One-dimensional flow with friction: Difference between revisions

Latest revision as of 18:26, 1 April 2026

Flow-station data

The starting point is the governing equations for one-dimensional steady-state flow

Continuity

$ρ_{1} u_{1} = ρ_{2} u_{2}$

Momentum

ρ_{1} u_{1}^{2} + p_{1} - {\bar{τ}}_{w} \frac{b L}{A} = ρ_{2} u_{2}^{2} + p_{2}

(Eq. 3.126)

where ${\bar{τ}}_{w}$ is the average wall-shear stress

{\bar{τ}}_{w} = \frac{1}{L} \int_{0}^{L} τ_{w} d x

(Eq. 3.127)

$b$ is the tube perimeter, and $L$ is the tube length. For circular cross sections

\frac{b L}{A} = {A = \frac{π D^{2}}{4}, b = π D} = \frac{4 L}{D}

(Eq. 3.128)

and thus

ρ_{1} u_{1}^{2} + p_{1} - \frac{4}{D} \int_{0}^{L} τ_{w} d x = ρ_{2} u_{2}^{2} + p_{2}

(Eq. 3.129)

Energy

h_{1} + \frac{1}{2} u_{1}^{2} = h_{2} + \frac{1}{2} u_{2}^{2}

(Eq. 3.130)

Differential Form

In order to remove the integral term in the momentum equation, the governing equations are written in differential form

Continuity

ρ_{1} u_{1} = ρ_{2} u_{2} = c o n s t \Rightarrow

(Eq. 3.131)

\frac{d}{d x} (ρ u) = 0

(Eq. 3.132)

Momentum

(ρ_{2} u_{2}^{2} + p_{2} - ρ_{1} u_{1}^{2} + p_{1}) = - \frac{4}{D} \int_{0}^{L} τ_{w} d x \Rightarrow

(Eq. 3.133)

\frac{d}{d x} (ρ u^{2} + p) = - \frac{4}{D} τ_{w}

(Eq. 3.134)

\frac{d}{d x} (ρ u^{2} + p) = ρ u \frac{d u}{d x} + u \frac{d}{d x} (ρ u) + \frac{d p}{d x} = {\frac{d}{d x} (ρ u) = 0} = ρ u \frac{d u}{d x} + \frac{d p}{d x}

(Eq. 3.135)

ρ u \frac{d u}{d x} + \frac{d p}{d x} = - \frac{4}{D} τ_{w}

(Eq. 3.136)

The wall shear stress is often approximated using a shear-stress factor, $f$ , according to

τ_{w} = f \frac{1}{2} ρ u^{2}

(Eq. 3.137)

and thus

ρ u \frac{d u}{d x} + \frac{d p}{d x} = - \frac{2}{D} f ρ u^{2}

(Eq. 3.138)

Energy

h_{1} + \frac{1}{2} u_{1}^{2} = h_{2} + \frac{1}{2} u_{2}^{2} = c o n s t

(Eq. 3.139)

h_{o_{1}} = h_{o_{2}} = c o n s t

(Eq. 3.140)

\frac{d}{d x} h_{o} = 0

(Eq. 3.141)

Summary

continuity:

\frac{d}{d x} (ρ u) = 0

(Eq. 3.142)

momentum:

ρ u \frac{d u}{d x} + \frac{d p}{d x} = - \frac{2}{D} f ρ u^{2}

(Eq. 3.143)

energy:

\frac{d}{d x} h_{o} = 0

(Eq. 3.144)

From chapter 3.9 we have the following expression for the momentum equation for one-dimensional flow with friction (equation (3.95))

d p + ρ u d u = - \frac{1}{2} ρ u^{2} \frac{4 f d x}{D}

(Eq. 3.145)

For cases dealing with calorically perfect gas, (3.95) can be recast completely in terms of Mach number using the following relations

speed of sound: $a^{2} = γ p / ρ$
the definition of Mach number: $M^{2} = u^{2} / a^{2}$
the ideal gas law for thermally perfect gas: $p = ρ R T$
the continuity equation: $ρ u = c o n s t$
the energy equation: $C_{p} T + u^{2} / 2 = c o n s t$

Continuity equation

We start with the continuity equation which for one-dimensional steady flows reads

ρ u = c o n s t

(Eq. 3.146)

Differentiating (\ref{eqn:cont:a}) gives

d (ρ u) = 0. \Leftrightarrow ρ d u + u d ρ = 0.

(Eq. 3.147)

If $u \neq 0.$ we can divide by $ρ u$ which gives us

\frac{d u}{u} + \frac{d ρ}{ρ} = 0.

(Eq. 3.148)

Now, if we divide and multiply the first term in (\ref{eqn:cont:b}) by $2 u$ and use the chain rule for derivatives we get

\frac{d (u^{2})}{2 u^{2}} + \frac{d ρ}{ρ} = 0.

(Eq. 3.149)

Energy equation

For an adiabatic one-dimensional flow we have that

C_{p} T + \frac{u^{2}}{2} = c o n s t

(Eq. 3.150)

If we differentiate (\ref{eqn:ttot:a}) we get

C_{p} d T + \frac{1}{2} d (u^{2}) = 0.

(Eq. 3.151)

We replace $C_{p}$ with $γ R / (γ - 1)$ and multiply and divide the first term with $T$ which gives us

\frac{γ R T}{(γ - 1)} \frac{d T}{T} + \frac{1}{2} d (u^{2}) = 0.

(Eq. 3.152)

Now, divide by $γ R T / (γ - 1)$ and multiply and divide the second term by $u^{2}$ gives

\frac{d T}{T} + \frac{(γ - 1)}{2} M^{2} \frac{d (u^{2})}{u^{2}} = 0.

(Eq. 3.153)

We want to remove the $d T / T$ -term in (\ref{eqn:ttot}). From the definition of Mach number we have that

a^{2} M^{2} = u^{2}

(Eq. 3.154)

which we can rewrite using the expression for speed of sound $(a^{2} = γ R T)$ according to

γ R T M^{2} = u^{2}

(Eq. 3.155)

Differentiating (\ref{eqn:Mach:b}) gives us

γ R M^{2} d T + γ R T d (M^{2}) = d (u^{2})

(Eq. 3.156)

Now, if we divide (\ref{eqn:Mach:c}) by $γ R T M^{2}$ and use $a^{2} = γ R T$ and $a^{2} M^{2} = u^{2}$ we get

\frac{d T}{T} + \frac{d (M^{2})}{M^{2}} = \frac{d (u^{2})}{u^{2}}

(Eq. 3.157)

Equation (\ref{eqn:Mach}) may now be used to replace the $d T / T$ -term in equation (\ref{eqn:ttot})

- \frac{d (M^{2})}{M^{2}} + \frac{d (u^{2})}{u^{2}} + \frac{(γ - 1)}{2} M^{2} \frac{d (u^{2})}{u^{2}} = 0.

(Eq. 3.158)

which can be rewritten according to

\frac{d (u^{2})}{u^{2}} = {[1 + \frac{(γ - 1)}{2} M^{2}]}^{- 1} \frac{d (M^{2})}{M^{2}}

(Eq. 3.159)

Using the chain rule for derivatives, the last term may be rewritten according to

\frac{d (M^{2})}{M^{2}} = 2 M \frac{d M}{M^{2}} = 2 \frac{d M}{M}

(Eq. 3.160)

which gives

\frac{d (u^{2})}{u^{2}} = 2 {[1 + \frac{(γ - 1)}{2} M^{2}]}^{- 1} \frac{d M}{M}

(Eq. 3.161)

The ideal gas law

For a perfect gas the ideal gas law reads

p = ρ R T

(Eq. 3.162)

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

d p = ρ R d T + R T d ρ

(Eq. 3.163)

If $p \neq 0.$ , we can divide (\ref{eqn:gaslaw:b}) by $p$ which gives

\frac{d p}{p} = \frac{d T}{T} + \frac{d ρ}{ρ}

(Eq. 3.164)

which can be rearranged according to

[\frac{d p}{p} - \frac{d ρ}{ρ}] = \frac{d T}{T}

(Eq. 3.165)

Now, inserting $d T / T$ from equation (\ref{eqn:ttot}) gives

[\frac{d p}{p} - \frac{d ρ}{ρ}] + \frac{(γ - 1)}{2} M^{2} \frac{d (u^{2})}{u^{2}} = 0.

(Eq. 3.166)

The $d ρ / ρ$ -term can be replaced using equation (\ref{eqn:cont})

\frac{d p}{p} + \frac{d (u^{2})}{2 u^{2}} + \frac{(γ - 1)}{2} M^{2} \frac{d (u^{2})}{u^{2}} = 0.

(Eq. 3.167)

Collect terms and rewrite gives

\frac{d p}{p} + [\frac{1 + (γ - 1) M^{2}}{2}] \frac{d (u^{2})}{u^{2}} = 0.

(Eq. 3.168)

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

For convenience equation (3.95) is written again here

d p + ρ u d u = - \frac{1}{2} ρ u^{2} \frac{4 f d x}{D}

(Eq. 3.169)

if $u \neq 0.$ , we can divide by $0.5 ρ u^{2}$ which gives

2 \frac{d p}{ρ u^{2}} + 2 \frac{ρ u d u}{ρ u^{2}} = - \frac{4 f d x}{D}

(Eq. 3.170)

using $M^{2} = u^{2} / a^{2}$ , $a^{2} = γ p / ρ$ and the chain rule in (\ref{eqn:mom:a}) gives

\frac{2}{γ M^{2}} \frac{d p}{p} + \frac{d (u^{2})}{u^{2}} = - \frac{4 f d x}{D}

(Eq. 3.171)

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

\frac{2}{γ M^{2}} {- [\frac{1 + (γ - 1) M^{2}}{2}] \frac{d (u^{2})}{u^{2}}} + \frac{d (u^{2})}{u^{2}} = - \frac{4 f d x}{D}

(Eq. 3.172)

collecting terms and rearranging gives

\frac{M^{2} - 1}{γ M^{2}} \frac{d (u^{2})}{u^{2}} = \frac{4 f d x}{D}

(Eq. 3.173)

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

\frac{4 f d x}{D} = \frac{2}{γ M^{2}} (1 - M^{2}) {[1 + \frac{(γ - 1)}{2} M^{2}]}^{- 1} \frac{d M}{M}

(Eq. 3.174)

Differential Relations

In analogy with the heat addition process discussed in the previous section, one-dimensional flow with heat addition is a continuous process. We will derive the differential relations for one-dimensional flow with friction, which will lead to trends for supersonic and supersonic flow with friction.

The continuity equation gives

d (ρ u) = u d ρ + ρ d u \Rightarrow

(Eq. 3.175)

\frac{d ρ}{ρ} = - \frac{d u}{u}

(Eq. 3.176)

The addition of friction does not affect total temperature and thus the total temperature is constant

T_{o} = T + \frac{u^{2}}{2 C_{p}} = c o n s t

(Eq. 3.177)

differentiating gives

d T_{o} = d T + \frac{1}{C p} u d u = 0

(Eq. 3.178)

with $u = M \sqrt{γ R T}$ , we get

\frac{d T}{T} = - (γ - 1) M^{2} \frac{d u}{u}

(Eq. 3.179)

A differential relation for pressure can be obtained from the ideal gas relation

p = ρ R T \Rightarrow d p = R (T d ρ + ρ d T) \Rightarrow

(Eq. 3.180)

\frac{d p}{p} = - (1 + (γ - 1) M^{2}) \frac{d u}{u}

(Eq. 3.181)

The entropy increase can be obtained from

d s = C_{v} \frac{d p}{p} - C_{p} \frac{d ρ}{ρ}

(Eq. 3.182)

and thus

d s = - R (1 - M^{2}) \frac{d u}{u}

(Eq. 3.183)

Finally, a relation describing the change in Mach number can be obtained from

M = \frac{u}{\sqrt{γ R T}} \Rightarrow d M = M \frac{d u}{u} - \frac{M}{2} \frac{d T}{T}

(Eq. 3.184)

which can be rewritten as

\frac{d M}{M} = (1 + (γ - 1) M^{2}) \frac{d u}{u}

(Eq. 3.185)

Eqns.~\ref{eq:fanno:drho} - \ref{eq:fanno:dM} are expressed as functions of $d u$ and in order to get a direct relation to the addition of friction caused by the increase in pipe length $d x$ , the equations are rewritten so that all variable changes are functions of the entropy increase $d s$ .

\frac{d ρ}{ρ} = - \frac{1}{R (1 - M^{2})} d s

(Eq. 3.186)

\frac{d T}{T} = - (γ - 1) M^{2} \frac{1}{R (1 - M^{2})} d s

(Eq. 3.187)

\frac{d p}{p} = - (1 + (γ - 1) M^{2}) \frac{1}{R (1 - M^{2})} d s

(Eq. 3.188)

\frac{d M}{M} = (1 + \frac{(γ - 1)}{2} M^{2}) \frac{1}{R (1 - M^{2})} d s

(Eq. 3.189)

\frac{d u}{u} = \frac{1}{R (1 - M^{2})} d s

(Eq. 3.190)

A relation for the change in total pressure can be obtained from

d s = C_{p} \frac{d T_{o}}{T_{o}} - R \frac{d p_{o}}{p_{o}}

(Eq. 3.191)

Since total temperature is constant the relation above gives

\frac{d p_{o}}{p_{o}} = - \frac{d s}{R}

(Eq. 3.192)

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}).

Friction Choking

Figure~\ref{fig:friction:Ts} shows the Fanno flow process in a $T s$ -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 ( $M = 1$ ). 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 $L^{*}$ , 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 $L^{*}$ for the new inlet conditions.

M_{1^{'}} = f (L^{*})

T_{1^{'}} = f (T_{o}, M_{1^{'}})

p_{1^{'}} = f (p_{o}, M_{1^{'}})

ρ_{1^{'}} = f (p_{1^{'}}, T_{1^{'}})

a_{1^{'}} = f (T_{1^{'}})

u_{1^{'}} = M_{1^{'}} a_{1^{'}}

For a choked supersonic flow, addition of more friction (increasing the length of the pipe such that $L > L^{*}$ ) may lead to the generation of a shock inside the pipe. In contrast to the one-dimensional flow with heat addition where a shock does not change $q^{*}$ , $L^{*}$ is increased over a shock. The internal shock will be generated in an axial location such that $L^{*}$ downstream of the shock equals the remaining pipe length at the shock location (see Figure~\ref{fig:friction:choking:sup}). As more length is added to the pipe, the shock will move further and further upstream in the pipe until it stands at the pipe entrance. If the pipe is longer than $L^{*}$ after o shock standing at the inlet, the shock will move to the upstream system and the pipe flow will be subsonic and the massflow will be adjusted such that $L = L^{*}$ according to the process described for subsonic choking above.

From prvevious derivations, we know that $L^{*}$ is a function of mach number according to

\frac{4 \bar{f} L^{*}}{D} = \frac{1 - M^{2}}{γ M^{2}} + (\frac{γ + 1}{2 γ}) \ln (\frac{(γ + 1) M^{2}}{2 + (γ - 1) M^{2}})

(Eq. 3.193)

by dividing both the numerator and denominator in the fractions by $M^{2}$ it is easy to see that the choking length (Figure~\ref{fig:friction:factor}) approaches a finite length for great Mach numbers and thus the upper limit for the choking length $L_{1}^{*}$ is given by

{\frac{4 \bar{f} L_{1}^{*}}{D} (M_{1}) |}_{M_{1} \to \infty} = - \frac{1}{γ} + (\frac{γ + 1}{2 γ}) \ln (\frac{γ + 1}{γ - 1})

(Eq. 3.194)

From the normal shock relations we know that the downstream Mach number approaches the finite value $\sqrt{(γ - 1) / 2 γ}$ large Mach numbers and thus the choking length downstream the shock is limited to

{\frac{4 \bar{f} L_{2}^{*}}{D} (M_{2}) |}_{M_{1} \to \infty} = (\frac{γ + 1}{γ (γ - 1)}) + (\frac{γ + 1}{2 γ}) \ln (\frac{(γ + 1) (γ - 1)}{4 γ + (γ - 1)^{2}})

(Eq. 3.195)

From the relations above we get

{(\frac{4 \bar{f} L_{2}^{*}}{D} (M_{2}) - \frac{4 \bar{f} L_{1}^{*}}{D} (M_{1})) |}_{M_{1} \to \infty} = (\frac{2}{γ - 1}) + (\frac{γ + 1}{2 γ}) \ln [(\frac{(γ + 1) (γ - 1)}{4 γ + (γ - 1)^{2}}) (\frac{γ - 1}{γ + 1})]

(Eq. 3.196)

Figure~\ref{fig:friction:factor:shock} shows the development of choking length $L_{1}^{*}$ in a supersonic flow as a function of Mach number in relation to the corresponding choking length $L_{2}^{*}$ 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.

One-dimensional flow with friction: Difference between revisions

Latest revision as of 18:26, 1 April 2026

Contents

Flow-station data

Continuity

Momentum

Energy

Differential Form

Continuity

Momentum

Energy

Summary

Continuity equation

Energy equation

The ideal gas law

Momentum equation

Differential Relations

Friction Choking

Navigation menu

@@ Line 1: / Line 1: @@
-[[Category:Compressible flow]]
+<!--
-[[Category:One-dimensional flow]]
+-->[[Category:Compressible flow]]<!--
-[[Category:Inviscid flow]]
+-->[[Category:One-dimensional flow]]<!--
-[[Category:Continuous solution]]
+-->[[Category:Inviscid flow]]<!--
+-->[[Category:Continuous solution]]<!--
+--><noinclude><!--
+-->[[Category:Compressible flow:Topic]]<!--
+--></noinclude><!--
-__TOC__
+--><nomobile><!--
+-->__TOC__<!--
+--></nomobile><!--
+--><noinclude><!--
+-->{{#vardefine:secno|3}}<!--
+-->{{#vardefine:eqno|125}}<!--
+--></noinclude><!--
+-->
 ==== Flow-station data ====
@@ Line 18: / Line 30: @@
 ===== Momentum =====
-<math display="block">
+{{NumEqn|<math>
 \rho_1 u_1^2+p_1-\bar{\tau}_w\frac{bL}{A}=\rho_2 u_2^2+p_2
-</math>
+</math>}}
 where <math>\bar{\tau}_w</math> is the average wall-shear stress
-<math display="block">
+{{NumEqn|<math>
 \bar{\tau}_w=\frac{1}{L}\int_0^L\tau_w dx
-</math>
+</math>}}
 <math>b</math> is the tube perimeter, and <math>L</math> is the tube length. For circular cross sections
-<math display="block">
+{{NumEqn|<math>
 \frac{bL}{A}=\left\{A=\frac{\pi D^2}{4}, b=\pi D\right\}=\frac{4L}{D}
-</math>
+</math>}}
 and thus
-<math display="block">
+{{NumEqn|<math>
 \rho_1 u_1^2+p_1-\frac{4}{D}\int_0^L\tau_w dx=\rho_2 u_2^2+p_2
-</math>
+</math>}}
 ===== Energy =====
-<math display="block">
+{{NumEqn|<math>
 h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2
-</math>
+</math>}}
 ==== Differential Form ====
@@ Line 52: / Line 64: @@
 ===== Continuity =====
-<math display="block">
+{{NumEqn|<math>
 \rho_1 u_1=\rho_2 u_2=const\Rightarrow
-</math>
+</math>}}
-<math display="block">
+{{NumEqn|<math>
 \frac{d}{dx}(\rho u)=0
-</math>
+</math>}}
 ===== Momentum =====
-<math display="block">
+{{NumEqn|<math>
 (\rho_2 u_2^2+p_2-\rho_1 u_1^2+p_1)=-\frac{4}{D}\int_0^L\tau_w dx\Rightarrow
-</math>
+</math>}}
-<math display="block">
+{{NumEqn|<math>
 \frac{d}{dx}(\rho u^2+p)=-\frac{4}{D}\tau_w
-</math>
+</math>}}
-<math display="block">
+{{NumEqn|<math>
 \frac{d}{dx}(\rho u^2+p)=\rho u\frac{du}{dx}+u\frac{d}{dx}(\rho u)+\frac{dp}{dx}=\left\{\frac{d}{dx}(\rho u)=0\right\}=\rho u\frac{du}{dx}+\frac{dp}{dx}
-</math>
+</math>}}
-<math display="block">
+{{NumEqn|<math>
 \rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{4}{D}\tau_w
-</math>
+</math>}}
 The wall shear stress is often approximated using a shear-stress factor, <math>f</math>, according to
-<math display="block">
+{{NumEqn|<math>
 \tau_w=f\frac{1}{2}\rho u^2
-</math>
+</math>}}
 and thus
-<math display="block">
+{{NumEqn|<math>
 \rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2
-</math>
+</math>}}
 ===== Energy =====
-<math display="block">
+{{NumEqn|<math>
 h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2=const
-</math>
+</math>}}
-<math display="block">
+{{NumEqn|<math>
 h_{o_1}=h_{o_2}=const
-</math>
+</math>}}
-<math display="block">
+{{NumEqn|<math>
 \frac{d}{dx}h_o=0
-</math>
+</math>}}
 ==== Summary ====
@@ Line 108: / Line 120: @@
 continuity:
-<math display="block">
+{{NumEqn|<math>
 \frac{d}{dx}(\rho u)=0
-</math>
+</math>}}
 momentum:
-<math display="block">
+{{NumEqn|<math>
 \rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2
-</math>
+</math>}}
 energy:
-<math display="block">
+{{NumEqn|<math>
 \frac{d}{dx}h_o=0
-</math>
+</math>}}
 From chapter 3.9 we have the following expression for the momentum equation for one-dimensional flow with friction (equation (3.95))
-<math display="block">
+{{NumEqn|<math>
 dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}
-</math>
+</math>}}
 For cases dealing with calorically perfect gas, (3.95) can be recast completely in terms of Mach number using the following relations
@@ Line 142: / Line 154: @@
 We start with the continuity equation which for one-dimensional steady flows reads
-<math display="block">
+{{NumEqn|<math>
 \rho u=const
-</math>
+</math>}}
 Differentiating (\ref{eqn:cont:a}) gives
-<math display="block">
+{{NumEqn|<math>
 d(\rho u)=0. \Leftrightarrow \rho du + ud\rho=0.
-</math>
+</math>}}
 If <math>u\neq 0.</math> we can divide by <math>\rho u</math> which gives us
-<math display="block">
+{{NumEqn|<math>
 \frac{du}{u}+\frac{d\rho}{\rho}=0.
-</math>
+</math>}}
 Now, if we divide and multiply the first term in (\ref{eqn:cont:b}) by <math>2u</math> and use the chain rule for derivatives we get
-<math display="block">
+{{NumEqn|<math>
 \frac{d(u^2)}{2u^2}+\frac{d\rho}{\rho}=0.
-</math>
+</math>}}
 ==== Energy equation ====
-\noindent For an adiabatic one-dimensional flow we have that \\
-\begin{equation}
+For an adiabatic one-dimensional flow we have that
-c_p T+\frac{u^2}{2}=const
-\label{eqn:ttot:a}
+{{NumEqn|<math>
-\end{equation}\\
+C_p T+\frac{u^2}{2}=const
+</math>}}
-\noindent If we differentiate (\ref{eqn:ttot:a}) we get\\
+If we differentiate (\ref{eqn:ttot:a}) we get
-\begin{equation}
+{{NumEqn|<math>
-c_p dT+\frac{1}{2}d(u^2)=0.
+C_p dT+\frac{1}{2}d(u^2)=0.
-\label{eqn:ttot:b}
+</math>}}
-\end{equation}\\
-\noindent We replace $c_p$ with $\gamma R/(\gamma-1)$ and multiply and divide the first term with $T$ which gives us\\
+We replace <math>C_p</math> with <math>\gamma R/(\gamma-1)</math> and multiply and divide the first term with <math>T</math> which gives us
-\begin{equation}
+{{NumEqn|<math>
 \frac{\gamma RT}{(\gamma-1)}\frac{dT}{T}+\frac{1}{2}d(u^2)=0.
-\label{eqn:ttot:c}
+</math>}}
-\end{equation}\\
-\noindent Now, divide by $\gamma RT/(\gamma-1)$ and multiply and divide the second term by $u^2$ gives\\
+Now, divide by <math>\gamma RT/(\gamma-1)</math> and multiply and divide the second term by <math>u^2</math> gives
-\begin{equation}
+{{NumEqn|<math>
 \frac{dT}{T}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
-\label{eqn:ttot}
+</math>}}
-\end{equation}\\
-\noindent We want to remove the $dT/T$-term in (\ref{eqn:ttot}). From the definition of Mach number we have that\\
+We want to remove the <math>dT/T</math>-term in (\ref{eqn:ttot}). From the definition of Mach number we have that
-\begin{equation}
+{{NumEqn|<math>
 a^2M^2=u^2
-\label{eqn:Mach:a}
+</math>}}
-\end{equation}\\
-\noindent which we can rewrite using the expression for speed of sound $(a^2=\gamma RT)$ according to\\
+which we can rewrite using the expression for speed of sound <math>(a^2=\gamma RT)</math> according to
-\begin{equation}
+{{NumEqn|<math>
 \gamma RTM^2=u^2
-\label{eqn:Mach:b}
+</math>}}
-\end{equation}\\
-\noindent Differentiating (\ref{eqn:Mach:b}) gives us\\
+Differentiating (\ref{eqn:Mach:b}) gives us
-\begin{equation}
+{{NumEqn|<math>
 \gamma RM^2 dT+\gamma RT d(M^2)=d(u^2)
-\label{eqn:Mach:c}
+</math>}}
-\end{equation}\\
-\noindent Now, if we divide (\ref{eqn:Mach:c}) by $\gamma RT M^2$ and use $a^2=\gamma RT$ and $a^2M^2=u^2$ we get\\
+Now, if we divide (\ref{eqn:Mach:c}) by <math>\gamma RT M^2</math> and use <math>a^2=\gamma RT</math> and <math>a^2M^2=u^2</math> we get
-\begin{equation}
+{{NumEqn|<math>
 \frac{dT}{T}+\frac{d(M^2)}{M^2}=\frac{d(u^2)}{u^2}
-\label{eqn:Mach}
+</math>}}
-\end{equation}\\
-\noindent Equation (\ref{eqn:Mach}) may now be used to replace the $dT/T$-term in equation (\ref{eqn:ttot})\\
+Equation (\ref{eqn:Mach}) may now be used to replace the <math>dT/T</math>-term in equation (\ref{eqn:ttot})
-\begin{equation}
+{{NumEqn|<math>
 -\frac{d(M^2)}{M^2}+\frac{d(u^2)}{u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.
-\label{eqn:ttot:Mach:a}
+</math>}}
-\end{equation}\\
-\noindent which can be rewritten according to\\
+which can be rewritten according to
-\begin{equation}
+{{NumEqn|<math>
 \frac{d(u^2)}{u^2}=\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{d(M^2)}{M^2}
-\label{eqn:ttot:Mach:b}
+</math>}}
-\end{equation}\\
-\noindent Using the chain rule for derivatives, the last term may be rewritten according to\\
+Using the chain rule for derivatives, the last term may be rewritten according to
-\[\frac{d(M^2)}{M^2}=2M\frac{dM}{M^2}=2\frac{dM}{M}\]\\
+{{NumEqn|<math>
+\frac{d(M^2)}{M^2}=2M\frac{dM}{M^2}=2\frac{dM}{M}
+</math>}}
-\noindent which gives\\
+which gives
-\begin{equation}
+{{NumEqn|<math>
 \frac{d(u^2)}{u^2}=2\left[1+\frac{(\gamma-1)}{2}M^2\right]^{-1}\frac{dM}{M}
-\label{eqn:ttot:Mach}
+</math>}}
-\end{equation}\\
+==== The ideal gas law ====
-\subsection{The ideal gas law}
+For a perfect gas the ideal gas law reads
-\noindent For a perfect gas the ideal gas law reads\\
-\begin{equation}
+{{NumEqn|<math>
 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}
+{{NumEqn|<math>
 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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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 ====
-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
-\[dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}\ (3.95)\]\\
+{{NumEqn|<math>
+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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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*}
+{{NumEqn|<math>
 \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 ====
-\noindent In analogy with the heat addition process discussed in the previous section, one-dimensional flow with heat addition is a continuous process. We will derive the differential relations for one-dimensional flow with friction, which will lead to trends for supersonic and supersonic flow with friction.
+In analogy with the heat addition process discussed in the previous section, one-dimensional flow with heat addition is a continuous process. We will derive the differential relations for one-dimensional flow with friction, which will lead to trends for supersonic and supersonic flow with friction.
+<!--
 \begin{figure}[ht!]
 \begin{center}
@@ Line 349: / Line 346: @@
 \label{fig:fanno:dx}
 \end{figure}
+-->
-\noindent The continuity equation gives
+The continuity equation gives
-\[d(\rho u)=ud\rho+\rho du \Rightarrow\]
+{{NumEqn|<math>
+d(\rho u)=ud\rho+\rho du \Rightarrow
+</math>}}
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{d\rho}{\rho}=-\dfrac{du}{u}
-\label{eq:fanno:drho}
+</math>}}
-\end{equation}
-\noindent The addition of friction does not affect total temperature and thus the total temperature is constant
+The addition of friction does not affect total temperature and thus the total temperature is constant
-\[T_o=T+\dfrac{u^2}{2C_p}=const\]
+{{NumEqn|<math>
+T_o=T+\dfrac{u^2}{2C_p}=const
+</math>}}
-\noindent differentiating gives
+differentiating gives
-\[dT_o=dT+\dfrac{1}{Cp}udu=0\]
+{{NumEqn|<math>
+dT_o=dT+\dfrac{1}{Cp}udu=0
+</math>}}
-\noindent with $u=M\sqrt{\gamma RT}$, we get
+with <math>u=M\sqrt{\gamma RT}</math>, we get
+{{NumEqn|<math>
-\begin{equation}
 \dfrac{dT}{T}=-(\gamma-1)M^2\dfrac{du}{u}
-\label{eq:fanno:dT}
+</math>}}
-\end{equation}\\
-\noindent A differential relation for pressure can be obtained from the ideal gas relation
+A differential relation for pressure can be obtained from the ideal gas relation
-\[p=\rho RT\Rightarrow dp=R(Td\rho+\rho dT)\Rightarrow\]
+{{NumEqn|<math>
+p=\rho RT\Rightarrow dp=R(Td\rho+\rho dT)\Rightarrow
+</math>}}
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}
-\label{eq:fanno:dp}
+</math>}}
-\end{equation}\\
-\noindent The entropy increase can be obtained from
+The entropy increase can be obtained from
-\[ds=C_v \dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}\]
+{{NumEqn|<math>
+ds=C_v \dfrac{dp}{p}-C_p\dfrac{d\rho}{\rho}
+</math>}}
-\noindent and thus
+and thus
-\begin{equation}
+{{NumEqn|<math>
 ds=-R(1-M^2)\dfrac{du}{u}
-\label{eq:fanno:ds}
+</math>}}
-\end{equation}\\
-\noindent Finally, a relation describing the change in Mach number can be obtained from \\
+Finally, a relation describing the change in Mach number can be obtained from
-\[M=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}\]
+{{NumEqn|<math>
+M=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}
+</math>}}
-\noindent which can be rewritten as
+which can be rewritten as
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{dM}{M}=\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}
-\label{eq:fanno:dM}
+</math>}}
-\end{equation}\\
-\noindent Eqns.~\ref{eq:fanno:drho} - \ref{eq:fanno:dM} are expressed as functions of $du$ and in order to get a direct relation to the addition of friction caused by the increase in pipe length $dx$, the equations are rewritten so that all variable changes are functions of the entropy increase $ds$.\\
+Eqns.~\ref{eq:fanno:drho} - \ref{eq:fanno:dM} are expressed as functions of <math>du</math> and in order to get a direct relation to the addition of friction caused by the increase in pipe length <math>dx</math>, the equations are rewritten so that all variable changes are functions of the entropy increase <math>ds</math>.
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{d\rho}{\rho}=-\dfrac{1}{R(1-M^2)}ds
-\label{eq:fanno:drho:b}
+</math>}}
-\end{equation}
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{dT}{T}=-(\gamma-1)M^2\dfrac{1}{R(1-M^2)}ds
-\label{eq:fanno:dT:b}
+</math>}}
-\end{equation}
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{1}{R(1-M^2)}ds
-\label{eq:fanno:dp:b}
+</math>}}
-\end{equation}
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{dM}{M}=\left(1+\dfrac{(\gamma-1)}{2}M^2\right)\dfrac{1}{R(1-M^2)}ds
-\label{eq:fanno:dp:b}
+</math>}}
-\end{equation}
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{du}{u}=\dfrac{1}{R(1-M^2)}ds
-\label{eq:fanno:drho:b}
+</math>}}
-\end{equation}\\
-\noindent A relation for the change in total pressure can be obtained from
+A relation for the change in total pressure can be obtained from
-\[ds=C_p\dfrac{dT_o}{T_o}-R\dfrac{dp_o}{p_o}\]\\
+{{NumEqn|<math>
+ds=C_p\dfrac{dT_o}{T_o}-R\dfrac{dp_o}{p_o}
+</math>}}
-\noindent Since total temperature is constant the relation above gives
+Since total temperature is constant the relation above gives
-\begin{equation}
+{{NumEqn|<math>
 \dfrac{dp_o}{p_o}=-\dfrac{ds}{R}
-\label{eq:fanno:dpo:b}
+</math>}}
-\end{equation}\\
-\noindent 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{subfigure}[b]{0.5\textwidth}
@@ Line 460: / Line 461: @@
 \label{fig:fanno:trends}
 \end{figure}
+-->
 ==== Friction Choking ====
+<!--
 \begin{figure}[ht!]
 \centering
@@ Line 469: / Line 472: @@
 \label{fig:friction:Ts}
 \end{figure}
+-->
-\noindent Figure~\ref{fig:friction:Ts} shows the Fanno flow process in a $Ts$-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 ($M=1$). 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 $L^\ast$, 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 $L^\ast$ 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.
-\[
+{{InfoBox|<math>
-\begin{aligned}
+M_{1^\prime} = f(L^\ast)
-M_{1'} & = f(L^\ast)\\
+</math><br><br><math>
-T_{1'} & = f(T_o, M_{1'})\\
+T_{1^\prime} = f(T_o,\ M_{1^\prime})
-p_{1'} & = f(p_o, M_{1'})\\
+</math><br><br><math>
-\rho_{1'} & = f(p_{1'}, T_{1'})\\
+p_{1^\prime} = f(p_o,\ M_{1^\prime})
-a_{1'} & = f(T_{1'})\\
+</math><br><br><math>
-u_{1'} & = M_{1'}a_{1'}\\
+\rho_{1^\prime} = f(p_{1^\prime},\ T_{1^\prime})
-\end{aligned}
+</math><br><br><math>
-\] \\
+a_{1^\prime} = f(T_{1^\prime})
+</math><br><br><math>
+u_{1^\prime} = M_{1^\prime}a_{1^\prime}
+</math>}}
+<!--
 \begin{figure}[ht!]
 \begin{subfigure}[b]{0.5\textwidth}
@@ Line 497: / Line 505: @@
 \label{fig:friction:choking:sub}
 \end{figure}
+-->
+<!--
 \begin{figure}[ht!]
 \centering
@@ Line 506: / Line 514: @@
 \label{fig:friction:choking:sup}
 \end{figure}
+-->
-\noindent For a choked supersonic flow, addition of more friction (increasing the length of the pipe such that $L>L^\ast$) may lead to the generation of a shock inside the pipe. In contrast to the one-dimensional flow with heat addition where a shock does not change $q^\ast$, $L^\ast$ is increased over a shock. The internal shock will be generated in an axial location such that $L^\ast$ downstream of the shock equals the remaining pipe length at the shock location (see Figure~\ref{fig:friction:choking:sup}). As more length is added to the pipe, the shock will move further and further upstream in the pipe until it stands at the pipe entrance. If the pipe is longer than $L^\ast$ after o shock standing at the inlet, the shock will move to the upstream system and the pipe flow will be subsonic and the massflow will be adjusted such that $L=L^\ast$ according to the process described for subsonic choking above.
+For a choked supersonic flow, addition of more friction (increasing the length of the pipe such that <math>L>L^\ast</math>) may lead to the generation of a shock inside the pipe. In contrast to the one-dimensional flow with heat addition where a shock does not change <math>q^\ast</math>, <math>L^\ast</math> is increased over a shock. The internal shock will be generated in an axial location such that <math>L^\ast</math> downstream of the shock equals the remaining pipe length at the shock location (see Figure~\ref{fig:friction:choking:sup}). As more length is added to the pipe, the shock will move further and further upstream in the pipe until it stands at the pipe entrance. If the pipe is longer than <math>L^\ast</math> after o shock standing at the inlet, the shock will move to the upstream system and the pipe flow will be subsonic and the massflow will be adjusted such that <math>L=L^\ast</math> according to the process described for subsonic choking above.
-\noindent From prvevious derivations, we know that $L^\ast$ is a function of mach number according to\\
+From prvevious derivations, we know that <math>L^\ast</math> is a function of mach number according to
-\[\dfrac{4\bar{f}L^\ast}{D}=\dfrac{1-M^2}{\gamma M^2}+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{(\gamma+1)M^2}{2+(\gamma-1)M^2}\right)\]\\
+{{NumEqn|<math>
+\dfrac{4\bar{f}L^\ast}{D}=\dfrac{1-M^2}{\gamma M^2}+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{(\gamma+1)M^2}{2+(\gamma-1)M^2}\right)
+</math>}}
-\noindent by dividing both the numerator and denominator in the fractions by $M^2$ it is easy to see that the choking length (Figure~\ref{fig:friction:factor}) approaches a finite length for great Mach numbers and thus the upper limit for the choking length $L^\ast_1$ is given by\\
+by dividing both the numerator and denominator in the fractions by <math>M^2</math> it is easy to see that the choking length (Figure~\ref{fig:friction:factor}) approaches a finite length for great Mach numbers and thus the upper limit for the choking length <math>L^\ast_1</math> is given by
-\[\left.\dfrac{4\bar{f}L_1^\ast}{D}(M_1)\right|_{M_1\rightarrow \infty}=-\dfrac{1}{\gamma}+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{\gamma+1}{\gamma-1}\right)\]\\
+{{NumEqn|<math>
+\left.\dfrac{4\bar{f}L_1^\ast}{D}(M_1)\right|_{M_1\rightarrow \infty}=-\dfrac{1}{\gamma}+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{\gamma+1}{\gamma-1}\right)
+</math>}}
+<!--
 \begin{figure}[ht!]
 \centering
@@ Line 523: / Line 537: @@
 \label{fig:friction:factor}
 \end{figure}
+-->
-\noindent From the normal shock relations we know that the downstream Mach number approaches the finite value $\sqrt{(\gamma-1)/2\gamma}$ large Mach numbers and thus the choking length downstream the shock is limited to \\
+From the normal shock relations we know that the downstream Mach number approaches the finite value <math>\sqrt{(\gamma-1)/2\gamma}</math> large Mach numbers and thus the choking length downstream the shock is limited to
-\[\left.\dfrac{4\bar{f}L_2^\ast}{D}(M_2)\right|_{M_1\rightarrow \infty}=\left(\dfrac{\gamma+1}{\gamma(\gamma-1)}\right)+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{(\gamma+1)(\gamma-1)}{4\gamma+(\gamma-1)^2}\right)\]\\
+{{NumEqn|<math>
+\left.\dfrac{4\bar{f}L_2^\ast}{D}(M_2)\right|_{M_1\rightarrow \infty}=\left(\dfrac{\gamma+1}{\gamma(\gamma-1)}\right)+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left(\dfrac{(\gamma+1)(\gamma-1)}{4\gamma+(\gamma-1)^2}\right)
+</math>}}
-\noindent From the relations above we get \\
+From the relations above we get
-\[\left.\left(\dfrac{4\bar{f}L_2^\ast}{D}(M_2)-\dfrac{4\bar{f}L_1^\ast}{D}(M_1)\right)\right|_{M_1\rightarrow \infty}=\left(\dfrac{2}{\gamma-1}\right)+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left[\left(\dfrac{(\gamma+1)(\gamma-1)}{4\gamma+(\gamma-1)^2}\right)\left(\dfrac{\gamma-1}{\gamma+1}\right)\right]\]\\
+{{NumEqn|<math>
+\left.\left(\dfrac{4\bar{f}L_2^\ast}{D}(M_2)-\dfrac{4\bar{f}L_1^\ast}{D}(M_1)\right)\right|_{M_1\rightarrow \infty}=\left(\dfrac{2}{\gamma-1}\right)+\left(\dfrac{\gamma+1}{2\gamma}\right)\ln\left[\left(\dfrac{(\gamma+1)(\gamma-1)}{4\gamma+(\gamma-1)^2}\right)\left(\dfrac{\gamma-1}{\gamma+1}\right)\right]
+</math>}}
-\noindent 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 $L_2^\ast$ 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.
+<!--
 \begin{figure}[ht!]
 \centering
@@ Line 540: / Line 560: @@
 \label{fig:friction:factor:shock}
 \end{figure}
+-->

One-dimensional flow with friction: Difference between revisions

Latest revision as of 18:26, 1 April 2026

Flow-station data

Continuity

Momentum

Energy

Differential Form

Continuity

Momentum

Energy

Summary

Continuity equation

Energy equation

The ideal gas law

Momentum equation

Differential Relations

Friction Choking

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