Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee

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1 Appendi to On the stability of a compressible aisymmetric rotating flow in a pipe By Z. Rusak & J. H. Lee Journal of Fluid Mechanics, vol. 5 4, pp. 5 4 This material has not been copy-edited or typeset by Cambridge University Press: its format is entirely the responsibility of the author. Appendi A. Derivation of perturbation equations 3 and 3 Elimination of pressure from 9 and by cross differentiation in terms of and y, respectively, followed by subtraction gives a relationship between, K and : +γm ω K 4y + yy t + y w y + yy yy + ωk K ytt d t yt =.A- Solving A- for K and substituting in linearized θ momentum equation results in: K t = ωk y + ωk + yy + + yy yy t ω +γm K ] 4y y w y ytt d t yt.a- Elimination of K from A- and A- by cross differentiation in terms of t and, respectively, followed by subtraction, and multiplying by ωk /, gives + yy + w t + yy + + yy tt ω K K y + w yy yy t [ ω = γm K 4y + t 4y t 3 ytt 3 yt yttt d w ] y tt y w w y. A-3 Differentiation of A-3 with respect to gives 3. Differentiating with respect to gives γm T t + T = γ γ [ ] P y u + γm P t + P T y u A-4.

2 From 6 we have T = P T. Substituting this in A-4 gives M P t + P γm T t + = γ γ P y T y u. A-5 Using to epress P t and P in A-5 and multiplying by gives [ γm T 3M w t + T M w 3M tt M = M y T y + γ P y M y γ ttt d ] M ytt + M y t M yt. A-6 Differentiating A-6 with respect to gives 3. Appendi B. Boundary conditions for 43 and 44 The substitution of 4 into 3-4 gives boundary conditions for and for and = γm, =,, / =, / B- T y, y = γm y, y + yy, y + ω K w 4y y σ, y + σ yy, y y [ ω K 4y = γm γ, y =, γm, y = y, y, σ, y w + ω K 4y y, y B- B-3 with, = y, =, B-4 ω K K y w yy, y + σ yy, y, y y, y σ y, y σ y, y + σ w yy, y +, y, y + σ, y σ yy w ], y 4σy, y 3σ y, y, B-5 3σ y, y σ w ] y, y y, y w w y, y, B-6 [ ] T σ 3M w, y + T M w, y 3σ M, y = σ y, y σ y, y σ w y, y, B-7, y =, σ, y +, y = B-8

3 for y /. Appendi C. Analysis of imaginary parts of 43 and 44 Substituting 6 into 43 and 44, collecting terms of the orders ɛ I, σ I, ɛ I σ R, ɛ I Ω and neglecting terms of the orders OσR, σ I, σ Rσ I, σ I ɛ R, σ I Ω and higher gives ɛ I I + Iyy + +σ I c +ɛ I σ R I + Iyy ɛ I I Ω K K y w yy w I + γm Ω K w 4y y I Iy + cyy w yy w c + γm Ω K 4 w y 4y w c 3 cy w yy w I + γm Ω K 4 w y 4y w I 3 Iy K K y +ɛ I Ω w I + γm K } 4y I =, C- M y T y + γ M Ω K 4y I M w Iy σ I T 3M + w T M w c y c + } γ γ cy + ɛ Iσ R T 3M w T M w I y I + } γ γ Iy γ K } +ɛ I Ω γt M w 4y I =. C- γm T M w From B--B-8, the boundary conditions for these equations are: for and I, =, I, / = I, / C-3 I, y =, γm I, y = Iy, y, T ɛ I Iy, y γm Ω K } } y 4y Iy, y σ I γm c, y } ɛ I σ R γm I, y ɛ I Ω γm K } 4y Iy, y = with I, = Iy, =, I, y Ω K K y ɛ I + Iyy, y + w yy w I, y +γm Ω K w 4y y I, y Iy, y Iyy, y } +ɛ I σ R γm wy I, y + Iy, y C-4 C-5 C-6 3

4 K K y +ɛ I Ω w I, y + γm K } 4y I, y =, C-7 I, y ɛ I + γm Ω K w 4y y I, y Iy, y +σ I c, y y + cyy, y +γm Ω K 4y w I, y +ɛ I σ R + Iyy, y y K +γm Ω K yy w 4 y c, y yy w I, y c, y 3 cy, y 4w 4y w y I, y 3 Iy, y } +ɛ I Ω γm 4y I, y =, C-8 σ I T 3M ɛ I I, y + w T M w c, y y c, y + } γ γ cy, y + ɛ Iσ R T 3M w T M w I, y y I, y + } γ γ Iy, y =, C-9 I, y =, σ I c, y + ɛ I I, y + ɛ I σ R I, y = C- for y /. Two integrations with respect to of C- and C- and the use of boundary conditions C-4 and C-7-C- result in I ɛ I + Ω K K y Iyy + w +σ I c +ɛ I σ R I + Iyy ɛ I I w yy I + γm Ω K w 4y y I Iy + cyy w yy w c + γm Ω K 4 w y 4y w c 3 cy w yy w I + γm Ω K 4 w y 4y w I 3 Iy K K y +ɛ I Ω w I + γm K } 4y I = ɛ I σ R f y, C- M y T y + γ M Ω K 4y I M w Iy σ I T 3M + w T M w c y c + } γ γ cy + ɛ Iσ R T 3M w T M w I y I + } γ γ Iy γm T M w γ K } +ɛ I Ω γt M w 4y I 4 = σ I f y + ɛ I σ R f 3 y, C-

5 Here f y = Iyy, y yy w f y = T M w f 3 y = T M w I, y + γm Ω K w ] y 4y w I, y Iy, y, M Sy + y Φy γ γ Φ yy, C-3 M I, y + y I, y γ γ Iy, y. Solution of C- for ɛ I I, substitution of the result in C-, multiplication by T M w /T and an additional integration with respect to gives 6. Appendi D. Analysis of real parts of 43 and 44 Substituting epansions 6 into 43 and 44 and neglecting terms of the orders OσR, σ I, σ Rɛ R, σ I ɛ I, σ R Ω and higher gives ɛ R R + Ryy + +σ R c ɛ R R Ω K K y w yy w R + γm Ω K w 4y y R Ry + cyy w yy w c + γm Ω K 4 w y 4y w c 3 cy K K y + Ω w c + γm K } 4y c =, D- M y T y + γ M Ω K 4y R M w Ry σ R T 3M + w T M w c y c + } γ γ cy γ K } Ω γt M w 4y c =. D- γm T M w From B--B-8, the boundary conditions for these equations are: for and R, =, R, / = R, / D-3 R, y =, γm R, y = Ry, y, T ɛ R Ry, y γm Ω K } } y 4y Ry, y σ R γm c, y = with R, = Ry, =, R, y Ω K K y ɛ R + Ryy, y + w yy w R, y +γm Ω K w 4y y R, y Ry, y D-4 D-5 D-6 5

6 ɛ R R, y K K y + Ω w c, y + γm K } 4y c, y =, D-7 + γm Ω K w 4y y R, y Ry, y +σ R c, y y + cyy, y Ω K yy c, y w +γm 4 w y 4y w c, y 3 cy, y =, D-8 ɛ R R, y σ R T 3M + w T M w c, y y c, y + } γ γ cy, y =, D-9 R, y =, σ R c, y + ɛ R R, y = D- for y /. Two integrations of D- and D- with respect to and the use of boundary conditions D-4, D-7-D- result in ɛ R R + Ryy + +σ R c ɛ R R Ω K K y w yy w R + γm Ω K w 4y y R Ry + cyy w yy w c + γm Ω K 4 w y 4y w c 3 cy K K y + Ω w c + γm K } 4y c =, D- M y T y + γ M Ω K 4y R M Ry σ R T 3M + w T M w c y c + } γ γ cy γ K } Ω γt M w 4y c = σ R f y. D- γm T M w Here f y is defined in C-3. Also, the conditions in D-5 and D-6 can be solved and show that ɛ R R, y = σ R γm π y y epαy [ gy ep αy dy ] dy, 4 γm R, y = σ R γm π y epαy [ gy ep αy dy ] D-3 4 where αy = y py dy, py = T T y K γm Ω, 4y T gy = Φy y y T y. Note that R, y = when M =. Also, note that in the general case R, / is now determined and may not be zero. For eample, in the case of a solid-body rotation 6

7 profile where K = and = T = we find py = γm Ω, αy = γm Ω y, and then R, y = γm π 4 y y epγm Ω y [ Φy y ep γm Ω y dy ] dy. D-4 Eamples of calculating R, y according to D-4 are shown in Fig. D- for various Mach numbers. It is clear that R, / is not zero. Solving D- for ɛ R R and substituting in D-, multiplying by T M w /T, and integrating again with respect to gives 63. 7

8 6 4 Mo=. Mo=.3 Mo=.5 4 Φ R r Figure : D-: Solutions of R, y for a solid-body rotation flow at various Mach numbers. 8