Shock Tube

The shock tube is a study tool with very little practical purposes. It is used in many cases to understand certain phenomena. Other situations can be examined and extended from these phenomena. A cylinder with two chambers connected by a diaphragm. On one side the pressure is high, while the pressure on the other side is low. When the diaphragm is ruptured the gas from the high pressure section flows into the low pressure section. When the pressure is high enough, a shock is created that it travels to the low pressure chamber. This is the same case as in the suddenly opened valve case described previously. At the back of the shock, expansion waves occur with a reduction of pressure. The temperature is known to reach several thousands degrees in a very brief period of time. The high pressure chamber is referred to in the literature is the driver section and the low section is referred to as the expansion section.

Initially, the gas from the driver section is coalescing from small shock waves into a large shock wave. In this analysis, it is assumed that this time is essentially zero. Zone 1 is an undisturbed gas and zone 2 is an area where the shock already passed. The assumption is that the shock is very sharp with zero width. On the other side, the expansion waves are moving into the high pressure chamber i.e. the driver section. The shock is moving at a supersonic speed (it depends on the definition, i.e., what reference temperature is being used) and the medium behind the shock is also moving but at a velocity, $U_2$ , which can be supersonic or subsonic in stationary coordinates. The velocities in the expansion chamber vary between three zones. In zone 3 is the original material that was in the high pressure chamber but is now the same pressure as zone 2. Zone 4 is where the gradual transition occurs between original high pressure to low pressure. The boundaries of zone 4 are defined by initial conditions. The expansion front is moving at the local speed of sound in the high pressure section. The expansion back front is moving at the local speed of sound velocity but the actual gas is moving in the opposite direction in $U_2$ . In fact, material in the expansion chamber and the front are moving to the left while the actual flow of the gas is moving to the right (refer to Figure (5.20)). In zone 5, the velocity is zero and the pressure is in its original value.

Figure 5.20: The shock tube schematic with a pressure "diagram."

The properties in the different zones have different relationships. The relationship between zone 1 and zone 2 is that of a moving shock into still medium (again, this is a case of sudden opened valve). The material in zone 2 and 3 is moving at the same velocity (speed) but the temperature and the entropy are different, while the pressure in the two zones are the same. The pressure, the temperature and their properties in zone 4 aren't constant and continuous between the conditions in zone 3 to the conditions in zone 5. The expansion front wave velocity is larger than the velocity at the back front expansion wave velocity. Zone 4 is expanding during the initial stage (until the expansion reaches the wall).

The shock tube is a relatively small length $1-2[m]$ and the typical velocity is in the range of the speed of sound, $c\sim \sqrt{340}$ thus the whole process takes only a few milliseconds or less. Thus, these kinds of experiments require fast recording devices (a relatively fast camera and fast data acquisition devices.). A typical design problem of a shock tube is finding the pressure to achieve the desired temperature or Mach number. The relationship between the different properties was discussed earlier and because it is a common problem, a review of the material is provided thus far.

The following equations were developed earlier and are repeated here for clarification. The pressure ratio between the two sides of the shock is

$${P_2 \over P_1 } = {k -1 \over k + 1} \left( {2k \over k -1 } {M_{s1}}^2 - 1 \right)$$ (5.82)

Rearranging equation (5.82) becomes

$${M_{s1}} = \sqrt{{k -1 \over 2k} + { k + 1 \over 2k } {P_2 \over P_1} }$$ (5.83)

Or expressing the velocity as

$$U_s = M_{s1} c_1 = c_1 \sqrt{ {k -1 \over 2k} + {k +1 \over 2k} {P_2 \over P_1} }$$ (5.84)

And the velocity ratio between the two sides of the shock is

$${U_1 \over U_2 } = {\rho_2 \over \rho_2} = { 1 + {k+1 \over k-1} {P_2 \over P_1} \over {k+1 \over k-1 } {P_2 \over P_1} }$$ (5.85)

The fluid velocity in zone 2 is the same

$${U_2}{'} = U_s - {U_2} = U_s \left( 1 - { U_2 \over U_s }\right)$$ (5.86)

From the mass conservation, it follows that

$${U_2 \over U_s} = {\rho_1 \over \rho_2}$$ (5.87)

$${U_2}^{'} = c_1 \sqrt{ {k-1 \over 2k} + {k+1 \over 2k} {P_2 \over... ...\over k -1 } + {P_2 \over P_1} \over 1 + {k +1 \over k -1 } {P_2 \over P_1} } }$$ (5.88)

After rearranging equation (5.88) the result is

$${U_2}^{'} = {c_1 \over k} \left( {P_2 \over P_1} - 1 \right) \sqrt{ {2k \over k+1 } \over {P_2 \over P_1} {k-1 \over 1+k} }$$ (5.89)

On the isentropic side, in zone 4, taking the derivative of the continuity equation, $d(\rho U) = 0$ , and dividing by the continuity equation the following is obtained:

$${d\rho \over \rho} = - {dU \over c}$$ (5.90)

Since the process in zone 4 is isentropic, applying the isentropic relationship ( $T \propto \rho^{k -1}$ ) yields

$${c \over c_5} = \sqrt{T \over T_5} = \left( \rho \over \rho_5\right)^{k-1 \over 2}$$ (5.91)

From equation (5.90) it follows that

$$dU = -c {d\rho \over \rho} = c_5 \left( \rho \over \rho_5\right)^ {k-1 \over 2} d\rho$$ (5.92)

Equation (5.92) can be integrated as follows:

$$\int_{U_5=0}^{U_3} dU = \int^{\rho_3}_{\rho_5} c_5 \left( \rho \over \rho_5\right)^ {k-1 \over 2} d\rho$$ (5.93)

The results of the integration are

$$U_3 = {2 c_5 \over k -1 } \left( 1 - \left(\rho_3 \over \rho_5 \right)^ {k-1 \over 2} \right)$$ (5.94)

Or in terms of the pressure ratio as

$$U_3 = {2 c_5 \over k -1 } \left( 1 - \left(P_3 \over P_5 \right)^ {k-1 \over 2k} \right)$$ (5.95)

As it was mentioned earlier the velocity at points $2^{'}$ and 3 are identical, hence equation (5.95) and equation (5.89) can be combined to yield

$${2 c_5 \over k -1 } \left( 1 - \left(P_3 \over P_5 \right)^ {k-1 ... ...P_1} - 1 \right) \sqrt{ {2k \over k+1 } \over {P_2 \over P_1} {k-1 \over 1+k} }$$ (5.96)

After some rearrangement, equation (5.96) is transformed into

$${P_5 \over P_1} = {P_2 \over P_1} \left( 1 - {(k -1 ) {c_1 \over ... ... ( k +1 ) \left( {P_2 \over P_1 } - 1\right)} } \right) ^ { - {2k \over k -1} }$$ (5.97)

Or in terms of the Mach number, $M_{s1}$

$${P_5 \over P_1} = { k_1 - 1 \over k+1 +1 } \left( {2 k \over k_1 ... ...er c_5} \left( {M_{s1}}^2 -1\right) \over M_{s1}}\right]^ { - {2k \over k -1} }$$ (5.98)

Using the Rankine-Hugoniot relationship and the perfect gas model, the following is obtained:

$${T_2 \over T_1} = { 1 + {k_1 - 1 \over k_1 + 1} { P_2 \over P_1 }\over 1 + {k_1 - 1 \over k_1 + 1} { P_1 \over P_2 } }$$ (5.99)

By utilizing the isentropic relationship for zone 3 to 5 results in

$${T_3 \over T_5} = \left( P_3 \over P_5\right)^{k_5 -1 \over k_5} = \left( {P_2 \over P_1} \over {P_5 \over P_1} \right)^{k_5 -1 \over k_5}$$ (5.100)

Solution

Not finished yet.