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Another fuel tube question?


Buckaroo

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2 hours ago, Ed Cesnalis said:

Wings level ball left is a skid not a slip fuel flow will favor burn from right tank and appear as though transferring from right to left. 

Ok I'm a simple man! Fuel will appear! What does that mean? Will the fuel follow the ball under any most typical situations?  Is it simple as the fuel follows the ball? 

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39 minutes ago, Buckaroo said:

Ok I'm a simple man! Fuel will appear! What does that mean? Will the fuel follow the ball under any most typical situations?  Is it simple as the fuel follows the ball? 

think 1/2 ball out vs 4 balls out.  one is reading a small amount of forces at work and the other represents a much larger amount.

If you use a typical small amount like Roger's 1/2 ball then you will appear to transfer from right to left in that over some time the imbalance will reduce or go away but because it was a small amount of head pressure differential you wouldn't be transferring fuel but simply burning more from the right then the left.

If you are 3 balls out maybe you now have enough lateral force going to reverse the flow on the left side. When this happens you would actually be transferring fuel from right to left as well as burning exclusively from the right.

Both cases can achieve balance, one slowly the other quickly but only in the more severe case would actual transfer from tank to tank take place.

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Remember you only have a 5/16" hose that has bends, fittings and has to go uphill as well as down. Transfer most likely will max out at a certain point. Movement with .5 & 1 ball out works well as we know from experience. Do you get  a lot more transfer with the ball farther out?  I don't know.

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Also remember that when you are trying to transfer fuel it will look like you have less fuel in the tank you are trying to transfer to, because the fuel in that tank moves outboard. It will only be after you bring the airplane back to a wings level straight flight that you will know if you have done any good.

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30 minutes ago, Roger Lee said:

Remember you only have a 5/16" hose that has bends, fittings and has to go uphill as well as down. Transfer most likely will max out at a certain point. Movement with .5 & 1 ball out works well as we know from experience. Do you get  a lot more transfer with the ball farther out?  I don't know.

I have done transfers with ball as much as two diameters out.  It does flow faster; I guess the higher head pressure with the wing up higher does have an effect.  It's probably not linear though, in other words two diameters out is not twice as fast as one out.  Maybe 30% faster.

  One ball out works better though, it's easier to predict and not over-transfer.

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1 hour ago, Tom Baker said:

Also remember that when you are trying to transfer fuel it will look like you have less fuel in the tank you are trying to transfer to, because the fuel in that tank moves outboard. It will only be after you bring the airplane back to a wings level straight flight that you will know if you have done any good.

Of course that is the other half of the story, you describe a slip to transfer. My re-write below describes both a slip and a skid to transfer.

Quote

when you are trying to transfer fuel it will look like you have less fuel in the tank you are trying to transfer to, because the fuel in that tank moves outboard. It will only be after you bring the airplane's nose back into the relative wind that you will know if you have done any good.

Tom's version should probably say wings level and ball centered instead of just wings level too.

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2 hours ago, Ed Cesnalis said:

Of course that is the other half of the story, you describe a slip to transfer. My re-write below describes both a slip and a skid to transfer.

Tom's version should probably say wings level and ball centered instead of just wings level too.

I said wings level and straight. If you are not turning, and the wings are level you will be pretty close to ball centered.

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If the "receiving" tank drain is unported, I'd say the transfer rate would be increased, as opposed to transferring fuel when the "receiving" port has fuel covering it. Less resistance in that line, because the drain line has absolutely no fuel pressure behind it. IOW, the tank transferring fuel is supplying two lines, one to the engine and the other to an "open" line (receiving tank), with nothing behind it.

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17 minutes ago, WmInce said:

If the "receiving" tank drain is unported, I'd say the transfer rate would be increased, as opposed to transferring fuel when the "receiving" port has fuel covering it. Less resistance in that line, because the drain line has absolutely no fuel pressure behind it. IOW, the tank transferring fuel is supplying two lines, one to the engine and the other to an "open" line (receiving tank), with nothing behind it.

I'm guess it doesn't come into play much. When very low fuel you might unport if you transfer but when very low fuel I want all of my remaining on one side where its easier to manage and I just have to keep that wing with its fuel at the root.

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Let's say the left tank is empty and you have almost none in the right. How do you  insure you keep fuel in the tube/wing root? I can see too much push to ball left could unport going to left side and of course pushing ball right would sling remaining fuel way out in right long tank! 

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13 minutes ago, Buckaroo said:

Let's say the left tank is empty and you have almost none in the right. How do you  insure you keep fuel in the tube/wing root? I can see too much push to ball left could unport going to left side and of course pushing ball right would sling remaining fuel way out in right long tank! 

If the left tank is empty it is already unported.

To keep small remaining amount in right wing at the root:

  • Trim your rudder while looking at your tube.
  • Better to set up a slip than a skid
  • Minimize turns ( left turns are slips right turns are skids - don't over think it just keep your right sight tube in your scan. If you begin a turn and the fuel begins to dissappear / slosh outboard then start over and complete the turn without loosing sight of your fuel.  Use rudder to keep fuel visible and bank to set direction and adjust rate of turn.)
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22 minutes ago, Buckaroo said:

Let's say the left tank is empty and you have almost none in the right. How do you  insure you keep fuel in the tube/wing root? I can see too much push to ball left could unport going to left side and of course pushing ball right would sling remaining fuel way out in right long tank! 

Answer:  Bank left a little (and I don't mean 'turn'!), until you see fuel in the right sight tube.

Like Ed said years go, "if your eyes can see fuel, so can your engine."

Just look up to the right. If there is fuel in that sight tube, it is feeding the engine.

Buckaroo, with all due respect, I think you are over complicating this.

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"Do you get  a lot more transfer with the ball farther out?  I don't know."

If a slip is the only maneuver used to transfer fuel, then the transfer rate increases as the deviation of the ball from center increases.  The transfer rate is proportional to the square root of the difference in the height of the fuel column in the two tanks.  So, for each doubling of the difference in the height of the fuel, the transfer rate increases by 1.414 (the square root of 2).  So, imagine that fuel transfers at some rate when (in a slip) the fuel in one tank is 3" higher than the fuel in the other tank.  If the bank is increased so that the fuel is then 12", the transfer rate will double (12 is four times greater than 3, the square root of four is two).  All this assumes that there are no yaw effects (i.e., the airplane is on a constant heading).  

Without doing some measurement, what I don't know is the relationship between the height of the fuel tanks and deviation of the ball.  Stated another way, how much bank angle corresponds to each ball diameter of ball movement?  Once that is known (and if it linear), then the rest can be calculated.  

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12 minutes ago, FredG said:

"Do you get  a lot more transfer with the ball farther out?  I don't know."

If a slip is the only maneuver used to transfer fuel, then the transfer rate increases as the deviation of the ball from center increases.  The transfer rate is proportional to the square root of the difference in the height of the fuel column in the two tanks.  So, for each doubling of the difference in the height of the fuel, the transfer rate increases by 1.414 (the square root of 2).  So, imagine that fuel transfers at some rate when (in a slip) the fuel in one tank is 3" higher than the fuel in the other tank.  If the bank is increased so that the fuel is then 12", the transfer rate will double (12 is four times greater than 3, the square root of four is two).  All this assumes that there are no yaw effects (i.e., the airplane is on a constant heading).  

Without doing some measurement, what I don't know is the relationship between the height of the fuel tanks and deviation of the ball.  Stated another way, how much bank angle corresponds to each ball diameter of ball movement?  Once that is known (and if it linear), then the rest can be calculated.  

Thanks Fred,

We don't have to assume 'no yaw or p-factor effects because the ball isn't indicating the result of the tank's height difference alone but instead all forces acting on it including your yaw forces.

My laymen's understanding wants to attribute transfer in a  slip to height differential or gravity. I see yaw forces taking over when you skid instead (wings level and ball out to same degree)

 

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3 hours ago, WmInce said:

Answer:  Bank left a little (and I don't mean 'turn'!), until you see fuel in the right sight tube.

Like Ed said years go, "if your eyes can see fuel, so can your engine."

Just look up to the right. If there is fuel in that sight tube, it is feeding the engine.

Buckaroo, with all due respect, I think you are over complicating this.

That's funny that I'm over complicating this with 118 discussions on my post about this simple subject!?

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