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The release process when relatively light gases are injected at low Mach speed in the atmosphere is governed by the forces of inertia, buoyancy and viscosity. Hence the following laws apply:

Law for interial forces $$ F_{r,inertia} \simeq \rho L^2 V^2 \Rightarrow \Pi_i = { {\rho L^2 V^2} \over F } $$ Law for gravitational forces $$ F_{r,gravitation} \simeq \Delta \rho L^3 g \Rightarrow \Pi_g = { {\Delta \rho g L^3 } \over F} $$ Law for viscous forces $$ \tau \simeq \mu { V \over L } \Rightarrow F_{r,viscous} \simeq \mu L V \Rightarrow \Pi_v = { {\mu L V } \over F } $$

As we don't have any external characteristic force F, only ratios of the above ratios will be of interest. In these ratios the fictional force F drops out and we get two well known \Pi-Numbers from the above three laws: The Reynolds number Re $$ Re = { \Pi_i \over \Pi_v } = { {L V } \over {\mu / \rho} } = { {L V } \over \nu } $$ and the Richardson number Ri $$ Ri = { \Pi_i \over \Pi_g } = { {\Delta \rho g L } \over \rho V^2 } $$ It is easy to see that the Richardson number is closely related to the Froude number Fr introduced in the Equation Approach description: $$ Ri = { {\Delta \rho} \over \rho } { 1 \over Fr^2 } = { 1 \over Fr^{\star} } $$ Where Fr^{\star} defines the densimetric Froude number.

In the model of the described experiments the flammable hydrogen is replaced by the inert gas helium. Helium is usually selected as its properties are close to hydrogen . However, for similarity the relevant dimensionless parameters, here the Re and Ri have to be identical, what can be expressed in the following scale factor requirements:

From Re^' = Re it follows $$ { { \hat{l} \hat{v} } \over \hat{\nu} } = 1 $$ and from Ri^' = Ri it follows $$ { { \hat{\Delta \rho} \hat{l} } \over {\hat{\rho} \hat{v}^2} } = 1 $$ where it has been assumed that also the experiments will be made on earth ( \hat{g} = 1 ) and use the same external gas, namely air ( \hat{\rho} = 1 ).

Because of the selected model gas the remaining two material scale factors are also fixed: $$ \hat{\nu} = {{\nu_{He}} \over {\nu_{H2}}} \approx 1.095 $$ $$ \hat{\Delta \rho} = {{\rho_{air}-\rho_{He}} \over {\rho_{air}-\rho_{H2}}} \approx 0.927 $$

The above requirements may be reformulated to yield the remaining velocity and size scale: $$ \hat{v} = \sqrt[3]{ {\hat{\nu} \hat{\Delta \rho}} \over {\hat{\rho}}} \approx 0.8 $$ $$ \hat{l} = \sqrt[3]{ {\hat{\rho} \hat{\nu}^2} \over \hat{\Delta \rho} } \approx 1.37 $$

So for similarity the velocities in the helium jet should be reduced by 20% and the geometry of the helium experiments should be enlarged by 37%, roughly a third.

With the same size in the model experiment and the prototype only velocities may be scaled only after further relaxation. Assuming that the buoyancy effects are stronger than the frictional effects the Ri -number provides the following orientation $$ \hat{v} = \sqrt{ \hat{\Delta \rho} \over \hat{\rho}} \approx 0.68 $$ Under these conditions the characteristic helium release velocities should be reduced by a third. Violating this will make a comparison of the release phases and all subsequent processes difficult.

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Page last modified on September 16, 2008, at 03:58 PM