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3/8 "is the outside diameter. The inside is @1/4"

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the inside is 5/16" on a 3/8" steel line

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The larger the hose diameter the less friction loss given the same hose length and discharge pressure.

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Sometimes it does

B U T ....a fire-hose is that traveling from a dead-stop to 60 mph in 2 seconds or so !

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Ok, lets do some math...

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Take the 3/8" diameter
So 3.1415 x .375"^2 x 100"x .0267lbs/in^3= 1.17 lbs mass

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1. Area = pi*D^2/4, your volume calculations are 4X too big. 2. You're using solid mechanics math on a fluid mechanics problem.

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There are two separate issues and some are confusing them. One is the acceleration of a fluid and I've explained that one previously, the size of the line isn't an issue. The other is the pressure loss from fluid flow through the lines and fittings. This is the area people should focus on. First find your actual max. flow rate and then calculate the velocity in the line and determine if you need a bigger line. See this page: http://en.wikipedia.org/wiki/Reynolds_number

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The larger the hose diameter the less friction loss given the same hose length and discharge pressure.

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B U T ....a fire-hose is that traveling from a dead-stop to 60 mph in 2 seconds or so !

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I guess you're missing the point Doc ? You're a Doctor of What ? Fiberglass ?

Fact is, the larger the diameter the less the friction loss is. A 1.5in line loses 25psi pressure per 100ft. A 2.5in line loses 14psi. A 4in line loses 4 psi....and so on.

A 5/16 line will have more friction loss than a 3/8in line....all things being equal. The velocity of the fluid has little to do with it....surface contact has everything to do with it. The larger the percentage of surface contact vs the total volume of fluid....the greater the friction loss.

All this is academic anyway....i agree that a 3/8-1/2 line is plenty for just about any application


Ron

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The larger the hose diameter the less friction loss given the same hose length and discharge pressure.

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B U T ....a fire-hose is that traveling from a dead-stop to 60 mph in 2 seconds or so !

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I guess you're missing the point Doc ? You're a Doctor of What ? Fiberglass ?

Fact is, the larger the diameter the less the friction loss is. A 1.5in line loses 25psi pressure per 100ft. A 2.5in line loses 14psi. A 4in line loses 4 psi....and so on.

A 5/16 line will have more friction loss than a 3/8in line....all things being equal. The velocity of the fluid has little to do with it....surface contact has everything to do with it. The larger the percentage of surface contact vs the total volume of fluid....the greater the friction loss.

All this is academic anyway....i agree that a 3/8-1/2 line is plenty for just about any application :thumbRon

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I have one stock 5/16 fuel line feeding both carbs. [Email]10.35@134.4[/Email] and 770HP on the dyno..

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Thats all I need know! Thanks Dave!

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I have one stock 5/16 fuel line feeding both carbs. [Email]10.35@134.4[/Email] and 770HP on the dyno..

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That's all I need know! Thanks Dave!

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There are a lot of possible explanations for all this but one is a car that falls on its face at 900-1000 feet due to fuel starvation.

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it's a FAST car...quite literally

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Can anyone on here verify if the fluid dynamics theory are based on stationary commercial applications? I'm thinking, that if that is so, then they, it, does not apply to fluid systems mounted onto accelerating vehicle like race cars, race boats, airplanes and rocket ships

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I've had a few tell me I may not have enough fuel for my car. My 74 cuda with 440 has 530ish Hp. I run a holley black pump with a regulator at carb which is an 850. no return. I doesnt stumble nor seem to fall on its face so i havent paid much thaught to it. My 1/4 mph is usually between 117-118mph.

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Looks like you are good to go. The most important consideration is line size from the tank/cell to the pump

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I have one stock 5/16 fuel line feeding both carbs. [Email]10.35@134.4[/Email] and 770HP on the dyno..

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Can anyone on here verify if the fluid dynamics theory are based on stationary commercial applications? I'm thinking, that if that is so, then they, it, does not apply to fluid systems mounted onto accelerating vehicle like race cars, race boats, airplanes and rocket ships

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Define "stationary', really, it's not as easy as you think. They say that the only truly stationary 'object' is the center of the universe, and that everything is moving away from it. Effectively, nothing is stationary and all motion is relative. The equation relating pressure of a column of fluid is from hydrostatics, absolutely. But that gravity is included means that an acceleration is involved, 32.2 ft/second^2. The more general form uses acceleration instead of gravity, any acceleration. A similar situation occurs with weight. Weight is just the force that results from a mass under the acceleration of gravity. Accelerate your car at 1G and the reaction force opposite your direction of travel equals the weight of the car/driver.

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Hmmmmm, most basic fluid dynamics equations seem leave gravity out I was told for what we did it was not really necessary....The fact that everything has an inward angular acceleration (or "rotates" as most say) from the gravitational pull of the sun really has little effect on anything. One can consider an object stationary if it is sitting on the earth with no velocity, the motions of the earth relative to the object you are examining have such a minimal effect that they need not be taken into consideration.
As far as the 1G thing goes it sounds like you are saying that if I were to accelerate at 32.2ft/s^s in the positive x-dir the weight of the car plus me would be on my in the negative x-dir. Pretty crushing force. As you sit on your computer right now you are opposing 1G, defined by Newton's Second Law, F=ma; F being basically what you see when you hop up on the scale in your bathroom, it IS your weight. I am 160lb, so if I were to accelerate at 32.2ft/s^2 that would be a 2G force, and the force acting on my body would be 320lb-force downward onto my seat, and as Newton's Third Law explains, the seat would react with a force equal in magnitude and opposite in direction, making you feel the force being put upon you.