Resulting system of equations for piecewise constant temperature For a composite cylinder consisting of n individual cylinders, we have to specify 2n constants of integration to determine the stress distribution in every concentric cylinder it contains. This is done 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 consists of concentric cylinders which each have uniform temperature T i. This is first done for systems consisting of solid cylinders and subsequently for boundaries where bladed cylinders are involved. Solid cylinders Using the temperature distribution T (r) = T i for r i 1 < r r i, the function Q i (r) for the i-th cylinder can be determined to be: Q i (r) = 1 2 T ir 2 (2.33) Substituting this expression in the displacement solution (2.17) and stress distributions (2.20) and (2.21) yields: 1 u i (r) = (1 ν)c i1 r + (1 + ν)c i2 r (1 + ν)αt ir (2.34) ( σ r,i (r) = E c i1 c i2 r 2 1 ) 2 αt i (2.35)

11 2.2 Composite cylinders 10 ( σ θ,i (r) = E c i1 + c i2 r 2 1 ) 2 αt i (2.36) As a first step in determining the constants, the free boundary conditions σ r (r 0 ) = σ r (r n ) = 0 from Section are applied to the radial stress expression (2.35), resulting in the following equations: c 11 c 12 r0 2 = 1 2 αt 1 (2.37) c n1 c n2 rn 2 = 1 2 αt n (2.38) Substitution of the radial stress expression (2.35) in (2.30) yields for the interface at r i : t i (c i1 c i2 r 2 i ) t j (c j1 c j2 ri 2 ) = 1 2 α(t it i t j T j ) (2.39) where j = i + 1. Finally, equation (2.32), after substitution of the displacement expression (2.34) reads: 1 ν 1 + ν (c i1 c j1 )r 2 i + (c i2 c j2 ) = 1 2 α(t j T i )r 2 i (2.40) We now have 2n equations to solve 2n constants of integration, enabling us to determine the stress distributions for composite cylinders these equations apply to. Calculation by hand easily becomes tedious for composite cylinders with n 3, so we let computer power take over at this point. The Matlab-script added in appendix A takes material parameters and vectors containing radius-, thickness- and temperature values, solves the above expressions for the integration constants and draws the graphs of the radial and tangential stress in red and green respectively. Bladed cylinders Around cylinders that consist of a (by approximation) axisymmetric set of blades as for example shown in Figure 2.2, the equations resulting from the interface boundary conditions are different than those for solid cylinders. We define the bladed cylinder at uniform temperature T i and thickness t i at r i 1 < r r i. With the temperature known, we can determine Q(r): Substituting this in the displacement expression (2.27) yields: The radial stress distribution is provided by (2.28) and becomes: Q(r) = rt j (2.41) u i (r) = (c i1 + αt i )r + c i2 (2.42) σ r (r) = Ec i1 (2.43) We can now use the boundary conditions to determine the constants, starting with the force equilibrium at the inner boundary of the bladed ring r = r i. Using j = i + 1, the force equilibrium equation (2.31) can be re-written to get: c i1 c i2 r 2 ma r i t i c j1 = 1 2 αt i (2.44) The displacement continuity equation (2.32) can be specified by substitution of the displacement functions (2.34) and (2.42): (1 ν)c i1 r i + (1 + ν) c ( i2 c j1 r i c j2 = αr i T j 1 ) r i 2 (1 + ν)t i (2.45)

12 2.2 Composite cylinders 11 Analogically, the displacement continuity at r = r j can be written as: (1 ν)c k1 r j + (1 + ν) c k2 r j c j1 r j c j2 = αr j ( T j 1 ) 2 (1 + ν)t k (2.46) This set of equations provides us a way of determining the constants of integration and therefore the exact stress distributions. This will be done in Chapter 3 for some imaginary situations to gain insight on thermo-elastic stress behavior of cylindrical objects. In Chapter 4, this system is applied to obtain stress distributions for a model of the PSR Jet System engine nozzle.

13 Chapter 3 Examples To provide a better understanding of thermo-elastic stresses predicted by the solutions that were derived in Chapter 2, the stress distributions in a number of typical cylindrical objects are evaluated in this chapter. First a single cylinder with a hole and a linear temperature distribution is analyzed. In the second example, stresses in two concentric cylinders with piecewise constant temperature distribution are computed. Finally, the assumption of piecewise constant temperature is used to numerically simulate the linear temperature profile of the first example cylinder. For all 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 calculated stresses to. Parameter Value E 190GP a ν 0.3 α K 1 760MP a σ y 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 temperature distribution between T = T 0 at the inner surface (r = r 0 ) and T = T 1 at the outer surface (r = ), as shown in the temperature profile of Figure 3.2. z r r0 r1 Figure 3.1: A single cylinder The temperature distribution in the material, i.e. for r 0 < r <, is thus: T (r) = T r r + T 0 (3.1)

14 3.1 Open cylinder with a linear temperature distribution 13 T T1 T(r) T0 r0 r r1 Figure 3.2: Linear temperature profile T (r) within the cylinder where T = T 1 T 0 and r = r 0. The primitive Q(r) of rt (r) can be determined as: Q(r) = 1 T 3 r r T 0r 2 (3.2) The radial stress as given by equation (2.20) becomes in this case: ( σ r (r) = E c αt 0 c 2 r 2 1 ) 3 α T r r Using the two free boundary conditions we can determine the constants c 1 and c 2. Starting with σ r (r 0 ) = 0 we obtain: (3.3) c αt 0 = c 2 r α T r r 0 (3.4) Substituting this result in (3.3) to eliminate c 1 yields: ( σ r (r) = E c 2 (r0 2 r 2 ) 1 ) 3 α T r (r r 0) The second boundary condition, σ r ( ) = 0, allows us to also eliminate c 2 as: so that we finally obtain: c 2 = σ r (r) = 1 3 αe T ( 1 ( r 0 r ) 2 1 ( r0 ) 2 (3.5) 1 3 α T 1 ( r0 ) 2 r2 0 (3.6) ) r r0 1 r0 This means that the radial stress in proportionally dependent on α, E and T. This third parameter reveals the cylinder is stress-free at a homogenous reference temperature (not strainfree). Furthermore, this stress distribution also implies that the radial stress has a hyperbolic shape, as shown in Figure 3.3 using the geometry- and temperature parameters shown in Table 3.2. This distribution can be understood as follows. As warmer material at the outside of the ring tends to expand more than that at the inside, a tensile stress exists in the material in the radial direction. This radial stress vanishes at the inner and outer surface to satisfy the stress-free boundary condition. The tangential stress can be obtained by substituting (3.1) and (3.2) in the general expression (2.21) to get: ( σ θ (r) = E c 1 + c 2 r αt 0 2 ) 3 α T r r (3.8) (3.7)

15 3.1 Open cylinder with a linear temperature distribution 14 r 0 T 0 T m 0.10 m 400 C 700 C Table 3.2: Properties of the single cylinder with linear temperature distribution. The given temperatures do not include the reference temperature radial stress [MPa] r [m] Figure 3.3: Radial stress profile σ r (r) in the cylinder with linear temperature distribution tangential stress [MPa] r [m] Figure 3.4: Tangential stress profile σ θ (r) in the cylinder with linear temperature distribution Then use expressions (3.4) and (3.6) to obtain: 2 0 r 2 ( σ θ (r) = αe T 1 ( r0 ) 2 2 r r0 1 r0 ) (3.9) As we can see in Figure 3.4, in which this distribution has been plotted, the tangential stress gradually changes from being positive at the inner surface to negative at the outer surface. This is also a result of increasing thermal expansion as we move outwards: the inner material partially follows the outer layers of material as they tend to expand more. As a result, inner layers are stretched tangentially. On the other hand, the outer part of the cylinder is restricted in its thermal expansion by the cooler inner part and therefore adopts a smaller radius than it would if unconstrained; a compressive tangential stress results. Figure 3.5 demonstrates that, for the cylinder to be in equilibrium, the net tangential force acting

16 3.2 Double concentric cylinder with piecewise constant temperatures 15 dz Figure 3.5: Tangential forces acting on a cross-section in a r-z-plane dr on an arbitrary cross-section in the r-z-plane must vanish. This tangential force balance means that the tangential stresses integrated over the area of this cross-section have to be zero. As we stated before that stresses are constant in z-direction and only depend on the radius r, this means the tangential stresses integrated over the radius have to be zero: r1 We can see that this relation indeed holds in Figure 3.4. r 0 σ θ (r) dr = 0 (3.10) 3.2 Double concentric cylinder with piecewise constant temperatures In this example we consider a composite cylinder which consists of two concentric uniform cylinders, as depicted in Figure 3.6. Both cylinders have their own constant uniform temperature T (r) = T i for r i 1 < r r i with i = 1, 2, as illustrated in Figure 3.7. z r t 1 t 2 r0 r1 r2 Figure 3.6: A composite of two concentric cylinders T T2 T(r) T1 r0 r r1 r2 Figure 3.7: Temperature profile T (r) within the double cylinder for T 1 < T 2 T0 EN T1 PLAATJE AANPASSEN General stress expressions for this temperature profile were derived in Section Also the equations resulting from the boundary conditions were

17 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: c αt 1 = c 12 r 2 0 (3.11) c αt 2 = c 22 r 2 2 (3.12) Substitution of these expressions in the radial stress distributions yields: σ r,1 (r) = c 12 E(r 2 0 r 2 ) (3.13) σ r,2 (r) = c 22 E(r 2 2 r 2 ) (3.14) To determine c 12 and c 22 we use the two interface conditions at r =. equilibrium (2.30) can be written as: The interface force c 12 = t 2 ( r0 t 1 r 2 ) 2 ( r0 ) 2 1 ( r0 ) 2 c 22 (3.15) Furthermore, the interface displacement continuity condition (2.40) can be expressed as: 1 ν 1 + ν (c 11 c 21 )r (c 12 c 22 ) = 1 2 α(t 2 T 1 )r 2 1 (3.16) With these four equations all the four integration constants can be expressed in terms of the given parameters. Substituting the results in the stress distributions (2.20) and (2.21) gives the stress expressions for this situation as: σ r,1 (r) = Eα T ( r0 ) 2 ( r0 r 2 ) 2 t 1 t 2 A + B ( 1 ( r0 ) ) 2 r (3.17) where σ r,2 (r) = Eα T 1 ( r0 ) 2 A + t2 t 1 B ( (r0 σ θ,1 (r) = Eα T ( r0 ) 2 ( r0 r 2 ) 2 t 1 t 2 A + B σ θ,2 (r) = Eα T 1 ( r0 ) 2 A + t2 t 1 B r ( (r0 ) ) 2 r 0 ( ) 2 r 2 ( 1 + r ( r0 ) ) 2 r ) ) 2 r 0 + ( ) 2 r 2 (3.18) (3.19) (3.20) A = (1 ( r 0 ) 2 )(1 ν)( r 0 r 2 ) 2 + (1 + ν)( r 0 ) 2 (3.21) B = (( r 0 ) 2 ( r 0 r 2 ) 2 )((1 ν) + (1 + ν)( r 0 ) 2 ) (3.22) Note that T = T 2 T 1 is now the temperature difference between the two cylinders. These quadratic functions are rather elaborate, containing many dimensionless products of geometry. In the following two subsections we will visualize these stress profiles and compare them with those generated by the Matlab script of Appendix A. This is first done for two cylinders of equal thicknesses and then for different thicknesses. The parameter values used (unless indicated otherwise) are given in Table 3.3.

18 3.2 Double concentric cylinder with piecewise constant temperatures 17 r 0 r 2 t 1 t 2 T 0 T m 0.10 m 0.15 m m 0.01 m 400 C 700 C Table 3.3: Properties of a double cylinder with piecewise constant temperature distribution Cylinders of equal thickness For t 1 = t 2 the dimensionless geometry terms in the stress distributions significantly simplify to: σ r,1 (r) = 1 2 Eα T r2 2 r1 2 ( ) r2 2 1 r2 0 r2 0 r 2 (3.23) σ r,2 (r) = 1 2 Eα T r2 1 r0 2 ( ) r 2 2 r2 2 r2 0 r 2 1 (3.24) σ θ,1 (r) = 1 2 Eα T r2 2 r 2 1 r 2 2 r2 0 ( 1 + r2 0 r 2 σ θ,2 (r) = 1 2 Eα T r2 1 r0 2 ( r 2 2 r2 2 r2 0 r ) ) (3.25) (3.26) By filling in the parameter values from Table 3.3, except for t 1, which is taken equal to t 2 = 0.10m, we can determine the stress distributions. The figures representing these distributions which are drawn here were provided by running the Matlab-file given in Appendix A with the used parameter values as entry, indicating the file works properly. z r r0 r1 r radial stress [MPa] r [m] Figure 3.8: Radial stress profile σ r (r) in the double cylinder of uniform thickness

19 3.2 Double concentric cylinder with piecewise constant temperatures tangential stress [MPa] r [m] Figure 3.9: Tangential stress profile σ θ (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 outer surfaces and reaches a maximum value at the interface r =. In contrast to the radial stress, the tangential stress is not a continuous function as Figure 3.9 demonstrates. For this system of a cold inner cylinder and a hot outer cylinder, the relatively cold inner cylinder resists the expansion of the hot outer cylinder, giving it a positive tangential stress. This is similar to the situation of linear temperature profile from Section 3.1, but now the temperature suddenly changes, resulting in a discontinuous change from positive to negative tangential stress at the interface. As can be concluded from the graph, the tangential force balance over the total cross-section nevertheless still applies: r2 r 0 σ θ (r) dr = r1 r 0 σ θ,1 (r) dr + r2 σ θ,2 (r) dr = 0 (3.27) Cylinders of different thickness For the more general case of two concentric cylinders with different thicknesses we take the parameter values given in Table 3.3. The radial and tangential stresses are as depicted in Figure 3.10 and Figure 3.11 respectively. Compared to Figure 3.8 for uniform thickness, the radial stress profile is no longer continuous at the interface. As the force balance (2.30) prescribes, it shows a jump by a factor which is the inverse of the thickness ratio t2 t 1 between 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 thickness difference into account, again confirming the situation to be in static equilibrium: t 1 r1 r2 σ θ,1 (r) dr = t 2 σ θ,2 (r) dr (3.28) r 0

20 3.3 Piecewise constant approximation of a linear temperature profile radial stress [MPa] r [m] Figure 3.10: Radial stress profile σ r (r) in the double cylinder of different thickness tangential stress [MPa] r [m] Figure 3.11: Tangential stress profile σ θ (r) in the double cylinder of different thickness 3.3 Piecewise constant approximation of a linear temperature profile To demonstrate the use of the piecewise constant temperature profile, the cylinder of Section 3.1 with 0.04m r 0.10m and 400 C T 700 C is subdivided in 12 parts. This means that every 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 and Figure 3.4 respectively. As becomes especially clear from the tangential stress distribution, a higher accuracy can be obtained by taking more, but smaller temperature steps as input.

21 3.3 Piecewise constant approximation of a linear temperature profile Temperature [K] r [ ] Figure 3.12: Temperature profile T (r) used in the piecewise constant approximation of a linear profile radial stress [MPa] r [m] Figure 3.13: Radial stress profile σ r (r) obtained from the piecewise constant temperature tangential stress [MPa] r [m] Figure 3.14: Tangential stress profile σ θ (r) obtained from the piecewise constant temperature

22 Chapter 4 Analysis of the Turbine Nozzle Now that it is possible to predict the stress distributions for many cylindrical geometries and temperature 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 following paragraphs 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 consisting of three concentric cylinders as shown in Figure 2.2. From the inside to the outside the cylinders represent: 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 to the planes of the other two cylinders. Together they then approximately form a cylinder of the same thickness of the cylinder s individual blades, see Figure 2.2. This cylinder is furthermore tangentially compliant, as individual blades are not able to transfer stresses in tangential direction onto one another. For the composite cylinder, the geometry given in Table 4.1 is used, more or less representing the actual sizes of the nozzle.

23 4.1 Operating conditions 22 The material that the nozzle consists of is Inconel 713, a nickel-based steel. The material properties are assumed to be constant at the values given in Table 3.1. The reference temperature, at which objects are assumed to be stress-free, is taken to be 20 C r i [mm] t i [mm] 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 set of equations that determine the constants of integration for both solid and bladed cylinders from Section Solving the resulting system of 6 equations for 6 integration constants and substituting 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 stress profiles 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 to be initiated under cooldown conditions, the stresses in this situation are examined subsequently in Section 4.2. Finally, the geometry of the inner and outer cylinder is adapted in the following three 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 to approximately 700 C, the outer nozzle wall to 650 C while the inner flange remains at 250 C as shown in Table T i [ C] 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 positive radial stress exists in every part of the nozzle, so the whole composite is under tension. The minimum values are, as they should be, at the radial-stress-free inner and outer surface of the nozzle. 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 close to 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 tangential stress. The relatively cold flange has a positive tangential stress, as it resist the expansion of the other two (hot) cylinders. In contrast, the outer cylinder has a negative tangential stress, as it is restricted in its expansion by the flange inside. Note that the maximum value of tangential stress, located near the inner surface, is higher than the maximum radial stress. From a physical perspective this is unexpected, as the cold inside of the flange is the least likely part of the nozzle to deform plastically. This result is probably due to the inaccurate geometrical modeling of the inner flange, but a more accurate explanation is yet to be found.

24 4.2 Cooldown radial stress [MPa] r [m] Figure 4.2: Radial stress profile σ r (r) of the PSR nozzle model under operating conditions tangential stress [MPa] r [m] Figure 4.3: Tangential stress profile σ θ (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 at around 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 the temperature of the blades changes to 450 C, yielding the temperature distribution shown in Table 4.3. This situation is expected to be the most critical for cracking, so the stress profile is examined

25 4.2 Cooldown 24 in this section T i [ C] Table 4.3: Temperature profile of the PSR nozzle model under cooldown conditions radial stress [MPa] r [m] Figure 4.4: Radial stress profile σ r (r) of the PSR nozzle model under cooldown conditions tangential stress [MPa] r [m] Figure 4.5: Tangential stress profile σ θ (r) of the PSR nozzle model under cooldown conditions

26 4.2 Cooldown 25 The radial stress depicted in Figure 4.4 shows a similar profile to that of the full operating conditions, but now with higher values. The maximum radial stress, located in the blades has increased from 628 MPa to 800 MPa. This means cracking is indeed more likely to occur in the blades in this cooldown situation than under operating conditions. The tangential stress profile of Figure 4.5 shows again how the hot nozzle wall s expansion is resisted by the flange and the tangentially weak bladed cylinder, this time with increased absolute values compared to the operating conditions. Based on the above observations, a logical way to avoid cracking in the nozzle s blades is to make the parts connected to the blades more compliant. This can be achieved by either making the flange more compliant, or the wall, or both. This will be done in the following subsections for the present cooldown situation. More compliance is introduced by reducing the radial dimensions of the solid cylinders Reduced flange stiffness To represent a more flexible flange, in this Section the center hole radius is imaginarily increased to r 0 = 45mm, thus reducing the radial thickness of the flange by 50%. All other parameters of the cooldown situation are still assumed to apply. The radial geometry for this situation is given in Table 4.4. i r i [mm] Table 4.4: Radial geometry of the cooldown PSR nozzle model with enlarged r radial stress [MPa] r [m] Figure 4.6: Radial stress profile σ r (r) of the cooldown PSR nozzle model with enlarged r 0 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

27 4.2 Cooldown tangential stress [MPa] r [m] Figure 4.7: Tangential stress profile σ θ (r) of the cooldown PSR nozzle model with enlarged r 0 decreased significantly; the maximum radial stress in the blades has dropped by 34%. The flange stiffness 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. Simultaneously, leaving out a part of the flange material causes the absolute tangential stress in the nozzle wall to decrease as the tangential force balance prescribes as given in the equation: t 1 r1 r 0 σ θ,1 (r) dr = t 3 where an increase in r 0 causes σ θ,3 to decrease Reduced nozzle wall stiffness r3 r 2 σ θ,3 (r) dr (4.1) Instead of reducing the flange stiffness, we can also weaken the nozzle wall. Again the radial thickness of the solid cylinder representing it is reduced by 50%, so that r 3 = 86.4mm, as shown in Table r i [mm] Table 4.5: Radial geometry of the cooldown PSR nozzle model with reduced r 3 Compared to the (dashed) reference stress distribution, the radial stress with reduced wall stiffness as given in Figure 4.8 has again decreased, this time by 28% in the blades. Removing the outer half of the wall therefore has a similar effect on the radial stresses as removing the inner half 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 Figure 4.9 shows.

28 4.2 Cooldown radial stress [MPa] r [m] Figure 4.8: Radial stress profile σ r (r) of the cooldown PSR nozzle model with reduced r tangential stress [MPa] r [m] Figure 4.9: Tangential stress profile σ θ (r) of the cooldown PSR nozzle model with reduced r Reduced flange and engine wall stiffness Knowing that splitting the flange or engine wall radially in half reduces the radial stresses in the blades by about one third, we will now see what happens if the two stiffness reductions are combined. Figure 4.10 shows how the radial stress values now have been reduced to 52% of their original value at cooldown. This 48% reduction is the same as reducing the stress by 34% and 28% suc-

29 4.2 Cooldown r i [mm] Table 4.6: Radial geometry of the cooldown PSR nozzle model with enlarged r 0 and reduced r radial stress [MPa] r [m] Figure 4.10: Radial stress profile σ r (r) of the cooldown PSR nozzle model with enlarged r 0 and reduced r 3 cessively, so the stiffness reductions from both sides add up.

30 Chapter 5 Conclusions The analytical model that is constructed int his paper showed that stressess in the material directly proportionally depend on temperature differences with a reference temperature. Other parameters on which the stresses depend directly proportional are the Youngs modulus and the thermal expansion coefficient, both properties that indicate the ability of the material to cope with strains and temperature gradients. For the turbine nozzle of the PSR Jet System, this meant that temperatures involved with turbine operation reached levels near the yield stress level, confirming the designers suspicions that cracking of the nozzle blades under prolonged operation is most likely caused by these temperatures. Compared to operating conditions, stress levels increased when the turbine is shut down so the nozzle blades cool rapidly while the adjacent material is still at high temperature. 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 reducing the radial thickness, proved to reduce stresses by significant amounts. The model composed and applied to analyse the thermo-mechanical behavior of the PSR Jet System nozzle has a number of limitations. Stress levels are not to be taken as very accurate, as the model is based on a set of assumptions that simplify stress results, for example the assumptions that the nozzle parts do not bend and are in a situation of plane stress. The single-cylinder modeling of the inner flange is another considerable simplification. What these results do provide is a global impression of the thermo-mechanical behavior of the turbine nozzle. It tells under which conditions cracking is probably caused and how it can be prevented. To make a more detailed analysis with accurate stress numbers, a numerical model of the turbine nozzle could be made. The analytical results from this paper can be used to locate areas of interest for numerical stress analysis of the nozzle, for example where the blades are attached to the flange and nozzle wall. Knowing that cooldown of the turbine is most likely to make the blades crack, a more detailed analysis of thermal-mechanical behavior during this (repeated) process can be done. Additionally, the flange can be modeled in more detail to examine the influence of compliance more thoroughly.

31 Appendix A Matlab-script to determine the stress profile for various solid composite cylinders clear all close all E = 1.9e5; % [MPa] v = 0.3; % [-] a = 16e-6; % [K^-1] Tref = 20; % [deg. C] % 3-cylinder PSR Jet Engine nozzle r0=0.03; rvec=[ ] ; % vector r1;r2;..;rn tvec=[ ] ; % vector t1;t2;..;tn Tvec=[ ] -Tref; % vector T1;T2;..;Tn N=length(rvec); % ENTER FREE BC S SYSTEM CSF=zeros(2,2*N+1); % BC r0: CSF(1,1)=1; CSF(1,2)=-1/(r0^2); CSF(1,2*N+1)=a*Tvec(1)/2; % BC 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 SYSTEM CS=zeros(N-1,2*N+1); % BC ri Rivec=[]; for i=1:(n-1) CS(i,2*i-1)=tvec(i);

32 31 end 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)); % ENTER INTERFACE DISPLACEMENT BC S SYSTEM CU=zeros(N-1,2*N+1); % BC ri for 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 CONSTANTS C=[CSF;CS;CU]; Cans=rref(C); % PLOT RADIAL STRESS GRAPHS figure rstep=0.0001; plot(r0:rstep:rvec(length(rvec)),0, k ) % plot black r-axis hold on % PLOT RADIAL STRESS GRAPH CYLINDER 1 r=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 GRAPHS for 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) end grid on xlabel( r [m] ) ylabel( radial stress [MPa] ) hold off print -depsc sr.eps % PLOT TANGENTIAL STRESS GRAPHS figure plot(r0:rstep:rvec(length(rvec)),0, k ) % plot black r-axis hold on % PLOT TANGENTIAL STRESS GRAPH CYLINDER 1 r=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 GRAPHS for i=2:n

33 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) end grid on xlabel( r [m] ) ylabel( tangential stress [MPa] ) hold off print -depsc st.eps

34 Appendix B Matlab-script to determine the stress profile for the 3-cylinder PSR Jet System nozzle model clear all close all E = 1.9e5; % [MPa] v = 0.3; % [-] a = 16e-6; % [K^-1] Tref = 20; % [deg. C] % 3-ring PSR-nozzle r0=0.03; rvec=[ ] ; tvec=[ ] ; Tvec=[ ] -Tref; % Enter free BC s system CSF=zeros(2,7); % BC r0: CSF(1,1)=1; % c11 CSF(1,2)=-1/(r0^2); % c12 CSF(1,7)=1/2*a*Tvec(1); % BC 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 system CS=zeros(2,7); % BC r1: CS(1,1)=1; % c11 CS(1,2)=-1/(rvec(1)^2); % c12 CS(1,3)=-tvec(2)/tvec(1); % c21 CS(1,7)=1/2*a*Tvec(1); % BC r2:

35 34 CS(2,3)=-rvec(1)*tvec(2)/(rvec(2)*tvec(3)); % c21 CS(2,5)=1; % c31 CS(2,6)=-1/(rvec(2)^2); % c32 CS(2,7)=1/2*a*Tvec(3); % Enter interface displacements BC s system CU=zeros(2,7); % BC r1 CU(1,1)=(1-v)*rvec(1)^2; % c11 CU(1,2)=(1+v); % c12 CU(1,3)=-rvec(1)^2; % c21 CU(1,4)=-rvec(1); % c22 CU(1,7)=a*(rvec(1)^2)*(Tvec(2)-1/2*(1+v)*Tvec(1)); % BC r2 CU(2,3)=-rvec(2)^2; % c21 CU(2,4)=-rvec(2); % c22 CU(2,5)=(1-v)*(rvec(2)^2); % c31 CU(2,6)=(1+v); % c32 CU(2,7)=a*(rvec(2)^2)*(Tvec(2)-1/2*(1+v)*Tvec(3)); % CREATE AND SOLVE TOTAL SYSTEM OF CONSTANTS C=[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 GRAPHS figure plot(r0:rstep:rvec(3),0, k ) % plot black r-axis hold 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 on xlabel( r [m] ) ylabel( radial stress [MPa] ) hold off % AXIS([ ]) print -depsc sr.eps