Clearly, when \rm p_r = p_0 (see Curve A), no flow can occur. If \rm p_r is slightly lower than \rm p_0, subsonic flow occurs (see Curve B). Lowering \rm p_r further results in a pressure distribution \rm p(x)/p_0 (see Curve C) where M = 1 in the throat. Now, a second particular \rm p_r – value generates again subsonic flow with M = 1 in the throat, i.e., Curve D, because there are two solutions to Eq. (3.57) for \rm M_{exit} or for that matter to Eq. (3.62).

\rm\frac{p_0}{p} =\left\lgroup1+\frac{k-1}{2}M^2 \right\rgroup ^{\frac{k}{k-1} }                                (3.57)

\rm\frac{A^*}{A} =M\left[\frac{k+1}{2+(k-1)M^2} \right] ^{\frac{k+1}{2(1-k)} }                              (3.62)

Given \rm A_{exit} / A_{throat} ≡ A / A^*= 40 /10 =4 and using Eq. (3.62), two \rm M_e – values can be obtained via trial-and-error, i.e., \rm M_e ≈ 0.147  and \rm M_e ≈ 2.94 . Employing Eq. (3.57), the corresponding exit pressures are

\rm p_e\equiv p_r\hat =p_c=0.985p_0~\text{and}~p_D=0.0298p_0

or with \rm p_0 = 500 kPa

\rm p_c=492.5\,kPa~\text{and}~p_D=15\,kPa

Graph:

Comments:
• To generate Curve C, the pressure drop is only \rm p_0 − p_c =7.5 kPa

• The associated exit temperature conditions are \rm T_c =292 K and \rm T_D = 107 K

• That allows computation of the gas exit velocities to

\rm v_c=M_c\sqrt{kRT_c}=50\,m/s

and

\rm v_D=M_D\sqrt{kRT_D}=610\,m/s

• Any outlet pressure in the range of \rm p_c < p_r < p_D , generating pressure distributions \rm p(x)/p_0 between Curves C and D, produce shock waves inside or outside the diverging part of the nozzle and the flow is generally non-isentropic.