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Next: Mass Flow Rate (Number) Up: Isentropic Converging-Diverging Flow in Previous: Relationship Between the Mach   Index

## Isentropic Flow Examples

Solution

The stagnation conditions at the reservoir will be maintained through out tube because the process is isentropic. Hence the stagnation temperature can be written and and both of them are known (the condition at the reservoir). For the point where the static pressure is known, the Mach number can be calculated utilizing the pressure ratio. With known Mach number, the temperature, and velocity can be calculated. Finally, the cross section can be calculated with all these information.

In the point where the static pressure is known

From Table (4.2) or from Figure (4.3) or by utilizing the enclosed program, Potto-GDC, or simply by using the equations that
Isentropic Flow Input: Pbar k = 1.4
M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*
0.886393 0.864201 0.694283 1.01155 0.6 0.606928 0.531054

With these values the static temperature and the density can be calculated.

The velocity at that point is

The tube area can be obtained from the mass conservation as

For a circular tube the diameter is about 1[cm].

Solution

With known Mach number at point A all the ratios of the static properties to total (stagnation) properties can be calculated. Therefore, the stagnation pressure at point A is known and stagnation temperature can be calculated.

At (supersonic flow) the ratios are

Isentropic Flow Input: M k = 1.4
M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*
2 0.555556 0.230048 1.6875 0.127805 0.21567 0.593093

With this information the pressure at Point B expressed

The corresponding Mach number for this pressure ratio is 1.8137788 and . The stagnation temperature can be bypassed'' to calculated the temperature at point

Solution

To obtain the Mach number at point B by finding the ratio of the area to the critical area. This relationship can be obtained by

With the value of from the Table (4.2) or from Potto-GDC two solutions can be obtained. The two possible solutions: the first supersonic M = 1.6265306 and second subsonic M = 0.53884934. Both solution are possible and acceptable. The supersonic branch solution is possible only if there where a transition at throat where M=1.
Isentropic Flow Input: A/A* k = 1.4
M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*
0.538865 0.945112 0.868378 1.27211 0.820715 1.04404 0.611863
1.62655 0.653965 0.345848 1.27211 0.226172 0.287717 0.563918

Next: Mass Flow Rate (Number) Up: Isentropic Converging-Diverging Flow in Previous: Relationship Between the Mach   Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21