THESE de DOCTORAT - cerfacs
THESE de DOCTORAT - cerfacs THESE de DOCTORAT - cerfacs
136 Chapter A: About the π ′ c = 0 assumption w 2 U v 2 w 1 v θ,2 v θ,1 v 1 U Figure A.1: Velocity triangles Considering now U 1 = U 2 = U for an axial compressor and ρ 1 u 1 = ρ 2 u 2 due to mass conservation, it is possible to express W c as function of U and v θ , W ρ 1 u 1 = h t,2 − h t,1 = U(v θ,2 − v θ,1 ) (A.5) and introducing now π T h t,1 (π T − 1) = U(v θ,2 − v θ,1 ) (A.6) Let us now define a parameter ζ = Uv θ h t . The variable π T is then expressed as h t,1 (π T − 1) = ζ 1 h t,1 − ζ 2 h t,2 = ζ 1 h t,1 − ζ 2 π T h t,1 (A.7) (π T − 1) = ζ 1 − ζ 2 π T (A.8) Finally, resolving for π T , it yields π T = ζ 1 + 1 ζ 2 + 1 (A.9) Applying the differential of the logarithm to Eq. (A.9) results in π ′ T ¯π T = ζ′ 1 ¯ζ 1 + 1 − ζ′ 2 ¯ζ 2 + 1 (A.10)
A.1 When is π T ′ equal to zero? 137 The fractional variation of ζ can in turn be found. It yields ζ ′ ¯ζ = v′ θ − h′ t v¯ θ ¯h t (A.11) From Eq. (A.10), it is then possible to state that π ′ T = 0 if both ζ′ 1 and ζ′ 2 are zero, or if ζ ′ 1 ¯ζ 1 +1 = ζ ′ 2 ¯ζ 2 +1 which is equivalent to ζ ′ 1 = ¯π T ζ ′ 2 (A.12) ( ) v ′ θ,1 − h′ t,1 v θ,1 ¯ ¯h t,1 ( ) v ′ θ,1 − h′ t,1 v θ,1 ¯ ¯h t,1 ( ) v ′ ¯ζ 1 = θ,2 ¯π T − h′ t,2 ¯ζ 2 v θ,2 ¯ ¯h t,2 ) ¯h t,2 ¯v θ,1 ( v ′ = θ,2 ¯π T − h′ t,2 ¯h t,1 v θ,2 ¯ ¯v θ,2 ¯h t,2 (A.13) (A.14) This relation can be splitted. Therefore π T ′ is zero if both v ′ θ,1 ¯v θ,1 ¯v θ,1 ¯h t,1 = ¯π T v ′ θ,2 ¯v θ,2 ¯v θ,2 ¯h t,2 (A.15) and h ′ t,1 ¯h t,1 ¯v θ,1 ¯h t,1 = ¯π T h ′ t,2 ¯h t,2 ¯v θ,2 ¯h t,2 (A.16) or if v ′ θ,1 = h′ t,1 v θ,1 ¯ ¯h t,1 or v ′ θ,2 = h′ t,2 v θ,2 ¯ ¯h t,2 (A.17) Equations (A.15) and (A.16) can be simplified in Eqs. (A.18) and (A.19) respectively. It results in v ′ θ,1 = v′ θ,2 (A.18)
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A.1 When is π<br />
T ′ equal to zero? 137<br />
The fractional variation of ζ can in turn be found. It yields<br />
ζ ′<br />
¯ζ = v′ θ<br />
− h′ t<br />
v¯<br />
θ<br />
¯h t<br />
(A.11)<br />
From Eq. (A.10), it is then possible to state that π ′ T = 0 if both ζ′ 1 and ζ′ 2 are zero, or if ζ ′ 1<br />
¯ζ 1 +1 =<br />
ζ ′ 2<br />
¯ζ 2 +1 which is equivalent to ζ ′ 1 = ¯π T ζ ′ 2 (A.12)<br />
( )<br />
v<br />
′<br />
θ,1<br />
− h′ t,1<br />
v θ,1 ¯ ¯h t,1<br />
( )<br />
v<br />
′<br />
θ,1<br />
− h′ t,1<br />
v θ,1 ¯ ¯h t,1<br />
( )<br />
v<br />
′<br />
¯ζ 1 = θ,2<br />
¯π T − h′ t,2 ¯ζ 2<br />
v θ,2 ¯ ¯h t,2<br />
)<br />
¯h t,2<br />
¯v θ,1<br />
(<br />
v<br />
′<br />
= θ,2<br />
¯π T − h′ t,2<br />
¯h t,1 v θ,2 ¯<br />
¯v θ,2<br />
¯h t,2<br />
(A.13)<br />
(A.14)<br />
This relation can be splitted. Therefore π<br />
T ′ is zero if both<br />
v ′ θ,1<br />
¯v θ,1<br />
¯v θ,1<br />
¯h t,1<br />
= ¯π T<br />
v ′ θ,2<br />
¯v θ,2<br />
¯v θ,2<br />
¯h t,2<br />
(A.15)<br />
and<br />
h ′ t,1<br />
¯h t,1<br />
¯v θ,1<br />
¯h t,1<br />
= ¯π T<br />
h ′ t,2<br />
¯h t,2<br />
¯v θ,2<br />
¯h t,2<br />
(A.16)<br />
or if<br />
v ′ θ,1<br />
= h′ t,1<br />
v θ,1 ¯ ¯h t,1<br />
or<br />
v ′ θ,2<br />
= h′ t,2<br />
v θ,2 ¯ ¯h t,2<br />
(A.17)<br />
Equations (A.15) and (A.16) can be simplified in Eqs. (A.18) and (A.19) respectively. It results<br />
in<br />
v ′ θ,1 = v′ θ,2<br />
(A.18)
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