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. 5.1)

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. 5.2)

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

(Eq. 5.3)

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

(Eq. 5.4)

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. 5.5)

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

(Eq. 5.6)

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

(Eq. 5.7)

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. 5.8)

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. 5.9)

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

(Eq. 5.10)

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

(Eq. 5.11)

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. 5.12)

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. 5.13)

$d(\rho uA)=0$

(Eq. 5.14)

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. 5.15)

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

(Eq. 5.16)

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

(Eq. 5.17)

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

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

(Eq. 5.18)

$dp=-\rho udu$

(Eq. 5.19)

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. 5.20)

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

(Eq. 5.21)

$dh+udu=0$

(Eq. 5.22)

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. 5.23)

divide by $\rho uA$ gives

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

(Eq. 5.24)

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. 5.25)

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

(Eq. 5.26)

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. 5.27)

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

(Eq. 5.28)

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

(Eq. 5.29)

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. 5.30)

or

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

(Eq. 5.31)

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. 5.32)

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. 5.33)

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

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

(Eq. 5.34)

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

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

(Eq. 5.35)

and thus

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

(Eq. 5.36)

Using the isentropic relations, we get

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

(Eq. 5.37)

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

(Eq. 5.38)

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. 5.39)

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. 5.40)

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. 5.41)

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

(Eq. 5.42)

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. 5.43)

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. 5.44)

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. 5.45)

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

(Eq. 5.46)

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

(Eq. 5.47)

which is the area-Mach-number relation.

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

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

(Eq. 5.48)

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

Choked flow

Geometric Choking

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

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

(Eq. 5.49)

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. 5.50)

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

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

(Eq. 5.51)

${\rho }^{*}$ 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. 5.52)

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

(Eq. 5.53)

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. 5.54)

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. 5.55)

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

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Diffusers

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