The Q1D equations

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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.