Template:OpenInfoBox

2026-04-01T18:50:14Z

Nian:

{{InfoBox|1={{{1|}}}|2={{{2|}}}|3={{{3|}}}|label={{{label|}}}|description={{{description|}}}|noborder=0|numw=6em}}

2026-04-01T18:43:24Z

Nian:

Template:OpenInfoBox

2026-04-01T18:41:20Z

2026-04-01T18:25:07Z

Nian: /* Thermal Choking */

[[Category:Compressible flow]][[Category:One-dimensional flow]][[Category:Inviscid flow]][[Category:Continuous solution]]<noinclude>[[Category:Compressible flow:Topic]]</noinclude><nomobile>__TOC__</nomobile><noinclude>{{#vardefine:secno|3}}{{#vardefine:eqno|70}}</noinclude>
==== Flow-station relations ====

The aim is to derive relations for pressure ratio and temperature ratio as a function of Mach numbers. We will do that starting from the momentum equation.

{{NumEqn|<math>
p_2-p_1=\rho_1 u_1^2 - \rho_2 u_2^2
</math>}}

Assuming calorically perfect gas

{{NumEqn|<math>
\rho u^2=\rho a^2 M^2=\rho \frac{\gamma p}{\rho} M^2=\gamma p M^2
</math>}}

which inserted in Eqn. \ref{eq:governing:mom} gives

{{NumEqn|<math>
p_2-p_1=\gamma p_1 M_1^2 - \gamma p_2 M_2^2
</math>}}

{{NumEqn|<math>
p_2(1+\gamma M_2^2)=p_1(1+\gamma M_1^2)
</math>}}

and thus

{{NumEqn|<math>
\frac{p_2}{p_1}=\frac{1+\gamma M_1^2}{1+\gamma M_2^2}
</math>}}

From the equation of state <math>p=\rho RT</math>, we get

{{NumEqn|<math>
\frac{T_2}{T_1}=\frac{p_2}{\rho_2 R}\frac{\rho_1 R}{p_1}=\frac{p_2}{p_1}\frac{\rho_1}{\rho_2}
</math>}}

Using the continuity equation, we can get <math>\rho_1/\rho_2</math>

{{NumEqn|<math>
\rho_1 u_1=\rho_2 u_2 \Rightarrow \frac{\rho_1}{\rho_2}=\frac{u_2}{u_1}
</math>}}

Inserted in Eqn. \ref{eq:tr:a} gives

{{NumEqn|<math>
\frac{T_2}{T_1}=\frac{p_2}{p_1}\frac{u_2}{u_1}
</math>}}

{{NumEqn|<math>
\frac{u_2}{u_1}=\frac{M_2a_2}{M_1a_1}=\frac{M_2}{M_1}\frac{\sqrt{\gamma RT_2}}{\sqrt{\gamma RT_1}}=\frac{M_2}{M_1}\sqrt{\frac{T_2}{T_1}}
</math>}}

Eqn. \ref{eq:tr:c} in Eqn. \ref{eq:tr:b} gives

{{NumEqn|<math>
\sqrt{\frac{T_2}{T_1}}=\frac{p_2}{p_1}\frac{M_2}{M_1}
</math>}}

With <math>p_2/p_1</math> from Eqn. \ref{eq:governing:mom:b}, Eqn \ref{eq:tr:d} becomes

{{NumEqn|<math>
\frac{T_2}{T_1}=\left(\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\right)^2\left(\frac{M_2}{M_1}\right)^2
</math>}}

==== Differential Relations ====

The equations presented in the previous section gives us the flow state after heat addition but since the heat addition, unlike the normal shock, is a continuous process, it is of interest to study the the heat addition from start to end. In order to do so we will now derive differential relations starting from the governing equations on differential form. We will start with converting the integral equation for conservation of mass for one-dimensional flows to differential form.

{{NumEqn|<math>
\rho_1 u_1 = \rho_2 u_2 = const \Rightarrow d(\rho u)=0
</math>}}

{{NumEqn|<math>
d(\rho u)=\rho du+ud\rho=0
</math>}}

Divide by <math>\rho u</math> gives

{{NumEqn|<math>
\dfrac{d\rho}{\rho}=\dfrac{du}{u}
</math>}}

The integral form of the conservation of momentum equation for one-dimensional flows is converted to differential form as follows.

{{NumEqn|<math>
p_1+\rho_1u_1^2=p_2+\rho_2u_2^2=const \Rightarrow d(p+\rho u^2)=0
</math>}}

{{NumEqn|<math>
dp+\rho udu+u\underbrace{d(\rho u)}_{=0}=0\Rightarrow dp=-\rho udu
</math>}}

with <math>\rho=\dfrac{p}{RT}</math> and <math>u^2=M^2a^2=M^2\gamma RT</math> in Eqn.~\ref{eq:governing:mom:diff:b}, we get

{{NumEqn|<math>
dp=-\dfrac{p}{RT}u^2\dfrac{du}{u}=-\dfrac{p}{RT}M^2\gamma RT\dfrac{du}{u}\Rightarrow
</math>}}

{{NumEqn|<math>
\dfrac{dp}{p}=-\gamma M^2\dfrac{du}{u}
</math>}}

which gives the relative change in pressure, <math>dp/p</math>, as a function of the relative change in flow velocity, <math>du/u</math>. The next equation to derive is an equation that describes the relative change in temperature, <math>dT/T</math>, as a function of the relative change in flow velocity, <math>du/u</math>. The starting point is the equation of state (the gas law).

{{NumEqn|<math>
p=\rho RT\Rightarrow dp = R(\rho dT+ Td\rho)\Rightarrow dT=\dfrac{1}{R\rho}dp-\dfrac{T}{\rho}d\rho
</math>}}

Divide by <math>T</math>

{{NumEqn|<math>
\dfrac{dT}{T}=\dfrac{1}{\rho RT}dp-\dfrac{1}{\rho}d\rho
</math>}}

substitute <math>dp</math> from Eqn.~\ref{eq:governing:mom:diff:c} and <math>d\rho</math> from Eqn.~\ref{eq:governing:cont:diff:b} gives

{{NumEqn|<math>
\dfrac{dT}{T}=(1-\gamma M^2)\dfrac{du}{u}
</math>}}

The entropy equation reads

{{NumEqn|<math>
ds=C_v\dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}
</math>}}

which after substituting <math>dp</math> from Eqn.~\ref{eq:governing:mom:diff:c} and <math>d\rho</math> from Eqn.~\ref{eq:governing:cont:diff:b} becomes

{{NumEqn|<math>
ds=C_v\gamma(1-M^2)\dfrac{du}{u}
</math>}}

From the definition of total temperature <math>T_o</math> we get

{{NumEqn|<math>
T_o=T+\dfrac{u^2}{2C_p}\Rightarrow dT_o=dT+\dfrac{1}{C_p}udu=dT+\dfrac{\gamma-1}{\gamma RT}T u^2\dfrac{du}{u}\Rightarrow
</math>}}

{{NumEqn|<math>
dT_o=dT+(\gamma-1)M^2 T\dfrac{du}{u}
</math>}}

Inserting <math>dT</math> from Eqn~\ref{eq:governing:temp:diff:c} in Eqn~\ref{eq:governing:To:diff:a} we get

{{NumEqn|<math>
dT_o=(1-\gamma M^2)T\dfrac{du}{u}+(\gamma-1)M^2 T\dfrac{du}{u}
</math>}}

or

{{NumEqn|<math>
dT_o=(1-M^2)T\dfrac{du}{u}
</math>}}

Dividing Eqn.~\ref{eq:governing:To:diff:b} by <math>T_o</math> and using

{{NumEqn|<math>
T_o=T\left(1+\dfrac{\gamma-1}{2}M^2\right)
</math>}}

we get

{{NumEqn|<math>
\dfrac{dT_o}{T_o}=\dfrac{1-M^2}{1+\dfrac{\gamma-1}{2}M^2}\dfrac{du}{u}
</math>}}

Finally, we will derive a differential relation that describes the change in Mach number.

{{NumEqn|<math>
M=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=\dfrac{1}{\sqrt{\gamma R}}(T^{1/2}du+ud(T^{-1/2}))=\dfrac{du}{\sqrt{\gamma RT}}-\dfrac{u}{2\sqrt{\gamma R}}T^{-3/2}dT\Rightarrow
</math>}}

{{NumEqn|<math>
dM=\dfrac{1}{\sqrt{\gamma RT}}\dfrac{du}{u}-\dfrac{1}{2}\dfrac{u}{\sqrt{\gamma RT}}\dfrac{dT}{T}=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}
</math>}}

Inserting <math>dT/T</math> from Eqn.~\ref{eq:governing:temp:diff:c}, we get

{{NumEqn|<math>
\dfrac{dM}{M}=\dfrac{1+\gamma M^2}{2}\dfrac{du}{u}
</math>}}

All the derived differential relations are expressed as functions of <math<du/u</math> but it would be more convenient to relate the changes in flow properties to the added heat or the change in total temperature, which can be related to the added heat through the energy equation.

{{NumEqn|<math>
dT_o=\dfrac{\delta q}{C_p}
</math>}}

From Eqn.~\ref{eq:governing:To:diff:c}, we get

{{NumEqn|<math>
\dfrac{du}{u}=\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
</math>}}

Now, we can substitute $du/u$ in all the above relations using Eqn.~\ref{eq:governing:du:diff:final}, we get the following relations

{{NumEqn|<math>
\dfrac{d\rho}{\rho}=-\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
</math>}}

{{NumEqn|<math>
\dfrac{dp}{p}=\gamma M^2\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
</math>}}

{{NumEqn|<math>
\dfrac{dT}{T}=(1-\gamma M^2)\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
</math>}}

{{NumEqn|<math>
\dfrac{dT}{T}=\left(\dfrac{1+\gamma M^2}{2}\right)\left(\dfrac{1+\dfrac{\gamma-1}{2}M^2}{1-M^2}\right)\dfrac{dT_o}{T_o}
</math>}}

{{NumEqn|<math>
\dfrac{dT}{T}=C_v\gamma \left(1+\dfrac{\gamma-1}{2}M^2\right)\dfrac{dT_o}{T_o}
</math>}}

==== Heat Addition Process ====

With the differential relations in place, we can now study the continuous change in flow quantities from the initial flow state to the flow state after the heat addition process by dividing the total amount of heat added to the flow, <math>q</math>, into small portions, <math>\delta q</math>, and calculate the change in flow properties for each of these heat additions, see Figure~\ref{fig:dq}.



Let's first examine the temperature change by rewriting Eqn.~\ref{eq:governing:dT:diff:final} as

{{NumEqn|<math>
dT=\dfrac{1-\gamma M^2}{1-M^2}dT_o\Leftrightarrow \dfrac{dT}{dT_o}=\dfrac{1-\gamma M^2}{1-M^2}
</math>}}

which is equivalent to

{{NumEqn|<math>
\dfrac{dh}{\delta q}=\dfrac{1-\gamma M^2}{1-M^2}
</math>}}

Form Eqn.~\ref{eq:governing:dT:diff:mod:a} we can make the following observation

{{NumEqn|<math>
\dfrac{dT}{dT_o}=0\Rightarrow \gamma M^2=1\Rightarrow M=\sqrt{1/\gamma}
</math>}}

which means that the maximum temperature will be reached when the Mach number is <math>\sqrt{1/\gamma}</math>. Since <math>\gamma</math> is a number greater than one for all gases, this implies that the maximum temperature can only be reached if the flow is subsonic. For air, this the maximum temperature will be reached at <math>M=0.845</math>.

If we evaluate Eqn.~\ref{eq:governing:dT:diff:mod:a} for sonic flow (<math>M=1</math>), we see that the derivative becomes infinite.

{{NumEqn|<math>
|M|\rightarrow 1.0 \Rightarrow \dfrac{dT}{dT_o}\rightarrow \pm \infty
</math>}}

Now, by specifying an initial subsonic flow state and dividing the heat addition corresponding to choked flow, <math>q^\ast</math>, into small portions <math>\delta q</math>, one can perform integration as indicated in Figure~\ref{fig:dq}. The result is presented in the in Figure~\ref{fig:TS:closeup}. The subsonic process corresponds to the upper line. As heat is added the Mach number is increased and at <math>M=\gamma^{-1/2}</math> the maximum temperature is reached. Adding more heat will reduce the temperature and increase the Mach number until sonic conditions are reached (<math>M=1.0</math>). As can be seen in Figure~\ref{fig:TS:closeup}, the lean of the subsonic branch of the Rayleigh line is lower than the isobars (gray lines), which means the increasing heat will reduce pressure. The lower part of the blue line in Figure~\ref{fig:TS:closeup} is the supersonic branch of the Rayleigh line, which is obtained in the same way starting from a supersonic flow condition. A flow state resulting in the same sonic conditions as for the subsonic case is calculated and used as a starting state. The corresponding $q^\ast$ is calculated and the same calculation of consecutive flow states in a step-wise manner is performed. As can be seen in Figure~\ref{fig:TS:closeup}, the lean of the supersonic part of the Rayleigh curve is steeper than the isobars (gray lines), which means that pressure increases as heat is added to the flow. As we saw from Eqn.~\ref{eq:governing:dT:diff:mod:b}, <math>dT/dT_o</math> becomes infinite when the flow approaches the sonic the sonic state. After the sonic state is reached, further heat addition is impossible without changing the upstream flow conditions. This will be made clearer in the next section.

Using the differential relations above, we can get a good picture of the development of flow variables as heat is continuously added to the flow (see Figure~\ref{fig:rayleigh:trends}).




==== Rayleigh Line ====

The continuity equation for steady-state, one-dimensional flow reads

{{NumEqn|<math>
\rho_1 u_1 = \rho_2 u_2 = C
</math>}}

where <math>C</math> is the massflow per square meter (massflow divided by area). Inserted in the momentum equation we get

{{NumEqn|<math>
p_1+\dfrac{C^2}{\rho_1}=p_2+\dfrac{C^2}{\rho_2} \Leftrightarrow p_1+\nu_1 C^2=p_2+\nu_2 C^2\Rightarrow \dfrac{p_2-p_1}{\nu_2-\nu_1}=-C^2
</math>}}

Eqn.~\ref{eq:governing:mom:b} tells us that any solution to the governing flow equations must lie along a line (a so-called Rayleigh line) in a <math>p\nu</math>-diagram. In Figure~\ref{fig:PV}, 1 corresponds to the flow state before heat addition and states 2 and 3 corresponds to the flow state after heat is added. If the flow in state 1 is subsonic, adding heat will change the flow state following the Rayleigh line to the right, i.e. towards flow state 2. If the initial flow state instead is supersonic, heat addition will move the flow state towards state 3.



Now we know in which direction we will move along the Rayleigh curve when heat is added but in order to find the flow state after heat addition we need to add the energy equation to the problem. If we draw a curve corresponding to the energy equation including the heat addition in the same <math>p\nu</math>-diagram, the intersection of this curve and the Rayleigh line corresponds to the downstream flow state (the flow state that fulfils the continuity, momentum, and energy equations). To be able to do this we will rewrite the energy equation such that it can be represented by a line in the <math>p\nu</math>-diagram.

The energy equation for one-dimensional flow with heat addition reads

{{NumEqn|<math>
h_1+\dfrac{1}{2}u_1^2+q=h_2+\dfrac{1}{2}u_2^2
</math>}}

Inserting the constant <math>C</math> from above (the massflow per <math>m^2</math>) and and and <math>h=C_pT</math>, we get

{{NumEqn|<math>
\dfrac{\gamma R}{\gamma - 1}T_1+\dfrac{1}{2}C^2\nu_1^2+q=\dfrac{\gamma R}{\gamma-1}+\dfrac{1}{2}C^2\nu_2^2
</math>}}

which may be rewritten as

{{NumEqn|<math>
\dfrac{p_2}{p_1}=\left(\dfrac{\nu_2}{\nu_1}-\dfrac{\gamma+1}{\gamma-1}-\dfrac{2q}{RT_1}\right)\left(1-\dfrac{\gamma+1}{\gamma-1}\dfrac{\nu_2}{\nu_1}\right)^{-1}
</math>}}









As you can see in the examples above (Figures~\ref{fig:TSPV:b} and \ref{fig:TSPV:d}), sonic conditions are reached when the Rayleigh line is tangent to the curve representing the energy equation in the <math>p\nu</math>-diagram. Adding more heat would move the energy equation line upwards and thus there can not be any solution after reaching this state unless the upstream conditions are changed such that the energy line intersects the Rayleigh line after further heat addition. Let's have a second look at the equations and see if it is possible to verify that the case where the Rayleigh line is a tangent to the energy-equation curve is in fact the sonic state.

Starting from Eqn.~\ref{eq:governing:energy:b}, it is easy to see that for any point along the energy equation curve the flow state may be expressed as a function of the initial flow state and the added heat <math>q</math> as

{{NumEqn|<math>
\dfrac{\gamma}{\gamma-1}p\nu+\dfrac{1}{2}C^2\nu^2=\dfrac{\gamma}{\gamma-1}p_1\nu_1+\dfrac{1}{2}C^2\nu_1^2+q=D
</math>}}

where <math>D</math> is a constant.

Now, let's different the Eqn.\ref{eq:governing:energy:d} with respect to <math>\nu</math>

{{NumEqn|<math>
\dfrac{\gamma}{\gamma-1}\left(\nu\dfrac{dp}{d\nu}+p\right)+C^2\nu=0\Rightarrow \dfrac{dp}{d\nu}=-\dfrac{\gamma}{\gamma-1}C^2-\dfrac{p}{\nu}
</math>}}

The Rayleigh line is a tangent to the energy equation curve when <math>dp/d\nu=-C^2</math> and thus

{{NumEqn|<math>
\dfrac{C^2}{\gamma}=\dfrac{p}{\nu}
</math>}}

By definition <math>C=\rho u</math> and <math>\nu=1/\rho</math>, which inserted in Eqn.~\ref{eq:governing:energy:f} gives

{{NumEqn|<math>
u=\sqrt{\dfrac{\gamma p}{\rho}}=a
</math>}}

==== Thermal Choking ====

When the heat addition reaches $q^\ast$ the flow becomes sonic and the flow is said to thermally choked. Thermal choking is illustrated in Figure~\ref{fig:TSPV:d}, where the curve representing the energy equation (the blue line in the <math>p\nu</math>-diagram) is tangent to the Rayleigh line and if more heat is added the blue line will move to the right of the Rayleigh line and thus there are no solutions for <math>q>q^\ast</math>. So what happens if more heat is added to the flow after thermal choking is reached. The answer is different if the flow is subsonic or supersonic. For a subsonic flow, the upstream flow will be adjusted such that the slope of the Rayleigh line changes and the energy equation curve becomes tangent to the Rayleigh line. This means that the massflow per unit area (<math>C</math>) is reduced and <math>q^\ast</math> is increased such that <math>q^\ast</math> equals the heat added to the flow. Note that the upstream total conditions will not be changed in this process (see Figure~\ref{fig:thermal:choking:sub}).

{{InfoBox|<math>
M_{1^\prime} = f(q^\ast)
</math><br><br><math>
T_{1^\prime} = f(T_o,\ M_{1^\prime})
</math><br><br><math>
p_{1^\prime} = f(p_o,\ M_{1^\prime})
</math><br><br><math>
\rho_{1^\prime} = f(p_{1^\prime},\ T_{1^\prime})
</math><br><br><math>
a_{1^\prime} = f(T_{1^\prime})
</math><br><br><math>
u_{1^\prime} = M_{1^\prime}a_{1^\prime}
</math>}}



In a choked supersonic flow, there is no possibility for pressure waves to travel upstream in the flow and thus the upstream flow conditions can not be changed as in the subsonic case. Moreover, since a normal shock is an adiabatic process (a jump between two points on the same Rayleigh line), the total temperature is not changed over a chock. From before we have

{{NumEqn|<math>
\dfrac{T_o}{T_o^\ast}=\dfrac{(\gamma+1)M_1^2}{(1+\gamma M_1^2)^2}(2+(\gamma-1)M_1^2)
</math>}}

Inserting the normal shock relation

{{NumEqn|<math>
M_2^2=\dfrac{2+(\gamma-1)M_1^2}{2\gamma M_1^2-(\gamma-1)}
</math>}}

one can show that

{{NumEqn|<math>
\dfrac{T_o}{T_o^\ast}=\dfrac{(\gamma+1)M_2^2}{(1+\gamma M_2^2)^2}(2+(\gamma-1)M_2^2)
</math>}}

and thus <math>T_o^\ast</math> is not changed by the normal shock and consequently <math>q^\ast</math> is unchanged if there is a normal shock between station 1 and 2. So, it is not possible to change the upstream static flow conditions and a normal shock will not make it possible to add more heat. The only possible solution is a normal shock upstream of station 1 and thus subsonic flow through the heat addition process.

Governing equations on differential form

2026-04-01T18:23:47Z

Nian:

[[Category:Compressible flow]][[Category:Governing equations]]<noinclude>[[Category:Compressible flow:Topic]]</noinclude><nomobile>__TOC__</nomobile><noinclude>{{#vardefine:secno|2}}{{#vardefine:eqno|28}}</noinclude>
=== The Differential Equations on Conservation Form ===

==== Conservation of Mass ====

The continuity equation on integral form reads

{{InfoBox|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
</math>}}

Apply Gauss's divergence theorem on the surface integral gives

{{NumEqn|<math>
\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV
</math>}}

Also, if <math>\Omega</math> is a fixed control volume

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV=\iiint_{\Omega} \frac{\partial \rho}{\partial t} dV
</math>}}

The continuity equation can now be written as a single volume integral.

{{NumEqn|<math>
\iiint_{\Omega} \left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})\right]dV=0
</math>}}

<math>\Omega</math> is an arbitrary control volume and thus

{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>|label=eq-cont-pde}}

which is the continuity equation on partial differential form.

==== Conservation of Momentum ====

The momentum equation on integral form reads

{{InfoBox|<math>
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
</math>}}

As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.

{{NumEqn|<math>
\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega} \nabla\cdot(\rho \mathbf{v}\mathbf{v})dV
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega} p\mathbf{n}dS=\iiint_{\Omega} \nabla pdV
</math>}}

Also, if <math>\Omega</math> is a fixed control volume

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV=\iiint_{\Omega} \frac{\partial}{\partial t}(\rho \mathbf{v}) dV
</math>}}

The momentum equation can now be written as one single volume integral

{{NumEqn|<math>
\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p - \rho \mathbf{f}\right]dV=0
</math>}}

<math>\Omega</math> is an arbitrary control volume and thus

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>|label=eq-mom-pde}}

which is the momentum equation on partial differential form

==== Conservation of Energy ====

The energy equation on integral form reads

{{InfoBox|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>}}

Gauss's divergence theorem applied to the surface integral term in the energy equation gives

{{NumEqn|<math>
\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\nabla\cdot(\rho h_o\mathbf{v})dV
</math>}}

Fixed control volume

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV=\iiint_{\Omega}\frac{\partial}{\partial t}(\rho e_o) dV
</math>}}

The energy equation can now be written as

{{NumEqn|<math>
\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) - \rho\mathbf{f}\cdot\mathbf{v} - \dot{q}\rho \right]dV=0
</math>}}

<math>\Omega</math> is an arbitrary control volume and thus

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|label=eq-energy-pde}}

which is the energy equation on partial differential form

==== Summary ====

The governing equations for compressible inviscid flow on partial differential form:

<div style="border: solid 1px;">
{{OpenInfoBox|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>|description=Continuity:}}

{{OpenInfoBox|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>|description=Momentum:}}

{{OpenInfoBox|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|description=Energy:}}
</div>

=== The Differential Equations on Non-Conservation Form ===

==== The Substantial Derivative ====

The substantial derivative operator is defined as

{{NumEqn|<math>
\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla
</math>|label=eq-cont-pde-non-cons}}

where the first term of the right hand side is the local derivative and the second term is the convective derivative.

==== Conservation of Mass ====

If we apply the substantial derivative operator to density we get

{{NumEqn|<math>
\frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho
</math>}}

From before we have the continuity equation on differential form as

{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>}}

which can be rewritten as

{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \rho(\nabla\cdot\mathbf{v}) + \mathbf{v}\cdot\nabla\rho=0
</math>}}

and thus

{{NumEqn|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
</math>|label=eq-pde-noncons-cont}}

{{EquationNote|label=eq-pde-noncons-cont|nopar=1}} says that the mass of a fluid element with a fixed set of fluid particles is constant as the element moves in space.

==== Conservation of Momentum ====

We start from the momentum equation on differential form derived above

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>}}

Expanding the first and the second terms gives

{{NumEqn|<math>
\rho\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v}\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla\mathbf{v} + \mathbf{v}(\nabla\cdot\rho\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>}}

Collecting terms, we can identify the substantial derivative operator applied to the velocity vector and the continuity equation.

{{NumEqn|<math>
\rho\underbrace{\left[\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right]}_{=\frac{D\mathbf{v}}{Dt}}+\mathbf{v}\underbrace{\left[\frac{\partial \rho}{\partial t}+\nabla\cdot\rho\mathbf{v}\right]}_{=0}+ \nabla p = \rho \mathbf{f}
</math>}}

which gives us the non-conservation form of the momentum equation

{{NumEqn|<math>
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
</math>|label=eq-mom-pde-non-cons}}

==== Conservation of Energy ====

The last equation on non-conservation differential form is the energy equation. We start by rewriting the energy equation on differential form {{EquationNote|label=eq-energy-pde}}, repeated here for convenience

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|nonumber=1}}

Total enthalpy, <math>h_o</math>, is replaced with total energy, <math>e_o</math>

{{NumEqn|<math>
h_o=e_o+\frac{p}{\rho}
</math>}}

which gives

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho e_o\mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}

Expanding the two first terms as

{{NumEqn|<math>
\rho\frac{\partial e_o}{\partial t} + e_o\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla e_o + e_o\nabla\cdot(\rho \mathbf{v}) + \nabla\cdot(p\mathbf{v})=</math><br><br><math>= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}

Collecting terms, we can identify the substantial derivative operator applied on total energy, <math>De_o/Dt</math> and the continuity equation

{{NumEqn|<math>
\rho\underbrace{\left[ \frac{\partial e_o}{\partial t} + \mathbf{v}\cdot\nabla e_o \right]}_{=\frac{De_o}{Dt}} + e_o\underbrace{\left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho \mathbf{v}) \right]}_{=0} + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}

and thus we end up with the energy equation on non-conservation differential form

{{NumEqn|<math>
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|label=eq-energy-pde-non-cons}}

==== Summary ====

<div style="border: solid 1px;">
{{OpenInfoBox|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
</math>|description=Continuity:}}

{{OpenInfoBox|<math>
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
</math>|description=Momentum:}}

{{OpenInfoBox|<math>
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|description=Energy:}}
</div>

=== Alternative Forms of the Energy Equation ===

==== Internal Energy Formulation ====

Total internal energy is defined as

{{NumEqn|<math>
e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}
</math>}}

Inserted in {{EquationNote|label=eq-energy-pde-non-cons|nopar=1}}, this gives

{{NumEqn|<math>
\rho\frac{De}{Dt} + \rho\mathbf{v}\cdot\frac{D \mathbf{v}}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}

Now, let's replace the substantial derivative <math>D\mathbf{v}/Dt</math> using the momentum equation on non-conservation form {{EquationNote|label=eq-mom-pde-non-cons}}.

{{NumEqn|<math>
\rho\frac{De}{Dt} -\mathbf{v}\cdot\nabla p + \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \nabla\cdot(p\mathbf{v}) = \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \dot{q}\rho
</math>}}

Now, expand the term <math>\nabla\cdot(p\mathbf{v})</math> gives

{{NumEqn|<math>
\rho\frac{De}{Dt} \cancel{-\mathbf{v}\cdot\nabla p} + \cancel{\mathbf{v}\cdot\nabla p} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\Rightarrow</math><br><br><math>\Rightarrow\rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho
</math>}}

Divide by <math>\rho</math>

{{NumEqn|<math>
\frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q}
</math>|label=eq-energy-pde-non-cons-b}}

Conservation of mass gives

{{NumEqn|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0\Rightarrow \nabla\cdot\mathbf{v} = -\frac{1}{\rho}\frac{D\rho}{Dt}
</math>}}

Insert in {{EquationNote|label=eq-energy-pde-non-cons-b|nopar=1}}

{{NumEqn|<math>
\frac{De}{Dt} - \frac{p}{\rho^2}\frac{D\rho}{Dt} = \dot{q}\Rightarrow \frac{De}{Dt} + p\frac{D}{Dt} \left(\frac{1}{\rho}\right)= \dot{q}
</math>}}

{{NumEqn|<math>
\frac{De}{Dt} + p\frac{D\nu}{Dt} = \dot{q}
</math>}}

Compare with the first law of thermodynamics: <math>de=\delta q-\delta w</math>

==== Enthalpy Formulation ====

{{NumEqn|<math>
h=e+\frac{p}{\rho}\Rightarrow \frac{Dh}{Dt}=\frac{De}{Dt}+\frac{1}{\rho}\frac{Dp}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho}\right)
</math>}}

with <math>De/Dt</math> from {{EquationNote|label=eq-energy-pde-non-cons-b|nopar=1}}

{{NumEqn|<math>
\frac{Dh}{Dt}=\dot{q} - \cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)} +\frac{1}{\rho}\frac{Dp}{Dt}+\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)}
</math>}}

{{NumEqn|<math>
\frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}
</math>|label=eq-energy-pde-non-cons-c}}

==== Total Enthalpy Formulation ====

{{NumEqn|<math>
h_o=h+\frac{1}{2}\mathbf{v}\mathbf{v}\Rightarrow\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\frac{D\mathbf{v}}{Dt}
</math>}}

From the momentum equation {{EquationNote|label=eq-mom-pde-non-cons}}

{{NumEqn|<math>
\frac{D\mathbf{v}}{Dt}=\mathbf{f}-\frac{1}{\rho}\nabla p
</math>}}

which gives

{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p
</math>}}

Inserting <math>Dh/Dt</math> from {{EquationNote|label=eq-energy-pde-non-cons-c|nopar=1}} gives

{{NumEqn|<math>
\frac{Dh_o}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p =</math><br><br><math>=\frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}
</math>}}

The substantial derivative operator applied to pressure

{{NumEqn|<math>
\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+\mathbf{v}\cdot\nabla p
</math>}}

and thus

{{NumEqn|<math>
\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p=\frac{\partial p}{\partial t}
</math>}}

which gives

{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t} + \dot{q} + \mathbf{v}\cdot\mathbf{f}
</math>}}

If we assume adiabatic flow without body forces

{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}
</math>}}

If we further assume the flow to be steady state we get

{{NumEqn|<math>
\frac{Dh_o}{Dt}=0
</math>}}

This means that in a steady-state adiabatic flow without body forces, total enthalpy is constant along a streamline.

Template:OpenInfoBox

2026-04-01T18:20:56Z

Nian: Created page with "{{InfoBox|1={{{1|}}}|2={{{2|}}}|3={{{3|}}}|label={{{label|}}}|description={{{description|}}}|noborder=1}}"

{{InfoBox|1={{{1|}}}|2={{{2|}}}|3={{{3|}}}|label={{{label|}}}|description={{{description|}}}|noborder=1}}

Template:InfoBox

2026-04-01T18:20:18Z

Nian:

{{NumEqn|1={{{1|}}}|2={{{2|}}}|3={{{3|}}}|label={{{label|}}}|nonumber=1|numw=2em|infobox=1|description={{{description|}}}|noborder={{{noborder|}}}}}

Template:InfoBox

2026-04-01T17:41:01Z

Nian: Created page with "{{NumEqn|1={{{1|}}}|2={{{2|}}}|3={{{3|}}}|label={{{label|}}}|nonumber=1|numw=2em|infobox=1|description={{{description|}}}}}"

Template:NumEqn

2026-04-01T16:01:05Z

Nian:

<includeonly>{{#if:{{{nonumber|}}}||{{#if:{{{number|}}}|{{#vardefine:leqno|{{{number}}}}}|{{#if:{{{3|}}}|{{#vardefine:leqno|{{{3}}}}}|{{#if:{{#var:eqno}}|{{#vardefine:eqno|{{#expr:{{#var:eqno}}+1}}}}|{{#vardefine:eqno|1}}}}{{#if:{{#var:secno}}|{{#vardefine:leqno|{{#var:secno}}.{{#var:eqno}}}}|{{#vardefine:leqno|{{#var:eqno}}}}}}}}}}{{#if:{{{noprefix|}}}||{{#if:{{{prefix|}}}|{{#vardefine:leqno|{{{prefix}}} {{#var:leqno}}}}|{{#if:{{#var:eqprefix}}|{{#vardefine:leqno|{{#var:eqprefix}} {{#var:leqno}}}}|{{#vardefine:leqno|Eq. {{#var:leqno}}}}}}}}}}}}<table style="width: 100%;{{#if:{{{noborder|}}}||{{#if:{{{border|}}}|border: 1px solid {{#if:{{{border-color|}}}|{{{border-color}}}|}};|{{#if:{{{infobox|}}}|border: 1px solid {{#if:{{{border-color|}}}|{{{border-color}}}|}};}}}}}}border-collapse: collapse;"><tr {{#if:{{{background-color|}}}|style="background-color: {{{background-color}}};"|{{#if:{{{infobox|}}}|style="background-color: whitesmoke;"}}}}><td style="width: {{#if:{{{numw|}}}|{{{numw}}}|5em}}; padding-left: 0.5em; {{#if:{{{inner-border|}}}|border: 1px solid {{#if:{{{border-color|}}}|{{{border-color}}}|}};}}"><span {{#if:{{{color|}}}|style="color: {{{color}}};"|{{#if:{{{infobox|}}}|style="color: steelblue;"}}}}>{{#if:{{{description|}}}|{{{description}}}}}</span></td><td style="padding: {{#if:{{{padding|}}}|{{{padding}}};|{{#if:{{{infobox|}}}|2.0em;|0.5em;}}}} text-align: {{#if:{{{align|}}}|{{{align}}}|center}};{{#if:{{{inner-border|}}}|border: 1px solid {{#if:{{{border-color|}}}|{{{border-color}}}|}};"}}><span {{#if:{{{color|}}}|style="color: {{{color}}};"|{{#if:{{{infobox|}}}|style="color: steelblue;"}}}}>{{{1}}}</span></td><td style="width: {{#if:{{{numw|}}}|{{{numw}}}|5em}}; padding-right: 0.5em; text-align: right; vertical-align: middle; {{#if:{{{inner-border|}}}|border: 1px solid {{#if:{{{border-color|}}}|{{{border-color}}}|}};}}"><span {{#if:{{{color|}}}|style="color: {{{color}}};"|{{#if:{{{infobox|}}}|style="color: steelblue;"}}}}>{{#if:{{{nonumber|}}}||{{#if:{{{label|}}}|{{#if:{{{number|}}}|{{EquationRef|{{{label}}}|{{#var:leqno}}|nopar={{{nopar|}}}|spar={{{spar|}}}|wpar={{{wpar|}}}}}|{{EquationRef|{{{label}}}|{{#var:leqno}}|nopar={{{nopar|}}}|spar={{{spar|}}}|wpar={{{wpar|}}}}}}}|{{#if:{{{number|}}}|{{EquationRef|{{#var:leqno}}|nopar={{{nopar|}}}|spar={{{spar|}}}|wpar={{{wpar|}}}}}|{{#if:{{{3|}}}|{{EquationRef|{{{2}}}|{{#var:leqno}}|nopar={{{nopar|}}}|spar={{{spar|}}}|wpar={{{wpar|}}}}}|{{#if:{{{2|}}}|{{EquationRef|{{{2}}}|{{#var:leqno}}|nopar={{{nopar|}}}|spar={{{spar|}}}|wpar={{{wpar|}}}}}|{{EquationRef|{{#var:leqno}}|nopar={{{nopar|}}}|spar={{{spar|}}}|wpar={{{wpar|}}}}}}}}}}}}}}}</span></td></tr></table></includeonly><noinclude>

</noinclude>

Governing equations on differential form

2026-04-01T15:31:45Z

Nian:

[[Category:Compressible flow]][[Category:Governing equations]]<noinclude>[[Category:Compressible flow:Topic]]</noinclude><nomobile>__TOC__</nomobile><noinclude>{{#vardefine:secno|2}}{{#vardefine:eqno|28}}</noinclude>
=== The Differential Equations on Conservation Form ===

==== Conservation of Mass ====

The continuity equation on integral form reads

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
</math>|nonumber=1|infobox=1|numw=2em}}

Apply Gauss's divergence theorem on the surface integral gives

{{NumEqn|<math>
\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV
</math>}}

Also, if <math>\Omega</math> is a fixed control volume

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV=\iiint_{\Omega} \frac{\partial \rho}{\partial t} dV
</math>}}

The continuity equation can now be written as a single volume integral.

{{NumEqn|<math>
\iiint_{\Omega} \left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})\right]dV=0
</math>}}

<math>\Omega</math> is an arbitrary control volume and thus

{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>|label=eq-cont-pde}}

which is the continuity equation on partial differential form.

==== Conservation of Momentum ====

The momentum equation on integral form reads

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
</math>|nonumber=1|infobox=1|numw=2em}}

As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.

{{NumEqn|<math>
\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega} \nabla\cdot(\rho \mathbf{v}\mathbf{v})dV
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega} p\mathbf{n}dS=\iiint_{\Omega} \nabla pdV
</math>}}

Also, if <math>\Omega</math> is a fixed control volume

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV=\iiint_{\Omega} \frac{\partial}{\partial t}(\rho \mathbf{v}) dV
</math>}}

The momentum equation can now be written as one single volume integral

{{NumEqn|<math>
\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p - \rho \mathbf{f}\right]dV=0
</math>}}

<math>\Omega</math> is an arbitrary control volume and thus

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>|label=eq-mom-pde}}

which is the momentum equation on partial differential form

==== Conservation of Energy ====

The energy equation on integral form reads

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>|nonumber=1|infobox=1|numw=2em}}

Gauss's divergence theorem applied to the surface integral term in the energy equation gives

{{NumEqn|<math>
\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\nabla\cdot(\rho h_o\mathbf{v})dV
</math>}}

Fixed control volume

{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV=\iiint_{\Omega}\frac{\partial}{\partial t}(\rho e_o) dV
</math>}}

The energy equation can now be written as

{{NumEqn|<math>
\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) - \rho\mathbf{f}\cdot\mathbf{v} - \dot{q}\rho \right]dV=0
</math>}}

<math>\Omega</math> is an arbitrary control volume and thus

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|label=eq-energy-pde}}

which is the energy equation on partial differential form

==== Summary ====

The governing equations for compressible inviscid flow on partial differential form:

<div style="border: solid 1px;">
{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>|infobox=1|description=Continuity:|nonumber=1|noborder=1}}

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>|infobox=1|description=Momentum:|nonumber=1|noborder=1}}

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|infobox=1|description=Energy:|nonumber=1|noborder=1}}
</div>

=== The Differential Equations on Non-Conservation Form ===

==== The Substantial Derivative ====

The substantial derivative operator is defined as

{{NumEqn|<math>
\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla
</math>|label=eq-cont-pde-non-cons}}

where the first term of the right hand side is the local derivative and the second term is the convective derivative.

==== Conservation of Mass ====

If we apply the substantial derivative operator to density we get

{{NumEqn|<math>
\frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho
</math>}}

From before we have the continuity equation on differential form as

{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>}}

which can be rewritten as

{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \rho(\nabla\cdot\mathbf{v}) + \mathbf{v}\cdot\nabla\rho=0
</math>}}

and thus

{{NumEqn|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
</math>|label=eq-pde-noncons-cont}}

{{EquationNote|label=eq-pde-noncons-cont|nopar=1}} says that the mass of a fluid element with a fixed set of fluid particles is constant as the element moves in space.

==== Conservation of Momentum ====

We start from the momentum equation on differential form derived above

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>}}

Expanding the first and the second terms gives

{{NumEqn|<math>
\rho\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v}\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla\mathbf{v} + \mathbf{v}(\nabla\cdot\rho\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>}}

Collecting terms, we can identify the substantial derivative operator applied to the velocity vector and the continuity equation.

{{NumEqn|<math>
\rho\underbrace{\left[\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right]}_{=\frac{D\mathbf{v}}{Dt}}+\mathbf{v}\underbrace{\left[\frac{\partial \rho}{\partial t}+\nabla\cdot\rho\mathbf{v}\right]}_{=0}+ \nabla p = \rho \mathbf{f}
</math>}}

which gives us the non-conservation form of the momentum equation

{{NumEqn|<math>
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
</math>|label=eq-mom-pde-non-cons}}

==== Conservation of Energy ====

The last equation on non-conservation differential form is the energy equation. We start by rewriting the energy equation on differential form {{EquationNote|label=eq-energy-pde}}, repeated here for convenience

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|nonumber=1}}

Total enthalpy, <math>h_o</math>, is replaced with total energy, <math>e_o</math>

{{NumEqn|<math>
h_o=e_o+\frac{p}{\rho}
</math>}}

which gives

{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho e_o\mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}

Expanding the two first terms as

{{NumEqn|<math>
\rho\frac{\partial e_o}{\partial t} + e_o\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla e_o + e_o\nabla\cdot(\rho \mathbf{v}) + \nabla\cdot(p\mathbf{v})=</math><br><br><math>= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}

Collecting terms, we can identify the substantial derivative operator applied on total energy, <math>De_o/Dt</math> and the continuity equation

{{NumEqn|<math>
\rho\underbrace{\left[ \frac{\partial e_o}{\partial t} + \mathbf{v}\cdot\nabla e_o \right]}_{=\frac{De_o}{Dt}} + e_o\underbrace{\left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho \mathbf{v}) \right]}_{=0} + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}

and thus we end up with the energy equation on non-conservation differential form

{{NumEqn|<math>
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|label=eq-energy-pde-non-cons}}

==== Summary ====

<div style="border: solid 1px;">
{{NumEqn|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
</math>|infobox=1|description=Continuity:|nonumber=1|noborder=1}}

{{NumEqn|<math>
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
</math>|infobox=1|description=Momentum:|nonumber=1|noborder=1}}

{{NumEqn|<math>
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|infobox=1|description=Energy:|nonumber=1|noborder=1}}
</div>

=== Alternative Forms of the Energy Equation ===

==== Internal Energy Formulation ====

Total internal energy is defined as

{{NumEqn|<math>
e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}
</math>}}

Inserted in {{EquationNote|label=eq-energy-pde-non-cons|nopar=1}}, this gives

{{NumEqn|<math>
\rho\frac{De}{Dt} + \rho\mathbf{v}\cdot\frac{D \mathbf{v}}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}

Now, let's replace the substantial derivative <math>D\mathbf{v}/Dt</math> using the momentum equation on non-conservation form {{EquationNote|label=eq-mom-pde-non-cons}}.

{{NumEqn|<math>
\rho\frac{De}{Dt} -\mathbf{v}\cdot\nabla p + \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \nabla\cdot(p\mathbf{v}) = \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \dot{q}\rho
</math>}}

Now, expand the term <math>\nabla\cdot(p\mathbf{v})</math> gives

{{NumEqn|<math>
\rho\frac{De}{Dt} \cancel{-\mathbf{v}\cdot\nabla p} + \cancel{\mathbf{v}\cdot\nabla p} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\Rightarrow</math><br><br><math>\Rightarrow\rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho
</math>}}

Divide by <math>\rho</math>

{{NumEqn|<math>
\frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q}
</math>|label=eq-energy-pde-non-cons-b}}

Conservation of mass gives

{{NumEqn|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0\Rightarrow \nabla\cdot\mathbf{v} = -\frac{1}{\rho}\frac{D\rho}{Dt}
</math>}}

Insert in {{EquationNote|label=eq-energy-pde-non-cons-b|nopar=1}}

{{NumEqn|<math>
\frac{De}{Dt} - \frac{p}{\rho^2}\frac{D\rho}{Dt} = \dot{q}\Rightarrow \frac{De}{Dt} + p\frac{D}{Dt} \left(\frac{1}{\rho}\right)= \dot{q}
</math>}}

{{NumEqn|<math>
\frac{De}{Dt} + p\frac{D\nu}{Dt} = \dot{q}
</math>}}

Compare with the first law of thermodynamics: <math>de=\delta q-\delta w</math>

==== Enthalpy Formulation ====

{{NumEqn|<math>
h=e+\frac{p}{\rho}\Rightarrow \frac{Dh}{Dt}=\frac{De}{Dt}+\frac{1}{\rho}\frac{Dp}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho}\right)
</math>}}

with <math>De/Dt</math> from {{EquationNote|label=eq-energy-pde-non-cons-b|nopar=1}}

{{NumEqn|<math>
\frac{Dh}{Dt}=\dot{q} - \cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)} +\frac{1}{\rho}\frac{Dp}{Dt}+\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)}
</math>}}

{{NumEqn|<math>
\frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}
</math>|label=eq-energy-pde-non-cons-c}}

==== Total Enthalpy Formulation ====

{{NumEqn|<math>
h_o=h+\frac{1}{2}\mathbf{v}\mathbf{v}\Rightarrow\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\frac{D\mathbf{v}}{Dt}
</math>}}

From the momentum equation {{EquationNote|label=eq-mom-pde-non-cons}}

{{NumEqn|<math>
\frac{D\mathbf{v}}{Dt}=\mathbf{f}-\frac{1}{\rho}\nabla p
</math>}}

which gives

{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p
</math>}}

Inserting <math>Dh/Dt</math> from {{EquationNote|label=eq-energy-pde-non-cons-c|nopar=1}} gives

{{NumEqn|<math>
\frac{Dh_o}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p =</math><br><br><math>=\frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}
</math>}}

The substantial derivative operator applied to pressure

{{NumEqn|<math>
\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+\mathbf{v}\cdot\nabla p
</math>}}

and thus

{{NumEqn|<math>
\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p=\frac{\partial p}{\partial t}
</math>}}

which gives

{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t} + \dot{q} + \mathbf{v}\cdot\mathbf{f}
</math>}}

If we assume adiabatic flow without body forces

{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}
</math>}}

If we further assume the flow to be steady state we get

{{NumEqn|<math>
\frac{Dh_o}{Dt}=0
</math>}}

This means that in a steady-state adiabatic flow without body forces, total enthalpy is constant along a streamline.

The Q1D equations

2026-04-01T15:22:04Z

Nian:

[[Category:Compressible flow]][[Category:Quasi-one-dimensional flow]][[Category:Inviscid flow]]<noinclude>[[Category:Compressible flow:Topic]]</noinclude><nomobile></nomobile><noinclude>{{#vardefine:secno|5}}{{#vardefine:eqno|0}}</noinclude>
=== Governing Equations ===



In the following quasi-one-dimensional flow will be assumed. That means that the cross-section is allowed to vary smoothly but flow quantities varies in one direction only. The equations that are derived will thus describe one-dimensional flow in axisymmetric tubes. Let's assume flow in the <math>x</math>-direction, which means that all flow quantities and the cross-section area will vary with the axial coordinate <math>x</math>.

{{NumEqn|<math>
A=A(x),\ \rho=\rho(x),\ u=u(x),\ p=p(x),\ ...
</math>}}

We will further assume steady-state flow, which means that unsteady terms will be zero.

The equations are derived with the starting point in the governing flow equations on integral form

==== Continuity Equation ====

Applying the integral form of the continuity equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives

{{NumEqn|<math>
\underbrace{\frac{d}{dt}\iiint_{\Omega}\rho d{V}}_{=0}+\iint_{\partial \Omega}\rho {\mathbf{v}}\cdot {\mathbf{n}}dS=0
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega}\rho {\mathbf{v}}\cdot {\mathbf{n}}dS=-\rho_1 u_1 A_1+\rho_2 u_2 A_2
</math>}}

{{NumEqn|<math>
\rho_1 u_1 A_1=\rho_2 u_2 A_2
</math>}}

==== Momentum Equation ====

Applying the integral form of the momentum equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives

{{NumEqn|<math>
\underbrace{\frac{d}{dt}\iiint_{\Omega}\rho{\mathbf{v}} d{V}}_{=0}+\iint_{\partial \Omega}\left[\rho ({\mathbf{v}}\cdot {\mathbf{n}}){\mathbf{v}}+p{\mathbf{n}}\right]dS=0
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega} \rho ({\mathbf{v}}\cdot {\mathbf{n}}){\mathbf{v}}dS=-\rho_1u_1^2A_1+\rho_2u_2^2A_2
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega} p{\mathbf{n}}dS=-p_1A_1+p_2A_2-\int_{A_1}^{A_2}pdA
</math>}}

collecting terms

{{NumEqn|<math>
\left(\rho_1u_1^2+p_1\right)A_1+\int_{A_1}^{A_2}pdA=\left(\rho_2u_2^2+p_2\right)A_2
</math>}}

==== Energy Equation ====

Applying the integral form of the energy equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives

{{NumEqn|<math>
\underbrace{\frac{d}{dt}\iiint_{\Omega}\rho e_o d{V}}_{=0}+\iint_{\partial \Omega}\left[\rho h_o ({\mathbf{v}}\cdot{\mathbf{n}})\right]dS=0
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega}\left[\rho h_o ({\mathbf{v}}\cdot{\mathbf{n}})\right]dS=-\rho_1u_1h_{o_1}A_1+\rho_2u_2h_{o_2}A_2
</math>}}

{{NumEqn|<math>
\rho_1u_1h_{o_1}A_1=\rho_2u_2h_{o_2}A_2
</math>}}

Now, using the continuity equation <math>\rho_1u_1A_1=\rho_2u_2A_2</math> gives

{{NumEqn|<math>
h_{o_1}=h_{o_2}
</math>}}

==== Differential Form ====

The integral term appearing the momentum equation is undesired and therefore the governing equations are converted to differential form.

The continuity equation (Eqn. \ref{eq:governing:cont:b}) is rewritten in differential form as

{{NumEqn|<math>
\rho_1u_1A_1=\rho_2u_2A_2=const
</math>}}

{{NumEqn|<math>
d(\rho uA)=0
</math>}}

The momentum equation (Eqn. \ref{eq:governing:mom:b}) is rewritten in differential form as

{{NumEqn|<math>
\left(\rho_1u_1^2+p_1\right)A_1+\int_{A_1}^{A_2}pdA=\left(\rho_2u_2^2+p_2\right)A_2\Rightarrow d\left[(\rho u^2+p)A\right]=pdA
</math>}}

{{NumEqn|<math>
d(\rho u^2A)+d(pA)=pdA
</math>}}

{{NumEqn|<math>
ud(\rho uA)+\rho uAdu+Adp+\cancel{pdA}=\cancel{pdA}
</math>}}

From the continuity equation we have <math>d(\rho uA)</math> and thus

{{NumEqn|<math>
\rho u\cancel{A}du+\cancel{A}dp=0\Rightarrow
</math>}}

{{NumEqn|<math>
dp=-\rho udu
</math>}}

which is the momentum equation on differential form. Also referred to as Euler's equation. Finally, the energy equation (Eqn. \ref{eq:governing:cont:b}) is rewritten in differential form as

{{NumEqn|<math>
h_{o_1}=h_{o_2}=const\Rightarrow dh_o=0
</math>}}

{{NumEqn|<math>
h_o=h+\frac{1}{2}u^2\Rightarrow dh+\frac{1}{2}d(u^2)=0
</math>}}

{{NumEqn|<math>
dh+udu=0
</math>}}

==== Summary ====

<div style="border: 1px solid;">
{{NumEqn|<math>
d(\rho uA)=0
</math>|nonumber=1|infobox=1|description=Continuity:|noborder=1}}

{{NumEqn|<math>
dp=-\rho udu
</math>|nonumber=1|infobox=1|description=Momentum:|noborder=1}}

{{NumEqn|<math>
dh+udu=0
</math>|nonumber=1|infobox=1|description=Energy:|noborder=1}}
</div>

The equations are valid for:

* quasi-one-dimensional flow
* steady state
* all gas models (no gas model assumptions made)
* inviscid flow

It should be noted that equations are exact but they are applied to a physical model that is approximate, i.e., the approximation that flow quantities varies in one dimension with a varying cross-section area. In reality, a variation of cross-section area would imply flow in three dimensions.

2026-04-01T13:39:32Z

Nian:

Choked flow

2026-04-01T13:39:13Z

Nian:

The Q1D equations

2026-04-01T13:38:57Z

Nian:

[[Category:Compressible flow]][[Category:Quasi-one-dimensional flow]][[Category:Inviscid flow]]<noinclude>[[Category:Compressible flow:Topic]]</noinclude><nomobile>__TOC__</nomobile><noinclude>{{#vardefine:secno|5}}{{#vardefine:eqno|0}}</noinclude>
=== Governing Equations ===



In the following quasi-one-dimensional flow will be assumed. That means that the cross-section is allowed to vary smoothly but flow quantities varies in one direction only. The equations that are derived will thus describe one-dimensional flow in axisymmetric tubes. Let's assume flow in the <math>x</math>-direction, which means that all flow quantities and the cross-section area will vary with the axial coordinate <math>x</math>.

{{NumEqn|<math>
A=A(x),\ \rho=\rho(x),\ u=u(x),\ p=p(x),\ ...
</math>}}

We will further assume steady-state flow, which means that unsteady terms will be zero.

The equations are derived with the starting point in the governing flow equations on integral form

==== Continuity Equation ====

Applying the integral form of the continuity equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives

{{NumEqn|<math>
\underbrace{\frac{d}{dt}\iiint_{\Omega}\rho d{V}}_{=0}+\iint_{\partial \Omega}\rho {\mathbf{v}}\cdot {\mathbf{n}}dS=0
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega}\rho {\mathbf{v}}\cdot {\mathbf{n}}dS=-\rho_1 u_1 A_1+\rho_2 u_2 A_2
</math>}}

{{NumEqn|<math>
\rho_1 u_1 A_1=\rho_2 u_2 A_2
</math>}}

==== Momentum Equation ====

Applying the integral form of the momentum equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives

{{NumEqn|<math>
\underbrace{\frac{d}{dt}\iiint_{\Omega}\rho{\mathbf{v}} d{V}}_{=0}+\iint_{\partial \Omega}\left[\rho ({\mathbf{v}}\cdot {\mathbf{n}}){\mathbf{v}}+p{\mathbf{n}}\right]dS=0
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega} \rho ({\mathbf{v}}\cdot {\mathbf{n}}){\mathbf{v}}dS=-\rho_1u_1^2A_1+\rho_2u_2^2A_2
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega} p{\mathbf{n}}dS=-p_1A_1+p_2A_2-\int_{A_1}^{A_2}pdA
</math>}}

collecting terms

{{NumEqn|<math>
\left(\rho_1u_1^2+p_1\right)A_1+\int_{A_1}^{A_2}pdA=\left(\rho_2u_2^2+p_2\right)A_2
</math>}}

==== Energy Equation ====

Applying the integral form of the energy equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives

{{NumEqn|<math>
\underbrace{\frac{d}{dt}\iiint_{\Omega}\rho e_o d{V}}_{=0}+\iint_{\partial \Omega}\left[\rho h_o ({\mathbf{v}}\cdot{\mathbf{n}})\right]dS=0
</math>}}

{{NumEqn|<math>
\iint_{\partial \Omega}\left[\rho h_o ({\mathbf{v}}\cdot{\mathbf{n}})\right]dS=-\rho_1u_1h_{o_1}A_1+\rho_2u_2h_{o_2}A_2
</math>}}

{{NumEqn|<math>
\rho_1u_1h_{o_1}A_1=\rho_2u_2h_{o_2}A_2
</math>}}

Now, using the continuity equation <math>\rho_1u_1A_1=\rho_2u_2A_2</math> gives

{{NumEqn|<math>
h_{o_1}=h_{o_2}
</math>}}

==== Differential Form ====

The integral term appearing the momentum equation is undesired and therefore the governing equations are converted to differential form.

The continuity equation (Eqn. \ref{eq:governing:cont:b}) is rewritten in differential form as

{{NumEqn|<math>
\rho_1u_1A_1=\rho_2u_2A_2=const
</math>}}

{{NumEqn|<math>
d(\rho uA)=0
</math>}}

The momentum equation (Eqn. \ref{eq:governing:mom:b}) is rewritten in differential form as

{{NumEqn|<math>
\left(\rho_1u_1^2+p_1\right)A_1+\int_{A_1}^{A_2}pdA=\left(\rho_2u_2^2+p_2\right)A_2\Rightarrow d\left[(\rho u^2+p)A\right]=pdA
</math>}}

{{NumEqn|<math>
d(\rho u^2A)+d(pA)=pdA
</math>}}

{{NumEqn|<math>
ud(\rho uA)+\rho uAdu+Adp+\cancel{pdA}=\cancel{pdA}
</math>}}

From the continuity equation we have <math>d(\rho uA)</math> and thus

{{NumEqn|<math>
\rho u\cancel{A}du+\cancel{A}dp=0\Rightarrow
</math>}}

{{NumEqn|<math>
dp=-\rho udu
</math>}}

which is the momentum equation on differential form. Also referred to as Euler's equation. Finally, the energy equation (Eqn. \ref{eq:governing:cont:b}) is rewritten in differential form as

{{NumEqn|<math>
h_{o_1}=h_{o_2}=const\Rightarrow dh_o=0
</math>}}

{{NumEqn|<math>
h_o=h+\frac{1}{2}u^2\Rightarrow dh+\frac{1}{2}d(u^2)=0
</math>}}

{{NumEqn|<math>
dh+udu=0
</math>}}

==== Summary ====

<div style="border: 1px solid;">
{{NumEqn|<math>
d(\rho uA)=0
</math>|nonumber=1|infobox=1|description=Continuity:|noborder=1}}

{{NumEqn|<math>
dp=-\rho udu
</math>|nonumber=1|infobox=1|description=Momentum:|noborder=1}}

{{NumEqn|<math>
dh+udu=0
</math>|nonumber=1|infobox=1|description=Energy:|noborder=1}}
</div>

The equations are valid for:

* quasi-one-dimensional flow
* steady state
* all gas models (no gas model assumptions made)
* inviscid flow

It should be noted that equations are exact but they are applied to a physical model that is approximate, i.e., the approximation that flow quantities varies in one dimension with a varying cross-section area. In reality, a variation of cross-section area would imply flow in three dimensions.

Diffusers

2026-04-01T13:38:31Z

Nian:

Area-velocity relation

2026-04-01T13:38:21Z

Nian:

Area-Mach relation

2026-04-01T13:37:54Z

Nian:

[[Category:Compressible flow]][[Category:Quasi-one-dimensional flow]][[Category:Inviscid flow]]<noinclude>[[Category:Compressible flow:Topic]]</noinclude><nomobile>__TOC__</nomobile><noinclude>{{#vardefine:secno|5}}{{#vardefine:eqno|31}}</noinclude>
=== The Area-Mach-Number Relation ===

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

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

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

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

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

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

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

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

and thus

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

Using the isentropic relations, we get

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

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

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

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

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

{{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}
</math>}}

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

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

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

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

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

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

{{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)}
</math>}}

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

{{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
</math>}}

{{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
</math>}}

{{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)}
</math>}}

which is the area-Mach-number relation.

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>.

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



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 <math>A^*</math> will change over the shock.