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 $T_{0} = constant$ and $P_{0} = constant$ 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

$$\bar{P} = {P \over P_0} = {3 [MPa] \over 5 [MPa] } = 0.6$$

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.

$$T = 0.86420338 \times (273+ 21) = 254.076 K$$

$$\rho = {\rho \over \rho_0} \overbrace{P_{0}\over R T_0}^{\rho_0}= 0.69... ...es {5 \times 10^6 [Pa] \over 287.0 \left[{J \over kg K}\right] \times 294 [K] }$$

$$= 41.1416 \left[{kg \over m^3 }\right]$$

The velocity at that point is

$$U = M \overbrace{\sqrt{kRT}}^{c} = 0.88638317 \times \sqrt { 1.4 \times 287 \times 294} = 304 [m /sec]$$

The tube area can be obtained from the mass conservation as

$$A = {\dot {m} \over \rho U} = 8.26 \times 10^{-5} [m^{3}]$$

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

^{4.3 4.4}

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 $M=2$ (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

$${P_{A} \over P_{0}} = \overbrace{P_... ...2}} \times {P_{A} \over P_{B}} = 0.12780453 \times {2.0 \over 1.5} = 0.17040604$$

The corresponding Mach number for this pressure ratio is 1.8137788 and $T_{B} = 0.60315132$ ${P_{B} \over P_{0} }= 0.17040879$ . The stagnation temperature can be ``bypassed'' to calculated the temperature at point $\mathbf{B}$

$$T_{B} = T_{A}\times \overbrace{T_{0} \over T_{A} }^{M=2} \times \... .....} = 250 [K] \times {1 \over 0.55555556} \times {0.60315132} \simeq 271.42 [K]$$

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

$${A_{B} \over A{*} } = {A _{B} \over A_{A} } \times {A_{A} \over A... ...4 }^{\hbox{from the Table \eqref{variableArea:tab:basicIsentropic}}} = 1.272112$$

With the value of ${A_{B} \over A{*} }$ 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