One-dimensional flow with friction - Revision history

Nian: /* Friction Choking */

2026-04-01T18:26:20Z

Friction Choking

← Older revision		Revision as of 18:26, 1 April 2026
Line 476:		Line 476:
	Figure~\ref{fig:friction:Ts} shows the Fanno flow process in a <math>Ts</math>-diagram. The dashed line represents the sonic temperature, which means that the flow states along the process line above the dashed line are subsonic flow states and the part of the line below the dashed line represents supersonic flow states. In both subsonic and supersonic flow addition of friction leads to a change in temperature in the direction towards the sonic temperature, i.e. the flow approaches sonic conditions (<math>M=1</math>). When the length of the pipe through which the fluid flows is equal to the length at which the flow is sonic, the flow is choked (friction choking) and further pipe length cannot be added without a change in the flow conditions. For an initially subsonic flow, a pipe longer than <math>L^\ast</math>, the change in flow conditions is analogous to the what happens for addition of heat to a subsonic flow that has reached sonic state discussed in the previous section. The inlet conditions will change such that the massflow is reduced without changing the inlet total conditions such as the pipe length is equal to <math>L^\ast</math> for the new inlet conditions.		Figure~\ref{fig:friction:Ts} shows the Fanno flow process in a <math>Ts</math>-diagram. The dashed line represents the sonic temperature, which means that the flow states along the process line above the dashed line are subsonic flow states and the part of the line below the dashed line represents supersonic flow states. In both subsonic and supersonic flow addition of friction leads to a change in temperature in the direction towards the sonic temperature, i.e. the flow approaches sonic conditions (<math>M=1</math>). When the length of the pipe through which the fluid flows is equal to the length at which the flow is sonic, the flow is choked (friction choking) and further pipe length cannot be added without a change in the flow conditions. For an initially subsonic flow, a pipe longer than <math>L^\ast</math>, the change in flow conditions is analogous to the what happens for addition of heat to a subsonic flow that has reached sonic state discussed in the previous section. The inlet conditions will change such that the massflow is reduced without changing the inlet total conditions such as the pipe length is equal to <math>L^\ast</math> for the new inlet conditions.

	{{~~NumEqn~~\|<math>		{{InfoBox\|<math>
	M_{1^\prime} = f(L^\ast)		M_{1^\prime} = f(L^\ast)
	</math><br><br><math>		</math><br><br><math>
Line 488:		Line 488:
	</math><br><br><math>		</math><br><br><math>
	u_{1^\prime} = M_{1^\prime}a_{1^\prime}		u_{1^\prime} = M_{1^\prime}a_{1^\prime}
	</math>~~\|nonumber=1\|infobox=1~~}}		</math>}}

	<!--		<!--

Nian at 08:57, 1 April 2026

2026-04-01T08:57:33Z

Show changes

Nian at 19:38, 20 March 2026

2026-03-20T19:38:37Z

← Older revision		Revision as of 19:38, 20 March 2026
Line 1:		Line 1:
	[[Category:Compressible flow]]		[[Category:Compressible flow]]
	~~[[Category:Topic]]~~
	[[Category:One-dimensional flow]]		[[Category:One-dimensional flow]]
	[[Category:Inviscid flow]]		[[Category:Inviscid flow]]
	[[Category:Continuous solution]]		[[Category:Continuous solution]]
			<noinclude>
			[[Category:Compressible flow:Topic]]
			</noinclude>

	__TOC__		__TOC__

Nian at 17:10, 20 March 2026

2026-03-20T17:10:56Z

← Older revision		Revision as of 17:10, 20 March 2026
Line 1:		Line 1:
	[[Category:Compressible flow]]		[[Category:Compressible flow]]
			[[Category:Topic]]
	[[Category:One-dimensional flow]]		[[Category:One-dimensional flow]]
	[[Category:Inviscid flow]]		[[Category:Inviscid flow]]

Nian: /* Friction Choking */

2026-03-17T12:59:37Z

Friction Choking

← Older revision		Revision as of 12:59, 17 March 2026
Line 549:		Line 549:
	</math>		</math>

	Figure~\ref{fig:friction:factor:shock} shows the development of choking length $L_1^\ast$ in a supersonic flow as a function of Mach number in relation to the corresponding choking length <math>L_2^\ast</math> downstream of a normal shock generated at the same Mach number. As can be seen from the figure, a normal shock will always increase the choking length.		Figure~\ref{fig:friction:factor:shock} shows the development of choking length <math>L_1^\ast</math> in a supersonic flow as a function of Mach number in relation to the corresponding choking length <math>L_2^\ast</math> downstream of a normal shock generated at the same Mach number. As can be seen from the figure, a normal shock will always increase the choking length.

	<!--		<!--

Nian: /* Friction Choking */

2026-03-17T12:58:00Z

Friction Choking

Show changes

Nian: /* Differential Relations */

2026-03-17T12:50:16Z

Differential Relations

← Older revision		Revision as of 12:50, 17 March 2026
Line 434:		Line 434:
	Using the differential relations above, we can get a good picture of the development of flow variables as friction is continuously added to the flow (see Figure~\ref{fig:fanno:trends}).		Using the differential relations above, we can get a good picture of the development of flow variables as friction is continuously added to the flow (see Figure~\ref{fig:fanno:trends}).

			<!--
	\begin{figure}[ht!]		\begin{figure}[ht!]
	\begin{subfigure}[b]{0.5\textwidth}		\begin{subfigure}[b]{0.5\textwidth}
Line 449:		Line 449:
	\label{fig:fanno:trends}		\label{fig:fanno:trends}
	\end{figure}		\end{figure}
			-->

	==== Friction Choking ====		==== Friction Choking ====

Nian: /* Differential Relations */

2026-03-17T12:49:30Z

Differential Relations

Show changes

Nian: /* Momentum equation */

2026-03-17T12:45:29Z

Momentum equation

← Older revision		Revision as of 12:45, 17 March 2026
Line 284:		Line 284:
	==== Momentum equation ====		==== Momentum equation ====

	By combining the above derived relations and the momentum equation on the form given by (3.95), we can get an expression where the friction force is a function of Mach number only\\		By combining the above derived relations and the momentum equation on the form given by (3.95), we can get an expression where the friction force is a function of Mach number only

	~~\noindent~~ For convenience equation (3.95) is written again here\\		For convenience equation (3.95) is written again here

	\[dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}~~\ (3.95)\]\\~~		<math display="block">
			dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}
			</math>

	~~\noindent~~ if $u\neq 0.$, we can divide by $0.5\rho u^2$ which gives\\		if <math>u\neq 0.</math>, we can divide by <math>0.5\rho u^2</math> which gives

	~~\begin{equation}~~		<math display="block">
	2\frac{dp}{\rho u^2}+2\frac{\rho u du}{\rho u^2}=-\frac{4 f dx}{D}		2\frac{dp}{\rho u^2}+2\frac{\rho u du}{\rho u^2}=-\frac{4 f dx}{D}
	~~\label{eqn:mom:a}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ using $M^2=u^2/a^2$, $a^2=\gamma p/\rho$ and the chain rule in (\ref{eqn:mom:a}) gives\\		using <math>M^2=u^2/a^2</math>, <math>a^2=\gamma p/\rho</math> and the chain rule in (\ref{eqn:mom:a}) gives

	~~\begin{equation}~~		<math display="block">
	\frac{2}{\gamma M^2}\frac{dp}{p}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}		\frac{2}{\gamma M^2}\frac{dp}{p}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}
	~~\label{eqn:mom:b}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ From equation (\ref{eqn:gaslaw}) we can get a relation that expresses the pressure derivative term, $dp/p$, in terms of Mach number and $d(u^2)/u^2$. Inserting this in (\ref{eqn:mom:b}) gives\\		From equation (\ref{eqn:gaslaw}) we can get a relation that expresses the pressure derivative term, <math>dp/p</math>, in terms of Mach number and <math>d(u^2)/u^2</math>. Inserting this in (\ref{eqn:mom:b}) gives

	~~\begin{equation}~~		<math display="block">
	\frac{2}{\gamma M^2}\left\{-\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}\right\}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}		\frac{2}{\gamma M^2}\left\{-\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}\right\}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}
	~~\label{eqn:mom:c}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ collecting terms and rearranging gives\\		collecting terms and rearranging gives

	~~\begin{equation}~~		<math display="block">
	\frac{M^2-1}{\gamma M^2}\frac{d(u^2)}{u^2}=\frac{4 f dx}{D}		\frac{M^2-1}{\gamma M^2}\frac{d(u^2)}{u^2}=\frac{4 f dx}{D}
	~~\label{eqn:mom:d}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ if we now use equation (\ref{eqn:ttot:Mach}) to get rid of the $d(u^2)/u^2$-term we end up with the following expression\\		if we now use equation (\ref{eqn:ttot:Mach}) to get rid of the <math>d(u^2)/u^2</math>-term we end up with the following expression

	~~\begin{equation*}~~		<math display="block">
	\frac{4 f dx}{D}=\frac{2}{\gamma M^2}(1-M^2)\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{dM}{M}		\frac{4 f dx}{D}=\frac{2}{\gamma M^2}(1-M^2)\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{dM}{M}
	~~\end{equation*}~~		</math>

	==== Differential Relations ====		==== Differential Relations ====

Nian: /* The ideal gas law */

2026-03-17T12:42:23Z

The ideal gas law

← Older revision		Revision as of 12:42, 17 March 2026
Line 238:		Line 238:
	</math>		</math>

	===== The ideal gas law =====		==== The ideal gas law ====
	~~\noindent For a perfect gas the ideal gas law reads\\~~

	~~\begin{equation}~~		For a perfect gas the ideal gas law reads

			<math display="block">
	p=\rho R T		p=\rho R T
	~~\label{eqn:gaslaw:a}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ Differentiating (\ref{eqn:gaslaw:a}) gives:\\		Differentiating (\ref{eqn:gaslaw:a}) gives:

	~~\begin{equation}~~		<math display="block">
	dp=\rho R dT+RT d\rho		dp=\rho R dT+RT d\rho
	~~\label{eqn:gaslaw:b}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ If $p\neq0.$, we can divide (\ref{eqn:gaslaw:b}) by $p$ which gives\\		If <math>p\neq0.</math>, we can divide (\ref{eqn:gaslaw:b}) by <math>p</math> which gives

	~~\begin{equation}~~		<math display="block">
	\frac{dp}{p}=\frac{dT}{T}+\frac{d\rho}{\rho}		\frac{dp}{p}=\frac{dT}{T}+\frac{d\rho}{\rho}
	~~\label{eqn:gaslaw:c}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ which can be rearranged according to\\		which can be rearranged according to

	~~\begin{equation}~~		<math display="block">
	\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]=\frac{dT}{T}		\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]=\frac{dT}{T}
	~~\label{eqn:gaslaw:d}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ Now, inserting $dT/T$ from equation (\ref{eqn:ttot}) gives\\		Now, inserting <math>dT/T</math> from equation (\ref{eqn:ttot}) gives

	~~\begin{equation}~~		<math display="block">
	\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.		\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
	~~\label{eqn:gaslaw:b}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ The $d\rho/\rho$-term can be replaced using equation (\ref{eqn:cont})\\		The <math>d\rho/\rho</math>-term can be replaced using equation (\ref{eqn:cont})

	~~\begin{equation}~~		<math display="block">
	\frac{dp}{p}+\frac{d(u^2)}{2u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.		\frac{dp}{p}+\frac{d(u^2)}{2u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
	~~\label{eqn:gaslaw:c}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ Collect terms and rewrite gives\\		Collect terms and rewrite gives

	~~\begin{equation}~~		<math display="block">
	\frac{dp}{p}+\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}=0.		\frac{dp}{p}+\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}=0.
	~~\label{eqn:gaslaw}~~		</math>
	~~\end{equation}\\~~

	==== Momentum equation ====		==== Momentum equation ====