One-dimensional flow with friction

Flow-station data

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

Continuity

$${\rho }_1{u}_1={\rho }_2{u}_2$$

Momentum

${\rho }_1{u}_1^2+p_1-{\overline{\tau }}_w\frac{bL}{A}={\rho }_2{u}_2^2+p_2$

(Eq. 3.126)

where ${\overline{\tau }}_w$ is the average wall-shear stress

${\overline{\tau }}_w=\frac{1}{L}\underset{0}{\overset{L}{\int }}{\tau }_{w}dx$

(Eq. 3.127)

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

$\frac{bL}{A}=\{A=\frac{\pi D^2}{4},b=\pi D\}=\frac{4L}{D}$

(Eq. 3.128)

and thus

${\rho }_1{u}_1^2+p_1-\frac{4}{D}\underset{0}{\overset{L}{\int }}{\tau }_{w}dx={\rho }_2{u}_2^2+p_2$

(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

${\rho }_1{u}_1={\rho }_2{u}_2=const\Rightarrow$

(Eq. 3.131)

$\frac{d}{dx}(\rho u)=0$

(Eq. 3.132)

Momentum

$({\rho }_2{u}_2^2+p_2-{\rho }_1{u}_1^2+p_1)=-\frac{4}{D}\underset{0}{\overset{L}{\int }}{\tau }_{w}dx\Rightarrow$

(Eq. 3.133)

$\frac{d}{dx}(\rho u^2+p)=-\frac{4}{D}{\tau }_w$

(Eq. 3.134)

$\frac{d}{dx}(\rho u^2+p)=\rho u\frac{du}{dx}+u\frac{d}{dx}(\rho u)+\frac{dp}{dx}=\{\frac{d}{dx}(\rho u)=0\}=\rho u\frac{du}{dx}+\frac{dp}{dx}$

(Eq. 3.135)

$\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{4}{D}{\tau }_w$

(Eq. 3.136)

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

${\tau }_w=f\frac{1}{2}\rho u^2$

(Eq. 3.137)

and thus

$\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2$

(Eq. 3.138)

Energy

$h_1+\frac{1}{2}{u}_1^2=h_2+\frac{1}{2}{u}_2^2=const$

(Eq. 3.139)

$h_{o_1}=h_{o_2}=const$

(Eq. 3.140)

$\frac{d}{dx}{h}_o=0$

(Eq. 3.141)

Summary

continuity:

$\frac{d}{dx}(\rho u)=0$

(Eq. 3.142)

momentum:

$\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2$

(Eq. 3.143)

energy:

$\frac{d}{dx}{h}_o=0$

(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))

$dp+\rho udu=-\frac{1}{2}\rho u^2\frac{4fdx}{D}$

(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: $a^2=\gamma p/\rho$
• the definition of Mach number: $M^2=u^2/a^2$
• the ideal gas law for thermally perfect gas: $p=\rho RT$
• the continuity equation: $\rho u=const$
• the energy equation: $C_{p}T+u^2/2=const$

Continuity equation

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

$\rho u=const$

(Eq. 3.146)

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

$d(\rho u)=0.\iff \rho du+ud\rho =0.$

(Eq. 3.147)

If $u\ne 0.$ we can divide by $\rho u$ which gives us

$\frac{du}{u}+\frac{d\rho }{\rho }=0.$

(Eq. 3.148)

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

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

(Eq. 3.149)

Energy equation

For an adiabatic one-dimensional flow we have that

$C_{p}T+\frac{u^2}{2}=const$

(Eq. 3.150)

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

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

(Eq. 3.151)

We replace $C_p$ with $\gamma R/(\gamma -1)$ and multiply and divide the first term with $T$ which gives us

$\frac{\gamma RT}{(\gamma -1)}\frac{dT}{T}+\frac{1}{2}d(u^2)=0.$

(Eq. 3.152)

Now, divide by $\gamma RT/(\gamma -1)$ and multiply and divide the second term by $u^2$ gives

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

(Eq. 3.153)

We want to remove the $dT/T$-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 $(a^2=\gamma RT)$ according to

$\gamma RT{M}^2=u^2$

(Eq. 3.155)

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

$\gamma R{M}^{2}dT+\gamma RTd(M^2)=d(u^2)$

(Eq. 3.156)

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

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

(Eq. 3.157)

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

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

(Eq. 3.158)

which can be rewritten according to

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

(Eq. 3.159)

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

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

(Eq. 3.160)

which gives

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

(Eq. 3.161)

The ideal gas law

For a perfect gas the ideal gas law reads

$p=\rho RT$

(Eq. 3.162)

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

$dp=\rho RdT+RTd\rho$

(Eq. 3.163)

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

$\frac{dp}{p}=\frac{dT}{T}+\frac{d\rho }{\rho }$

(Eq. 3.164)

which can be rearranged according to

$[\frac{dp}{p}-\frac{d\rho }{\rho }]=\frac{dT}{T}$

(Eq. 3.165)

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

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

(Eq. 3.166)

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

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

(Eq. 3.167)

Collect terms and rewrite gives

$\frac{dp}{p}+\left[\frac{1+(\gamma -1)M^2}{2}\right]\frac{d(u^2)}{u^2}=0.$

(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

$dp+\rho udu=-\frac{1}{2}\rho u^2\frac{4fdx}{D}$

(Eq. 3.169)

if $u\ne 0.$, we can divide by $0.5\rho u^2$ which gives

$2\frac{dp}{\rho u^2}+2\frac{\rho udu}{\rho u^2}=-\frac{4fdx}{D}$

(Eq. 3.170)

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

$\frac{2}{\gamma M^2}\frac{dp}{p}+\frac{d(u^2)}{u^2}=-\frac{4fdx}{D}$

(Eq. 3.171)

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

$\frac{2}{\gamma M^2}\{-\left[\frac{1+(\gamma -1)M^2}{2}\right]\frac{d(u^2)}{u^2}\}+\frac{d(u^2)}{u^2}=-\frac{4fdx}{D}$

(Eq. 3.172)

collecting terms and rearranging gives

$\frac{M^2-1}{\gamma M^2}\frac{d(u^2)}{u^2}=\frac{4fdx}{D}$

(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

$\frac{4fdx}{D}=\frac{2}{\gamma M^2}(1-M^2){[1+\frac{(\gamma -1)}{2}{M}^2]}^{-1}\frac{dM}{M}$

(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

$d(\rho u)=ud\rho +\rho du\Rightarrow$

(Eq. 3.175)

$\frac{d\rho }{\rho }=-\frac{du}{u}$

(Eq. 3.176)

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

$T_o=T+\frac{u^2}{2C_p}=const$

(Eq. 3.177)

differentiating gives

$d{T}_o=dT+\frac{1}{Cp}udu=0$

(Eq. 3.178)

with $u=M\sqrt{\gamma RT}$, we get

$\frac{dT}{T}=-(\gamma -1)M^2\frac{du}{u}$

(Eq. 3.179)

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

$p=\rho RT\Rightarrow dp=R(Td\rho +\rho dT)\Rightarrow$

(Eq. 3.180)

$\frac{dp}{p}=-(1+(\gamma -1)M^2)\frac{du}{u}$

(Eq. 3.181)

The entropy increase can be obtained from

$ds=C_v\frac{dp}{p}-C_p\frac{d\rho }{\rho }$

(Eq. 3.182)

and thus

$ds=-R(1-M^2)\frac{du}{u}$

(Eq. 3.183)

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

$M=\frac{u}{\sqrt{\gamma RT}}\Rightarrow dM=M\frac{du}{u}-\frac{M}{2}\frac{dT}{T}$

(Eq. 3.184)

which can be rewritten as

$\frac{dM}{M}=(1+(\gamma -1)M^2)\frac{du}{u}$

(Eq. 3.185)

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 $ds$.

$\frac{d\rho }{\rho }=-\frac{1}{R(1-M^2)}ds$

(Eq. 3.186)

$\frac{dT}{T}=-(\gamma -1)M^2\frac{1}{R(1-M^2)}ds$

(Eq. 3.187)

$\frac{dp}{p}=-(1+(\gamma -1)M^2)\frac{1}{R(1-M^2)}ds$

(Eq. 3.188)

$\frac{dM}{M}=(1+\frac{(\gamma -1)}{2}{M}^2)\frac{1}{R(1-M^2)}ds$

(Eq. 3.189)

$\frac{du}{u}=\frac{1}{R(1-M^2)}ds$

(Eq. 3.190)

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

$ds=C_p\frac{d{T}_o}{T_o}-R\frac{d{p}_o}{p_o}$

(Eq. 3.191)

Since total temperature is constant the relation above gives

$\frac{d{p}_o}{p_o}=-\frac{ds}{R}$

(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 $Ts$-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.

$M_{1^{\prime }}=f(L^{\ast })$ $T_{1^{\prime }}=f(T_o,M_{1^{\prime }})$ $p_{1^{\prime }}=f(p_o,M_{1^{\prime }})$ ${\rho }_{1^{\prime }}=f(p_{1^{\prime }},T_{1^{\prime }})$ $a_{1^{\prime }}=f(T_{1^{\prime }})$ $u_{1^{\prime }}=M_{1^{\prime }}{a}_{1^{\prime }}$

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.

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

$\frac{4\overline{f}{L}^{\ast }}{D}=\frac{1-M^2}{\gamma M^2}+\left(\frac{\gamma +1}{2\gamma }\right)\ln \left(\frac{(\gamma +1)M^2}{2+(\gamma -1)M^2}\right)$

(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 $L_1^{\ast }$ is given by

${\frac{4\overline{f}{L}_1^{\ast }}{D}(M_1)|}_{M_1\to \mathrm{\infty }}=-\frac{1}{\gamma }+\left(\frac{\gamma +1}{2\gamma }\right)\ln \left(\frac{\gamma +1}{\gamma -1}\right)$

(Eq. 3.194)

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

${\frac{4\overline{f}{L}_2^{\ast }}{D}(M_2)|}_{M_1\to \mathrm{\infty }}=\left(\frac{\gamma +1}{\gamma (\gamma -1)}\right)+\left(\frac{\gamma +1}{2\gamma }\right)\ln \left(\frac{(\gamma +1)(\gamma -1)}{4\gamma +(\gamma -1{)}^2}\right)$

(Eq. 3.195)

From the relations above we get

${(\frac{4\overline{f}{L}_2^{\ast }}{D}(M_2)-\frac{4\overline{f}{L}_1^{\ast }}{D}(M_1))|}_{M_1\to \mathrm{\infty }}=\left(\frac{2}{\gamma -1}\right)+\left(\frac{\gamma +1}{2\gamma }\right)\ln \left[\left(\frac{(\gamma +1)(\gamma -1)}{4\gamma +(\gamma -1{)}^2}\right)\left(\frac{\gamma -1}{\gamma +1}\right)\right]$

(Eq. 3.196)

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.