Compressible flow

Thermodynamics

Specific heat

For thermally perfect and calorically perfect gases

$\begin{array}{cc}& C_p=\frac{dh}{dT}\\ & C_v=\frac{de}{dT}\end{array}$

(Eq. 1)

From the definition of enthalpy and the equation of state $p=\rho RT$

$h=e+\frac{p}{\rho }=e+RT$

(Eq. 2)

Differentiate (Eq. 2) with respect to temperature gives

$\frac{dh}{dT}=\frac{de}{dT}+\frac{d(RT)}{dT}$

(Eq. 3)

Inserting the specific heats gives

$C_p=C_v+R$

(Eq. 4)

Dividing (Eq. 4) by $C_v$ gives

$\frac{C_p}{C_v}=1+\frac{R}{C_v}$

(Eq. 5)

Introducing the ratio of specific heats defined as

$\gamma =\frac{C_p}{C_v}$

(Eq. 6)

Now, inserting (Eq. 6) in Eqn. \ref{eq:specificheat:c} gives

$C_v=\frac{R}{\gamma -1}$

(Eq. 7)

In the same way, dividing (Eq. 4) with $C_p$ gives

$1=\frac{C_v}{C_p}+\frac{R}{C_p}=\frac{1}{\gamma }+\frac{R}{C_p}$

(Eq. 8)

and thus

$C_p=\frac{\gamma R}{\gamma -1}$

(Eq. 9)

Isentropic relations

First law of thermodynamics

First law of thermodynamics:

$de=\delta q-\delta w$

(Eq. 10)

For a reversible process: $\delta w=pd(1/\rho )$ and $\delta q=Tds$

$de=Tds-pd\left(\frac{1}{\rho }\right)$

(Eq. 11)

Enthalpy is defined as: $h=e+p/\rho$ and thus

$dh=de+pd\left(\frac{1}{\rho }\right)+\left(\frac{1}{\rho }\right)dp$

(Eq. 12)

Eliminate $de$ in (Eq. 11) using (Eq. 12)

$Tds=dh-\cancel{pd\left(\frac{1}{\rho }\right)}-\left(\frac{1}{\rho }\right)dp+\cancel{pd\left(\frac{1}{\rho }\right)}$

(Eq. 13)

$ds=\frac{dh}{T}-\frac{dp}{\rho T}$

(Eq. 14)

Using $dh=C_{p}T$ and the equation of state $p=\rho RT$, we get

$ds=C_p\frac{dT}{T}-R\frac{dp}{p}$

(Eq. 15)

Integrating (Eq. 15) gives

$s_2-s_1=\underset{1}{\overset{2}{\int }}{C}_p\frac{dT}{T}-R\ln \left(\frac{p_2}{p_1}\right)$

(Eq. 16)

For a calorically perfect gas, $C_p$ is constant (not a function of temperature) and can be moved out from the integral and thus

$s_2-s_1=C_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{p_2}{p_1}\right)$

(Eq. 17)

An alternative form of (Eq. 17) is obtained by using $de=C_{v}dT$ in (Eq. 11), which gives

$s_2-s_1=\underset{1}{\overset{2}{\int }}{C}_v\frac{dT}{T}-R\ln \left(\frac{{\rho }_2}{{\rho }_1}\right)$

(Eq. 18)

Again, for a calorically perfect gas, we get

$s_2-s_1=C_v\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{{\rho }_2}{{\rho }_1}\right)$

(Eq. 19)

Isentropic Relations

Adiabatic and reversible processes, i.e., isentropic processes implies $ds=0$ and thus (Eq. 17) reduces to

$\frac{C_p}{R}\ln \left(\frac{T_2}{T_1}\right)=\ln \left(\frac{p_2}{p_1}\right)$

(Eq. 20)

$$\frac{C_p}{R}=\frac{\gamma }{\gamma -1}$$

$$\frac{\gamma }{\gamma -1}\ln \left(\frac{T_2}{T_1}\right)=\ln \left(\frac{p_2}{p_1}\right)\Rightarrow$$

$\frac{p_2}{p_1}={\left(\frac{T_2}{T_1}\right)}^{\gamma /(\gamma -1)}$

(Eq. 21)

In the same way, (Eq. 19) gives

$\frac{{\rho }_2}{{\rho }_1}={\left(\frac{T_2}{T_1}\right)}^{1/(\gamma -1)}$

(Eq. 22)

Eqn. (Eq. 21) and Eqn. (Eq. 22) constitutes the isentropic relations

$\frac{p_2}{p_1}={\left(\frac{{\rho }_2}{{\rho }_1}\right)}^{\gamma }={\left(\frac{T_2}{T_1}\right)}^{\gamma /(\gamma -1)}$

(Eq. 23)

Thermodynamic processes

$ds=C_v\frac{dT}{T}+R\frac{d\nu }{\nu }$

(Eq. 24)

$d\nu =\frac{\nu }{R}ds-C_v\frac{\nu }{RT}dT=\frac{\nu }{R}ds-\frac{C_v}{p}dT$

(Eq. 25)

for an isentropic process ($ds=0$), $d\nu <0$ for positive values of $dT$.

$ds=C_p\frac{dT}{T}-R\frac{dp}{p}$

(Eq. 26)

$dp=-\frac{p}{R}ds+C_p\frac{p}{RT}dT=-\frac{p}{R}ds+C_p\rho dT$

(Eq. 27)

for an isentropic process ($ds=0$), $dp>0$ for positive values of $dT$.

Since $\nu$ decreases with temperature and pressure increases with temperature for an isentropic process, we can see from (Eq. 25) that $d\nu$ will be greater at lower temperatures and thus isochores (lines of constant specific volume) will be closely spaced at low temperatures and more sparse at higher temperatures. For an isochore $dv=0$ which implies

$0=\frac{\nu }{R}(ds-C_v\frac{dT}{T})\Rightarrow \frac{dT}{ds}=\frac{T}{C_v}$

(Eq. 28)

and thus we can see that the slope of an isochore in a $T-s$-diagram is positive and that the slope increases with temperature.

In analogy, we can see that an isobar ($dp=0$) leads to the following relation

$0=\frac{p}{R}(C_p\frac{dT}{T}-ds)\Rightarrow \frac{dT}{ds}=\frac{T}{C_p}$

(Eq. 29)

and consequently isobars will also have a positive slope that increases with temperature in a $T-s$-diagram. Moreover, isobars are less steep than ischores as $C_p>C_v$.

Governing equations

Governing equations on integral form

The governing equations stems from mass conservation, conservation of momentum and conservation of energy

The Continuity Equation

"Mass can be neither created nor destroyed, which implies that mass is conserved"

The net massflow into the control volume $\Omega$ in Fig. \ref{fig:generic:cv} is obtained by integrating mass flux over the control volume surface $\partial \Omega$

$-\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS$

(Eq. 30)

Now, let's consider a small infinitesimal volume $dV$ inside $\Omega$. The mass of $dV$ is $\rho dV$. Thus, the mass enclosed within $\Omega$ can be calculated as

$\underset{\Omega }{\iiint }\rho dV$

(Eq. 31)

The rate of change of mass within $\Omega$ is obtained as

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV$

(Eq. 32)

Mass is conserved, which means that the rate of change of mass within $\Omega$ must equal the net flux over the control volume surface.

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV=-\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS$

(Eq. 33)

or

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV+\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=0$

(Eq. 34)

which is the integral form of the continuity equation.

The Momentum Equation

"The time rate of change of momentum of a body equals the net force exerted on it"

$\frac{d}{dt}(m𝐯)=𝐅$

(Eq. 35)

What type of forces do we have?

• Body forces acting on the fluid inside $\Omega$
  • gravitation
  • electromagnetic forces
  • Coriolis forces
• Surface forces: pressure forces and shear forces

Body forces inside $\Omega$:

$\underset{\Omega }{\iiint }\rho 𝐟dV$

(Eq. 36)

Surface force on $\partial \Omega$:

$-\underset{\partial \Omega }{\iint }p𝐧dS$

(Eq. 37)

Since we are considering inviscid flow, there are no shear forces and thus we have the net force as

$𝐅=\underset{\Omega }{\iiint }\rho 𝐟dV-\underset{\partial \Omega }{\iint }p𝐧dS$

(Eq. 38)

The fluid flowing through $\Omega$ will carry momentum and the net flow of momentum out from $\Omega$ is calculated as

$\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧dS)𝐯=\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧)𝐯dS$

(Eq. 39)

Integrated momentum inside $\Omega$

$\underset{\Omega }{\iiint }\rho 𝐯dV$

(Eq. 40)

Rate of change of momentum due to unsteady effects inside $\Omega$

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV$

(Eq. 41)

Combining the rate of change of momentum, the net momentum flux and the net forces we get

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV+\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧)𝐯dS=\underset{\Omega }{\iiint }\rho 𝐟dV-\underset{\partial \Omega }{\iint }p𝐧dS$

(Eq. 42)

combining the surface integrals, we get

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV+\underset{\partial \Omega }{\iint }[(\rho 𝐯\cdot 𝐧)𝐯+p𝐧]dS=\underset{\Omega }{\iiint }\rho 𝐟dV$

(Eq. 43)

which is the momentum equation on integral form.

The Energy Equation

"Energy can be neither created nor destroyed; it can only change in form"

$$E_1+E_2=E_3$$

$E_1$ Rate of heat added to the fluid in $\Omega$ from the surroundings

heat transfer

radiation

$E_2$ Rate of work done on the fluid in $\Omega$

$E_3$ Rate of change of energy of the fluid as it flows through $\Omega$

$E_1=\underset{\Omega }{\iiint }\dot{q}\rho dV$

(Eq. 44)

where $\dot{q}$ is the rate of heat added per unit mass

The rate of work done on the fluid in $\Omega$ due to pressure forces is obtained from the pressure force term in the momentum equation.

$E_{2_{pressure}}=-\underset{\partial \Omega }{\iint }(p𝐧dS)\cdot 𝐯=-\underset{\partial \Omega }{\iint }p𝐯\cdot 𝐧dS$

(Eq. 45)

The rate of work done on the fluid in $\Omega$ due to body forces is

$E_{2_{bodyforces}}=\underset{\Omega }{\iiint }(\rho 𝐟dV)\cdot 𝐯=\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV$

(Eq. 46)

$E_2=E_{2_{pressure}}+E_{2_{bodyforces}}=-\underset{\partial \Omega }{\iint }p𝐯\cdot 𝐧dS+\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV$

(Eq. 47)

The energy of the fluid per unit mass is the sum of internal energy $e$ (molecular energy) and the kinetic energy $V^2/2$ and the net energy flux over the control volume surface is calculated by the following integral

$\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧dS)(e+\frac{V^2}{2})$

(Eq. 48)

Analogous to mass and momentum, the total amount of energy of the fluid in $\Omega$ is calculated as

$\underset{\Omega }{\iiint }\rho (e+\frac{V^2}{2})dV$

(Eq. 49)

The time rate of change of the energy of the fluid in $\Omega$ is obtained as

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho (e+\frac{V^2}{2})dV$

(Eq. 50)

Now, $E_3$ is obtained as the sum of the time rate of change of energy of the fluid in $\Omega$ and the net flux of energy carried by fluid passing the control volume surface.

$E_3=\frac{d}{dt}\underset{\Omega }{\iiint }\rho (e+\frac{V^2}{2})dV+\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧dS)(e+\frac{V^2}{2})$

(Eq. 51)

With all elements of the energy equation defined, we are now ready to finally compile the full equation

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho (e+\frac{V^2}{2})dV+\underset{\partial \Omega }{\iint }[\rho (e+\frac{V^2}{2})(𝐯\cdot 𝐧)+p𝐯\cdot 𝐧]dS=$ $\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV+\underset{\Omega }{\iiint }\dot{q}\rho dV$

(Eq. 52)

The surface integral in the energy equation may be rewritten as

$\underset{\partial \Omega }{\iint }[\rho (e+\frac{V^2}{2})(𝐯\cdot 𝐧)+p𝐯\cdot 𝐧]dS=$ $\underset{\partial \Omega }{\iint }\rho [e+\frac{p}{\rho }+\frac{V^2}{2}](𝐯\cdot 𝐧)dS$

(Eq. 53)

and with the definition of enthalpy $h=e+p/\rho$, we get

$\underset{\partial \Omega }{\iint }\rho [h+\frac{V^2}{2}](𝐯\cdot 𝐧)dS$

(Eq. 54)

Furthermore, introducing total internal energy $e_o$ and total enthalpy $h_o$ defined as

$e_o=e+\frac{1}{2}{V}^2$

(Eq. 55)

and

$h_o=h+\frac{1}{2}{V}^2$

(Eq. 56)

the energy equation is written as

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho e_{o}dV+\underset{\partial \Omega }{\iint }\rho h_o(𝐯\cdot 𝐧)dS=$ $\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV+\underset{\Omega }{\iiint }\dot{q}\rho dV$

(Eq. 57)

Summary

The integral form of the governing equations for inviscid compressible flow has been derived

Continuity:

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV+\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=0$

Momentum:

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV+\underset{\partial \Omega }{\iint }[(\rho 𝐯\cdot 𝐧)𝐯+p𝐧]dS=\underset{\Omega }{\iiint }\rho 𝐟dV$

Energy:

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho e_{o}dV+\underset{\partial \Omega }{\iint }\rho h_o(𝐯\cdot 𝐧)dS=$ $\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV+\underset{\Omega }{\iiint }\dot{q}\rho dV$

Governing equations on differential form

The Differential Equations on Conservation Form

Conservation of Mass

The continuity equation on integral form reads

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV+\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=0$

Apply Gauss's divergence theorem on the surface integral gives

$\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=\underset{\Omega }{\iiint }\mathrm{\nabla }\cdot (\rho 𝐯)dV$

(Eq. 58)

Also, if $\Omega$ is a fixed control volume

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV=\underset{\Omega }{\iiint }\frac{\partial \rho }{\partial t}dV$

(Eq. 59)

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

$\underset{\Omega }{\iiint }[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)]dV=0$

(Eq. 60)

$\Omega$ is an arbitrary control volume and thus

$\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)=0$

(Eq. 61)

which is the continuity equation on partial differential form.

Conservation of Momentum

The momentum equation on integral form reads

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV+\underset{\partial \Omega }{\iint }[(\rho 𝐯\cdot 𝐧)𝐯+p𝐧]dS=\underset{\Omega }{\iiint }\rho 𝐟dV$

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

$\underset{\partial \Omega }{\iint }(\rho 𝐯\cdot 𝐧)𝐯dS=\underset{\Omega }{\iiint }\mathrm{\nabla }\cdot (\rho 𝐯𝐯)dV$

(Eq. 62)

$\underset{\partial \Omega }{\iint }p𝐧dS=\underset{\Omega }{\iiint }\mathrm{\nabla }pdV$

(Eq. 63)

Also, if $\Omega$ is a fixed control volume

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV=\underset{\Omega }{\iiint }\frac{\partial }{\partial t}(\rho 𝐯)dV$

(Eq. 64)

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

$\underset{\Omega }{\iiint }[\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\cdot (\rho 𝐯𝐯)+\mathrm{\nabla }p-\rho 𝐟]dV=0$

(Eq. 65)

$\Omega$ is an arbitrary control volume and thus

$\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\cdot (\rho 𝐯𝐯)+\mathrm{\nabla }p=\rho 𝐟$

(Eq. 66)

which is the momentum equation on partial differential form

Conservation of Energy

The energy equation on integral form reads

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho e_{o}dV+\underset{\partial \Omega }{\iint }\rho h_o(𝐯\cdot 𝐧)dS=$ $\underset{\Omega }{\iiint }\rho 𝐟\cdot 𝐯dV+\underset{\Omega }{\iiint }\dot{q}\rho dV$

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

$\underset{\partial \Omega }{\iint }\rho h_o(𝐯\cdot 𝐧)dS=\underset{\Omega }{\iiint }\mathrm{\nabla }\cdot (\rho h_o𝐯)dV$

(Eq. 67)

Fixed control volume

$\frac{d}{dt}\underset{\Omega }{\iiint }\rho e_{o}dV=\underset{\Omega }{\iiint }\frac{\partial }{\partial t}(\rho e_o)dV$

(Eq. 68)

The energy equation can now be written as

$\underset{\Omega }{\iiint }[\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho h_o𝐯)-\rho 𝐟\cdot 𝐯-\dot{q}\rho ]dV=0$

(Eq. 69)

$\Omega$ is an arbitrary control volume and thus

$\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho h_o𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 70)

which is the energy equation on partial differential form

Summary

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

Continuity:

$\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)=0$

Momentum:

$\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\cdot (\rho 𝐯𝐯)+\mathrm{\nabla }p=\rho 𝐟$

Energy:

$\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho h_o𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

The Differential Equations on Non-Conservation Form

The Substantial Derivative

The substantial derivative operator is defined as

$\frac{D}{Dt}=\frac{\partial }{\partial t}+𝐯\cdot \mathrm{\nabla }$

(Eq. 71)

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

$\frac{D\rho }{Dt}=\frac{\partial \rho }{\partial t}+𝐯\cdot \mathrm{\nabla }\rho$

(Eq. 72)

From before we have the continuity equation on differential form as

$\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)=0$

(Eq. 73)

which can be rewritten as

$\frac{\partial \rho }{\partial t}+\rho (\mathrm{\nabla }\cdot 𝐯)+𝐯\cdot \mathrm{\nabla }\rho =0$

(Eq. 74)

and thus

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0$

(Eq. 75)

Eq. 75 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

$\frac{\partial }{\partial t}(\rho 𝐯)+\mathrm{\nabla }\cdot (\rho 𝐯𝐯)+\mathrm{\nabla }p=\rho 𝐟$

(Eq. 76)

Expanding the first and the second terms gives

$\rho \frac{\partial 𝐯}{\partial t}+𝐯\frac{\partial \rho }{\partial t}+\rho 𝐯\cdot \mathrm{\nabla }𝐯+𝐯(\mathrm{\nabla }\cdot \rho 𝐯)+\mathrm{\nabla }p=\rho 𝐟$

(Eq. 77)

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

$\rho \underset{=\frac{D𝐯}{Dt}}{\underbrace{[\frac{\partial 𝐯}{\partial t}+𝐯\cdot \mathrm{\nabla }𝐯]}}+𝐯\underset{=0}{\underbrace{[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot \rho 𝐯]}}+\mathrm{\nabla }p=\rho 𝐟$

(Eq. 78)

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

$\frac{D𝐯}{Dt}+\frac{1}{\rho }\mathrm{\nabla }p=𝐟$

(Eq. 79)

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 (Eq. 70), repeated here for convenience

$\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho h_o𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

Total enthalpy, $h_o$, is replaced with total energy, $e_o$

$h_o=e_o+\frac{p}{\rho }$

(Eq. 80)

which gives

$\frac{\partial }{\partial t}(\rho e_o)+\mathrm{\nabla }\cdot (\rho e_o𝐯)+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 81)

Expanding the two first terms as

$\rho \frac{\partial e_o}{\partial t}+e_o\frac{\partial \rho }{\partial t}+\rho 𝐯\cdot \mathrm{\nabla }{e}_o+e_o\mathrm{\nabla }\cdot (\rho 𝐯)+\mathrm{\nabla }\cdot (p𝐯)=$ $=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 82)

Collecting terms, we can identify the substantial derivative operator applied on total energy, $D{e}_o/Dt$ and the continuity equation

$\rho \underset{=\frac{D{e}_o}{Dt}}{\underbrace{[\frac{\partial e_o}{\partial t}+𝐯\cdot \mathrm{\nabla }{e}_o]}}+e_o\underset{=0}{\underbrace{[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho 𝐯)]}}+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 83)

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

$\rho \frac{D{e}_o}{Dt}+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 84)

Summary

Continuity:

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0$

Momentum:

$\frac{D𝐯}{Dt}+\frac{1}{\rho }\mathrm{\nabla }p=𝐟$

Energy:

$\rho \frac{D{e}_o}{Dt}+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

Alternative Forms of the Energy Equation

Internal Energy Formulation

Total internal energy is defined as

$e_o=e+\frac{1}{2}𝐯\cdot 𝐯$

(Eq. 85)

Inserted in Eq. 84, this gives

$\rho \frac{De}{Dt}+\rho 𝐯\cdot \frac{D𝐯}{Dt}+\mathrm{\nabla }\cdot (p𝐯)=\rho 𝐟\cdot 𝐯+\dot{q}\rho$

(Eq. 86)

Now, let's replace the substantial derivative $D𝐯/Dt$ using the momentum equation on non-conservation form (Eq. 79).

$\rho \frac{De}{Dt}-𝐯\cdot \mathrm{\nabla }p+\cancel{\rho 𝐟\cdot 𝐯}+\mathrm{\nabla }\cdot (p𝐯)=\cancel{\rho 𝐟\cdot 𝐯}+\dot{q}\rho$

(Eq. 87)

Now, expand the term $\mathrm{\nabla }\cdot (p𝐯)$ gives

$\rho \frac{De}{Dt}\cancel{-𝐯\cdot \mathrm{\nabla }p}+\cancel{𝐯\cdot \mathrm{\nabla }p}+p(\mathrm{\nabla }\cdot 𝐯)=\dot{q}\rho \Rightarrow$ $\Rightarrow \rho \frac{De}{Dt}+p(\mathrm{\nabla }\cdot 𝐯)=\dot{q}\rho$

(Eq. 88)

Divide by $\rho$

$\frac{De}{Dt}+\frac{p}{\rho }(\mathrm{\nabla }\cdot 𝐯)=\dot{q}$

(Eq. 89)

Conservation of mass gives

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0\Rightarrow \mathrm{\nabla }\cdot 𝐯=-\frac{1}{\rho }\frac{D\rho }{Dt}$

(Eq. 90)

Insert in Eq. 89

$\frac{De}{Dt}-\frac{p}{{\rho }^2}\frac{D\rho }{Dt}=\dot{q}\Rightarrow \frac{De}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho }\right)=\dot{q}$

(Eq. 91)

$\frac{De}{Dt}+p\frac{D\nu }{Dt}=\dot{q}$

(Eq. 92)

Compare with the first law of thermodynamics: $de=\delta q-\delta w$

Enthalpy Formulation

$h=e+\frac{p}{\rho }\Rightarrow \frac{Dh}{Dt}=\frac{De}{Dt}+\frac{1}{\rho }\frac{Dp}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho }\right)$

(Eq. 93)

with $De/Dt$ from Eq. 89

$\frac{Dh}{Dt}=\dot{q}-\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho }\right)}+\frac{1}{\rho }\frac{Dp}{Dt}+\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho }\right)}$

(Eq. 94)

$\frac{Dh}{Dt}=\dot{q}+\frac{1}{\rho }\frac{Dp}{Dt}$

(Eq. 95)

Total Enthalpy Formulation

$h_o=h+\frac{1}{2}𝐯𝐯\Rightarrow \frac{D{h}_o}{Dt}=\frac{Dh}{Dt}+𝐯\cdot \frac{D𝐯}{Dt}$

(Eq. 96)

From the momentum equation (Eq. 79)

$\frac{D𝐯}{Dt}=𝐟-\frac{1}{\rho }\mathrm{\nabla }p$

(Eq. 97)

which gives

$\frac{D{h}_o}{Dt}=\frac{Dh}{Dt}+𝐯\cdot 𝐟-\frac{1}{\rho }𝐯\cdot \mathrm{\nabla }p$

(Eq. 98)

Inserting $Dh/Dt$ from Eq. 95 gives

$\frac{D{h}_o}{Dt}=\dot{q}+\frac{1}{\rho }\frac{Dp}{Dt}+𝐯\cdot 𝐟-\frac{1}{\rho }𝐯\cdot \mathrm{\nabla }p=$ $=\frac{1}{\rho }[\frac{Dp}{Dt}-𝐯\cdot \mathrm{\nabla }p]+\dot{q}+𝐯\cdot 𝐟$

(Eq. 99)

The substantial derivative operator applied to pressure

$\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+𝐯\cdot \mathrm{\nabla }p$

(Eq. 100)

and thus

$\frac{Dp}{Dt}-𝐯\cdot \mathrm{\nabla }p=\frac{\partial p}{\partial t}$

(Eq. 101)

which gives

$\frac{D{h}_o}{Dt}=\frac{1}{\rho }\frac{\partial p}{\partial t}+\dot{q}+𝐯\cdot 𝐟$

(Eq. 102)

If we assume adiabatic flow without body forces

$\frac{D{h}_o}{Dt}=\frac{1}{\rho }\frac{\partial p}{\partial t}$

(Eq. 103)

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

$\frac{D{h}_o}{Dt}=0$

(Eq. 104)

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

The entropy equation

From the second law of thermodynamics

$\frac{De}{Dt}=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)$

(Eq. 105)

From the energy equation on differential non-conservation form internal energy formulation

$\frac{De}{Dt}=\dot{q}-\frac{p}{\rho }(\mathrm{\nabla }\cdot 𝐯)$

(Eq. 106)

The continuity equation on differential non-conservation form

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0\Rightarrow \mathrm{\nabla }\cdot 𝐯=-\frac{1}{\rho }\frac{D\rho }{Dt}$

(Eq. 107)

and thus

$\frac{De}{Dt}=\dot{q}+\frac{p}{{\rho }^2}\frac{D\rho }{Dt}$

(Eq. 108)

$\frac{D\rho }{Dt}=-\frac{1}{{\nu }^2}\frac{D\nu }{Dt}$

(Eq. 109)

$\rho \frac{De}{Dt}=\rho \dot{q}-\frac{p}{\rho {\nu }^2}\frac{D\nu }{Dt}=\rho \dot{q}-\rho p\frac{D\nu }{Dt}$

(Eq. 110)

$\rho [\frac{De}{Dt}+p\frac{D\nu }{Dt}-\dot{q}]=0\Rightarrow \frac{De}{Dt}=\dot{q}-p\frac{D\nu }{Dt}$

(Eq. 111)

Insert $De/Dt$ in Eqn. \ref{eq:second:law}

$\dot{q}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)\Rightarrow$

(Eq. 112)

$T\frac{Ds}{Dt}=-\dot{q}$

(Eq. 113)

Adiabatic flow:

$T\frac{Ds}{Dt}=0$

(Eq. 114)

In an adiabatic, steady-state, inviscid flow, entropy is constant along a streamline.

Crocco's equation

The momentum equation without body forces

$\rho \frac{D𝐯}{Dt}=-\mathrm{\nabla }p$

(Eq. 115)

Expanding the substantial derivative

$\rho \frac{\partial 𝐯}{\partial t}+\rho 𝐯\cdot \mathrm{\nabla }𝐯=-\mathrm{\nabla }p$

(Eq. 116)

The first and second law of thermodynamics gives

$T\mathrm{\nabla }s=\mathrm{\nabla }h-\frac{\mathrm{\nabla }p}{\rho }$

(Eq. 117)

Insert $\mathrm{\nabla }p$ from the momentum equation

$T\mathrm{\nabla }s=\mathrm{\nabla }h+\frac{\partial 𝐯}{\partial t}+𝐯\cdot \mathrm{\nabla }𝐯$

(Eq. 118)

Definition of total enthalpy ($h_o$)

$h_o=h+\frac{1}{2}𝐯\cdot 𝐯\Rightarrow \mathrm{\nabla }h=\mathrm{\nabla }{h}_o-\mathrm{\nabla }(\frac{1}{2}𝐯\cdot 𝐯)$

(Eq. 119)

The last term can be rewritten as

$\mathrm{\nabla }(\frac{1}{2}𝐯\cdot 𝐯)=𝐯\times (\mathrm{\nabla }\times 𝐯)+𝐯\cdot \mathrm{\nabla }𝐯$

(Eq. 120)

which gives

$\mathrm{\nabla }h=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)-𝐯\cdot \mathrm{\nabla }𝐯$

(Eq. 121)

Insert $\mathrm{\nabla }h$ in the entropy equation gives

$T\mathrm{\nabla }s=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)-\cancel{𝐯\cdot \mathrm{\nabla }𝐯}+\frac{\partial 𝐯}{\partial t}+\cancel{𝐯\cdot \mathrm{\nabla }𝐯}$

(Eq. 122)

$T\mathrm{\nabla }s=\mathrm{\nabla }{h}_o-𝐯\times (\mathrm{\nabla }\times 𝐯)+\frac{\partial 𝐯}{\partial t}$

(Eq. 123)

One-dimensional flow

One-dimensional flow

Two-dimensional flow

Two-dimensional flow

Quasi-one-dimensional flow

The Q1D equations

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 $x$-direction, which means that all flow quantities and the cross-section area will vary with the axial coordinate $x$.

$A=A(x),\rho =\rho (x),u=u(x),p=p(x),\mathrm{...}$

(Eq. 124)

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

$\underset{=0}{\underbrace{\frac{d}{dt}\underset{\Omega }{\iiint }\rho dV}}+\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=0$

(Eq. 125)

$\underset{\partial \Omega }{\iint }\rho 𝐯\cdot 𝐧dS=-{\rho }_1{u}_1{A}_1+{\rho }_2{u}_2{A}_2$

(Eq. 126)

${\rho }_1{u}_1{A}_1={\rho }_2{u}_2{A}_2$

(Eq. 127)

Momentum Equation

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

$\underset{=0}{\underbrace{\frac{d}{dt}\underset{\Omega }{\iiint }\rho 𝐯dV}}+\underset{\partial \Omega }{\iint }[\rho (𝐯\cdot 𝐧)𝐯+p𝐧]dS=0$

(Eq. 128)

$\underset{\partial \Omega }{\iint }\rho (𝐯\cdot 𝐧)𝐯dS=-{\rho }_1{u}_1^2{A}_1+{\rho }_2{u}_2^2{A}_2$

(Eq. 129)

$\underset{\partial \Omega }{\iint }p𝐧dS=-p_1{A}_1+p_2{A}_2-\underset{A_1}{\overset{A_2}{\int }}pdA$

(Eq. 130)

collecting terms

$({\rho }_1{u}_1^2+p_1)A_1+\underset{A_1}{\overset{A_2}{\int }}pdA=({\rho }_2{u}_2^2+p_2)A_2$

(Eq. 131)

Energy Equation

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

$\underset{=0}{\underbrace{\frac{d}{dt}\underset{\Omega }{\iiint }\rho e_{o}dV}}+\underset{\partial \Omega }{\iint }[\rho h_o(𝐯\cdot 𝐧)]dS=0$

(Eq. 132)

$\underset{\partial \Omega }{\iint }[\rho h_o(𝐯\cdot 𝐧)]dS=-{\rho }_1{u}_1{h}_{o_1}{A}_1+{\rho }_2{u}_2{h}_{o_2}{A}_2$

(Eq. 133)

${\rho }_1{u}_1{h}_{o_1}{A}_1={\rho }_2{u}_2{h}_{o_2}{A}_2$

(Eq. 134)

Now, using the continuity equation ${\rho }_1{u}_1{A}_1={\rho }_2{u}_2{A}_2$ gives

$h_{o_1}=h_{o_2}$

(Eq. 135)

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

${\rho }_1{u}_1{A}_1={\rho }_2{u}_2{A}_2=const$

(Eq. 136)

$d(\rho uA)=0$

(Eq. 137)

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

$({\rho }_1{u}_1^2+p_1)A_1+\underset{A_1}{\overset{A_2}{\int }}pdA=({\rho }_2{u}_2^2+p_2)A_2\Rightarrow d[(\rho u^2+p)A]=pdA$

(Eq. 138)

$d(\rho u^{2}A)+d(pA)=pdA$

(Eq. 139)

$ud(\rho uA)+\rho uAdu+Adp+\cancel{pdA}=\cancel{pdA}$

(Eq. 140)

From the continuity equation we have $d(\rho uA)$ and thus

$\rho u\cancel{A}du+\cancel{A}dp=0\Rightarrow$

(Eq. 141)

$dp=-\rho udu$

(Eq. 142)

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

$h_{o_1}=h_{o_2}=const\Rightarrow d{h}_o=0$

(Eq. 143)

$h_o=h+\frac{1}{2}{u}^2\Rightarrow dh+\frac{1}{2}d(u^2)=0$

(Eq. 144)

$dh+udu=0$

(Eq. 145)

Summary

Continuity:

$d(\rho uA)=0$

Momentum:

$dp=-\rho udu$

Energy:

$dh+udu=0$

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.

Area-velocity relation

The Area-Velocity Relation

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

$d(\rho uA)=0\Rightarrow \rho udA+\rho Adu+uAd\rho =0$

(Eq. 146)

divide by $\rho uA$ gives

$\frac{d\rho }{\rho }+\frac{du}{u}+\frac{dA}{A}=0$

(Eq. 147)

As the name suggests, the area-velocity relation is a relation including the area and the flow velocity. Therefore, the next step is to replace the density terms.

This can be achieved using the momentum equation (Eqn. \ref{eq:governing:mom})

$dp=-\rho udu\iff \frac{dp}{\rho }=-udu$

(Eq. 148)

$\frac{dp}{\rho }=\frac{dp}{d\rho }\frac{d\rho }{\rho }=-udu$

(Eq. 149)

If we assume adiabatic and reversible flow processes, i.e., isentropic flow

$\frac{dp}{d\rho }={\left(\frac{dp}{d\rho }\right)}_s=a^2\Rightarrow a^2\frac{d\rho }{\rho }=-udu$

(Eq. 150)

$a^2\frac{d\rho }{\rho }=-udu=-u^2\frac{du}{u}$

(Eq. 151)

$\frac{d\rho }{\rho }=-M^2\frac{du}{u}$

(Eq. 152)

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

$-M^2\frac{du}{u}+\frac{du}{u}+\frac{dA}{A}=0$

(Eq. 153)

or

$\frac{dA}{A}=(M^2-1)\frac{du}{u}$

(Eq. 154)

which is the area-velocity relation.

From the area-velocity relation (Eqn. \ref{eq:governing:av}), we can learn that in a subsonic flow, the flow will accelerate if the cross-section area is decreased and decelerate if the cross-section area is increased. It can also be seen that for supersonic flow, the relation between flow velocity and cross-section area will be the opposite of that for subsonic flows, see Fig. \ref{fig:areavelocity}. For sonic flow, $M=1$, the relation shows that $dA=0$, which means that sonic flow can only occur at a cross-section area maximum or minimum. From the subsonic versus supersonic flow discussion, it can be understood that sonic flow at the minimum cross section area is the only valid option (see Fig. \ref{fig:sonic}).

Area-Mach relation

The Area-Mach-Number Relation

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

$d(\rho uA)=0\Rightarrow \rho uA=const$

(Eq. 155)

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

$\rho uA={\rho }^{*}{u}^{*}{A}^{*}=\{u^{*}=a^{*}\}={\rho }^{*}{a}^{*}{A}^{*}$

(Eq. 156)

divide by $\rho u{A}^{*}$ gives

$\frac{{\rho }^{*}}{\rho }\frac{a^{*}}{u}=\frac{A}{A^{*}}$

(Eq. 157)

$a^{*}/u=1/M^{*}$ but ${\rho }^{*}/\rho$ is unknown

$\frac{{\rho }^{*}}{\rho }=\frac{{\rho }^{*}}{{\rho }_o}\frac{{\rho }_o}{\rho }$

(Eq. 158)

and thus

$\frac{{\rho }^{*}}{{\rho }_o}\frac{{\rho }_o}{\rho }\frac{1}{M^{*}}=\frac{A}{A^{*}}$

(Eq. 159)

Using the isentropic relations, we get

$\frac{{\rho }^{*}}{{\rho }_o}=\frac{1}{{[\frac{1}{2}(\gamma -1)]}^{1/(\gamma -1)}}$

(Eq. 160)

$\frac{{\rho }_o}{\rho }={[1+\frac{1}{2}(\gamma +1)M^2]}^{1/(\gamma -1)}$

(Eq. 161)

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

$\frac{A}{A^{*}}=\frac{1}{M^{*}}{\left[\frac{2+(\gamma -1)M^2}{\gamma +1}\right]}^{1/(\gamma -1)}$

(Eq. 162)

What remains now is to replace $M^{*}$

${M^{*}}^2=\frac{u^2}{{a^{*}}^2}=\frac{u^2}{a^2}\frac{a^2}{{a^{*}}^2}=\frac{u^2}{a^2}\frac{a^2}{a_o^2}\frac{a_o^2}{{a^{*}}^2}=M^2\frac{a^2}{a_o^2}\frac{a_o^2}{{a^{*}}^2}$

(Eq. 163)

For a calorically perfect gas $a=\sqrt{\gamma RT}$, which gives

$\frac{a^2}{a_o^2}=\frac{T}{T_o}={[1+\frac{1}{2}(\gamma -1)M^2]}^{-1}$

(Eq. 164)

$\frac{a_o^2}{{a^{*}}^2}=\frac{T_o}{T^{*}}=\frac{1}{2}(\gamma +1)$

(Eq. 165)

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

${M^{*}}^2=\frac{(\gamma +1)M^2}{2+(\gamma -1)M^2}$

(Eq. 166)

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

${\left(\frac{A}{A^{*}}\right)}^2=\frac{1}{{M^{*}}^2}{\left[\frac{2+(\gamma -1)M^2}{\gamma +1}\right]}^{2/(\gamma -1)}$

(Eq. 167)

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

${\left(\frac{A}{A^{*}}\right)}^2=\frac{2+(\gamma -1)M^2}{(\gamma +1)M^2}{\left[\frac{2+(\gamma -1)M^2}{\gamma +1}\right]}^{2/(\gamma -1)}\Rightarrow$

(Eq. 168)

${\left(\frac{A}{A^{*}}\right)}^2=\frac{1}{M^2}{\left[\frac{2+(\gamma -1)M^2}{\gamma +1}\right]}^{1+2/(\gamma -1)}\Rightarrow$

(Eq. 169)

${\left(\frac{A}{A^{*}}\right)}^2=\frac{1}{M^2}{\left[\frac{2+(\gamma -1)M^2}{\gamma +1}\right]}^{(\gamma +1)/(\gamma -1)}$

(Eq. 170)

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, $A^{*}$.

$M=f\left(\frac{A}{A^{*}}\right)$

(Eq. 171)

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.

Choked flow

Geometric Choking

For steady-state nozzle flow, the massflow is obtained as

$\dot{m}=\rho uA=const$

(Eq. 172)

Eqn. \ref{eq:massflow:a} can be evaluated at any location inside the nozzle and if evaluated at sonic conditions we get

$\dot{m}={\rho }^{*}{u}^{*}{A}^{*}$

(Eq. 173)

By definition $u^{*}=a^{*}$ and thus

$\dot{m}={\rho }^{*}{a}^{*}{A}^{*}$

(Eq. 174)

${\rho }^{*}$ and $a^{*}$ can be obtained using the ratios ${\rho }^{*}/{\rho }_o$ and $a^{*}/a_o$

${\rho }^{*}=\left(\frac{{\rho }^{*}}{{\rho }_o}\right){\rho }_o=\frac{p_o}{R{T}_o}{\left(\frac{2}{\gamma +1}\right)}^{1/(\gamma -1)}$

(Eq. 175)

$a^{*}=\left(\frac{a^{*}}{a_o}\right)a_o=a_o{\left(\frac{2}{\gamma +1}\right)}^{1/2}=\sqrt{\gamma R{T}_o}{\left(\frac{2}{\gamma +1}\right)}^{1/2}$

(Eq. 176)

Eqns. \ref{eq:as} and \ref{eq:rhos} in Eqn. \ref{eq:massflow:b} gives

$\dot{m}=\frac{p_o}{R{T}_o}{\left(\frac{2}{\gamma +1}\right)}^{1/(\gamma -1)}\sqrt{\gamma R{T}_o}{\left(\frac{2}{\gamma +1}\right)}^{1/2}{A}^{*}$

(Eq. 177)

which can be rewritten as

$\dot{m}=\frac{p_o{A}^{*}}{\sqrt{T_o}}\sqrt{\frac{\gamma }{R}{\left(\frac{2}{\gamma +1}\right)}^{(\gamma +1)/(\gamma -1)}}$

(Eq. 178)

Eqn. \ref{eq:massflow:c} valid for:

• quasi-one-dimensional flow
• steady state
• inviscid flow
• calorically perfect gas

It should be noted that the choked massflow can be calculated using Eqn. \ref{eq:massflow:c} even for cases with shocks downstream of the throat.

Nozzle flow

add description of nozzle flows here...

Diffusers

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