Now that we have taken a look at a greatly simplified standard vented gas tank while in flight, it is time to look at the so called “uniflow tank” and illuminate it's operation.

Before I begin I need to take a moment to mention that I am again simplifying things greatly. This is because there are entire college courses devoted to subjects such as ours, they are called Physics, Dynamics, Fluid Dynamics, Classic Mechanics. So, I will only be looking at acceleration in one direction, out of the circle. Not acceleration due to gravity, or due to the plane speeding up as a result of increase engine power (such as a 2-4 break). I fully well realize that this is not the best way to address this topic, but it is beyond the scope of this series of articles to educate the reader in fields beyond our subject-Control Line modeling. Enough of this digression, and on to the Uniflow!

Looking back at my last installment, we found that at the beginning of a flight there was 1/2 inch of hydraulic head pressure (head pressure), when sitting on the ground motionless. As the plane begins it flight, and accelerates to a angular velocity of 5 seconds per lap, there is now (assuming 3Gs of centripetal acceleration, it is actually 3.34Gs) 1-1/2 inch of positive head pressure, and at the end 4-1/2 inches of negative head pressure. Note I have ignored discussing pressure in the tank, because it is the same from the beginning to the end of the run -- one atmosphere. This is simply not the case with a uniflow tank as we will now see.

Using the same tank a standard 1x2x4 inch stunt tank, while the pane is sitting on the ground and not moving, the pressure inside the tank is one atmosphere, minus 1/2 inch (depth of the fuel to the vent opening inside the tank) times the density of the fuel. Stated concisely yields:

.033lbs/in³ x 1/2 in = 0.0165lbs/in²

So at the beginning of the engine run we have inside the tank-one atmosphere minus 0.0165 lbs per square inch. Assuming an atmosphere is 14.7 lbs per square inch gives:

14.7lbs/in² - 0.0165lbs/in2 = 14.6835lbs/in²

Now I am going to digress for a moment to talk about atmospheric pressure, as I have yet to define what I mean by this rather ambiguous but important term. I stated rather boldly that one atmosphere was 14.7 lbs per square inch without so much as batting an eyelash. This was the number used in my past science education, and it was always understood that, that number was for mean average sea level, at 70 degrees Fahrenheit, and 20% relative humidity. Atmospheric pressure is the weight (force) of the column of air, above you all the way to the near perfect vacuum of space. So a 1 inch by 1 inch column of fairly dry air, all the way to space, while you are at sea level, weighs 14.7 pounds, on a 70 degree day. For our purposes it is not all that important that we know what the actual atmospheric pressure is, but for some CL flyers this is a very critical number. Most serious speed flyers will carry a portable weather station with them that gives temperature, pressure, and relative humidity, as well as such things a wind speed, due point, and altitude. Most of us are accustomed to hearing the weather man say the pressure is, “30 inches” and dropping. Since most weather broadcasts are limited in time they have dropped the proper units of pressure and should state: 30 inches of mercury (760mm in Europe), and dropping. This unit for our purposes is not that useful as it depends on a physical constant we do not have, the density of mercury. Suffice to say that the weight of the column of air, is sufficient to raise 30 inches of mercury into a tube. Oh, it was no accident that I chose 30 inches, (760mm or Torricelli if you prefer) as my number in this example, as this is equivalent to 14.7 pounds per square inch. If you fly at Delta Park in Portland, Ore., this number is a pretty good approximation of what your model will encounter, as the circle is at 3.5 feet above mean sea level. Yes this is correct. It is behind a dike that Interstate 5 runs atop, the Columbia River near Delta Park is not actually a river, but an estuary, and flows with the force of the tide, clear up to Bonneville dam. I can tell you for sure, the altitude at Eugene, Roseburg, or Salem are much higher, and have typically less atmospheric pressure. (EDITOR'S NOTE: See end of column for elevations of some Northwest flying sites.) I must apologize for getting off the topic, back to our tank.

So just to make fuel flow at the beginning of the run there must be a pressure differential of at least .0165 pounds per square inch, between the venturi and the tank. Further knowing the tank has one atmosphere minus .0165 pounds per square inch, then the venturi must have at least .0165 lbs/in² less to induce flow. Now let's look at the plane flying five second laps in the level upright configuration. There is now 3.34 times the acceleration due to gravity out of the circle, the fuel level is now half an inch above the spray bar, and there is one and a half inches to the bottom of the tank. In our last discussion it was found that the fuel will in a standard tank try and run downhill and the engine will go rich. However, this is not a standard vented tank, it has the interior vent at the bottom of the tank, so how does this affect the pressure inside the tank? In order for a bubble to reach the bottom of the tank, it must overcome 2 inches of fuel head, at 3.34 Gs, the calculation is as follows:

033lbs/in³ x 2 in x 3.34 = 0.22lbs/in²

So before a bubble of air can reach the bottom of the tank there must be at least 0.22lbs/in² of pressure differential between the atmosphere and the venturi. Note, however that we still get a boost from the half inch of hydraulic head at the beginning of the tank. Using the same numbers as from above and our scalar gives us a pressure of:

.033lbs/in3 x 1.5in x 3.34 = .165lbs/in2

Subtracting this pressure from the pressure in the tank, will give us the amount of pressure in the tank:

14.7lbs/in² - 0.165lbs/in² = 14.535lbs/in²

Now let's extend this same logic to the end of the run, with the uniflow exposed and the tank basically empty, only having the fuel in the pickup tube. In this case the depth to the uniflow is zero, so the tank has one atmosphere of pressure in it, fuel must now flow 1.5 inches inboard against 3.34Gs of acceleration, and there must be at least .165lbs/in² of pressure differential, before the fuel can reach the venturi. So at the beginning and at the end of the run, there needs to be at least .165 lbs/in² of pressure differential between the tank and the venturi. These full and empty tank conditions are identical! So the fuel will feed to the venturi at the same rate of flow when the tank is full as it will when it is empty-hence the name, UNIFLOW.

Now, I have not gotten into what happens when a loop is performed and there is no centripetal acceleration, or what happens when the plane is dead overhead when there is 2.34Gs at the bottom of the tank. These calculations are at this point academic and will be left to the reader, to ponder. Also I did not consider any acceleration except that out of the circle. To do so here would involve a rather detailed discussion of Vectors, which is a study better left to the reader.

Next time, I will introduce the fascinating subject of Material Strength and how it applies to your chosen line set.

FLYING SITE ALTITUDES: In Oregon, Eugene Airport is 374 feet, Roseburg is 529 feet and Salem is 214 feet. In Washington, Arlington Airport is 137 feet, Auburn Airport is 57 feet, Chehalis Airport is 177 feet, Harvey Field (Snohomish) is 22 feet, and Pierce County Airport (Thun Field) is 538 feet).

More information on uniflow tanks, including plans for building your own, can be found at the Aeromaniacs website.