Now that we have made a rudimentary study of line tension, or more precisely "Centripetal Acceleration" and the effect it has on an object traveling on a circular path, it is time to look at what happens inside a fuel tank on a control line model, while in flight.

As with our previous discussion, a complete and exhaustive study of classical mechanics, dynamics or fluid mechanics is way beyond the scope of this article, so we will be looking at the aircraft, engine and tank in a very much simplified way.  I will be looking at it for just a brief instant, so that I can treat the situation in more of a static (motionless), manner.  The end result of, increasing our understanding of fuel tanks, will be similar, and we will not need to enroll in a college level math, engineering, or physics course.

Having flown control line models (now) for the better part of 28 years, I have heard virtually every explanation for how a fuel tank works, and what happens from the transition from sitting on the ground with the engine running, to flying level upright, inverted, looping and flying overhead.  Most involve some terns like suction, standard tank, uniflow, pressure, prop loading/unloading, and the like.  To begin it bears mention that all fuel systems currently in use on model planes are pressure systems, whether this is muffler, bladder, crank case, or atmospheric, fuel is fed to the needle valve and into the venturi by pressure.

The first illusion that every modeler must overcome is that an engine equipped with a "hard" or "clunk" tank, and whose vents are open to the atmosphere, runs on suction.  The layperson calls vacuum or partial vacuum as in this case -- suction.  The simple truth is that a perfect vacuum is by definition nothing, and I think we can all agree that nothing cannot do something, in this case work of delivering fuel to your engines; so the something in a fuel tank that is bringing fuel to the engine is -- pressure.  So to restate the hard tank fuel situation more correctly: Fuel flow is induced by pressure inside the fuel tank, pushing fuel into the partial vacuum of the venturi, in an attempt to equalize this difference in pressure.  Before I make even the most rudimentary calculations, it bears mention that in every pressure system, that the fluid(s) will flow from an area of high pressure to an area of low pressure.  Wind blows from the high pressure into the low pressure area.  That is just how things work, and we all need to get use to that fact.  The second fact that will come to play here is that, when a fluid accelerates it pressure drops, or decreases.  I think we all have heard about the air on the top of an airfoil speeding up, and creating a low pressure, which lifts the aircraft off the ground (once there is adequate pressure differential, induced by forward motion).  This is exactly the same effect which occurs inside our engines venturis.        

Now let's look at a simple example of how pressure induces a fluid to flow, in this case I will use water.  Suppose you take a three pound coffee can and punch a hole in it at the very bottom just above the crimped lip.  Next you plug it off, and then fill it with water to a depth of 10 inches above the hole.  Now you set the can on your garden wall and pull the plug.  At first the water will flow out with considerable vigor, and will reach quite a way from the can/wall.  As water leaves the can, the depth of the top surface of the water to hole will decrease, and as this happens the water will progressively hit the ground closer to the can/wall.  At the end as the water is at the level of the hole it will just trickle out.  If you have not observed this, first hand, I suggest this exact experiment, if you have done this before then read on.   The 10 inch level in this example is what scientists call hydraulic head, or head pressure.  Very often as with water, or air, just a simple distance will be given, as it assumed the fluid is homogenous, and so, its weight density is the same throughout.  In our example the depth of water is 10 inches; the density of water is 62.39 pounds per cubic foot, or .58 ounces per cubic inch.  Let's look at this calculation:

 

62.38 lbs/cubic foot x 1cubic foot/1728 cubic inches.  1728 is the value of 12 to the third power, or 12x12x12.

62.38 lbs/ft³ x 1ft/1728 in = .036 lbs/in³

 

And as we all know there are 16 ounces in a pound, so: 

.036 lbs/in³ x 16oz/1lb = .5775 oz/in³ or .58oz/in³

 

So at the bottom of the can the pressure is: 

.58oz/in³ x 10 in = 5.8oz/in²

 

Now this is not the typical unit people are used to seeing so let me do the same calculation with pounds per square inch:

.036lbs/in³ x 10in = .36lbs/in²

 

Now that is better, pounds per square inch is what most of us are used to seeing, and it makes this unit the proverbial ÒPSIÓ (psi) we see everywhere.  So it is of interest to note that for the rest of this example we will use the very simple formula of, depth times density, equals pressure.  And in looking at the above calculation the units show this.  

The above example assumes the acceleration that water is under is that of gravity, or one "G."  As we learned in my last article, a control line aircraft is not exclusively flying only under the acceleration due to gravity, but is also acted upon by centripetal acceleration.  Using the "stunter" in my last article we have:   

  "So let's find the centripetal acceleration of a 48 oz (three pounds) model that has a five (5.00) second lap time, and 70 foot lines.  The angular velocity for this model is given by: 

W = 2p/5.00 seconds

W = 1.26rad/sec.

 

Now that we have the angular velocity we need to square it and multiply it by the line length:

A_c= (1.26)²x70

A_c=111.13 ft/sec² = centripetal acceleration"

 

So we have an outward acceleration of 111.13 ft/sec² (feet per second squared), which is approximately three times that of gravity.  So if we look at the coffee can example, what would the pressure be at the bottom of the can (tank) if the acceleration were that of our example "stunter"?  For this example we could take the mass of a given volume of water, and then use NewtonÕs F=ma, and come up with a new weight density, and then multiply by the depth to the hole ... But you know there is a much easier way.  If we assume the acceleration due to gravity to be 32.4 ft/sec² all we have to do is make a ratio out of the two accelerations and multiply the 1G pressure to get our result.

 

111.13ft/sec² divided by 32.4ft/sec² gives:

3.43 (which is our scalar quantity)

Note I have called this a "scalar."  This is because it is exactly what it is, a dimensionless number that is used to scale an existing quantity to a new value which is more useful at that time.  So then getting back to our pressure calculation:

.36lbs/in² x 3.43 = 1.23 lbs/in²

 

So we are looking at more than three times the pressure at the bottom of the coffee can when the acceleration is that of the "stunter" in my last article.  This is of paramount importance to us in this article because this is what the fuel in a tank is experiencing when the plane is flying level. 

         Before I can go into a lot of detail about head pressures and the like, I will need a weight density for a given volume of fuel.  Assuming 10% nitro methane, and 20 percent castor oil (Sig Champion 10/20), I measure this to be:   .033 lbs /in³  Now let's look at an example using a standard (non uniflow) stunt tank, that measures 1 inch x 2 inches x 4 inches.  We will start with a profile plane, the tank is sitting horizontal, the tank is mounted so the apex of the wedge is exactly even with the center line of the venturi/cylinder, it is mounted .1" behind the back plate of the Fox 35 engine.  Also, note that the needle valve of the engine is .5 inches above the motor mounts, and it is 1.6 inches to the needle valve from the back plate (pan head screws are needed to get this dimension).   Oh, inside this tank the fuel pick up is assumed to be .1" ahead of the rear end cap, and it is mounded on/at the apex of the wedge (on the centerline of the tank).

         So the plane is sitting on the ground the tank is full and the engine is not yet running.  The fuel will, if not stopped, drain into the venturi, just under the force of gravity.  Most (modelers) in this situation will pinch the fuel line with a hemo-stat, to stop this.  Why does the fuel want to flow—it has 0.5" of hydraulic head pressure to the pickup.   So let's look at the actual amount of pressure in this rather frustrating situation:

 

.033 lbs/in³ x 0.5 in = .0165 lbs/in²

To state this, pounds per cubic inch, times depth in inches, gives pounds per square inch. 

Remember I picked a non-uniflow tank, and I had the plane sitting motionless on the ground.  Now let's suppose the plane gets rolling and lifts off and flies level while doing the above mentioned 5 second laps, on the same 70 foot lines.  As the plane picks up speed the fuel inside the tank will migrate to the outside edge of the tank, and will be held to the outside by centripetal acceleration.  In this example 111.13 ft/sec² of centripetal acceleration, or about three Gs.

         So at the beginning of the tank, the air space has one atmosphere of pressure (this is an assumption, for the sake of simplicity, on my part it may be more or less depending on how the vents are made), or 14.7 pound per square inch, acting on it.  Then there is the centripetal acceleration times the depth of the fuel to the needle valve.  Note I do not need to know the pressure at the bottom of the tank, as the needle is 0.5Ó from the motor mounts, or one half inch outboard.  So what is the pressure of the fuel at this half inch depth? 

 

0.5 in x (.033 lbs/in³ x 3.34) = .055 lbs/in³

 

The 3.34 is the scalar from the earlier example and makes what would seem like a paltry .0165 lbs / in² into a sizable sum which cannot be ignored.  Please note, as this is a non uni-flow tank the air column acting on the fuel is the same from the beginning to the end, one atmosphere.  So, we now know that the fuel flowing to the venturi has positive pressure for the first half an inch of fuel.  What happens when the fuel is consumed and flows to a depth of half an inch?  At that point there is no longer any hydraulic head, and only pressure differential between the inside of the tank (1 atmosphere) and the venturi, induce fuel flow.  Also at this point the fuel levels are equal, both in the tank and the fuel line going to the engine.  As fuel is used let's say another half an inch, then the pressure differential (between the tank and venturi) has to work against the centripetal acceleration of the model and fuel density to deliver fuel to the engine.  So we need not make a calculation for this, as it is the same problem as above.  Just from a different perspective.  I would state it thusly:  There must be at least .055 lbs / in²  of differential for the fuel to reach the venturi.  Lastly what happens when the fuel is just about out, and the level in the tank is 1.5 inches below the venturi?

 

1.5 in x .055 lbs / in² = .165lbs/in²

 

So at the start the fuel in the tank has a surplus (head pressure) of .055 lbs /in² and at the end it has a deficit of .165lbs/in².  To put it differently it has to draw .165 psi (of pressure) to just stay running.  Some would in making this calculations assign a "sign" to the numbers, my convention would be to have all distances above the needle/venturi position to be positive, and those below to be negative.  So the numbers would be:  .055 psi at the start, and -.165 psi at the end of the tank.  The total differential for this system is the absolute value of both values summed:  .055 psi + .165 psi = .22 psi total change.

         This example perfectly explains the tendency of a non-uniflow tank to start with a very rich setting, and to lean out throughout the run, and end the flight so lean the engine is cycling. 

         It bears mention that I went to quite a lot of trouble measuring the weight density of glow fuel, for this article.  This entire example could be done solely with just the linear distances from the tank sides, and the needle valve, as the density of the fuel is the same throughout the system.  This would be the way a hydrologist would do things.  So The above Fox 35 example becomes 0.5" of head at the beginning of the run, on the ground, transitioning to 1.67" in the air (actual distance multiplied by the G force scalar) to an ending deficient of -5.01" (-1.5 x 3.34)  at the end of the run; and the total head difference would be:  6.68".   Only went through this example to help assign a number to the actual pressures the engine and tank are seeing.  It was for me, purely academic. 

         Lastly, I need to make a disclaimer for this entire dissertation:  If the leadouts are not in exactly the right place, then the fuel feed may be running "up or down hill" from the tank, which will affect the head pressure, there may be changes in the actual atmospheric pressure as the plane goes from sitting still to flying around at some speed; there may be some compaction, or bulk change in the fuel as it experiences 3 times the force of gravity, the temperature may change as air blows over the tank; there may even be bubbling of the fuel caused by the vibration of the engine.  I accounted for none of this, simply because I can't say what they are, and I have no way to measure them. 

         Next Time I will delve into the fascinating subject of the Uniflow Tank.