3.2 Governing Equations on Integral FormNiklas Andersson
Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
Figure 3.1: Generic control volume
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 Ω \Omega Ω Figure 3.1 ∂ Ω \partial \Omega ∂ Ω
− ∯ ∂ Ω ρ v ⋅ n d S -\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS − ∬ ∂ Ω ρ v ⋅ n d S Now, let’s consider a small infinitesimal volume d V d\mathscr{V} d V Ω \Omega Ω d V d\mathscr{V} d V ρ d V \rho d\mathscr{V} ρ d V Ω \Omega Ω
∭ Ω ρ d V \iiint_{\Omega} \rho d\mathscr{V} ∭ Ω ρ d V The rate of change of mass within Ω \Omega Ω
d d t ∭ Ω ρ d V \frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V} d t d ∭ Ω ρ d V Mass is conserved, which means that the rate of change of mass within Ω \Omega Ω
d d t ∭ Ω ρ d V = − ∯ ∂ Ω ρ v ⋅ n d S \frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V}=-\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS d t d ∭ Ω ρ d V = − ∬ ∂ Ω ρ v ⋅ n d S or
d d t ∭ Ω ρ d V + ∯ ∂ Ω ρ v ⋅ n d S = 0 \frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V}+\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0 d t d ∭ Ω ρ d V + ∬ ∂ Ω ρ v ⋅ n d S = 0 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
d d t ( m v ) = F \frac{d}{dt}(m\mathbf{v})=\mathbf{F} d t d ( m v ) = F What type of forces do we have?
Body forces inside Ω \Omega Ω
∭ Ω ρ f d V \iiint_{\Omega}\rho \mathbf{f}d\mathscr{V} ∭ Ω ρ f d V Surface force on ∂ Ω \partial \Omega ∂ Ω
− ∯ ∂ Ω p n d S -\oiint_{\partial \Omega} p\mathbf{n}dS − ∬ ∂ Ω p n d S Since we are considering inviscid flow, there are no shear forces and thus we have the net force as
F = ∭ Ω ρ f d V − ∯ ∂ Ω p n d S \mathbf{F}=\iiint_{\Omega}\rho \mathbf{f}d\mathscr{V}-\oiint_{\partial \Omega} p\mathbf{n}dS F = ∭ Ω ρ f d V − ∬ ∂ Ω p n d S The fluid flowing through Ω \Omega Ω Ω \Omega Ω
∯ ∂ Ω ( ρ v ⋅ n d S ) v = ∯ ∂ Ω ( ρ v ⋅ n ) v d S \oiint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n}dS)\mathbf{v}=\oiint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS ∬ ∂ Ω ( ρ v ⋅ n d S ) v = ∬ ∂ Ω ( ρ v ⋅ n ) v d S Integrated momentum inside Ω \Omega Ω
∭ Ω ρ v d V \iiint_{\Omega} \rho \mathbf{v} d\mathscr{V} ∭ Ω ρ v d V Rate of change of momentum due to unsteady effects inside Ω \Omega Ω
d d t ∭ Ω ρ v d V \frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V} d t d ∭ Ω ρ v d V Combining the rate of change of momentum, the net momentum flux and the net forces we get
d d t ∭ Ω ρ v d V + ∯ ∂ Ω ( ρ v ⋅ n ) v d S = ∭ Ω ρ f d V − ∯ ∂ Ω p n d S \frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V}+\oiint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega}\rho \mathbf{f}d\mathscr{V}-\oiint_{\partial \Omega} p\mathbf{n}dS d t d ∭ Ω ρ v d V + ∬ ∂ Ω ( ρ v ⋅ n ) v d S = ∭ Ω ρ f d V − ∬ ∂ Ω p n d S combining the surface integrals, we get
d d t ∭ Ω ρ v d V + ∯ ∂ Ω [ ( ρ v ⋅ n ) v + p n ] d S = ∭ Ω ρ f d V \frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V}+\oiint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}d\mathscr{V} d t d ∭ Ω ρ v d V + ∬ ∂ Ω [ ( ρ v ⋅ n ) v + p n ] d S = ∭ Ω ρ f d V 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+E_2=E_3 E 1 + E 2 = E 3 E 1 E_1 E 1 Ω \Omega Ω
E 2 E_2 E 2 Ω \Omega Ω
E 3 E_3 E 3 Ω \Omega Ω
E 1 = ∭ Ω q ˙ ρ d V E_1=\iiint_{\Omega} \dot{q}\rho d\mathscr{V} E 1 = ∭ Ω q ˙ ρ d V where q ˙ \dot{q} q ˙
The rate of work done on the fluid in Ω \Omega Ω
E 2 p r e s s u r e = − ∯ ∂ Ω ( p n d S ) ⋅ v = − ∯ ∂ Ω p v ⋅ n d S E_{2_{pressure}}=-\oiint_{\partial \Omega}(p\mathbf{n}dS)\cdot\mathbf{v}=-\oiint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS E 2 p ress u re = − ∬ ∂ Ω ( p n d S ) ⋅ v = − ∬ ∂ Ω p v ⋅ n d S The rate of work done on the fluid in Ω \Omega Ω
E 2 b o d y f o r c e s = ∭ Ω ( ρ f d V ) ⋅ v = ∭ Ω ρ f ⋅ v d V E_{2_{body\ forces}}=\iiint_{\Omega}(\rho\mathbf{f}d\mathscr{V})\cdot\mathbf{v}=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V} E 2 b o d y f orces = ∭ Ω ( ρ f d V ) ⋅ v = ∭ Ω ρ f ⋅ v d V E 2 = E 2 p r e s s u r e + E 2 b o d y f o r c e s = − ∯ ∂ Ω p v ⋅ n d S + ∭ Ω ρ f ⋅ v d V E_2=E_{2_{pressure}}+E_{2_{body\ forces}}=-\oiint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS+\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V} E 2 = E 2 p ress u re + E 2 b o d y f orces = − ∬ ∂ Ω p v ⋅ n d S + ∭ Ω ρ f ⋅ v d V The energy of the fluid per unit mass is the sum of internal energy e e e V 2 / 2 V^2/2 V 2 /2
∯ ∂ Ω ( ρ v ⋅ n d S ) ( e + V 2 2 ) \oiint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right) ∬ ∂ Ω ( ρ v ⋅ n d S ) ( e + 2 V 2 ) Analogous to mass and momentum, the total amount of energy of the fluid in Ω \Omega Ω
∭ Ω ρ ( e + V 2 2 ) d V \iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V} ∭ Ω ρ ( e + 2 V 2 ) d V The time rate of change of the energy of the fluid in Ω \Omega Ω
d d t ∭ Ω ρ ( e + V 2 2 ) d V \frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V} d t d ∭ Ω ρ ( e + 2 V 2 ) d V Now, E 3 E_3 E 3 Ω \Omega Ω
E 3 = d d t ∭ Ω ρ ( e + V 2 2 ) d V + ∯ ∂ Ω ( ρ v ⋅ n d S ) ( e + V 2 2 ) E_3=\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}+\oiint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right) E 3 = d t d ∭ Ω ρ ( e + 2 V 2 ) d V + ∬ ∂ Ω ( ρ v ⋅ n d S ) ( e + 2 V 2 ) With all elements of the energy equation defined, we are now ready to finally compile the full equation
d d t ∭ Ω ρ ( e + V 2 2 ) d V + ∯ ∂ Ω [ ρ ( e + V 2 2 ) ( v ⋅ n ) + p v ⋅ n ] d S = ∭ Ω ρ f ⋅ v d V + ∭ Ω q ˙ ρ d V \frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}+\oiint_{\partial \Omega}\left[\rho\left(e+\frac{V^2}{2}\right)(\mathbf{v}\cdot\mathbf{n}) + p\mathbf{v}\cdot\mathbf{n}\right]dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}+\iiint_{\Omega} \dot{q}\rho d\mathscr{V} d t d ∭ Ω ρ ( e + 2 V 2 ) d V + ∬ ∂ Ω [ ρ ( e + 2 V 2 ) ( v ⋅ n ) + p v ⋅ n ] d S = ∭ Ω ρ f ⋅ v d V + ∭ Ω q ˙ ρ d V The surface integral in the energy equation may be rewritten as
∯ ∂ Ω [ ρ ( e + V 2 2 ) ( v ⋅ n ) + p v ⋅ n ] d S = ∯ ∂ Ω ρ [ e + p ρ + V 2 2 ] ( v ⋅ n ) d S \oiint_{\partial \Omega}\left[\rho\left(e+\frac{V^2}{2}\right)(\mathbf{v}\cdot\mathbf{n}) + p\mathbf{v}\cdot\mathbf{n}\right]dS=\oiint_{\partial \Omega}\rho\left[e+\frac{p}{\rho}+\frac{V^2}{2}\right](\mathbf{v}\cdot\mathbf{n})dS ∬ ∂ Ω [ ρ ( e + 2 V 2 ) ( v ⋅ n ) + p v ⋅ n ] d S = ∬ ∂ Ω ρ [ e + ρ p + 2 V 2 ] ( v ⋅ n ) d S and with the definition of enthalpy h = e + p / ρ h=e+p/\rho h = e + p / ρ
∯ ∂ Ω ρ [ h + V 2 2 ] ( v ⋅ n ) d S \oiint_{\partial \Omega}\rho\left[h+\frac{V^2}{2}\right](\mathbf{v}\cdot\mathbf{n})dS ∬ ∂ Ω ρ [ h + 2 V 2 ] ( v ⋅ n ) d S Furthermore, introducing total internal energy e o e_o e o h o h_o h o
e o = e + 1 2 V 2 e_o=e+\frac{1}{2}V^2 e o = e + 2 1 V 2 and
h o = h + 1 2 V 2 h_o=h+\frac{1}{2}V^2 h o = h + 2 1 V 2 the energy equation is written as
d d t ∭ Ω ρ e o d V + ∯ ∂ Ω ρ h o ( v ⋅ n ) d S = ∭ Ω ρ f ⋅ v d V + ∭ Ω q ˙ ρ d V \frac{d}{dt}\iiint_{\Omega}\rho e_o d\mathscr{V}+\oiint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}+\iiint_{\Omega} \dot{q}\rho d\mathscr{V} d t d ∭ Ω ρ e o d V + ∬ ∂ Ω ρ h o ( v ⋅ n ) d S = ∭ Ω ρ f ⋅ v d V + ∭ Ω q ˙ ρ d V