Compressible flow outline

Collection page linking to compressible flow subpages

• Unsteady waves
• Quasi-one-dimensional flow
• Isentropic relations
• Specific heat
• Thermodynamic processes
• Compressible flow
• Governing equations
• Thermodynamics
• Oblique shocks and expansion waves
• Shock-tube relations
• Moving shock waves
• Moving expansion waves
• Finite non-linear waves
• Acoustic theory
• Nozzle flow
• Choked flow
• The Q1D equations
• Diffusers
• Area-velocity relation
• Area-Mach relation
• Shock-expansion theory
• Expansion waves
• Oblique shocks
• One-dimensional flow with heat addition
• One-dimensional flow with friction
• The entropy equation
• Crocco's equation
• Compressible flow
• Governing equations on differential form
• Governing equations on integral form
• Reference flow states
• Normal-shock relations
• Shock waves
• Acoustic waves
• One-dimensional inviscid flow

Reference flow states

Stagnation Flow Properties

$h_o=h+\frac{1}{2}{u}^2\Rightarrow Cp{T}_o=C_{p}T+\frac{1}{2}{u}^2\Rightarrow \frac{T_o}{T}=1+\frac{1}{2C_p}{M}^2\gamma RT=1+\frac{\gamma -1}{2}{M}^2$

(Eq. 1)

Acoustic waves

In Fig. \ref{fig:soundwave}, station 1 represents the flow state ahead of the sound wave and station 2 the flow state behind the sound wave. Set up the continuity equation for one-dimensional flows between 1 and 2. If we could change frame of reference and follow the sound wave, we would see fluid approaching the wave with the propagation speed of the wave, $a$, and behind the wave, the fluid would have a slightly modified speed, $a+da$. There would also be a slight in all other flow properties. Let's apply the one-dimensional continuity equation between station 1 and station 2.

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

(Eq. 2)

$\rho a=(\rho +d\rho )(a+da)$

(Eq. 3)

$\cancel{\rho a}=\cancel{\rho a}+\rho da+ad\rho +\underset{\sim 0}{\underbrace{d\rho da}}\Rightarrow$

(Eq. 4)

$a=-\rho \frac{da}{d\rho }$}

(Eq. 5)

The one-dimensional momentum equation between station 1 and station 2 gives

${\rho }_1{u}_1^2+p_1={\rho }_2{u}_2^2+p_2$

(Eq. 6)

$\rho a^2+p=(\rho +d\rho )(a+da{)}^2+(p+dp)$

(Eq. 7)

$\cancel{\rho a^2}+\cancel{p}=\cancel{\rho a^2}+2\rho ada+\underset{\sim 0}{\underbrace{\rho d{a}^2}}+d\rho a^2+\underset{\sim 0}{\underbrace{2d\rho ada}}+\underset{\sim 0}{\underbrace{d\rho d{a}^2}}+\cancel{p}+dp\Rightarrow$

(Eq. 8)

$dp=-2\rho ada-d\rho a^2\Rightarrow$

(Eq. 9)

$da=-\frac{dp+d\rho a^2}{2\rho a}=-\frac{d\rho }{2a\rho }(\frac{dp}{d\rho }+a^2)\Rightarrow$

(Eq. 10)

$\frac{da}{d\rho }=-\frac{1}{2a\rho }(\frac{dp}{d\rho }+a^2)$

(Eq. 11)

Eq. 11 in Eq. 5 gives

$a=\frac{1}{2a}(\frac{dp}{d\rho }+a^2)\Rightarrow$

(Eq. 12)

$a^2=\frac{dp}{d\rho }$

(Eq. 13)

Sound wave:

• gradients are small
• irreversible (dissipative effects are negligible)
• no heat addition

Thus, the change of flow properties as the sound wave passes can be assumed to be an isentropic process

$a^2={\left(\frac{dp}{d\rho }\right)}_s$

(Eq. 14)

$a=\sqrt{{\left(\frac{dp}{d\rho }\right)}_s}=\sqrt{\frac{1}{\rho {\tau }_s}}$

(Eq. 15)

where ${\tau }_s$ is the compressibility of the gas. Eq. 14 is valid for all gases. It can be seen from the equation, that truly incompressible flow (${\tau }_s=0$) would imply infinite speed of sound.

Since the process is isentropic, we can use the isentropic relations if we also assume the gas to be calorically perfect

$\frac{p_2}{p_1}={\left(\frac{{\rho }_2}{{\rho }_1}\right)}^{\gamma }\Rightarrow p=C{\rho }^{\gamma }$

(Eq. 16)

$a^2={\left(\frac{dp}{d\rho }\right)}_s=\gamma C{\rho }^{\gamma -1}=\gamma \underset{=p}{\underbrace{\left[C{\rho }^{\gamma }\right]}}{\rho }^{-1}=\frac{\gamma p}{\rho }\Rightarrow$

(Eq. 17)

$a=\sqrt{\frac{\gamma p}{\rho }}$

(Eq. 18)

or

$a=\sqrt{\gamma RT}$

(Eq. 19)

From the relation above, it is obvious that the local speed of sound is related to the temperature of the flow, which in turn is a measure of the motion of elementary particles (atoms and/or molecules) of the fluid at a specific location. This stems from the fact that sound waves are propagated via interaction of these elementary particles. Since information in a flow is propagated via molecular interaction the relation between the speed at which this information is conveyed and the speed of the flow has important physical implications. Figure~\ref{fig:speed:of:sound} compares three sound wave patterns generated by a a beacon. In the left picture, the sound transmitter is stationary and thus the acoustic waves are centered around the transmitter. In the middle image, the transmitter is moving to the left at a speed less than the speed of sound and thus the transmitter will always be within all sound wave circles but it will be off-centered with a bias in the direction that the transmitter is moving. In the right image the transmitter is moving faster than the speed of sound and thus it will always be located outside of all acoustic waves. In a supersonic flow, no information can travel upstream and therefore there is no way for the flow to adjust to downstream obstacles. This is compensated for by the introduction of shocks in the flow. Over a shock flow properties changes discontinuity. An example is given in Figure~\ref{fig:supersonic:flow}.

Shock waves

The starting point is to set up the governing equations for one-dimensional steady compressible flow over a control volume enclosing the normal shock (Fig. \ref{fig:shock:cv}).

continuity:

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

(Eq. 20)

momentum:

${\rho }_1{u}_1^2+p_1={\rho }_2{u}_2^2+p_2$

(Eq. 21)

energy:

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

(Eq. 22)

Divide the momentum equation by ${\rho }_1{u}_1$

$\frac{1}{{\rho }_1{u}_1}({\rho }_1{u}_1^2+p_1)=\frac{1}{{\rho }_1{u}_1}({\rho }_2{u}_2^2+p_2)=\{{\rho }_1{u}_1={\rho }_2{u}_2\}=\frac{1}{{\rho }_2{u}_2}({\rho }_2{u}_2^2+p_2)\Rightarrow$

(Eq. 23)

$\frac{p_1}{{\rho }_1{u}_1}-\frac{p_2}{{\rho }_2{u}_2}=u_2-u_1$

(Eq. 24)

For a calorically perfect gas $a=\sqrt{\gamma p/\rho }$, which if implemented in Eqn. \ref{eq:governing:mom:b} gives

$\frac{a_1^2}{\gamma u_1}-\frac{a_2^2}{\gamma u_2}=u_2-u_1$

(Eq. 25)

The energy equation (Eqn. \ref{eq:governing:energy}) with $h=C_{p}T$

$C_p{T}_1+\frac{1}{2}{u}_1^2=C_p{T}_2+\frac{1}{2}{u}_2^2$

(Eq. 26)

Replacing $C_p$ with $\gamma R/(\gamma -1)$ gives

$\frac{\gamma R{T}_1}{\gamma -1}+\frac{1}{2}{u}_1^2=\frac{\gamma R{T}_2}{\gamma -1}+\frac{1}{2}{u}_2^2$

(Eq. 27)

With $a=\sqrt{\gamma RT}$ this becomes

$\frac{a_1^2}{\gamma -1}+\frac{1}{2}{u}_1^2=\frac{a_2^2}{\gamma -1}+\frac{1}{2}{u}_2^2$

(Eq. 28)

Eqn. \ref{eq:governing:energy:d} can be set up between any two points in the flow. Specifically, we can use the relation to relate the flow velocity, $u$, and speed of sound, $a$, in any point to the corresponding flow properties at sonic conditions ($u=a=a^{*}$).

$\frac{a^2}{\gamma -1}+\frac{1}{2}{u}^2=\frac{\gamma +1}{2(\gamma -1)}{a^{*}}^2$

(Eq. 29)

If Eqn. \ref{eq:governing:energy:e} is evaluated in locations 1 and 2, we get

$\begin{array}{cc}& a_1^2=\frac{\gamma +1}{2}{a^{*}}^2-\frac{\gamma -1}{2}{u}_1^2\\ & a_2^2=\frac{\gamma +1}{2}{a^{*}}^2-\frac{\gamma -1}{2}{u}_2^2\end{array}$

(Eq. 30)

Since the change in flow conditions over the shock is adiabatic (no heat is added inside the shock), critical properties will be constant over the shock. Especially $a^{*}$ will be constant.

Eqn. \ref{eq:governing:energy:f} inserted in \ref{eq:governing:mom:c} gives\\

$\frac{1}{\gamma u_1}(\frac{\gamma +1}{2}{a^{*}}^2-\frac{\gamma -1}{2}{u}_1^2)-\frac{1}{\gamma u_2}(\frac{\gamma +1}{2}{a^{*}}^2-\frac{\gamma -1}{2}{u}_2^2)=u_2-u_1\Rightarrow$

(Eq. 31)

$\left(\frac{\gamma +1}{2\gamma }\right){a^{*}}^2(\frac{1}{u_1}-\frac{1}{u_2})=\left(\frac{\gamma +1}{2\gamma }\right)(u_2-u_1)\Rightarrow$

(Eq. 32)

${a^{*}}^2(\frac{1}{u_1}-\frac{1}{u_2})=(u_2-u_1)\Rightarrow$

(Eq. 33)

${a^{*}}^2(\frac{u_2}{u_1{u}_2}-\frac{u_1}{u_1{u}_2})=(u_2-u_1)\Rightarrow$

(Eq. 34)

$\frac{1}{u_1{u}_2}{a^{*}}^2(u_2-u_1)=(u_2-u_1)\Rightarrow$

(Eq. 35)

${a^{*}}^2=u_1{u}_2$

(Eq. 36)

Eqn. \ref{eq:prandtl} is sometimes referred to as the Prandtl relation. Divide the Prandtl relation by ${a^{*}}^2$ on both sides gives

$1=\frac{u_1}{a^{*}}\frac{u_2}{a^{*}}=M_1^{*}{M}_2^{*}$

(Eq. 37)

or

$M_2^{*}=\frac{1}{M_1^{*}}$

(Eq. 38)

The relation between $M^{*}$ and $M$ is given by

${M^{*}}^2=\frac{(\gamma +1)M^2}{2+(\gamma -1)M^2}$

(Eq. 39)

from which is can be seen that $M^{*}$ will follow the Mach number $M$ in the sense that

• $M=1\Rightarrow M^{*}=1$
• $M<1\Rightarrow M^{*}<1$
• $M>1\Rightarrow M^{*}>1$

Eqn. \ref{eq:MachStar} inserted in Eqn. \ref{eq:NormalMach} gives

$\frac{(\gamma +1)M_1^2}{2+(\gamma -1)M_1^2}=\frac{2+(\gamma -1)M_2^2}{(\gamma +1)M_2^2}$

(Eq. 40)

$M_2^2=\frac{1+[(\gamma -1)/2]M_1^2}{\gamma M_1^2-(\gamma -1)/2}$

(Eq. 41)

The Mach number relations above effectively show that if the Mach number upstream of the shock is greater than one, the downstream Mach number must be less than one and vice versa. We can also see that a sonic upstream flow $M_1=1.0$ gives sonic flow downstream of the shock. So, apparently the relation as such holds for both supersonic and subsonic upstream flow mathematically. The question is if it is also physically correct. For a supersonic upstream flow we will get a discontinuous compression and if the flow upstream of the control volume is subsonic we will instead get a discontinuous expansion inside the control volume but, again, is this physically correct? We will get the answer by analyzing the entropy change over the control volume.

Analyzing the energy equation and the second law of thermodynamics shows that there is a direct relation between entropy increase and total pressure drop.

$s_2-s_1=C_p\ln \frac{T_2}{T_1}-R\ln \frac{p_2}{p_1}$

(Eq. 42)

$s_2-s_1=C_p\ln \frac{T_2}{T_{o_2}}\frac{T_{o_1}}{T_1}\frac{T_{o_2}}{T_{o_1}}-R\ln \frac{p_2}{p_{o_2}}\frac{p_{o_1}}{p_1}\frac{p_{o_2}}{p_{o_1}}$

(Eq. 43)

using the isentropic relations we get

$s_2-s_1=C_p\ln \frac{T_{o_2}}{T_{o_1}}-R\ln \frac{p_{o_2}}{p_{o_1}}$

(Eq. 44)

and since the process is adiabatic and thus $T_{o_2}=T_{o_1}$ the change in entropy is directly related to the change in total pressure as

$s_2-s_1=-R\ln \frac{p_{o_2}}{p_{o_1}}$

(Eq. 45)

or

$\frac{p_{o_2}}{p_{o_1}}=e^{-(s_2-s_1)/R}$

(Eq. 46)

Figure~\ref{fig:shock:entropy} shows the entropy change over a normal shock. As can be seen in the figure, a subsonic upstream Mach number leads to a reduction of entropy, which once and for all rules out all such solutions as non-physical and thus the question about the upstream conditions can now be considered answered. This in turn implies that the Mach number downstream of a normal shock will always be subsonic, which can be seen in Fig~\ref{fig:shock:downstream:Mach} below.

By rewriting the right-hand side of Eqn.\ref{eq:NormalMach:b}, it is easy to realize that the downstream Mach number $M_2$ approaches a finite value for large values of the upstream Mach number, $M_1$.

${M_2^2|}_{M_1\to \mathrm{\infty }}={\frac{2/M_1^2+(\gamma -1)}{2\gamma -(\gamma -1)/M_1^2}|}_{M_1\to \mathrm{\infty }}=\frac{\gamma -1}{2\gamma }$

(Eq. 47)

Normal-shock relations

Rewriting the continuity equation (Eqn. \ref{eq:governing:cont})

$\frac{{\rho }_2}{{\rho }_1}=\frac{u_1}{u_2}=\frac{u_1^2}{u_1{u}_2}=\{{a^{*}}^2=u_1{u}_2\}=\frac{u_1^2}{{a^{*}}^2}={M_1^{*}}^2$

(Eq. 48)

Eqn. \ref{eq:MachStar} in Eqn. \ref{eq:Normal:density:a} gives

$\frac{{\rho }_2}{{\rho }_1}=\frac{(\gamma +1)M_1^2}{2+(\gamma -1)M_1^2}$

(Eq. 49)

To get a corresponding relation for the pressure ratio over the shock, we go back to the momentum equation (Eqn. \ref{eq:governing:mom})

$p_2-p_1={\rho }_1{u}_1^2-{\rho }_2{u}_2^2=\{{\rho }_1{u}_1={\rho }_2{u}_1\}={\rho }_1{u}_1(u_1-u_2)={\rho }_1{u}_1^2(1-\frac{u_2}{u_1})$

(Eq. 50)

$\frac{p_2-p_1}{p_1}=\frac{{\rho }_1{u}_1^2}{p_1}(1-\frac{u_2}{u_1})=\{a_1=\sqrt{\frac{\gamma p_1}{{\rho }_1}}\}=\gamma \frac{u_1^2}{a_1^2}(1-\frac{u_2}{u_1})=\gamma M_1^2(1-\frac{u_2}{u_1})$

(Eq. 51)

$\frac{p_2}{p_1}-1=\gamma M_1^2(1-\frac{u_2}{u_1})=\{\frac{u_2}{u_1}=\frac{{\rho }_1}{{\rho }_2}\}=\gamma M_1^2(1-\frac{2+(\gamma -1)M_1^2}{(\gamma +1)M_1^2})$

(Eq. 52)

$\frac{p_2}{p_1}=1+\frac{2\gamma }{\gamma +1}(M_1^2-1)$

(Eq. 53)

Figure~\ref{fig:shock:pressure:ratio} shows that the pressure must increase over the shock due to the fact that, based on the discussion above, the upstream Mach number must be greater than one and thus the shock is a discontinuous compression process.

The temperature ratio over the shock can be obtained using the already derived relations for pressure ratio and density ratio together with the equation of state $p=\rho RT$

$\frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)\left(\frac{{\rho }_1}{{\rho }_2}\right)$

(Eq. 54)

$\frac{T_2}{T_1}=[1+\frac{2\gamma }{\gamma +1}(M_1^2-1)]\left[\frac{(\gamma +1)M_1^2}{2+(\gamma -1)M_1^2}\right]$

(Eq. 55)

Figure~\ref{fig:normal:shock:relations} below shows how different flow properties change over a normal shock as a function of upstream Mach number.

Now, one question remains. How come that we by analyzing the control volume using the upstream and downstream states get the normal shock relations. There is no way that the governing equations could have known about the fact that we assumed that there would be a shock inside of the control volume, or is it? The answer is that we have assumed that there will be a change in flow properties from upstream to downstream. We have further assumed that the flow is adiabatic (we are using the adiabatic energy equation) so there is no heat exchange. We are, however, allowing for irreversibilities in the flow. The only way to accomplish a change in flow properties under those constraints is a formation of a normal shock (a discontinuity in flow properties - a sudden flow compression) between station 1 and station 2.

The Hugoniot equation

The Hugoniot equation is an alternative normal shock relation based on thermodynamic quantities only. It is derived from the governing equations and relates the change in energy to the change in pressure and specific volume. The starting point of the derivation of the Hugoniot equation is the governing equations (Eqns~\ref{eq:governing:cont} - \ref{eq:governing:energy}).

The continuity equation is rewritten and inserted into the momentum equation

$u_1=\left(\frac{{\rho }_2}{{\rho }_1}\right)u_2$

(Eq. 56)

Replace $u_1$ in Eqn. \ref{eq:governing:mom} using Eqn. \ref{eq:governing:cont:b}

${\rho }_1{\left(\frac{{\rho }_2}{{\rho }_1}\right)}^2{u}_2^2+p_1={\rho }_2{u}_2^2+p_2$

(Eq. 57)

$u_2^2({\rho }_1{\left(\frac{{\rho }_2}{{\rho }_1}\right)}^2-{\rho }_2)=(p_2-p_1)$

(Eq. 58)

$u_2^2\left(\left(\frac{{\rho }_2}{{\rho }_1}\right)({\rho }_2-{\rho }_1)\right)=(p_2-p_1)$

(Eq. 59)

$u_2^2=\left(\frac{{\rho }_1}{{\rho }_2}\right)\frac{p_2-p_1}{{\rho }_2-{\rho }_1}$

(Eq. 60)

Eqn. \ref{eq:governing:cont:b} and \ref{eq:governing:mom:b} gives

$u_1^2=\left(\frac{{\rho }_2}{{\rho }_1}\right)\frac{p_2-p_1}{{\rho }_2-{\rho }_1}$

(Eq. 61)

Eqn. \ref{eq:governing:mom:b} and Eqn. \ref{eq:governing:mom:c} inserted in the energy equation (Eqn. \ref{eq:governing:energy}) gives

$h_1+\frac{1}{2}\left(\frac{{\rho }_2}{{\rho }_1}\right)\left(\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\right)=h_2+\frac{1}{2}\left(\frac{{\rho }_1}{{\rho }_2}\right)\left(\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\right)$

(Eq. 62)

$h_2-h_1=\frac{p_2-p_1}{2}[\left(\frac{{\rho }_2}{{\rho }_1}\right)\left(\frac{1}{{\rho }_2-{\rho }_1}\right)-\left(\frac{{\rho }_1}{{\rho }_2}\right)\left(\frac{1}{{\rho }_2-{\rho }_1}\right)]$

(Eq. 63)

$h_2-h_1=\frac{p_2-p_1}{2}\left[\frac{{\rho }_2^2-{\rho }_1^2}{{\rho }_1{\rho }_2({\rho }_2-{\rho }_1)}\right]=\frac{p_2-p_1}{2}\left[\frac{{\rho }_2+{\rho }_1}{{\rho }_1{\rho }_2}\right]$

(Eq. 64)

$h_2-h_1=\frac{p_2-p_1}{2}(\frac{1}{{\rho }_1}+\frac{1}{{\rho }_2})$

(Eq. 65)

Now, replacing the enthalpies with internal energies using $h=e+p/\rho$ gives

$e_2-e_1=\frac{p_1}{{\rho }_1}-\frac{p_2}{{\rho }_2}+\frac{p_2-p_1}{2}(\frac{1}{{\rho }_1}+\frac{1}{{\rho }_2})$

(Eq. 66)

which after some rewriting becomes the Hugoniot equation

$e_2-e_1=\frac{p_2+p_1}{2}(\frac{1}{{\rho }_1}-\frac{1}{{\rho }_2})=\frac{p_2+p_1}{2}({\nu }_1-{\nu }_2)$

(Eq. 67)

To give an idea about how the normal shock relates to an isentropic compression (a flow compression process without losses) the change in flow density as a function of pressure ratio is compared in Figure~\ref{fig:normal:shock:compression:vs:isentropic}. One can see that the normal-shock compression is more effective but less efficient than the corresponding isentropic process.

Introducing $C$ as the massflow per unit area (which is a constant)

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

(Eq. 68)

Inserted into the momentum equation this gives

$p_1+\frac{C^2}{{\rho }_1}=p_2+\frac{C^2}{{\rho }_2}$

(Eq. 69)

or

$\frac{p_2-p_1}{{\nu }_2-{\nu }_1}=-C^2$

(Eq. 70)

which implies that all possible solutions to the governing equations must be located on a line in $p\nu$-space (the so-called Rayleigh line). If we add the Hugoniot relation to this we will find that there are two possible solutions, the upstream condition and the condition downstream of the normal shock and the flow cannot be in any of the intermediate stages. The normal-process is a so-called wave solution to the governing equations where the flow state must jump directly from one flow state to another without passing the intermediate conditions. If we add heat or friction to the problem we will instead get continuous solutions as we will see in the following sections. Figures \ref{fig:shock:pv} and \ref{fig:shock:Ts} shows a normal shock process in a $p\nu$- and $Ts$-diagram, respectively. Note that the flow passes the characteristic conditions as it is going through the shock, which means that the flow goes from supersonic to subsonic.

One-dimensional flow with heta addition

Flow-station relations

The aim is to derive relations for pressure ratio and temperature ratio as a function of Mach numbers. We will do that starting from the momentum equation.

$p_2-p_1={\rho }_1{u}_1^2-{\rho }_2{u}_2^2$

(Eq. 71)

Assuming calorically perfect gas

$\rho u^2=\rho a^2{M}^2=\rho \frac{\gamma p}{\rho }{M}^2=\gamma p{M}^2$

(Eq. 72)

which inserted in Eqn. \ref{eq:governing:mom} gives

$p_2-p_1=\gamma p_1{M}_1^2-\gamma p_2{M}_2^2$

(Eq. 73)

$p_2(1+\gamma M_2^2)=p_1(1+\gamma M_1^2)$

(Eq. 74)

and thus

$\frac{p_2}{p_1}=\frac{1+\gamma M_1^2}{1+\gamma M_2^2}$

(Eq. 75)

From the equation of state $p=\rho RT$, we get

$\frac{T_2}{T_1}=\frac{p_2}{{\rho }_{2}R}\frac{{\rho }_{1}R}{p_1}=\frac{p_2}{p_1}\frac{{\rho }_1}{{\rho }_2}$

(Eq. 76)

Using the continuity equation, we can get ${\rho }_1/{\rho }_2$

${\rho }_1{u}_1={\rho }_2{u}_2\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{u_2}{u_1}$

(Eq. 77)

Inserted in Eqn. \ref{eq:tr:a} gives

$\frac{T_2}{T_1}=\frac{p_2}{p_1}\frac{u_2}{u_1}$

(Eq. 78)

$\frac{u_2}{u_1}=\frac{M_2{a}_2}{M_1{a}_1}=\frac{M_2}{M_1}\frac{\sqrt{\gamma R{T}_2}}{\sqrt{\gamma R{T}_1}}=\frac{M_2}{M_1}\sqrt{\frac{T_2}{T_1}}$

(Eq. 79)

Eqn. \ref{eq:tr:c} in Eqn. \ref{eq:tr:b} gives

$\sqrt{\frac{T_2}{T_1}}=\frac{p_2}{p_1}\frac{M_2}{M_1}$

(Eq. 80)

With $p_2/p_1$ from Eqn. \ref{eq:governing:mom:b}, Eqn \ref{eq:tr:d} becomes

$\frac{T_2}{T_1}={\left(\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\right)}^2{\left(\frac{M_2}{M_1}\right)}^2$

(Eq. 81)

Differential Relations

The equations presented in the previous section gives us the flow state after heat addition but since the heat addition, unlike the normal shock, is a continuous process, it is of interest to study the the heat addition from start to end. In order to do so we will now derive differential relations starting from the governing equations on differential form. We will start with converting the integral equation for conservation of mass for one-dimensional flows to differential form.

${\rho }_1{u}_1={\rho }_2{u}_2=const\Rightarrow d(\rho u)=0$

(Eq. 82)

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

(Eq. 83)

Divide by $\rho u$ gives

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

(Eq. 84)

The integral form of the conservation of momentum equation for one-dimensional flows is converted to differential form as follows.

$p_1+{\rho }_1{u}_1^2=p_2+{\rho }_2{u}_2^2=const\Rightarrow d(p+\rho u^2)=0$

(Eq. 85)

$dp+\rho udu+u\underset{=0}{\underbrace{d(\rho u)}}=0\Rightarrow dp=-\rho udu$

(Eq. 86)

with $\rho =\frac{p}{RT}$ and $u^2=M^2{a}^2=M^2\gamma RT$ in Eqn.~\ref{eq:governing:mom:diff:b}, we get

$dp=-\frac{p}{RT}{u}^2\frac{du}{u}=-\frac{p}{RT}{M}^2\gamma RT\frac{du}{u}\Rightarrow$

(Eq. 87)

$\frac{dp}{p}=-\gamma M^2\frac{du}{u}$

(Eq. 88)

which gives the relative change in pressure, $dp/p$, as a function of the relative change in flow velocity, $du/u$. The next equation to derive is an equation that describes the relative change in temperature, $dT/T$, as a function of the relative change in flow velocity, $du/u$. The starting point is the equation of state (the gas law).

$p=\rho RT\Rightarrow dp=R(\rho dT+Td\rho )\Rightarrow dT=\frac{1}{R\rho }dp-\frac{T}{\rho }d\rho$

(Eq. 89)

Divide by $T$

$\frac{dT}{T}=\frac{1}{\rho RT}dp-\frac{1}{\rho }d\rho$

(Eq. 90)

substitute $dp$ from Eqn.~\ref{eq:governing:mom:diff:c} and $d\rho$ from Eqn.~\ref{eq:governing:cont:diff:b} gives

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

(Eq. 91)

The entropy equation reads

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

(Eq. 92)

which after substituting $dp$ from Eqn.~\ref{eq:governing:mom:diff:c} and $d\rho$ from Eqn.~\ref{eq:governing:cont:diff:b} becomes

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

(Eq. 93)

From the definition of total temperature $T_o$ we get

$T_o=T+\frac{u^2}{2C_p}\Rightarrow d{T}_o=dT+\frac{1}{C_p}udu=dT+\frac{\gamma -1}{\gamma RT}T{u}^2\frac{du}{u}\Rightarrow$

(Eq. 94)

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

(Eq. 95)

Inserting $dT$ from Eqn~\ref{eq:governing:temp:diff:c} in Eqn~\ref{eq:governing:To:diff:a} we get

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

(Eq. 96)

or

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

(Eq. 97)

Dividing Eqn.~\ref{eq:governing:To:diff:b} by $T_o$ and using

$T_o=T(1+\frac{\gamma -1}{2}{M}^2)$

(Eq. 98)

we get

$\frac{d{T}_o}{T_o}=\frac{1-M^2}{1+\frac{\gamma -1}{2}{M}^2}\frac{du}{u}$

(Eq. 99)

Finally, we will derive a differential relation that describes the change in Mach number.

$M=\frac{u}{\sqrt{\gamma RT}}\Rightarrow dM=\frac{1}{\sqrt{\gamma R}}(T^{1/2}du+ud(T^{-1/2}))=\frac{du}{\sqrt{\gamma RT}}-\frac{u}{2\sqrt{\gamma R}}{T}^{-3/2}dT\Rightarrow$

(Eq. 100)

$dM=\frac{1}{\sqrt{\gamma RT}}\frac{du}{u}-\frac{1}{2}\frac{u}{\sqrt{\gamma RT}}\frac{dT}{T}=M\frac{du}{u}-\frac{M}{2}\frac{dT}{T}$

(Eq. 101)

Inserting $dT/T$ from Eqn.~\ref{eq:governing:temp:diff:c}, we get

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

(Eq. 102)

All the derived differential relations are expressed as functions of <math<du/u</math> but it would be more convenient to relate the changes in flow properties to the added heat or the change in total temperature, which can be related to the added heat through the energy equation.

$d{T}_o=\frac{\delta q}{C_p}$

(Eq. 103)

From Eqn.~\ref{eq:governing:To:diff:c}, we get

$\frac{du}{u}=\left(\frac{1+\frac{\gamma -1}{2}{M}^2}{1-M^2}\right)\frac{d{T}_o}{T_o}$

(Eq. 104)

Now, we can substitute $du/u$ in all the above relations using Eqn.~\ref{eq:governing:du:diff:final}, we get the following relations

$\frac{d\rho }{\rho }=-\left(\frac{1+\frac{\gamma -1}{2}{M}^2}{1-M^2}\right)\frac{d{T}_o}{T_o}$

(Eq. 105)

$\frac{dp}{p}=\gamma M^2\left(\frac{1+\frac{\gamma -1}{2}{M}^2}{1-M^2}\right)\frac{d{T}_o}{T_o}$

(Eq. 106)

$\frac{dT}{T}=(1-\gamma M^2)\left(\frac{1+\frac{\gamma -1}{2}{M}^2}{1-M^2}\right)\frac{d{T}_o}{T_o}$

(Eq. 107)

$\frac{dT}{T}=\left(\frac{1+\gamma M^2}{2}\right)\left(\frac{1+\frac{\gamma -1}{2}{M}^2}{1-M^2}\right)\frac{d{T}_o}{T_o}$

(Eq. 108)

$\frac{dT}{T}=C_v\gamma (1+\frac{\gamma -1}{2}{M}^2)\frac{d{T}_o}{T_o}$

(Eq. 109)

Heat Addition Process

With the differential relations in place, we can now study the continuous change in flow quantities from the initial flow state to the flow state after the heat addition process by dividing the total amount of heat added to the flow, $q$, into small portions, $\delta q$, and calculate the change in flow properties for each of these heat additions, see Figure~\ref{fig:dq}.

Let's first examine the temperature change by rewriting Eqn.~\ref{eq:governing:dT:diff:final} as

$dT=\frac{1-\gamma M^2}{1-M^2}d{T}_o\iff \frac{dT}{d{T}_o}=\frac{1-\gamma M^2}{1-M^2}$

(Eq. 110)

which is equivalent to

$\frac{dh}{\delta q}=\frac{1-\gamma M^2}{1-M^2}$

(Eq. 111)

Form Eqn.~\ref{eq:governing:dT:diff:mod:a} we can make the following observation

$\frac{dT}{d{T}_o}=0\Rightarrow \gamma M^2=1\Rightarrow M=\sqrt{1/\gamma }$

(Eq. 112)

which means that the maximum temperature will be reached when the Mach number is $\sqrt{1/\gamma }$. Since $\gamma$ is a number greater than one for all gases, this implies that the maximum temperature can only be reached if the flow is subsonic. For air, this the maximum temperature will be reached at $M=0.845$.

If we evaluate Eqn.~\ref{eq:governing:dT:diff:mod:a} for sonic flow ($M=1$), we see that the derivative becomes infinite.

$|M|\to 1.0\Rightarrow \frac{dT}{d{T}_o}\to \pm \mathrm{\infty }$

(Eq. 113)

Now, by specifying an initial subsonic flow state and dividing the heat addition corresponding to choked flow, $q^{\ast }$, into small portions $\delta q$, one can perform integration as indicated in Figure~\ref{fig:dq}. The result is presented in the in Figure~\ref{fig:TS:closeup}. The subsonic process corresponds to the upper line. As heat is added the Mach number is increased and at $M={\gamma }^{-1/2}$ the maximum temperature is reached. Adding more heat will reduce the temperature and increase the Mach number until sonic conditions are reached ($M=1.0$). As can be seen in Figure~\ref{fig:TS:closeup}, the lean of the subsonic branch of the Rayleigh line is lower than the isobars (gray lines), which means the increasing heat will reduce pressure. The lower part of the blue line in Figure~\ref{fig:TS:closeup} is the supersonic branch of the Rayleigh line, which is obtained in the same way starting from a supersonic flow condition. A flow state resulting in the same sonic conditions as for the subsonic case is calculated and used as a starting state. The corresponding $q^\ast$ is calculated and the same calculation of consecutive flow states in a step-wise manner is performed. As can be seen in Figure~\ref{fig:TS:closeup}, the lean of the supersonic part of the Rayleigh curve is steeper than the isobars (gray lines), which means that pressure increases as heat is added to the flow. As we saw from Eqn.~\ref{eq:governing:dT:diff:mod:b}, $dT/d{T}_o$ becomes infinite when the flow approaches the sonic the sonic state. After the sonic state is reached, further heat addition is impossible without changing the upstream flow conditions. This will be made clearer in the next section.

Using the differential relations above, we can get a good picture of the development of flow variables as heat is continuously added to the flow (see Figure~\ref{fig:rayleigh:trends}).

Rayleigh Line

The continuity equation for steady-state, one-dimensional flow reads

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

(Eq. 114)

where $C$ is the massflow per square meter (massflow divided by area). Inserted in the momentum equation we get

$p_1+\frac{C^2}{{\rho }_1}=p_2+\frac{C^2}{{\rho }_2}\iff p_1+{\nu }_1{C}^2=p_2+{\nu }_2{C}^2\Rightarrow \frac{p_2-p_1}{{\nu }_2-{\nu }_1}=-C^2$

(Eq. 115)

Eqn.~\ref{eq:governing:mom:b} tells us that any solution to the governing flow equations must lie along a line (a so-called Rayleigh line) in a $p\nu$-diagram. In Figure~\ref{fig:PV}, 1 corresponds to the flow state before heat addition and states 2 and 3 corresponds to the flow state after heat is added. If the flow in state 1 is subsonic, adding heat will change the flow state following the Rayleigh line to the right, i.e. towards flow state 2. If the initial flow state instead is supersonic, heat addition will move the flow state towards state 3.

Now we know in which direction we will move along the Rayleigh curve when heat is added but in order to find the flow state after heat addition we need to add the energy equation to the problem. If we draw a curve corresponding to the energy equation including the heat addition in the same $p\nu$-diagram, the intersection of this curve and the Rayleigh line corresponds to the downstream flow state (the flow state that fulfils the continuity, momentum, and energy equations). To be able to do this we will rewrite the energy equation such that it can be represented by a line in the $p\nu$-diagram.

The energy equation for one-dimensional flow with heat addition reads

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

(Eq. 116)

Inserting the constant $C$ from above (the massflow per $m^2$) and and and $h=C_{p}T$, we get

$\frac{\gamma R}{\gamma -1}{T}_1+\frac{1}{2}{C}^2{\nu }_1^2+q=\frac{\gamma R}{\gamma -1}+\frac{1}{2}{C}^2{\nu }_2^2$

(Eq. 117)

which may be rewritten as

$\frac{p_2}{p_1}=(\frac{{\nu }_2}{{\nu }_1}-\frac{\gamma +1}{\gamma -1}-\frac{2q}{R{T}_1}){(1-\frac{\gamma +1}{\gamma -1}\frac{{\nu }_2}{{\nu }_1})}^{-1}$

(Eq. 118)

As you can see in the examples above (Figures~\ref{fig:TSPV:b} and \ref{fig:TSPV:d}), sonic conditions are reached when the Rayleigh line is tangent to the curve representing the energy equation in the $p\nu$-diagram. Adding more heat would move the energy equation line upwards and thus there can not be any solution after reaching this state unless the upstream conditions are changed such that the energy line intersects the Rayleigh line after further heat addition. Let's have a second look at the equations and see if it is possible to verify that the case where the Rayleigh line is a tangent to the energy-equation curve is in fact the sonic state.

Starting from Eqn.~\ref{eq:governing:energy:b}, it is easy to see that for any point along the energy equation curve the flow state may be expressed as a function of the initial flow state and the added heat $q$ as

$\frac{\gamma }{\gamma -1}p\nu +\frac{1}{2}{C}^2{\nu }^2=\frac{\gamma }{\gamma -1}{p}_1{\nu }_1+\frac{1}{2}{C}^2{\nu }_1^2+q=D$

(Eq. 119)

where $D$ is a constant.

Now, let's different the Eqn.\ref{eq:governing:energy:d} with respect to $\nu$

$\frac{\gamma }{\gamma -1}(\nu \frac{dp}{d\nu }+p)+C^2\nu =0\Rightarrow \frac{dp}{d\nu }=-\frac{\gamma }{\gamma -1}{C}^2-\frac{p}{\nu }$

(Eq. 120)

The Rayleigh line is a tangent to the energy equation curve when $dp/d\nu =-C^2$ and thus

$\frac{C^2}{\gamma }=\frac{p}{\nu }$

(Eq. 121)

By definition $C=\rho u$ and $\nu =1/\rho$, which inserted in Eqn.~\ref{eq:governing:energy:f} gives

$u=\sqrt{\frac{\gamma p}{\rho }}=a$

(Eq. 122)

Thermal Choking

When the heat addition reaches $q^\ast$ the flow becomes sonic and the flow is said to thermally choked. Thermal choking is illustrated in Figure~\ref{fig:TSPV:d}, where the curve representing the energy equation (the blue line in the $p\nu$-diagram) is tangent to the Rayleigh line and if more heat is added the blue line will move to the right of the Rayleigh line and thus there are no solutions for $q>q^{\ast }$. So what happens if more heat is added to the flow after thermal choking is reached. The answer is different if the flow is subsonic or supersonic. For a subsonic flow, the upstream flow will be adjusted such that the slope of the Rayleigh line changes and the energy equation curve becomes tangent to the Rayleigh line. This means that the massflow per unit area ($C$) is reduced and $q^{\ast }$ is increased such that $q^{\ast }$ equals the heat added to the flow. Note that the upstream total conditions will not be changed in this process (see Figure~\ref{fig:thermal:choking:sub}).

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

In a choked supersonic flow, there is no possibility for pressure waves to travel upstream in the flow and thus the upstream flow conditions can not be changed as in the subsonic case. Moreover, since a normal shock is an adiabatic process (a jump between two points on the same Rayleigh line), the total temperature is not changed over a chock. From before we have

$\frac{T_o}{T_o^{\ast }}=\frac{(\gamma +1)M_1^2}{(1+\gamma M_1^2{)}^2}(2+(\gamma -1)M_1^2)$

(Eq. 123)

Inserting the normal shock relation

$M_2^2=\frac{2+(\gamma -1)M_1^2}{2\gamma M_1^2-(\gamma -1)}$

(Eq. 124)

one can show that

$\frac{T_o}{T_o^{\ast }}=\frac{(\gamma +1)M_2^2}{(1+\gamma M_2^2{)}^2}(2+(\gamma -1)M_2^2)$

(Eq. 125)

and thus $T_o^{\ast }$ is not changed by the normal shock and consequently $q^{\ast }$ is unchanged if there is a normal shock between station 1 and 2. So, it is not possible to change the upstream static flow conditions and a normal shock will not make it possible to add more heat. The only possible solution is a normal shock upstream of station 1 and thus subsonic flow through the heat addition process.

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. 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. 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. 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. 129)

Energy

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

(Eq. 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. 131)

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

(Eq. 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. 133)

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

(Eq. 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. 135)

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

(Eq. 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. 137)

and thus

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

(Eq. 138)

Energy

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

(Eq. 139)

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

(Eq. 140)

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

(Eq. 141)

Summary

continuity:

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

(Eq. 142)

momentum:

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

(Eq. 143)

energy:

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

(Eq. 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. 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. 146)

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

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

(Eq. 147)

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

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

(Eq. 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. 149)

Energy equation

For an adiabatic one-dimensional flow we have that

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

(Eq. 150)

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

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

(Eq. 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. 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. 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. 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. 155)

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

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

(Eq. 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. 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. 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. 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. 160)

which gives

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

(Eq. 161)

The ideal gas law

For a perfect gas the ideal gas law reads

$p=\rho RT$

(Eq. 162)

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

$dp=\rho RdT+RTd\rho$

(Eq. 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. 164)

which can be rearranged according to

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

(Eq. 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. 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. 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. 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. 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. 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. 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. 172)

collecting terms and rearranging gives

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

(Eq. 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. 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. 175)

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

(Eq. 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. 177)

differentiating gives

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

(Eq. 178)

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

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

(Eq. 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. 180)

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

(Eq. 181)

The entropy increase can be obtained from

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

(Eq. 182)

and thus

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

(Eq. 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. 184)

which can be rewritten as

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

(Eq. 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. 186)

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

(Eq. 187)

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

(Eq. 188)

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

(Eq. 189)

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

(Eq. 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. 191)

Since total temperature is constant the relation above gives

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

(Eq. 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. 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. 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. 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. 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.