Governing equations on differential form - Revision history

Nian at 18:23, 1 April 2026

2026-04-01T18:23:47Z

← Older revision		Revision as of 18:23, 1 April 2026
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	The continuity equation on integral form reads		The continuity equation on integral form reads

	{{~~NumEqn~~\|<math>		{{InfoBox\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0		\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
	</math>~~\|nonumber=1\|infobox=1\|numw=2em~~}}		</math>}}

	Apply Gauss's divergence theorem on the surface integral gives		Apply Gauss's divergence theorem on the surface integral gives
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	The momentum equation on integral form reads		The momentum equation on integral form reads

	{{~~NumEqn~~\|<math>		{{InfoBox\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV		\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
	</math>~~\|nonumber=1\|infobox=1\|numw=2em~~}}		</math>}}

	As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.		As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
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	The energy equation on integral form reads		The energy equation on integral form reads

	{{~~NumEqn~~\|<math>		{{InfoBox\|<math>
	\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV		\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
	</math>~~\|nonumber=1\|infobox=1\|numw=2em~~}}		</math>}}

	Gauss's divergence theorem applied to the surface integral term in the energy equation gives		Gauss's divergence theorem applied to the surface integral term in the energy equation gives
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	<div style="border: solid 1px;">		<div style="border: solid 1px;">
	{{~~NumEqn~~\|<math>		{{OpenInfoBox\|<math>
	\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0		\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
	</math>~~\|infobox=1~~\|description=Continuity:~~\|nonumber=1\|noborder=1~~}}		</math>\|description=Continuity:}}

	{{~~NumEqn~~\|<math>		{{OpenInfoBox\|<math>
	\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}		\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
	</math>~~\|infobox=1~~\|description=Momentum:~~\|nonumber=1\|noborder=1~~}}		</math>\|description=Momentum:}}

	{{~~NumEqn~~\|<math>		{{OpenInfoBox\|<math>
	\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>~~\|infobox=1~~\|description=Energy:~~\|nonumber=1\|noborder=1~~}}		</math>\|description=Energy:}}
	</div>		</div>

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	<div style="border: solid 1px;">		<div style="border: solid 1px;">
	{{~~NumEqn~~\|<math>		{{OpenInfoBox\|<math>
	\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0		\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
	</math>~~\|infobox=1~~\|description=Continuity:~~\|nonumber=1\|noborder=1~~}}		</math>\|description=Continuity:}}

	{{~~NumEqn~~\|<math>		{{OpenInfoBox\|<math>
	\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}		\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
	</math>~~\|infobox=1~~\|description=Momentum:~~\|nonumber=1\|noborder=1~~}}		</math>\|description=Momentum:}}

	{{~~NumEqn~~\|<math>		{{OpenInfoBox\|<math>
	\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>~~\|infobox=1~~\|description=Energy:~~\|nonumber=1\|noborder=1~~}}		</math>\|description=Energy:}}
	</div>		</div>

Nian at 15:31, 1 April 2026

2026-04-01T15:31:45Z

← Older revision		Revision as of 15:31, 1 April 2026
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0		\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
	</math>\|nonumber=1\|~~border~~=1~~\|background-color=whitesmoke\|padding=2em\|color=steelblue~~\|numw=2em}}		</math>\|nonumber=1\|infobox=1\|numw=2em}}

	Apply Gauss's divergence theorem on the surface integral gives		Apply Gauss's divergence theorem on the surface integral gives
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV		\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
	</math>\|nonumber=1\|~~border~~=1~~\|background-color=whitesmoke\|padding=2em\|color=steelblue~~\|numw=2em}}		</math>\|nonumber=1\|infobox=1\|numw=2em}}

	As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.		As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV		\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
	</math>\|nonumber=1\|~~border~~=1~~\|background-color=whitesmoke\|padding=2em\|color=steelblue~~\|numw=2em}}		</math>\|nonumber=1\|infobox=1\|numw=2em}}

	Gauss's divergence theorem applied to the surface integral term in the energy equation gives		Gauss's divergence theorem applied to the surface integral term in the energy equation gives
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0		\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
	</math>\|~~background-color~~=~~whitesmoke\|color=steelblue~~\|description=Continuity:\|nonumber=1\|~~padding~~=~~2em~~}}		</math>\|infobox=1\|description=Continuity:\|nonumber=1\|noborder=1}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}		\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
	</math>\|~~background-color~~=~~whitesmoke\|color=steelblue~~\|description=Momentum:\|nonumber=1\|~~padding~~=~~2em~~}}		</math>\|infobox=1\|description=Momentum:\|nonumber=1\|noborder=1}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>\|~~background-color~~=~~whitesmoke\|color=steelblue~~\|description=Energy:\|nonumber=1\|~~padding~~=~~2em~~}}		</math>\|infobox=1\|description=Energy:\|nonumber=1\|noborder=1}}
	</div>		</div>

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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0		\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
	</math>\|~~background-color~~=~~whitesmoke\|color=steelblue~~\|description=Continuity:\|nonumber=1\|~~padding~~=~~2em~~}}		</math>\|infobox=1\|description=Continuity:\|nonumber=1\|noborder=1}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}		\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
	</math>\|~~background-color~~=~~whitesmoke\|color=steelblue~~\|description=Momentum:\|nonumber=1\|~~padding~~=~~2em~~}}		</math>\|infobox=1\|description=Momentum:\|nonumber=1\|noborder=1}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>\|~~background-color~~=~~whitesmoke\|color=steelblue~~\|description=Energy:\|nonumber=1\|~~padding~~=~~2em~~}}		</math>\|infobox=1\|description=Energy:\|nonumber=1\|noborder=1}}
	</div>		</div>

Nian at 05:15, 1 April 2026

2026-04-01T05:15:08Z

← Older revision		Revision as of 05:15, 1 April 2026
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	--><noinclude><!--		--><noinclude><!--
	-->{{#vardefine:secno\|2}}<!--		-->{{#vardefine:secno\|2}}<!--
	-->{{#vardefine:eqno\|31}}<!--		-->{{#vardefine:eqno\|28}}<!--
	--></noinclude><!--		--></noinclude><!--

Nian at 20:20, 31 March 2026

2026-03-31T20:20:25Z

← Older revision		Revision as of 20:20, 31 March 2026
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	The governing equations for compressible inviscid flow on partial differential form:		The governing equations for compressible inviscid flow on partial differential form:

			<div style="border: solid 1px;">
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0		\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
	</math>}}		</math>\|background-color=whitesmoke\|color=steelblue\|description=Continuity:\|nonumber=1\|padding=2em}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}		\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
	</math>}}		</math>\|background-color=whitesmoke\|color=steelblue\|description=Momentum:\|nonumber=1\|padding=2em}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>}}		</math>\|background-color=whitesmoke\|color=steelblue\|description=Energy:\|nonumber=1\|padding=2em}}
			</div>

	=== The Differential Equations on Non-Conservation Form ===		=== The Differential Equations on Non-Conservation Form ===
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	==== Summary ====		==== Summary ====

	~~Continuity~~:		<div style="border: solid 1px;">

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0		\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
	</math>}}		</math>\|background-color=whitesmoke\|color=steelblue\|description=Continuity:\|nonumber=1\|padding=2em}}

	~~Momentum:~~

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}		\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
	</math>}}		</math>\|background-color=whitesmoke\|color=steelblue\|description=Momentum:\|nonumber=1\|padding=2em}}

	~~Energy:~~

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>}}		</math>\|background-color=whitesmoke\|color=steelblue\|description=Energy:\|nonumber=1\|padding=2em}}
			</div>

	=== Alternative Forms of the Energy Equation ===		=== Alternative Forms of the Energy Equation ===

Nian at 19:13, 31 March 2026

2026-03-31T19:13:31Z

← Older revision		Revision as of 19:13, 31 March 2026
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0		\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
	</math>}}		</math>\|label=eq-cont-pde}}

	which is the continuity equation on partial differential form.		which is the continuity equation on partial differential form.
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}		\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
	</math>}}		</math>\|label=eq-mom-pde}}

	which is the momentum equation on partial differential form		which is the momentum equation on partial differential form
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>}}		</math>\|label=eq-energy-pde}}

	which is the energy equation on partial differential form		which is the energy equation on partial differential form
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla		\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla
	</math>}}		</math>\|label=eq-cont-pde-non-cons}}

	where the first term of the right hand side is the local derivative and the second term is the convective derivative.		where the first term of the right hand side is the local derivative and the second term is the convective derivative.
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}		\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
	</math>}}		</math>\|label=eq-mom-pde-non-cons}}

	==== Conservation of Energy ====		==== Conservation of Energy ====

	The last equation on non-conservation differential form is the energy equation. We start by rewriting the energy equation on differential form ~~(Eqn. \ref~~{eq~~:governing:~~energy:pde}), repeated here for convenience		The last equation on non-conservation differential form is the energy equation. We start by rewriting the energy equation on differential form {{EquationNote\|label=eq-energy-pde}}, repeated here for convenience

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>}}		</math>\|nonumber=1}}

	Total enthalpy, <math>h_o</math>, is replaced with total energy, <math>e_o</math>		Total enthalpy, <math>h_o</math>, is replaced with total energy, <math>e_o</math>
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\rho\frac{\partial e_o}{\partial t} + e_o\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla e_o + e_o\nabla\cdot(\rho \mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\rho\frac{\partial e_o}{\partial t} + e_o\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla e_o + e_o\nabla\cdot(\rho \mathbf{v}) + \nabla\cdot(p\mathbf{v})=</math><br><br><math>= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>}}		</math>}}

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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho		\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
	</math>}}		</math>\|label=eq-energy-pde-non-cons}}

	==== Summary ====		==== Summary ====
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	</math>}}		</math>}}

	Inserted in ~~Eqn. \ref~~{eq~~:governing:~~energy:non}, this gives		Inserted in {{EquationNote\|label=eq-energy-pde-non-cons\|nopar=1}}, this gives

	{{NumEqn\|<math>		{{NumEqn\|<math>
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	</math>}}		</math>}}

	Now, let's replace the substantial derivative <math>D\mathbf{v}/Dt</math> using the momentum equation on non-conservation form ~~(Eqn. \ref~~{eq~~:governing:~~mom:non}).		Now, let's replace the substantial derivative <math>D\mathbf{v}/Dt</math> using the momentum equation on non-conservation form {{EquationNote\|label=eq-mom-pde-non-cons}}.

	{{NumEqn\|<math>		{{NumEqn\|<math>
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\rho\frac{De}{Dt} \cancel{-\mathbf{v}\cdot\nabla p} + \cancel{\mathbf{v}\cdot\nabla p} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\Rightarrow \rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho		\rho\frac{De}{Dt} \cancel{-\mathbf{v}\cdot\nabla p} + \cancel{\mathbf{v}\cdot\nabla p} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\Rightarrow</math><br><br><math>\Rightarrow\rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho
	</math>}}		</math>}}

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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q}		\frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q}
	</math>}}		</math>\|label=eq-energy-pde-non-cons-b}}

	Conservation of mass gives		Conservation of mass gives
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	</math>}}		</math>}}

	Insert in ~~Eqn. \ref~~{eq~~:governing:~~energy:non:b}		Insert in {{EquationNote\|label=eq-energy-pde-non-cons-b\|nopar=1}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
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	</math>}}		</math>}}

	with <math>De/Dt</math> from ~~Eqn. \ref~~{eq~~:governing:~~energy:non:b}		with <math>De/Dt</math> from {{EquationNote\|label=eq-energy-pde-non-cons-b\|nopar=1}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}		\frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}
	</math>}}		</math>\|label=eq-energy-pde-non-cons-c}}

	==== Total Enthalpy Formulation ====		==== Total Enthalpy Formulation ====
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	</math>}}		</math>}}

	From the momentum equation ~~(Eqn. \ref~~{eq~~:governing:~~mom:non})		From the momentum equation {{EquationNote\|label=eq-mom-pde-non-cons}}

	{{NumEqn\|<math>		{{NumEqn\|<math>
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	</math>}}		</math>}}

	Inserting <math>Dh/Dt</math> from ~~Eqn. \ref~~{eq~~:governing:~~energy:non:c} gives		Inserting <math>Dh/Dt</math> from {{EquationNote\|label=eq-energy-pde-non-cons-c\|nopar=1}} gives

	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{Dh_o}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p = \frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}		\frac{Dh_o}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p =</math><br><br><math>=\frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}
	</math>}}		</math>}}

Nian at 18:40, 31 March 2026

2026-03-31T18:40:51Z

← Older revision		Revision as of 18:40, 31 March 2026
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0		\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
	</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue\|numw=2em}}

	Apply Gauss's divergence theorem on the surface integral gives		Apply Gauss's divergence theorem on the surface integral gives
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	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV		\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
	</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue\|numw=2em}}

	As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.		As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
Line 96:		Line 96:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV		\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
	</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue\|numw=2em}}

	Gauss's divergence theorem applied to the surface integral term in the energy equation gives		Gauss's divergence theorem applied to the surface integral term in the energy equation gives

Nian at 18:38, 31 March 2026

2026-03-31T18:38:36Z

← Older revision		Revision as of 18:38, 31 March 2026
Line 24:		Line 24:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0		\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
	</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue}}

	Apply Gauss's divergence theorem on the surface integral gives		Apply Gauss's divergence theorem on the surface integral gives
Line 58:		Line 58:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV		\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
	</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue}}

	As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.		As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
Line 96:		Line 96:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV		\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
	</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em\|color=steelblue}}

	Gauss's divergence theorem applied to the surface integral term in the energy equation gives		Gauss's divergence theorem applied to the surface integral term in the energy equation gives

Nian at 17:38, 31 March 2026

2026-03-31T17:38:37Z

← Older revision		Revision as of 17:38, 31 March 2026
Line 24:		Line 24:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0		\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
	</math>\|nonumber=1}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em}}

	Apply Gauss's divergence theorem on the surface integral gives		Apply Gauss's divergence theorem on the surface integral gives
Line 30:		Line 30:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV		\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV
	</math>~~\|number=5.34\|spar=1\|prefix=ekv.\|border=1\|padding=2em~~}}		</math>}}


	Also, if <math>\Omega</math> is a fixed control volume		Also, if <math>\Omega</math> is a fixed control volume
Line 59:		Line 58:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV		\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
	</math>\|nonumber=1}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em}}

	As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.		As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
Line 97:		Line 96:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV		\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
	</math>\|nonumber=1}}		</math>\|nonumber=1\|border=1\|background-color=whitesmoke\|padding=2em}}

	Gauss's divergence theorem applied to the surface integral term in the energy equation gives		Gauss's divergence theorem applied to the surface integral term in the energy equation gives

Nian at 14:49, 31 March 2026

2026-03-31T14:49:57Z

← Older revision		Revision as of 14:49, 31 March 2026
Line 30:		Line 30:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV		\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV
	</math>}}		</math>\|number=5.34\|spar=1\|prefix=ekv.\|border=1\|padding=2em}}


	Also, if <math>\Omega</math> is a fixed control volume		Also, if <math>\Omega</math> is a fixed control volume

Nian at 07:54, 31 March 2026

2026-03-31T07:54:21Z

← Older revision		Revision as of 07:54, 31 March 2026
Line 20:		Line 20:
	==== Conservation of Mass ====		==== Conservation of Mass ====

	Apply Gauss's divergence theorem on the surface integral ~~in Eqn. \ref{eq:governing:cont:int}~~ gives		The continuity equation on integral form reads

			{{NumEqn\|<math>
			\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
			</math>\|nonumber=1}}

			Apply Gauss's divergence theorem on the surface integral gives

	{{NumEqn\|<math>		{{NumEqn\|<math>
Line 47:		Line 53:

	==== Conservation of Momentum ====		==== Conservation of Momentum ====

			The momentum equation on integral form reads

			{{NumEqn\|<math>
			\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
			</math>\|nonumber=1}}

	As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.		As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
Line 80:		Line 92:
	==== Conservation of Energy ====		==== Conservation of Energy ====

	Gauss's divergence theorem applied to the surface integral term in the energy equation ~~(Eqn. \ref{eq:governing:energy:int})~~ gives		The energy equation on integral form reads

			{{NumEqn\|<math>
			\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
			</math>\|nonumber=1}}

			Gauss's divergence theorem applied to the surface integral term in the energy equation gives

	{{NumEqn\|<math>		{{NumEqn\|<math>
Line 158:		Line 176:
	{{NumEqn\|<math>		{{NumEqn\|<math>
	\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0		\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
	</math>}}		</math>\|label=eq-pde-noncons-cont}}

	~~Eqn. \ref~~{eq~~:governing:~~cont~~:non~~} says that the mass of a fluid element with a fixed set of fluid particles is constant as the element moves in space.		{{EquationNote\|label=eq-pde-noncons-cont\|nopar=1}} says that the mass of a fluid element with a fixed set of fluid particles is constant as the element moves in space.

	==== Conservation of Momentum ====		==== Conservation of Momentum ====