One-dimensional flow with friction: Difference between revisions

==== Flow-station data ====

===== Momentum =====

\rho_1 u_1^2+p_1-\bar{\tau}_w\frac{bL}{A}=\rho_2 u_2^2+p_2

where <math>\bar{\tau}_w</math> is the average wall-shear stress

\bar{\tau}_w=\frac{1}{L}\int_0^L\tau_w dx

<math>b</math> is the tube perimeter, and <math>L</math> is the tube length. For circular cross sections

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

and thus

\rho_1 u_1^2+p_1-\frac{4}{D}\int_0^L\tau_w dx=\rho_2 u_2^2+p_2

===== Energy =====

h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2

==== Differential Form ====

===== Continuity =====

\rho_1 u_1=\rho_2 u_2=const\Rightarrow

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

===== Momentum =====

(\rho_2 u_2^2+p_2-\rho_1 u_1^2+p_1)=-\frac{4}{D}\int_0^L\tau_w dx\Rightarrow

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

\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}

\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{4}{D}\tau_w

The wall shear stress is often approximated using a shear-stress factor, <math>f</math>, according to

\tau_w=f\frac{1}{2}\rho u^2

and thus

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

===== Energy =====

h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2=const

h_{o_1}=h_{o_2}=const

\frac{d}{dx}h_o=0

==== Summary ====

continuity:

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

momentum:

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

energy:

\frac{d}{dx}h_o=0

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

dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}

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

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

\rho u=const

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

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

If <math>u\neq 0.</math> we can divide by <math>\rho u</math> which gives us

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

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

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

==== Energy equation ====

For an adiabatic one-dimensional flow we have that

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

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

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

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

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

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

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

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

a^2M^2=u^2

which we can rewrite using the expression for speed of sound <math>(a^2=\gamma RT)</math> according to

\gamma RTM^2=u^2

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

\gamma RM^2 dT+\gamma RT d(M^2)=d(u^2)

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

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

Equation (\ref{eqn:Mach}) may now be used to replace the <math>dT/T</math>-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.

which can be rewritten according to

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

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}

which gives

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

==== The ideal gas law ====

For a perfect gas the ideal gas law reads

p=\rho R T

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

dp=\rho R dT+RT d\rho

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

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

which can be rearranged according to

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

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

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

The <math>d\rho/\rho</math>-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.

Collect terms and rewrite gives

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

==== Momentum equation ====

For convenience equation (3.95) is written again here

dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}

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

2\frac{dp}{\rho u^2}+2\frac{\rho u du}{\rho u^2}=-\frac{4 f dx}{D}

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

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

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

\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}

collecting terms and rearranging gives

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

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

\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}

==== Differential Relations ====

The continuity equation gives

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

\dfrac{d\rho}{\rho}=-\dfrac{du}{u}

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

differentiating gives

dT_o=dT+\dfrac{1}{Cp}udu=0

with <math>u=M\sqrt{\gamma RT}</math>, we get

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

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

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

\dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}

The entropy increase can be obtained from

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

and thus

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

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}

which can be rewritten as

\dfrac{dM}{M}=\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}

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>.

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

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

\dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{1}{R(1-M^2)}ds

\dfrac{dM}{M}=\left(1+\dfrac{(\gamma-1)}{2}M^2\right)\dfrac{1}{R(1-M^2)}ds

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

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}

Since total temperature is constant the relation above gives

\dfrac{dp_o}{p_o}=-\dfrac{ds}{R}

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}).

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.

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}

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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)

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)

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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)

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]

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.