Question 15.12: Supersonic Diffuser with Normal Shock Air at 10 kPa approach...

The three principal states of the flow, shown in Figure 15.44, are redrawn in the h-s diagram of Figure 15.45, along with the corresponding total states. Evaluate the flow states, State-1 and State-2, before and after the normal shock. While there is a drop in total pressure from State-1 to State-2 ( p_{t 2}<p_{t 1} ), caused by the normal shock, the flow is isentropic thereafter, so that p_{t 2}=p_{3} . Do an isentropic analysis for the flow between State-2 and State-3 where the diverging section works as a subsonic diffuser.

Assumptions
One-dimensional, isentropic flow of a perfect gas after the shock.
Analysis
The total pressure at the inlet, State-1, can be found from the isentropic table:

h_{t 1}=h_{t 2} ; \quad p_{t 1}=p_{1}\left(\frac{p_{1}}{p_{t 1}}\right)^{-1}=p_{1}\left[\left(\frac{p}{p_{t}}\right)_{@ M_{1}=3}\right]^{-1}=10(0.02722)^{-1}=367.38 kPa

Using the shock table or the gas dynamics TESTcalc, State-2 can be obtained:

p_{t 2}=p_{t 1} \frac{p_{t 2}}{p_{t 1}}=p_{t 1}\left(\frac{p_{t e}}{p_{t i}}\right)_{@ M_{1}=3}(367.38)(0.3282)=120.57 kPa ; \quad M_{2}=0.475

The flow is isentropic downstream of State-2. Using the mass equation for an isentropic flow between State-2 and -3, A_{* 3}=A_{* 2} , we obtain:

\left(\frac{A}{A_{*}}\right)_{@ M_{3}}=\frac{A_{3}}{A_{* 3}}=\frac{A_{3}}{A_{* 2}}=\frac{A_{3}}{A_{2}} \frac{A_{2}}{A_{* 2}}=\frac{A_{3}}{A_{1}} \frac{A_{2}}{A_{* 2}}=(3)\left(\frac{A}{A_{*}}\right)_{@ M_{1}=0.475}=3(1.39)=4.17

Using the isentropic table we get:

M_{3}=0.14 ; \quad \text { and } \quad p_{3}=p_{t 3} \frac{p_{3}}{p_{t 3}}=p_{t 2}\left(\frac{p}{p_{t}}\right)_{@ M_{3}}=(120.57)(0.986)=118.9 kPa

Therefore,  F_{p} \equiv \frac{p_{t 3}}{p_{t 1}}=\frac{p_{t 2}}{p_{t 1}}=\frac{120.57}{367.38}=0.328

TEST Solution
Launch the gas dynamics TESTcalcs and select Air as the working fluid. Calculate State-1 partially from p1, M1, and A1=1 m^2 (assumed). Obtain M2 and p_t2/p_t1 from the shock table in the table panel. Calculate State-2 with M2, p_t2 = 0.328*p_t1, and A2 = A1. Finally, evaluate State-3 with A3 = 3*A2, p_t3 = p_t2 (entropy equation), and Astar3 = Astar2 (mass equation). See TEST-code (posted in TEST > TEST-codes) for details.

Discussion
With the pressure recovery factor evaluated, the isentropic efficiency can be calculated from Eq. (15.46). Obviously, the presence of the shock is the main reason behind the poor performance of this diffuser.

\eta_{\text {diffuser }}=\frac{\left[1+M_{i}^{2}(k-1) / 2\right]\left(F_{p}\right)^{(k-1) / k}-1}{M_{i}^{2}(k-1) / 2}                     (15.46)