Area-Mach relation: Difference between revisions

From Flowpedia
Jump to navigation Jump to search
← Older editNewer edit →
Nian (talk | contribs)
Bureaucrats, Interface administrators, Administrators
580 edits
VisualWikitext
No edit summary
No edit summary
Line 1: Line 1:
[[Category:Compressible flow]]
<!--
[[Category:Quasi-one-dimensional flow]]
-->[[Category:Compressible flow]]<!--
[[Category:Inviscid flow]]
-->[[Category:Quasi-one-dimensional flow]]<!--
<noinclude>
-->[[Category:Inviscid flow]]<!--
[[Category:Compressible flow:Topic]]
--><noinclude><!--
</noinclude>
-->[[Category:Compressible flow:Topic]]<!--
--></noinclude><!--


__TOC__
--><nomobile><!--
__TOC__<!--
--></nomobile><!--


--><noinclude><!--
-->{{#vardefine:secno|5}}<!--
-->{{#vardefine:eqno|31}}<!--
--></noinclude><!--
-->
=== The Area-Mach-Number Relation ===
=== The Area-Mach-Number Relation ===


Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):
Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):


<math display="block">
{{NumEqn|<math>
d(\rho uA)=0 \Rightarrow \rho u A=const
d(\rho uA)=0 \Rightarrow \rho u A=const
</math>
</math>}}


This applies everywhere in the nozzle and therefore the sonic conditions can be used as a reference
This applies everywhere in the nozzle and therefore the sonic conditions can be used as a reference


<math display="block">
{{NumEqn|<math>
\rho uA=\rho^*u^*A^*=\left\{u^*=a^*\right\}=\rho^*a^*A^*
\rho uA=\rho^*u^*A^*=\left\{u^*=a^*\right\}=\rho^*a^*A^*
</math>
</math>}}


divide by <math>\rho uA^*</math> gives
divide by <math>\rho uA^*</math> gives


<math display="block">
{{NumEqn|<math>
\frac{\rho^*}{\rho}\frac{a^*}{u}=\frac{A}{A^*}
\frac{\rho^*}{\rho}\frac{a^*}{u}=\frac{A}{A^*}
</math>
</math>}}


<math>a^*/u=1/M^*</math> but <math>\rho^*/\rho</math> is unknown
<math>a^*/u=1/M^*</math> but <math>\rho^*/\rho</math> is unknown


<math display="block">
{{NumEqn|<math>
\frac{\rho^*}{\rho}=\frac{\rho^*}{\rho_o}\frac{\rho_o}{\rho}
\frac{\rho^*}{\rho}=\frac{\rho^*}{\rho_o}\frac{\rho_o}{\rho}
</math>
</math>}}


and thus
and thus


<math display="block">
{{NumEqn|<math>
\frac{\rho^*}{\rho_o}\frac{\rho_o}{\rho}\frac{1}{M^*}=\frac{A}{A^*}
\frac{\rho^*}{\rho_o}\frac{\rho_o}{\rho}\frac{1}{M^*}=\frac{A}{A^*}
</math>
</math>}}


Using the isentropic relations, we get
Using the isentropic relations, we get


<math display="block">
{{NumEqn|<math>
\frac{\rho^*}{\rho_o}=\frac{1}{\left[\dfrac{1}{2}(\gamma-1)\right]^{1/(\gamma-1)}}
\frac{\rho^*}{\rho_o}=\frac{1}{\left[\dfrac{1}{2}(\gamma-1)\right]^{1/(\gamma-1)}}
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
\frac{\rho_o}{\rho}=\left[1+\frac{1}{2}(\gamma+1)M^2\right]^{1/(\gamma-1)}
\frac{\rho_o}{\rho}=\left[1+\frac{1}{2}(\gamma+1)M^2\right]^{1/(\gamma-1)}
</math>
</math>}}


Eqns. \ref{eq:rho:a} and \ref{eq:rho:b} in Eqn. \ref{eq:areamach:a} gives
Eqns. \ref{eq:rho:a} and \ref{eq:rho:b} in Eqn. \ref{eq:areamach:a} gives


<math display="block">
{{NumEqn|<math>
\frac{A}{A^*}=\frac{1}{M^*}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{1/(\gamma-1)}
\frac{A}{A^*}=\frac{1}{M^*}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{1/(\gamma-1)}
</math>
</math>}}


What remains now is to replace <math>M^*</math>
What remains now is to replace <math>M^*</math>


<math display="block">
{{NumEqn|<math>
{M^*}^2=\frac{u^2}{{a^*}^2}=\frac{u^2}{a^2}\frac{a^2}{{a^*}^2}=\frac{u^2}{a^2}\frac{a^2}{a_o^2}\frac{a_o^2}{{a^*}^2}=M^2\frac{a^2}{a_o^2}\frac{a_o^2}{{a^*}^2}
{M^*}^2=\frac{u^2}{{a^*}^2}=\frac{u^2}{a^2}\frac{a^2}{{a^*}^2}=\frac{u^2}{a^2}\frac{a^2}{a_o^2}\frac{a_o^2}{{a^*}^2}=M^2\frac{a^2}{a_o^2}\frac{a_o^2}{{a^*}^2}
</math>
</math>}}


For a calorically perfect gas <math>a=\sqrt{\gamma R T}</math>, which gives
For a calorically perfect gas <math>a=\sqrt{\gamma R T}</math>, which gives


<math display="block">
{{NumEqn|<math>
\frac{a^2}{a_o^2}=\frac{T}{T_o}=\left[1+\frac{1}{2}(\gamma-1)M^2\right]^{-1}
\frac{a^2}{a_o^2}=\frac{T}{T_o}=\left[1+\frac{1}{2}(\gamma-1)M^2\right]^{-1}
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
\frac{a_o^2}{{a^*}^2}=\frac{T_o}{T^*}=\frac{1}{2}(\gamma+1)
\frac{a_o^2}{{a^*}^2}=\frac{T_o}{T^*}=\frac{1}{2}(\gamma+1)
</math>
</math>}}


Eqns. \ref{eq:a:a} and \ref{eq:a:b} in Eqn. \ref{eq:mstar:a} gives
Eqns. \ref{eq:a:a} and \ref{eq:a:b} in Eqn. \ref{eq:mstar:a} gives


<math display="block">
{{NumEqn|<math>
{M^*}^2=\frac{(\gamma+1)M^2}{2+(\gamma-1)M^2}
{M^*}^2=\frac{(\gamma+1)M^2}{2+(\gamma-1)M^2}
</math>
</math>}}


Now, rewrite Eqn. \ref{eq:areamach:b} as
Now, rewrite Eqn. \ref{eq:areamach:b} as


<math display="block">
{{NumEqn|<math>
\left(\frac{A}{A^*}\right)^2=\frac{1}{{M^*}^2}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{2/(\gamma-1)}
\left(\frac{A}{A^*}\right)^2=\frac{1}{{M^*}^2}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{2/(\gamma-1)}
</math>
</math>}}


and insert <math>{M^*}^2</math> from Eqn. \ref{eq:mstar:b}
and insert <math>{M^*}^2</math> from Eqn. \ref{eq:mstar:b}


<math display="block">
{{NumEqn|<math>
\left(\frac{A}{A^*}\right)^2=\frac{2+(\gamma-1)M^2}{(\gamma+1)M^2}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{2/(\gamma-1)} \Rightarrow
\left(\frac{A}{A^*}\right)^2=\frac{2+(\gamma-1)M^2}{(\gamma+1)M^2}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{2/(\gamma-1)} \Rightarrow
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
\left(\frac{A}{A^*}\right)^2=\frac{1}{M^2}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{1+2/(\gamma-1)} \Rightarrow
\left(\frac{A}{A^*}\right)^2=\frac{1}{M^2}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{1+2/(\gamma-1)} \Rightarrow
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
\left(\frac{A}{A^*}\right)^2=\frac{1}{M^2}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{(\gamma+1)/(\gamma-1)}
\left(\frac{A}{A^*}\right)^2=\frac{1}{M^2}\left[\frac{2+(\gamma-1)M^2}{\gamma+1}\right]^{(\gamma+1)/(\gamma-1)}
</math>
</math>}}


which is the area-Mach-number relation.
which is the area-Mach-number relation.
Line 102: Line 110:
For a nozzle flow, the area-Mach-number relation gives the Mach number, <math>M</math>, at any location inside the nozzle as a function of the ratio between the local cross-section area, <math>A</math>, and the throat area at choked conditions, <math>A^*</math>.
For a nozzle flow, the area-Mach-number relation gives the Mach number, <math>M</math>, at any location inside the nozzle as a function of the ratio between the local cross-section area, <math>A</math>, and the throat area at choked conditions, <math>A^*</math>.


<math display="block">
{{NumEqn|<math>
M=f\left(\frac{A}{A^*}\right)
M=f\left(\frac{A}{A^*}\right)
</math>
</math>}}


<!--
<!--

Revision as of 11:08, 1 April 2026

The Area-Mach-Number Relation

Starting point - the continuity equation (Eqn. \ref{eq:governing:cont}):

d(ρuA)=0⇒ρuA=const(Eq. 5.32)

This applies everywhere in the nozzle and therefore the sonic conditions can be used as a reference

ρuA=ρ*u*A*={u*=a*}=ρ*a*A*(Eq. 5.33)

divide by ρuA* gives

ρ*ρa*u=AA*(Eq. 5.34)

a*/u=1/M* but ρ*/ρ is unknown

ρ*ρ=ρ*ρoρoρ(Eq. 5.35)

and thus

ρ*ρoρoρ1M*=AA*(Eq. 5.36)

Using the isentropic relations, we get

ρ*ρo=1[12(γ−1)]1/(γ−1)(Eq. 5.37)
ρoρ=[1+12(γ+1)M2]1/(γ−1)(Eq. 5.38)

Eqns. \ref{eq:rho:a} and \ref{eq:rho:b} in Eqn. \ref{eq:areamach:a} gives

AA*=1M*[2+(γ−1)M2γ+1]1/(γ−1)(Eq. 5.39)

What remains now is to replace M*

M*2=u2a*2=u2a2a2a*2=u2a2a2ao2ao2a*2=M2a2ao2ao2a*2(Eq. 5.40)

For a calorically perfect gas a=γRT, which gives

a2ao2=TTo=[1+12(γ−1)M2]−1(Eq. 5.41)
ao2a*2=ToT*=12(γ+1)(Eq. 5.42)

Eqns. \ref{eq:a:a} and \ref{eq:a:b} in Eqn. \ref{eq:mstar:a} gives

M*2=(γ+1)M22+(γ−1)M2(Eq. 5.43)

Now, rewrite Eqn. \ref{eq:areamach:b} as

(AA*)2=1M*2[2+(γ−1)M2γ+1]2/(γ−1)(Eq. 5.44)

and insert M*2 from Eqn. \ref{eq:mstar:b}

(AA*)2=2+(γ−1)M2(γ+1)M2[2+(γ−1)M2γ+1]2/(γ−1)⇒(Eq. 5.45)
(AA*)2=1M2[2+(γ−1)M2γ+1]1+2/(γ−1)⇒(Eq. 5.46)
(AA*)2=1M2[2+(γ−1)M2γ+1](γ+1)/(γ−1)(Eq. 5.47)

which is the area-Mach-number relation.

For a nozzle flow, the area-Mach-number relation gives the Mach number, M, at any location inside the nozzle as a function of the ratio between the local cross-section area, A, and the throat area at choked conditions, A*.

M=f(AA*)(Eq. 5.48)


Due to the assumptions made in the derivation, the area-Mach-number relation is only valid for isentropic flows of calorically perfect gases. This means that it cannot be used throughout the divergent part of a convergent-divergent nozzle in case there is a shock within the nozzle. It can, however, be used both upstream and downstream of the shock. Note that A* will change over the shock.


Retrieved from "https://wiki.g3dflow.com/index.php?title=Area-Mach_relation&oldid=522"