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General Velocities Issues

When a valve or membrane is suddenly opened, a shock is created and propagates downstream. With the exception of close proximity to the valve, the shock moves in a constant velocity (5.12(a)). Using a coordinates system which moves with the shock results in a stationary shock and the flow is moving to the left see Figure (5.12(b)). The ``upstream'' will be on the right (see Figure (5.12(b))).

**Figure:** A shock moves into a still medium as a result of a sudden and complete opening of a valve
stationary coordinates moving coordinates

Similar definitions of the right side and the left side of the shock Mach numbers can be utilized. It has to be noted that the ``upstream'' and ``downstream'' are the reverse from the previous case. The ``upstream'' Mach number is

$\displaystyle {M_x} = {U_s \over c_x} = M_{sx }$

(5.59)

The ``downstream'' Mach number is

$\displaystyle {M_y} = {U_s - {U_y}^{'} \over c_y} = M_{sy} - {M_y}^{'}$

(5.60)

Note that in this case the stagnation temperature in stationary coordinates changes (as in the previous case) whereas the thermal energy (due to pressure difference) is converted into velocity. The stagnation temperature (of moving coordinates) is

$\displaystyle {T_0}_y - {T_0}_x = T_y \left( 1 + {k -1 \over 2} \left(M_{sy} - ... ...ht)^2 \right) - T_x \left( 1 + {k -1 \over 2} \left( M_{x} \right)^2 \right) =0$

(5.61)

A similar rearrangement to the previous case results in

$\displaystyle {{T_0}_y}^{'} - {{T_0}_x}^{'} = T_y \left( 1 + {k -1 \over 2} \left(-2M_{sy}M_y + {M_{y}}^2 \right)^2 \right)$

(5.62)

**Figure 5.13:** The number of iterations to achieve convergence.
${M_y}^{'} = 0.3$ ${M_y}^{'}=1.3$

The same question that was prominent in the previous case appears now, what will be the shock velocity for a given upstream Mach number? Again, the relationship between the two sides is

$\displaystyle M_{sy} = {M_y}^{'} + \sqrt{ \left(M_{sx}\right)^{2} + {2 \over k -1} \over {2k \over k -1} \left( M_{sx}\right)^{2} - 1 }$

(5.63)

Since M_sx can be represented by M_sy theoretically equation (5.63) can be solved. It is common practice to solve this equation by numerical methods. One such methods is ``successive substitutions.'' This method is applied by the following algorithm:

Assume that M_x = 1.0.
\label{shock:item:openValve} Calculate the Mach number M_y by utilizing the tables or Potto--GDC.
Utilizing M_x = sqrt(T_y / T_x) ( M_y + M_y′ ) calculate the new ``improved'' M_x.
Check the new and improved M_x against the old one. If it is satisfactory, stop or return to stage \eqref{shock:item:openValve}.

To illustrate the convergence of the procedure, consider the case of M_y′ =0.3 and M_y′ =1.3. The results show that the convergence occurs very rapidly (see Figure (5.13)). The larger the value of M_y′, the larger number of the iterations required to achieve the same accuracy. Yet, for most practical purposes, sufficient results can be achieved after 3-4 iterations.

Subsections

Piston Velocity

Next: Piston Velocity Up: Moving Shock into Stationary Previous: Moving Shock into Stationary Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21