Diffuser Thrust Force Illustration

[larry cottrill]

Here's a simple illustration of how a diffuser with air flowing rearward through it can generate a force in the thrust direction. This is an oversimplification, because we won't look at how the outside air flowing around the engine might be affecting things, but it will be very close to reality. Also, keep in mind that this won't give us the net engine thrust, because the anti-thrust internal effects of the exhaust nozzle aren't covered. Referring to the drawing below:

Assume a basic ramjet diffuser layout as shown at the top of the drawing. The half-angle of the conic sheet was arbitrarily chosen as 6 degrees. Air enters at the left and then slows as it makes the transition into larger and larger cross-sectional areas. As it slows, it assumes higher static pressure. The gauges are just a hokey way to illustrate the pressure change - PSIg simply means we are indicating only pressure greater than the surrounding air; this greatly simplifies the illustration. If there is a slight 'venturi' type entrance (fairly typical), there will be an initial speed gain and pressure drop at the narrow section, but the air will reach a point where the open area is equal to the entrance area, and the static pressure will be restored (0 PSIg), as shown. We'll assume that we can get a pressure rise of at least 1.5 PSI over atmospheric, a not unreasonable value. This is shown by the gauge at the right end, near the widest part of the engine. Detail A is simply a blow-up of a section of the engine shell in this area.

Air does not produce low pressure just because it is moving rapidly along a surface. Thrust force (by this I mean any forward-directed force applied to the structure) on the diffuser results from the fact that, even though the air is moving along the diffuser wall, its static pressure still acts upon that wall. All of the static pressure gain will be applied to the diffuser as force acting in the normal direction ("normal" meaning, perpendicular to any point on the surface). However, because of the conical contour of the wall, this pressure is not simply contained by the mechanical resistance of the opposing walls of the structure, and the effect can be illustrated as follows:

Referring to Detail A, imagine a "slice" or "band" of the diffuser 1 inch wide, taken at the point where we are able to measure 1.5 PSI over atmpospheric. Now imagine 1 square inch at some point on that band. The blue vector shows the force f1 created by this air pressure on that square inch, i.e. 1.5 pounds outward force, acting normal to the surface. The resistance to this force can be resolved into two vectors, one transverse to the diffuser centerline (red) and one parallel to the centerline (olive green). The red vector represent the tensile and shear forces (summed as f2) in the band preventing our 1-square-inch section from blowing outward. The value of this vector is the cosine of 6 degrees x f1, or 1.49 lb. The longitudinal vector represents force f3, the thrust force attempting to propel our square inch forward. It is the sine of 6 degrees x f1 = 0.157 lb.

Now, 0.157 lb isn't a lot of thrust, but that's just one inch width of a single 1-inch slice of the whole diffuser. If this was a little engine like Steve's TLAR ramjet, there might be 10 such square inches around the perimeter at this pressure level - 1.57 pounds from only a single inch segment of the diffuser length! If this were from a big engine, say a 1-inch slice that was 31 inches ID, we'd have 100 square inches and our 1-inch slice would be netting us 15.7 pounds! It doesn't take too many slices of this to come up with a pretty decent thrust force, though of course we have to understand that each square inch is worth less and less as we move toward the intake, and we also have fewer square inches in each slice!

An interesting question is whether this force is the same as we would get by "projecting" the "rear view" of the inside surface of the conical band rearward as a transverse "plate" (really sort of a skinny washer) that the static pressure would work against (since static pressure will create a force in any orientation of material we want to use). Our band is 1 inch wide, so the increase in radius from front to rear will be the tangent of 6 degrees x 1 inch = 0.105 inch. On the big engine example I mentioned earlier, we had 100 inches of circumference, hence A = 100 x .105 = 10.5 square inches. To get the "flat plate thrust force" for this skinny washer, we multiply area times pressure: 10.5 sq.in x 1.5 PSIg = 15.7 lb! So, it is, in fact equivalent (mathematically) to the analysis we just saw. To sum the whole diffuser, we would simply calculate the area AND the static pressure applicable to each "washer" and sum them.

So, a ramjet diffuser really IS a front plate, though it is not one that takes a uniform pressure! And again, remember that we aren't calculating the whole engine net thrust here - to get that, we would have to adjust for the effects of the exhaust nozzle, i.e. the forces created by changing pressures as the heated gases exit.

L Cottrill