Quote:
Originally Posted by JayUtah
(Post 12668812)
So in a rocket thrust chamber, is there a difference in ambient pressure between the top (injector end) and the bottom (the nozzle throat)?
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Or more accurately, let's imagine a sealed cylindrical container in a vacuum, in the absence of gravity. In this container has been placed a small amount of propellant that will magically spontaneously ignite after a time. The container is otherwise a vacuum. The propellant ignites, creating a mixture of gas, liquids, and solids. We have made the cylinder from a material robust enough to completely contain the effects of this combustion without leakage.
If the gas has any static pressure whatsoever, the difference in pressure between the interior of the cylinder and the ambient vacuum exerts mechanical force on the walls of the cylinder. The cylinder responds by incurring stress and deforming, even if that's only by an unnoticed amount. The tensile properties of any solid material confirm that in the absence of fracture, stress is transmitted from one part of the cylinder to another.
The question is
in what direction these forces act? Consider a slice of the cylindrical wall. At each point, the gas pressure acts in the direction normal to the wall as defined at that point. Mechanically this has two important effects. First, all the points on the perimeter have a vector sum of zero. An equal pressure operating along each of those vectors has a vector sum of zero. The net force reckoned this way is zero, but it is a sum of actual non-zero vectors acting in the plane through which we took our slice. Second, the material of the cylinder is pushed outward uniformly along the perimeter, resulting in stress through the wall material. If the force were to exceed the tensile strength of the material, it would fracture along a line whose direction we could predict to be at right angles to the vectors of stress.
If we took our slice adjacent to the circular endcaps, part of the stress would be transmitted to the cap and produce a stress in it acting collectively in a way that wants the cap to have a greater diameter. (This is not the only stress the cap must withstand; read on.) We could predict the fracture line there too, knowing what forces are present and how the material responds.
We mention stresses and possible fracture to illustrate that the cylinder walls and the contained gas are mechanically coupled. The tensile properties of the cylinder and the gas properties of the gas reach a mechanical equilibrium that is a predictable result of known mechanical interaction. Gas exerts pressure either statically or as the result of fluid flow, and maintains pressure when mechanically constrained. Solids resist deformation when stressed.
Now consider the endcaps. In what direction does the gas exert pressure? As before, in the normal direction at each point where the gas applies. Except here the endcaps are planar. One endcap receives a force in the direction away from the center of the cylinder, of a magnitude equal to the unit gas pressure multiplied by the surface area of the cap. The cap responds to the pressure again by deformation, which may be miniscule and unnoticeable. But more importantly, it also responds by wanting to separate from the cylinder wall at the edge between the cap and the wall. The tensile effect there is that the right-angle joint between the wall material and the cap material is being stressed in a direction parallel to the cylinder axis.
The opposite endcap receives the same attention, but in the opposite direction. Thus the net force resulting from gas pressure is again zero, but for geometrically different reasons than apply to the cylinder wall. There you had a uniform pressure acting against the interior of a circle equally in all places. Here you have two parallel planes that are mechanically coupled by being attached to opposite ends of the same tube, and against each of which a force is acting in opposite directions with equal force. The vector sum here is zero: F and -F.
Also we have a new stress mode. If you envision the cylinder laying sideways, the left endcap is trying very hard to go to the left, pressed in that direction by the force of gas inside. The right endcap is trying equally hard to go to the right. The strain that results from this tug-of-war manifests itself as a tendency for the tube to elongate.
All this occurs because the system is responding to the effect of pressure versus and ambient background. If the ambient pressure were the same as the pressure of the exhaust from the spent propellant, there would be no net forces across the cylinder walls, no stress, and no danger of yield or rupture from gas pressure. The degree of mechanical responses and effects increase in proportion to the difference in pressure across the chamber walls.
Now imagine I have a button which, when pressed, causes the left endcap simply to cease to exist. I press it, and let's freeze time at that very instant. What can we say about the system now?
Ignore that the gas will later begin to expand into the newly-exposed vacuum to the left. We've frozen the system at Time 0. The left endcap is no longer there, and therefore no longer receiving any mechanical force from the pressure of the gas against it. More importantly, it's no longer transmitting that force to the cylinder wall that it is no longer connected to.
But the right endcap still is. The gas is still all in place, and the gas pressure is still acting against the right endcap. And that endcap is still transmitting tension -- i.e., pulling on -- the right edge of the tube. There is no balancing force on the left edge, so the tube wall now reacts differently. It doesn't want to elongate anymore, in response to opposing equal forces. The only force now acting parallel to the cylinder axis is the force applied by the right endcap.
We implied at the beginning that the cylinder was freely positioned in its environment, i.e. not connected to anything. If a force is applied to a stationary object preferentially in one direction, what do the laws of motion say will happen? Acceleration, of course. We would divide the force of the gas pressure against the right end cap by the mass of the cylinder, and from this we could accurately predict the momentary acceleration at that precise instant. (Momentary acceleration and other such values were an important innovation from Sir Isaac, and the reason he had to invent calculus to make everything work.)
The situation with the tube wall hasn't changed. The "contained" gas is still acting uniformly against the cylinder walls with zero net force. Our magic button changed only the geometry of the end caps.
Now advance the clock one microsecond. Some of the gas has escaped, but not all of it. Ignoring viscous flow and temperature effects for now, let's say that half the original gas mass is still present in the cylinder, and half of it escaped into the vacuum. It's gone, irrelevant. The half of the gas still remaining in what remains of the cylinder is still exerting pressure against the tube walls and against the remaining endcap -- but obviously only half as much as at Time Zero. We stipulate that to accommodate the escaped gas, the remnant has repositioned itself to be uniformly distributed.
We now have a finite, non-zero amount of time over which a force has acted continuously in the unbalanced way we described above at Time Zero. At Time 0 the instantaneous acceleration was endcap-pressure divided by cylinder mass. At T+1μs, instantaneous acceleration is half the original endcap pressure divided by cylinder mass. Pressure has been non-zero during that time, and therefore so has acceleration. If we have non-zero acceleration over non-zero time on a free body, at this point there
must have occurred a change in velocity.
Notice that we haven't considered in the least what happened to the gas that left, or speculated in any way what it might have "pushed against." That's because that's not the pressure gradient we care about. The pressure gradient we care about is that across the endcap that remains. That's what's producing motive force. If I have a formulation that results in motion and which doesn't incorporate any "push-off" effect, I have refuted the notion that such an effect is needed.
At a certain point in this formulation, of course, all the gas will have left the cylinder at one rate or another, and no more unbalanced force will be applied against the inside of the right endcap. But given the specifics of the gas properties and the cylinder, we can accurately compute how much acceleration was imparted to the cylinder over the time in which a portion of the gas was still present and still exerting static pressure on the various parts of the cylinder.
Now for more magic, pretend that we have an apparatus that will cause more propellants magically to appear and combust inside the cylinder, just enough to replace in one microsecond the amount that escapes out the left in that same microsecond. If the lead time (to allow for the time combustion takes to occur) is carefully calibrated, I should be able to maintain a constant static pressure inside the cylinder, and therefore a constant pressure on the left endcap, and therefore a constant force to apply to acceleration.
And that would be a rocket.