12.3.2: Geometrical Explanation
- Page ID
- 856
Fig. 12.23 The schematic of the turning flow.
The change in the flow direction is assume to be result of the change in the tangential component. Hence, the total Mach number increases. Therefore, the Mach angle increase and result in a change in the direction of the flow. The velocity component in the direction of the Mach line is assumed to be constant to satisfy the assumption that the change is a result of the contour only. Later, this assumption will be examined. The typical simplifications for geometrical functions are used:
\[
\begin{array}{rl}
d\nu & \sim \sin (d\nu) ; \
\cos (d\nu) & \sim 1
\end{array}
\label{pm:eq:angle}
\]
\[
dx = (U + dU) \cos\nu -U = dU
\label{pm:eq:horizontalLine}
\]
In the same manner, the velocity perpendicular to the flow, \(dy\), is
\[
dy = (U + dU) \sin(d\nu) = U d\nu
\label{pm:eq:verticalLine}
\]
The \(\tan \mu\) is the ratio of \(dy/dx\) (see Figure ??)
\[
\tan \mu = \dfrac{dx }{ dy} = \dfrac{ dU }{ U\, d\nu }
\label{pm:eq:tanMU}
\]
The ratio \(dU/U\) was shown to be
\[
\dfrac{dU }{ U } = \dfrac{ dM^2 }{ 2\,M^2
\left( 1 + \dfrac{k -1 }{ 2} M^2 \right) }
\label{pm:eq:adiabatic}
\]
Combining equations (6) and (7) transforms it into
\[
d\nu = - \dfrac{ \sqrt{M^2 - 1} dM^2 }
{ 2\,M^2 \left( 1 + \dfrac{k -1 }{ 2} M^2 \right) }
\label{pm:eq:d-Nu}
\]
After integration of equation (8) becomes
Turnning Angle
\[
\begin{array}{c}
\nu (M) = -\sqrt{\dfrac{k+1 }{ k-1 } }
\tan^{-1} \sqrt{ \dfrac{k-1 }{ k+1} \left( M^2 -1\right)} \\
+ \tan^{-1} \sqrt{ \left( M^2 -1\right)} + \text{constant}
\end{array}
\label{pm:eq:nu}
\]
The constant can be chosen in a such a way that \(\nu=0\) at \(M=1\).
Alternative Approach to Governing Equations
Fig. 12.24 The schematic of the coordinate based on the mathematical description.
In the previous section, a simplified version was derived based on geometrical arguments. In this section, a more rigorous explanation is provided. It must be recognized that here the cylindrical coordinates are advantageous because the flow turns around a single point. For this coordinate system, the mass conservation can be written as
\[
\dfrac{\partial \left( \rho\, r\, U_r \right) }{ \partial r} +
\dfrac{\partial \left( \rho \, U_\theta \right) }{ \partial \theta} = 0
\label{pm:eq:mass}
\]
The momentum equations are expressed as
\[
U_r \, \dfrac{\partial U_r }{ \partial r } +
\dfrac{U_\theta }{ r}\, \dfrac{\partial U_r }{ \partial \theta } -
\dfrac{{U_\theta}^2 }{ r} = - \dfrac{ 1 }{ \rho}
\dfrac{\partial P }{ \partial r} =
- \dfrac{ c^2 }{ \rho} \, \dfrac{\partial \rho }{ \partial r}
\label{pm:eq:mom1}
\]
\[
U_r \,\dfrac{\partial U_\theta }{ \partial r } +
\dfrac{U_\theta }{ r} \, \dfrac{\partial U_\theta }{ \partial \theta } -
\dfrac{{U_\theta} U_r }{ r} = - \dfrac{ 1 }{r\, \rho}
\dfrac{\partial P }{ \partial \theta } =
- \dfrac{ c^2 }{ r\, \rho}\, \dfrac{\partial \rho }{ \partial \theta}
\label{pm:eq:mom2}
\]
If the assumption is that the flow isn't a function of the radius, \(r\), then all the derivatives with respect to the radius will vanish. One has to remember that when \(r\) enters to the function, like the first term in the mass equation, the derivative isn't zero. Hence, the mass equation is reduced to
\[
\rho U_r +
\dfrac{\partial \left( \rho U_\theta \right) }{ \partial \theta} = 0
\label{pm:eq:massTheta1}
\]
Equation (13) can be rearranged as transformed into
\[
- \dfrac{1 }{ U_\theta} \left( U_r +
\dfrac{\partial U_\theta }{ \partial \theta} \right) =
\dfrac{1 }{ \rho} \dfrac{\partial \rho }{ \partial \theta}
\label{pm:eq:massTheta}
\]
The momentum equations now obtain the form of
\[
\begin{array}{c}
\dfrac{U_\theta }{ r} \,\dfrac{\partial U_r }{ \partial \theta } -
\dfrac{{U_\theta}^2 }{ r} = 0 \\
U_\theta \,\left( \dfrac{\partial U_r }{ \partial \theta }
- {U_\theta} \right) =0
\end{array}
\label{pm:eq:mom1Theta}
\]
\[
\begin{array}{c}
\dfrac{U_\theta }{ r} \,\dfrac{\partial U_\theta }{ \partial \theta } -
\dfrac{{U_\theta} \,U_r }{ r} =
- \dfrac{ c^2 }{ r \rho}\, \dfrac{\partial \rho }{ \partial \theta} \\
{U_\theta }\, \left( \dfrac{\partial U_\theta }{ \partial \theta } -
U_r \right) =
- \dfrac{ c^2 }{\rho} \,\dfrac{\partial \rho }{ \partial \theta}
\end{array}
\label{pm:eq:mom2Theta1}
\]
Substituting the term \({1 \over \rho}{ \partial \rho \over \partial \theta}\) from equation (??) into equation (16) results in
\[
U_\theta \left( {\partial U_\theta \over \partial \theta } -
{U_r } \right) = {c^2 \over U_\theta} \left( U_r +
{\partial U_\theta \over \partial \theta} \right)
\label{pm:eq:mom2Theta}
\]
or
\[
{U_\theta}^2
\left( U_r + {\partial U_\theta \over \partial \theta} \right)
= { c^2} \left( U_r +
{\partial U_\theta \over \partial \theta} \right)
\label{pm:eq:massMom}
\]
And an additional rearrangement results in
\[
\left( c^2 - {U_\theta}^2 \right)
\left( U_r + \dfrac{\partial U_\theta }{ \partial \theta} \right) = 0
\label{pm:eq:massMom1}
\]
From equation (19) it follows that
\[
U_\theta = c
\label{pm:eq:tangialVelocity}
\]
It is remarkable that the tangential velocity at every turn is at the speed of sound! It must be pointed out that the total velocity isn't at the speed of sound, but only the tangential component. In fact, based on the definition of the Mach angle, the component shown in Figure 12.23 under \(U_y\) is equal to the speed of sound, \(M=1\). After some additional rearrangement, equation (15) becomes
\[
\dfrac{U_\theta }{ r}
\left( \dfrac{\partial U_r }{ \partial \theta} - U_\theta \right)
= 0
\label{pm:eq:dUrUtheta}
\]
If \(r\) isn't approaching infinity, \(\infty\) and since \(U_\theta \neq 0\) leads to
\[
\dfrac{\partial U_r }{ \partial \theta } =
{U_\theta}
\label{pm:eq:Uthetac}
\]
In the literature, these results are associated with the characteristic line. This analysis can be also applied to the same equation when they are normalized by Mach number. However, the non–dimensionalization can be applied at this stage as well. The energy equation for any point on a stream line is
\[
h(\theta) + \dfrac{{U_\theta}^2 + {U_r}^2 }{ 2} = h_0
\label{pm:eq:energy}
\]
Enthalpy in perfect gas with a constant specific heat, \(k\), is
\[
h(\theta) = C_p\, T = C_p\,\dfrac{ R }{ R } \,T =
\dfrac{1 }{ (k-1)}
\overbrace{ \underbrace{\dfrac{C_p }{ C_v} }_k
\,R\,T}^{c(\theta)^2 }
= \dfrac{ c^2 }{ k-1}
\label{pm:eq:bernolliSound}
\]
and substituting this equality, equation (24), into equation (23) results in
\[
\dfrac{ c^2 }{ k-1} + \dfrac{{U_\theta}^2 + {U_r}^2 }{ 2} = h_0
\label{pm:eq:energyDE0}
\]
Utilizing equation (20) for the speed of sound and substituting equation (22) which is the radial velocity transforms equation (25) into
\[
\dfrac{ {\left(\dfrac{\partial U_r }{ \partial \theta} \right)}^2 }{ k-1}
+ \dfrac{\left(\dfrac{\partial U_r }{ \partial \theta} \right)^2
+ {U_r}^2 }{ 2} = h_0
\label{pm:eq:energyDE}
\]
After some rearrangement, equation (26) becomes
\[
\dfrac{ k+1 }{ k-1} \,
\left( \dfrac{\partial U_r }{ \partial \theta} \right)^2 + {U_r}^2 = 2 h_0
\label{pm:eq:energyDEa}
\]
Note that \(U_r\) must be positive. The solution of the differential equation (27) incorporating the constant becomes
\[
U_r = \sqrt{2h_0} \sin \left( \theta \sqrt{ \dfrac{ k-1 }{ k+1 } } \right)
\label{pm:eq:energySolution}
\]
which satisfies equation (27) because \(\sin^2\theta + \cos^2\theta = 1\). The arbitrary constant in equation (28) is chosen such that \(U_r (\theta=0) =0\). The tangential velocity obtains the form
\[
U_\theta = c = {\partial U_r \over \partial \theta} =
\sqrt{k-1 \over k+1 } \sqrt{2\;h_0} \;\;\cos
\left( \theta \sqrt{k-1 \over k+1} \right)
\label{pm:eq:veloctyRadious}
\]
The Mach number in the turning area is
\[
M^2 = {{U_\theta}^2 + {U_r}^2 \over c^2} =
{{U_\theta}^2 + {U_r}^2 \over {U_\theta}^2 } =
1 + \left( {U_r} \over U_\theta \right) ^2
\label{pm:eq:Mach}
\]
Now utilizing the expression that was obtained for \(U_r\) and \(U_{\theta}\) equations (29) and (28) results for the Mach number is
\[
M^2 = 1 + {k+1 \over k-1 } \tan^2
\left( \theta \sqrt{k-1 \over k+1} \right)
\label{m:eq:Mtheta}
\]
or the reverse function for \(\theta\) is
Reversed Angle
\[
\label{pm:eq:reverseTheta}
\theta = \sqrt{\dfrac{ k+1 }{ k-1 } }\, \tan^{-1}
\left( \sqrt{\dfrac{k-1 }{ k+1} } \, \left( M^2 -1 \right) \right)
\]
What happens when the upstream Mach number is not 1? That is when the initial condition for the turning angle doesn't start with \(M=1\) but is already at a different angle. The upstream Mach number is denoted in this segment as \(M_{starting}\). For this upstream Mach number (see Figure (12.22))
\[
\tan \nu = \sqrt{{M_{starting}}^2 - 1}
\label{pm:eq:nuInfty}
\]
\[
\nu (M) = \theta(M) - \theta(M_{starting})
\label{pm:eq:nuTheta1}
\]
Deflection Angle
\[
\label{pm:eq:nuTheta}
\nu (M) = \sqrt{k+1\over k-1}
\tan^{-1} \left( \sqrt{k-1\over k+1} \sqrt{ M^2 -1}\right)
- \tan^{-1} \sqrt{ M^2 -1}
\]
These relationships are plotted in Figure 12.26.
Comparison And Limitations between the Two Approaches
The two models produce exactly the same results, but the assumptions for the construction of these models are different. In the geometrical model, the assumption is that the velocity change in the radial direction is zero. In the rigorous model, it was assumed that radial velocity is only a function of \(\theta\). The statement for the construction of the geometrical model can be improved by assuming that the frame of reference is moving radially in a constant velocity. Regardless of the assumptions that were used in the construction of these models, the fact remains that there is a radial velocity at \(U_r(r=0)= constant\). At this point (\(r=0\)) these models fail to satisfy the boundary conditions and something else happens there. On top of the complication of the turning point, the question of boundary layer arises. For example, how did the gas accelerate to above the speed of sound when there is no nozzle (where is the nozzle?)? These questions are of interest in engineering but are beyond the scope of this book (at least at this stage). Normally, the author recommends that this function be used everywhere beyond 2-4 the thickness of the boundary layer based on the upstream length. In fact, analysis of design commonly used in the industry and even questions posted to students show that many assume that the turning point can be sharp. At a small Mach number, (\(1+\epsilon\)) the radial velocity is small \(\epsilon\). However, an increase in the Mach number can result in a very significant radial velocity. The radial velocity is "fed'' through the reduction of the density. Aside from its close proximity to turning point, mass balance is maintained by the reduction of the density. Thus, some researchers recommend that, in many instances, the sharp point should be replaced by a smoother transition.
Contributors and Attributions
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.