6.2 Governing EquationsNiklas Andersson
Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
Figure 6.1: Quasi-one-dimensional flow - control volume
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 x x x x x
A = A ( x ) , ρ = ρ ( x ) , u = u ( x ) , p = p ( x ) , . . . A=A(x),\ \rho=\rho(x),\ u=u(x),\ p=p(x),\ ... A = A ( x ) , ρ = ρ ( x ) , u = u ( x ) , p = p ( x ) , ... 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. Figure 6.1
d d t ∭ Ω ρ d V ⏟ = 0 + ∯ ∂ Ω ρ v ⋅ n d S = 0 \underbrace{\frac{d}{dt}\iiint_{\Omega}\rho d{\mathscr{V}}}_{=0}+\oiint_{\partial \Omega}\rho {\mathbf{v}}\cdot {\mathbf{n}}dS=0 = 0 d t d ∭ Ω ρ d V + ∬ ∂ Ω ρ v ⋅ n d S = 0 ∯ ∂ Ω ρ v ⋅ n d S = − ρ 1 u 1 A 1 + ρ 2 u 2 A 2 \oiint_{\partial \Omega}\rho {\mathbf{v}}\cdot {\mathbf{n}}dS=-\rho_1 u_1 A_1+\rho_2 u_2 A_2 ∬ ∂ Ω ρ v ⋅ n d S = − ρ 1 u 1 A 1 + ρ 2 u 2 A 2 ρ 1 u 1 A 1 = ρ 2 u 2 A 2 \rho_1 u_1 A_1=\rho_2 u_2 A_2 ρ 1 u 1 A 1 = ρ 2 u 2 A 2 Momentum Equation ¶ Applying the integral form of the momentum equation on the quasi-one-dimensional flow control volume (Fig. Figure 6.1
d d t ∭ Ω ρ v d V ⏟ = 0 + ∯ ∂ Ω [ ρ ( v ⋅ n ) v + p n ] d S = 0 \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 = 0 d t d ∭ Ω ρ v d V + ∬ ∂ Ω [ ρ ( v ⋅ n ) v + p n ] d S = 0 ∯ ∂ Ω ρ ( v ⋅ n ) v d S = − ρ 1 u 1 2 A 1 + ρ 2 u 2 2 A 2 \oiint_{\partial \Omega} \rho ({\mathbf{v}}\cdot {\mathbf{n}}){\mathbf{v}}dS=-\rho_1u_1^2A_1+\rho_2u_2^2A_2 ∬ ∂ Ω ρ ( v ⋅ n ) v d S = − ρ 1 u 1 2 A 1 + ρ 2 u 2 2 A 2 ∯ ∂ Ω p n d S = − p 1 A 1 + p 2 A 2 − ∫ A 1 A 2 p d A \oiint_{\partial \Omega} p{\mathbf{n}}dS=-p_1A_1+p_2A_2-\int_{A_1}^{A_2}pdA ∬ ∂ Ω p n d S = − p 1 A 1 + p 2 A 2 − ∫ A 1 A 2 p d A collecting terms
( ρ 1 u 1 2 + p 1 ) A 1 + ∫ A 1 A 2 p d A = ( ρ 2 u 2 2 + p 2 ) 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 ( ρ 1 u 1 2 + p 1 ) A 1 + ∫ A 1 A 2 p d A = ( ρ 2 u 2 2 + p 2 ) A 2 Energy Equation ¶ Applying the integral form of the energy equation on the quasi-one-dimensional flow control volume (Fig. Figure 6.1
d d t ∭ Ω ρ e o d V ⏟ = 0 + ∯ ∂ Ω [ ρ h o ( v ⋅ n ) ] d S = 0 \underbrace{\frac{d}{dt}\iiint_{\Omega}\rho e_o d{\mathscr{V}}}_{=0}+\oiint_{\partial \Omega}\left[\rho h_o ({\mathbf{v}}\cdot{\mathbf{n}})\right]dS=0 = 0 d t d ∭ Ω ρ e o d V + ∬ ∂ Ω [ ρ h o ( v ⋅ n ) ] d S = 0 ∯ ∂ Ω [ ρ h o ( v ⋅ n ) ] d S = − ρ 1 u 1 h o 1 A 1 + ρ 2 u 2 h o 2 A 2 \oiint_{\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 ∬ ∂ Ω [ ρ h o ( v ⋅ n ) ] d S = − ρ 1 u 1 h o 1 A 1 + ρ 2 u 2 h o 2 A 2 ρ 1 u 1 h o 1 A 1 = ρ 2 u 2 h o 2 A 2 \rho_1u_1h_{o_1}A_1=\rho_2u_2h_{o_2}A_2 ρ 1 u 1 h o 1 A 1 = ρ 2 u 2 h o 2 A 2 Now, using the continuity equation ρ 1 u 1 A 1 = ρ 2 u 2 A 2 \rho_1u_1A_1=\rho_2u_2A_2 ρ 1 u 1 A 1 = ρ 2 u 2 A 2
h o 1 = h o 2 h_{o_1}=h_{o_2} h o 1 = h o 2 The integral term appearing the momentum equation is undesired and therefore the governing equations are converted to differential form.
The continuity equation (Eqn. (6.4)
ρ 1 u 1 A 1 = ρ 2 u 2 A 2 = c o n s t \rho_1u_1A_1=\rho_2u_2A_2=const ρ 1 u 1 A 1 = ρ 2 u 2 A 2 = co n s t d ( ρ u A ) = 0 d(\rho uA)=0 d ( ρ u A ) = 0 The momentum equation (Eqn. (6.8)
( ρ 1 u 1 2 + p 1 ) A 1 + ∫ A 1 A 2 p d A = ( ρ 2 u 2 2 + p 2 ) A 2 ⇒ d [ ( ρ u 2 + p ) A ] = p d A \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 ( ρ 1 u 1 2 + p 1 ) A 1 + ∫ A 1 A 2 p d A = ( ρ 2 u 2 2 + p 2 ) A 2 ⇒ d [ ( ρ u 2 + p ) A ] = p d A d ( ρ u 2 A ) + d ( p A ) = p d A d(\rho u^2A)+d(pA)=pdA d ( ρ u 2 A ) + d ( p A ) = p d A u d ( ρ u A ) + ρ u A d u + A d p + p d A = p d A ud(\rho uA)+\rho uAdu+Adp+\cancel{pdA}=\cancel{pdA} u d ( ρ u A ) + ρ u A d u + A d p + p d A = p d A From the continuity equation we have d ( ρ u A ) d(\rho uA) d ( ρ u A )
ρ u A d u + A d p = 0 ⇒ \rho u\cancel{A}du+\cancel{A}dp=0\Rightarrow ρ u A d u + A d p = 0 ⇒ d p = − ρ u d u dp=-\rho udu d p = − ρ u d u which is the momentum equation on differential form. Also referred to as Euler’s equation. Finally, the energy equation (Eqn. (6.4)
h o 1 = h o 2 = c o n s t ⇒ d h o = 0 h_{o_1}=h_{o_2}=const\Rightarrow dh_o=0 h o 1 = h o 2 = co n s t ⇒ d h o = 0 h o = h + 1 2 u 2 ⇒ d h + 1 2 d ( u 2 ) = 0 h_o=h+\frac{1}{2}u^2\Rightarrow dh+\frac{1}{2}d(u^2)=0 h o = h + 2 1 u 2 ⇒ d h + 2 1 d ( u 2 ) = 0 Continuity:
d ( ρ u A ) = 0 d(\rho uA)=0 d ( ρ u A ) = 0 Momentum:
d p = − ρ u d u dp=-\rho udu d p = − ρ u d u Energy:
The equations are valid for:
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.