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4.8 One-Dimensional Flow with Friction

Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics

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

Continuity:

ρ1u1=ρ2u2\rho_1 u_1=\rho_2 u_2

Momentum:

ρ1u12+p1τˉwbLA=ρ2u22+p2\rho_1 u_1^2+p_1-\bar{\tau}_w\frac{bL}{A}=\rho_2 u_2^2+p_2

where τˉw\bar{\tau}_w is the average wall-shear stress

τˉw=1L0Lτwdx\bar{\tau}_w=\frac{1}{L}\int_0^L\tau_w dx

bb is the tube perimeter, and LL is the tube length. For circular cross sections

bLA={A=πD24,b=πD}=4LD\frac{bL}{A}=\left\{A=\frac{\pi D^2}{4}, b=\pi D\right\}=\frac{4L}{D}

and thus

ρ1u12+p14D0Lτwdx=ρ2u22+p2\rho_1 u_1^2+p_1-\frac{4}{D}\int_0^L\tau_w dx=\rho_2 u_2^2+p_2

Energy:

h1+12u12=h2+12u22h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2

Differential Form

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

Continuity:

ρ1u1=ρ2u2=const\rho_1 u_1=\rho_2 u_2=const\Rightarrow
ddx(ρu)=0\frac{d}{dx}(\rho u)=0

Momentum:

(ρ2u22+p2ρ1u12+p1)=4D0Lτwdx(\rho_2 u_2^2+p_2-\rho_1 u_1^2+p_1)=-\frac{4}{D}\int_0^L\tau_w dx\Rightarrow
ddx(ρu2+p)=4Dτw\frac{d}{dx}(\rho u^2+p)=-\frac{4}{D}\tau_w
ddx(ρu2+p)=ρududx+uddx(ρu)+dpdx={ddx(ρu)=0}=ρududx+dpdx\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}
ρududx+dpdx=4Dτw\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, ff, according to

τw=f12ρu2\tau_w=f\frac{1}{2}\rho u^2

and thus

ρududx+dpdx=2Dfρu2\rho u\frac{du}{dx}+\frac{dp}{dx}=-\frac{2}{D}f\rho u^2

Energy:

h1+12u12=h2+12u22=consth_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2=const
ho1=ho2=consth_{o_1}=h_{o_2}=const
ddxho=0\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+ρudu=12ρu24fdxD (3.95)dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}\ (3.95)

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

Continuity equation

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

ρu=const\rho u=const

Differentiating (4.176) gives

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

If u0.u\neq 0. we can divide by ρu\rho u which gives us

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

Now, if we divide and multiply the first term in (4.178) by 2u2u and use the chain rule for derivatives we get

d(u2)2u2+dρρ=0.\frac{d(u^2)}{2u^2}+\frac{d\rho}{\rho}=0.

Energy equation

For an adiabatic one-dimensional flow we have that

cpT+u22=constc_p T+\frac{u^2}{2}=const

If we differentiate (4.180) we get

cpdT+12d(u2)=0.c_p dT+\frac{1}{2}d(u^2)=0.

We replace cpc_p with γR/(γ1)\gamma R/(\gamma-1) and multiply and divide the first term with TT which gives us

γRT(γ1)dTT+12d(u2)=0.\frac{\gamma RT}{(\gamma-1)}\frac{dT}{T}+\frac{1}{2}d(u^2)=0.

Now, divide by γRT/(γ1)\gamma RT/(\gamma-1) and multiply and divide the second term by u2u^2 gives

dTT+(γ1)2M2d(u2)u2=0.\frac{dT}{T}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.

We want to remove the dT/TdT/T-term in (4.183). From the definition of Mach number we have that

a2M2=u2a^2M^2=u^2

which we can rewrite using the expression for speed of sound (a2=γRT)(a^2=\gamma RT) according to

γRTM2=u2\gamma RTM^2=u^2

Differentiating (4.185) gives us

γRM2dT+γRTd(M2)=d(u2)\gamma RM^2 dT+\gamma RT d(M^2)=d(u^2)

Now, if we divide (4.186) by γRTM2\gamma RT M^2 and use a2=γRTa^2=\gamma RT and a2M2=u2a^2M^2=u^2 we get

dTT+d(M2)M2=d(u2)u2\frac{dT}{T}+\frac{d(M^2)}{M^2}=\frac{d(u^2)}{u^2}

Equation (4.187) may now be used to replace the dT/TdT/T-term in equation (4.183)

d(M2)M2+d(u2)u2+(γ1)2M2d(u2)u2=0.-\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

d(u2)u2=[1+(γ1)2M2]1d(M2)M2\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

d(M2)M2=2MdMM2=2dMM\frac{d(M^2)}{M^2}=2M\frac{dM}{M^2}=2\frac{dM}{M}

which gives

d(u2)u2=2[1+(γ1)2M2]1dMM\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=ρRTp=\rho R T

Differentiating (4.192) gives:

dp=ρRdT+RTdρdp=\rho R dT+RT d\rho

If p0.p\neq0., we can divide (4.193) by pp which gives

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

which can be rearranged according to

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

Now, inserting dT/TdT/T from equation (4.183) gives

[dppdρρ]+(γ1)2M2d(u2)u2=0.\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0.

The dρ/ρd\rho/\rho-term can be replaced using equation (4.179)

dpp+d(u2)2u2+(γ1)2M2d(u2)u2=0.\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

dpp+[1+(γ1)M22]d(u2)u2=0.\frac{dp}{p}+\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}=0.

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+ρudu=12ρu24fdxD (3.95)dp+\rho u du=-\frac{1}{2}\rho u^2 \frac{4 f dx}{D}\ (3.95)

if u0.u\neq 0., we can divide by 0.5ρu20.5\rho u^2 which gives

2dpρu2+2ρuduρu2=4fdxD2\frac{dp}{\rho u^2}+2\frac{\rho u du}{\rho u^2}=-\frac{4 f dx}{D}

using M2=u2/a2M^2=u^2/a^2, a2=γp/ρa^2=\gamma p/\rho and the chain rule in (4.200) gives

2γM2dpp+d(u2)u2=4fdxD\frac{2}{\gamma M^2}\frac{dp}{p}+\frac{d(u^2)}{u^2}=-\frac{4 f dx}{D}

From equation (4.198) we can get a relation that expresses the pressure derivative term, dp/pdp/p, in terms of Mach number and d(u2)/u2d(u^2)/u^2. Inserting this in (4.201) gives

2γM2{[1+(γ1)M22]d(u2)u2}+d(u2)u2=4fdxD\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

M21γM2d(u2)u2=4fdxD\frac{M^2-1}{\gamma M^2}\frac{d(u^2)}{u^2}=\frac{4 f dx}{D}

if we now use equation (4.191) to get rid of the d(u2)/u2d(u^2)/u^2-term we end up with the following expression

4fdxD=2γM2(1M2)[1+(γ1)2M2]1dMM\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

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.

infinitesimal pipe segment

Figure 4.28:Change in flow quantities due to the addition of an infinitesimal pipe segment with the length dxdx

The continuity equation gives

d(ρu)=udρ+ρdud(\rho u)=ud\rho+\rho du \Rightarrow
dρρ=duu\dfrac{d\rho}{\rho}=-\dfrac{du}{u}

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

To=T+u22Cp=constT_o=T+\dfrac{u^2}{2C_p}=const

differentiating gives

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

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

dTT=(γ1)M2duu\dfrac{dT}{T}=-(\gamma-1)M^2\dfrac{du}{u}

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

p=ρRTdp=R(Tdρ+ρdT)p=\rho RT\Rightarrow dp=R(Td\rho+\rho dT)\Rightarrow
dpp=(1+(γ1)M2)duu\dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}

The entropy increase can be obtained from

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

and thus

ds=R(1M2)duuds=-R(1-M^2)\dfrac{du}{u}

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

M=uγRTdM=MduuM2dTTM=\dfrac{u}{\sqrt{\gamma RT}}\Rightarrow dM=M\dfrac{du}{u}-\dfrac{M}{2}\dfrac{dT}{T}

which can be rewritten as

dMM=(1+(γ1)M2)duu\dfrac{dM}{M}=\left(1+(\gamma-1)M^2\right)\dfrac{du}{u}

Eqns. (4.206) - (4.215) are expressed as functions of dudu and in order to get a direct relation to the addition of friction caused by the increase in pipe length dxdx, the equations are rewritten so that all variable changes are functions of the entropy increase dsds.

dρρ=1R(1M2)ds\dfrac{d\rho}{\rho}=-\dfrac{1}{R(1-M^2)}ds
dTT=(γ1)M21R(1M2)ds\dfrac{dT}{T}=-(\gamma-1)M^2\dfrac{1}{R(1-M^2)}ds
dpp=(1+(γ1)M2)1R(1M2)ds\dfrac{dp}{p}=-\left(1+(\gamma-1)M^2\right)\dfrac{1}{R(1-M^2)}ds
dMM=(1+(γ1)2M2)1R(1M2)ds\dfrac{dM}{M}=\left(1+\dfrac{(\gamma-1)}{2}M^2\right)\dfrac{1}{R(1-M^2)}ds
duu=1R(1M2)ds\dfrac{du}{u}=\dfrac{1}{R(1-M^2)}ds

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

ds=CpdToToRdpopods=C_p\dfrac{dT_o}{T_o}-R\dfrac{dp_o}{p_o}

Since total temperature is constant the relation above gives

dpopo=dsR\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 Figures Figure 4.29 and Figure 4.30).

Friction Choking

Fanno flow process illustrated in a Ts-diagram

Figure 4.31:The Fanno flow process illustrated in a TsTs-diagram. The dashed line shows the critical temperature, TT^\ast, the blue line is the Fanno flow process, subsonic above TT^\ast and supersonic below TT^\ast. The gray lines are isobars.

Figure Figure 4.31 shows the Fanno flow process in a TsTs-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=1M=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 LL^\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 LL^\ast for the new inlet conditions.

M1=f(L)T1=f(To,M1)p1=f(po,M1)ρ1=f(p1,T1)a1=f(T1)u1=M1a1\begin{aligned} M_{1'} & = f(L^\ast)\\ T_{1'} & = f(T_o, M_{1'})\\ p_{1'} & = f(p_o, M_{1'})\\ \rho_{1'} & = f(p_{1'}, T_{1'})\\ a_{1'} & = f(T_{1'})\\ u_{1'} & = M_{1'}a_{1'}\\ \end{aligned}
change of the inlet static conditions on subsonic over choking

Figure 4.32:For friction-choked subsonic flow, further increase of the length of the pipe/duct will lead to an update of the inlet static conditions such that the massflow per unit area is changed and the length corresponding to friction choking LL^\ast is increased.

change of the inlet static conditions on subsonic over choking (close up)

Figure 4.33:change of the inlet static conditions on subsonic over choking (close up)

over choking shock

Figure 4.34:For friction-choked supersonic flow, further increase of the length of the pipe/duct may lead to the generation of a shock inside of the pipe. The location of the shock will be such that LL^\ast downstream of the shock equals the remaining length of the pipe at the shock location.

For a choked supersonic flow, addition of more friction (increasing the length of the pipe such that L>LL>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 qq^\ast, LL^\ast is increased over a shock. The internal shock will be generated in an axial location such that LL^\ast downstream of the shock equals the remaining pipe length at the shock location (see Figure Figure 4.34. 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 LL^\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=LL=L^\ast according to the process described for subsonic choking above.

From prvevious derivations, we know that LL^\ast is a function of mach number according to

4fˉLD=1M2γM2+(γ+12γ)ln((γ+1)M22+(γ1)M2)\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 M2M^2 it is easy to see that the choking length (Figure Figure 4.35) approaches a finite length for great Mach numbers and thus the upper limit for the choking length L1L^\ast_1 is given by

4fˉL1D(M1)M1=1γ+(γ+12γ)ln(γ+1γ1)\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)
L_star as a function of Mach number

Figure 4.35:LL^\ast as a function of Mach number. For high supersonic Mach numbers, the choking length approaches a finite value.

From the normal shock relations we know that the downstream Mach number approaches the finite value (γ1)/2γ\sqrt{(\gamma-1)/2\gamma} large Mach numbers and thus the choking length downstream the shock is limited to

4fˉL2D(M2)M1=(γ+1γ(γ1))+(γ+12γ)ln((γ+1)(γ1)4γ+(γ1)2)\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

(4fˉL2D(M2)4fˉL1D(M1))M1=(2γ1)+(γ+12γ)ln[((γ+1)(γ1)4γ+(γ1)2)(γ1γ+1)]\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 Figure 4.36 shows the development of choking length L1L_1^\ast in a supersonic flow as a function of Mach number in relation to the corresponding choking length L2L_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.

choking length change over shock

Figure 4.36:LL^\ast as a function of M1M_1 (Mach number in station 1) for initially supersonic flow. L1L_1^\ast is the choking length corresponding to M1M_1 and L2L_2^\ast the choking length downstream of a normal shock at station 1. A normal shock will always increase the choking length.