Stream thrust averaging

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In fluid dynamics, stream thrust averaging is a process used to convert three-dimensional flow through a duct into one-dimensional uniform flow. It makes the assumptions that the flow is mixed adiabatically and without friction. However, due to the mixing process, there is a net increase in the entropy of the system. Although there is an increase in entropy, the stream thrust averaged values are more representative of the flow than a simple average as a simple average would violate the second law of thermodynamics.

Equations for a perfect gas

Stream thrust:

F = ∫ ( ρ V ⋅ d A ) V ⋅ f + ∫ p d A ⋅ f . {\displaystyle F=\int \left(\rho \mathbf {V} \cdot d\mathbf {A} \right)\mathbf {V} \cdot \mathbf {f} +\int pd\mathbf {A} \cdot \mathbf {f} .} {\displaystyle F=\int \left(\rho \mathbf {V} \cdot d\mathbf {A} \right)\mathbf {V} \cdot \mathbf {f} +\int pd\mathbf {A} \cdot \mathbf {f} .}

Mass flow:

m ˙ = ∫ ρ V ⋅ d A . {\displaystyle {\dot {m}}=\int \rho \mathbf {V} \cdot d\mathbf {A} .} {\displaystyle {\dot {m}}=\int \rho \mathbf {V} \cdot d\mathbf {A} .}

Stagnation enthalpy:

H = 1 m ˙ ∫ ( ρ V ⋅ d A ) ( h + | V | 2 2 ) , {\displaystyle H={1 \over {\dot {m}}}\int \left({\rho \mathbf {V} \cdot d\mathbf {A} }\right)\left(h+{|\mathbf {V} |^{2} \over 2}\right),} {\displaystyle H={1 \over {\dot {m}}}\int \left({\rho \mathbf {V} \cdot d\mathbf {A} }\right)\left(h+{|\mathbf {V} |^{2} \over 2}\right),}
U ¯ 2 ( 1 − R 2 C p ) − U ¯ F m ˙ + H R C p = 0. {\displaystyle {\overline {U}}^{2}\left({1-{R \over 2C_{p}}}\right)-{\overline {U}}{F \over {\dot {m}}}+{HR \over C_{p}}=0.} {\displaystyle {\overline {U}}^{2}\left({1-{R \over 2C_{p}}}\right)-{\overline {U}}{F \over {\dot {m}}}+{HR \over C_{p}}=0.}

Solutions

Solving for U ¯ {\displaystyle {\overline {U}}} {\displaystyle {\overline {U}}} yields two solutions. They must both be analyzed to determine which is the physical solution. One will usually be a subsonic root and the other a supersonic root. If it is not clear which value of velocity is correct, the second law of thermodynamics may be applied.

ρ ¯ = m ˙ U ¯ A , {\displaystyle {\overline {\rho }}={{\dot {m}} \over {\overline {U}}A},} {\displaystyle {\overline {\rho }}={{\dot {m}} \over {\overline {U}}A},}
p ¯ = F A − ρ ¯ U ¯ 2 , {\displaystyle {\overline {p}}={F \over A}-{{\overline {\rho }}{\overline {U}}^{2}},} {\displaystyle {\overline {p}}={F \over A}-{{\overline {\rho }}{\overline {U}}^{2}},}
h ¯ = p ¯ C p ρ ¯ R . {\displaystyle {\overline {h}}={{\overline {p}}C_{p} \over {\overline {\rho }}R}.} {\displaystyle {\overline {h}}={{\overline {p}}C_{p} \over {\overline {\rho }}R}.}

Second law of thermodynamics:

∇ s = C p ln ⁡ ( T ¯ T 1 ) + R ln ⁡ ( p ¯ p 1 ) . {\displaystyle \nabla s=C_{p}\ln({{\overline {T}} \over T_{1}})+R\ln({{\overline {p}} \over p_{1}}).} {\displaystyle \nabla s=C_{p}\ln({{\overline {T}} \over T_{1}})+R\ln({{\overline {p}} \over p_{1}}).}

The values T 1 {\displaystyle T_{1}} {\displaystyle T_{1}} and p 1 {\displaystyle p_{1}} {\displaystyle p_{1}} are unknown and may be dropped from the formulation. The value of entropy is not necessary, only that the value is positive.

∇ s = C p ln ⁡ ( T ¯ ) + R ln ⁡ ( p ¯ ) . {\displaystyle \nabla s=C_{p}\ln({\overline {T}})+R\ln({\overline {p}}).} {\displaystyle \nabla s=C_{p}\ln({\overline {T}})+R\ln({\overline {p}}).}

One possible unreal solution for the stream thrust averaged velocity yields a negative entropy. Another method of determining the proper solution is to take a simple average of the velocity and determining which value is closer to the stream thrust averaged velocity.

References