Supersonic wind tunnel with total pressure loss? I am reading "I am reading "Fundamentals of Aerodynamics" 5th edition, J.D.Anderson. If you have the book, go to chapter 10: Compressible Flow
through Nozzles, Diffusers, and Wind Tunnels".
In order to produce Mach 2.5 uniform flow in a laboratory, we can build a wind tunnel like this:

The back pressure equals atmospheric pressure, the Mach 2.5 flow passes into the surroundings as a free jet. But we need high-pressure air supply at 17.09 atm which can be expensive, so we build this wind tunnel instead:

A normal shock-wave stand right at the end of the wind tunnel. The reservoir with a pressure of only 2.4 atm which reducing the cost compared with the first wind tunnel. 
Then the author said: A normal shock is the strongest possible shock, hence creating the largest total pressure loss. If we could replace the normal shock with a weaker shock, the total pressure loss would be less, and the required
reservoir pressure $p_0$ would be less than 2.4 atm. So again, we build this instead:

with reflected oblique shocks behind the test section.
The objective is to have $p_o$ as low as possible and the exit pressure must be equal to atmosperic pressure (= 1 atm). But why is the $p_o$ in case 3 is smaller than $p_o$ in normal shock case ? You might think due to total pressure loss is smaller, but the final total pressures are not the same in two case, so how could we compare the $p_o$.
 A: If we want the outflow pressure to be 1 atm, then we know the final pressure.  We know for an oblique shock that the change in pressure across the shock is given by:
$$
\frac{P_{2}}{P_{1}} = 1 + \frac{ 2 \ \gamma }{ \gamma + 1 } \left( M_{1}^{2} \ \sin^{2} \beta - 1 \right) \tag{1}
$$
where the subscripts $1$ and $2$ correspond to the upstream and downstream regions, respectively, $\beta$ is the angle of the shock plane from the incident flow direction, $\gamma$ is the ratio of specific heats, $M_{j}$ is the Mach number in the $j$th region, and $P_{j}$ is the average pressure in the $j$th region.  In the limit as $\beta \rightarrow 90^{\circ}$, i.e., a normal shock, we approach the following:
$$
\frac{P_{2}}{P_{1}} = 1 + \frac{ 2 \ \gamma }{ \gamma + 1 } \left( M_{1}^{2} - 1 \right) \tag{2}
$$
which is the standard pressure ratio change.

But why is the po in case 3 is smaller than po in normal shock case?

If you look at Equation 1 above, you will see that as $\beta \rightarrow 0$, the pressure ratio of the downstream to upstream drops considerably.  That is, the maximum of Equation 1 is given by Equation 2, i.e., the case where the shock plane is orthogonal (or normal) to the incident flow.
This is why the 2nd Po can be smaller than in the first case.  The final/third example relates to Equation 1.  You can generate the same Me with a smaller pressure change by forcing the shocks to be oblique or you can reduce Me to reduce the pressure change.
The 2nd nozzel/throat is a converging channel, which will kill the reflecting shocks.  This is because shock waves are unstable in a converging channel, as I discussed at https://physics.stackexchange.com/a/137842/59023.  This will further reduce the necessary pressure change, i.e., the combination of oblique shocks and a converging channel.
