The Q1D equations: Difference between revisions

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[[Category:Compressible flow]]
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=== Governing Equations ===
=== Governing Equations ===


Line 20: Line 30:
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>.  
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>.  


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


We will further assume steady-state flow, which means that unsteady terms will be zero.
We will further assume steady-state flow, which means that unsteady terms will be zero.
Line 32: Line 42:
Applying the integral form of the continuity equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives
Applying the integral form of the continuity equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives


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


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


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


==== Momentum Equation ====
==== Momentum Equation ====
Line 48: Line 58:
Applying the integral form of the momentum equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives
Applying the integral form of the momentum equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives


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


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


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


collecting terms
collecting terms


<math display="block">
{{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
\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>
</math>}}


==== Energy Equation ====
==== Energy Equation ====
Line 70: Line 80:
Applying the integral form of the energy equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives
Applying the integral form of the energy equation on the quasi-one-dimensional flow control volume (Fig. \ref{fig:cv}) gives


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


<math display="block">
{{NumEqn|<math>
\varoiint_{\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
\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>
</math>}}


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


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


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


==== Differential Form ====
==== Differential Form ====
Line 94: Line 104:
The continuity equation (Eqn. \ref{eq:governing:cont:b}) is rewritten in differential form as
The continuity equation (Eqn. \ref{eq:governing:cont:b}) is rewritten in differential form as


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


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


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


<math display="block">
{{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
\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>
</math>}}


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


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


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


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


<math display="block">
{{NumEqn|<math>
dp=-\rho udu
dp=-\rho udu
</math>
</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
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


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


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


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


==== Summary ====
==== Summary ====


Continuity:
<div style="border: 1px solid;">
 
{{OpenInfoBox|<math>
<math display="block">
d(\rho uA)=0
d(\rho uA)=0
</math>
</math>|description=Continuity:}}
 
Momentum:


<math display="block">
{{OpenInfoBox|<math>
dp=-\rho udu
dp=-\rho udu
</math>
</math>|description=Momentum:}}


Energy:
{{OpenInfoBox|<math>
 
<math display="block">
dh+udu=0
dh+udu=0
</math>
</math>|description=Energy:}}
</div>


The equations are valid for:
The equations are valid for:


* quasi-one-dimensional flow
* quasi-one-dimensional flow

Latest revision as of 18:53, 1 April 2026

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), ρ=ρ(x), u=u(x), p=p(x), ...(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

ddtΩρdV=0+Ωρ𝐯𝐧dS=0(Eq. 5.2)
Ωρ𝐯𝐧dS=ρ1u1A1+ρ2u2A2(Eq. 5.3)
ρ1u1A1=ρ2u2A2(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

ddtΩρ𝐯dV=0+Ω[ρ(𝐯𝐧)𝐯+p𝐧]dS=0(Eq. 5.5)
Ωρ(𝐯𝐧)𝐯dS=ρ1u12A1+ρ2u22A2(Eq. 5.6)
Ωp𝐧dS=p1A1+p2A2A1A2pdA(Eq. 5.7)

collecting terms

(ρ1u12+p1)A1+A1A2pdA=(ρ2u22+p2)A2(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

ddtΩρeodV=0+Ω[ρho(𝐯𝐧)]dS=0(Eq. 5.9)
Ω[ρho(𝐯𝐧)]dS=ρ1u1ho1A1+ρ2u2ho2A2(Eq. 5.10)
ρ1u1ho1A1=ρ2u2ho2A2(Eq. 5.11)

Now, using the continuity equation ρ1u1A1=ρ2u2A2 gives

ho1=ho2(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

ρ1u1A1=ρ2u2A2=const(Eq. 5.13)
d(ρuA)=0(Eq. 5.14)

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

(ρ1u12+p1)A1+A1A2pdA=(ρ2u22+p2)A2d[(ρu2+p)A]=pdA(Eq. 5.15)
d(ρu2A)+d(pA)=pdA(Eq. 5.16)
ud(ρuA)+ρuAdu+Adp+pdA=pdA(Eq. 5.17)

From the continuity equation we have d(ρuA) and thus

ρuAdu+Adp=0(Eq. 5.18)
dp=ρudu(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

ho1=ho2=constdho=0(Eq. 5.20)
ho=h+12u2dh+12d(u2)=0(Eq. 5.21)
dh+udu=0(Eq. 5.22)

Summary

Continuity:d(ρuA)=0
Momentum:dp=ρudu
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.