# UQx Hypers301x 3.5.5.2 Oblique Shockwave Example Problem Part 2

So this is where we finished off at the end of our last video. We’d formulated the problem and we’d got our flow conditions. The first step we’re going to take is we are going to look at what happens to the flow as it passes across the first shock wave. We’ll calculate the conditions at state 2 and then, for the rest of the problem, we’ll calculate across the second shock wave to get to state 3, then across third shock wave to get to state 4. We’ll start off now by looking at the first shock wave, S1. Let me just slide the screen down. So we’re interested in shock wave, S1. Let’s just do a sketch of its geometry. There’s the shock wave and its at angle, β1, to the oncoming flow. There’s the oncoming Mach number, M1, and the Mach number downstream of the shock wave, M2. From the geometry, the angle here is β1, this angle in here is θ1, so this angle in here is β1 – θ1. And we’ll put in our normal components of Mach number … so M_n1 and M_n2. So our oncoming Mach number is M1 is equal to 6.8 and the angle through which the flow is deflected, θ1, is 8 degrees. The first thing we need to do is work out what this shock deflection angle, β1, is. So we need to find β1 and to do that, recall this expression here,
which was a relationship between the flow deflection angle, θ, the shock wave angle, β, and the normal component of Mach number, M_n1, with the ratio of specific heats in it. The problem here is that we don’t have an explicit expression for finding β1. So we are going to have to find a method to determine (what that flow) what that shock wave angle, β, is. So, in order to get the shock wave angle from the flow deflection angle and the Mach number, we can use one of these θ-β-Mach number plots. So this was the plot that we had (in our) when we were looking at the results from the Drummond Tunnel. So we’ve got flow deflection angle here going from 0 to 50 degrees and the shock wave angle going from 0 to 90 degrees. And the curve were interested in here is the curve for Mach 6.8. And we’re interested here in a flow deflection angle of 8 degrees. So we’ve got to work out what’s the shock wave angle for that 8 degree deflection angle. Let’s zoom in a little bit on that plot. So here I’ve plotted it from 0 to just over 14 degrees flow deflection angle. The shock wave angle is now going from 0 to 30 degrees and we’re going to be interested in what is the shock wave angle for a deflection angle of 8 degrees. So this is the one here. So when we take that across, we see that it’s just under 15 degrees. So, let’s say, around about 14.7 degrees is the shock wave angle for a Mach 6.8 flow with a deflection angle of 8 degrees. So, from our θ-β-Mach number plot, we get that the shock wave angle, β1, is approximately 14.7 degrees. I’ll just slide this screen down a little bit. And now we have enough information to start doing our calculations of how things change across the shock wave. So the first thing (we have is or) we can calculate is what the Mach number component normal to the shock wave is upstream of the shock, so M_n1. And that’s going to be M1 times sin(β1). So we can put in those numbers … our Mach number, M1 is 6.8 and we now have our β1, which is 14.7 degrees. And that gives us a component of M1 normal to the shock wave of 1.73. And let’s now try and calculate what the static pressure is downstream of the shock wave. When we did our oblique shock wave analysis, we obtained this expression here. We said we could calculate how the static pressure changed across the oblique shock wave if we knew the component of the Mach number normal to the shock wave, upstream of the shock, and we knew the ratio of specific heats. So we can put in our numbers here … So p2/p1 is going to be 2 times γ, which is 1.4, times the component of Mach number upstream of the shock wave, 1.73^2 – 1.4 + 1 over 2.4. And when we do that calculation, we get the pressure increases by a factor of 3.31 across that first shock wave. Therefore our static pressure downstream of that shock, p2, is going to be 3.31 times the static pressure upstream of the shock, which was 1000 Pa. And that comes out to be 3.31 kPa. Next, we’ll look what the static density is downstream of the shock wave. And we had this expression for the static density change across an oblique shock wave, if we knew the component of the Mach number normal to the shock, upstream of the shock. So therefore our ρ2/ρ1 is going to be equal to γ+1, which is 2.4, times the component of Mach number upstream of the shock wave … the component is 1.73 … squared, over 2 plus (γ-1), 0.4, times 1.73^2. That comes out to be 2.24. So therefore, our ρ2 is going to be 2.24 times our density upstream of the shock wave, which if you look further up, is 0.01584. And that gives us 0.0355 kg/m^3. I’ll just slide the screen down again. Next we’ll calculate the static temperature downstream of the shock wave. Recall that this was the expression we had for the static temperature change across a shock wave. This one looks pretty horrible but we can do it a little more easily than that if we note that we can relate the static temperature, T2, to the static pressure, p2, the gas constant, R, and the (get it right!) the static density, ρ2. So therefore our T2 is our p2, which is our 3.31 x 10^3 Pa, over R, which is our 287, ρ2, which was 0.0355 kg/m^3, which comes out to be 325 K. So we’ve got our pressure, density and temperature downstream of the shock. The next thing we want to calculate is what’s the Mach number downstream of the shock wave and we have this expression to calculate what the component of the Mach number downstream of the shock wave, normal to the shock wave, was. So we can calculate M_n2 as being the square root of 1 plus (γ-1)/2, which is 0.2, times our 1.73^2, over γ,1.4, times our 1.73^2 again minus (γ-1)/2, which is 0.2, and all of that to the 1/2. And that comes out to be 0.634. So we have the component of Mach number, normal to the shock wave, downstream of the shock, but we want what the Mach number is. But we know that M_n2 is going to be M2 sin (β1 – θ1), which gives us that M2 is equal to our M_n2, 0.634, over sin (β1 – θ1) … so that’s going to be sin(β1), which was 14.7 degrees, minus θ1, which was 8 degrees. And we do that calculation and we find that M2 is 5.43. So I’ll just slide the screen down and summarize what we have so far. So this is it. We’ve got the Mach number, static pressure, static density and static temperature downstream of the first shock wave. And we’re ready now to go on to look at the second shock wave and what happens across that. But before we do that, what I want to do is I want to redo those calculations using a compressible flow calculator.
So we’ll do that in the next video.

## 2 thoughts on “UQx Hypers301x 3.5.5.2 Oblique Shockwave Example Problem Part 2”

1. ThatRosco458 says:

2. Gary Pal says: