It's okay--because I never played magic...though I do watch star trek frequently.
It will be interesting to see what the detailed numbers say, but I'm certain the difference in pressure across a realistic (are we talking about the realistic case now or the imaginary, crazy long case?) set of cheeses will be relatively small. Here's a back of the envelope analysis.
The energy involved shouldn't be enormous. The work, W, being done by the press on the cheeses is equal to the force, F, being applied times the distance, d, the "piston" travels through, or, in nerd: W =Fd (this is the equation for constant F across the entire length of d, which we will assume for simplicity). If we want 6 psi across a 6 in wheel, we'll need ~ 170 lbs ( 3in^2 * pi * 6 psi). Let's say our molds hold 9 inches of curd that will be compressed down to 3 inches (that seems like a lot to me--so our number should be higher than we find in reality). That means for every cheese we have in the line, we'll compress 6 inches. So, if we have 10 cheese, we expect to compress 5 feet (10 * 6 in). Using our formula for work, we will do W = 5 ft * 170 lbs = 850 ft lbs of work. I don't do well with ft lbs of work, so let's convert those to joules. There are 1.36 joules in a ft lb, so we have 1156 joules of work being done. While not all of this work will go into heat, let's pretend it does and see how much heat that is. This amount of energy would raise 1 liter of water 0.28 degrees C--so not very much energy.
Considering again the relative motion of the cheeses in the stack, the first one being pressed moving the fastest, and the last one in the stack not really moving at all, the difference they experience in pressure will be directly related to the difference in force across the cheese. This force will be easiest to determine if we could measure the acceleration of the cheeses. Again, the first cheese will experience the most significant difference in pressure since it moves the fastest (thus, accelerates the most). However, I don't have access to an industrial, horizontal cheese press (it's on the wish list
). Let's be a little crazy and pretend that the cheese accelerates at 1/10 the acceleration of gravity. Just to help think about this, if something accelerates at that rate for 1 second, it will be traveling at a velocity = acceleration * time = 1 m/s^2 (really 0.981 m/s^2...) * 1 second = 1 m/s ~ 2.2 miles/hour. People walk at about 3 miles/hour, so this is slightly slower than you walk, and, I presume, really stinking fast for a press to be pressing cheese. Now, let's assume the cheeses we are pressing are 2 kg each. The difference in force across this cheese, then, if F = ma = 2 kg * 1 m/s^2 = 2 Newtons. In fact, in the very fist instance, there is no reaction from the second cheese on the first cheese, so the 2 Newton "difference" is really all the force that the press is providing. However, as soon as the first cheese "engages" the second cheese, the second cheese provides a reaction, and the press must provide more than 2 Newtons. A similar phenomenon happens for each cheese in the stack--the second cheese has an instantaneous acceleration before being engaged by the third cheese, and so on. In this first instant, we find the most significant difference in pressures the cheeses experience. Since the last cheese is engaged last, it experiences 0 force, and thus 0 pressure while all the other cheeses are being pressed. Once all the cheeses are engaged, the force provided by the press continues to go up. If we allow the cheeses to continue to accelerate at the same constant rate of 1 m/s^2 (remember, an unrealistically fast acceleration), we will continue to have a 2 Newton difference across the first cheese. However the press will be providing a greater force than just those 2 Newtons because it has to "counter" all of the reactions from the entire stack of cheeses that have now been engaged. Let's suppose, that after 10 seconds, the press will have reached 1/2 of it's final pressure. If we're looking at a 6 inch cheese pressed at 6 psi, then we'll need about 750 Netwons of final pressing force, half of that, then puts us at 375 Newtons. At that point, our 2 Newton difference between the first cheese and the last cheese (which has almost no force differential across it) is just over 1/2% difference in force (and thus in pressure). When we get up to full pressure, the difference will be closer to 1/4%. However, what we've ignored is that when we get to full pressure, the acceleration of the first cheese will be essentially 0, thus the difference in force across the first cheese will be 0, and the difference in pressure between the first and the last cheese will be 0.
From this analysis (which is exaggerated to make the differences bigger) we find that: the first cheese in the press experiences more pressure than the last cheese, the difference lasts only so long as the cheeses are compressing, the difference is proportional to the rate at which the cheeses are accelerating, and, in an exaggerated scenario, this difference in force represents an extremely small difference in pressure.