Fuel Flow Calculation

If you Google Fuel flow calculation you'll get some help.

Don't speed.

I can't find anything useful. The car doesn't see the road much. Throttle is tempting and dangerous.

Here try this. GPM=0.0408X (Pipe Dia inch)^2 X ft/min (velocity) (for water)

1 gal=231 in^3

K=0.110 gals/hp K GPM in this case was for a different calculation, but it says thats how many gallons you need per hp.

Cheers

Thanks. I found calculations like that or the same but they include velocity. I know what the pump flow is, but its not going to be the same in the line is it? The pump flows 900 lbs/hr.

Could I somehow convert lbs/in^2 to velocity?

what do you think will be the item that determines fuel flow? The pump can flow 900 lbs per hr. At what pressure? What size are the fittings on the fuel pump? if the y are bigger than the hose, the hose may be a restriction. What about the needle valve(es) on the carburetor? A .110, or.120"would be common on a Holley. Duel inlet? Multiply by 2. Dual quads? Multiply by 4.

The limiting factor for you must be the engine's fuel consumption demands. Holley sells high performance fuel pumps. (among many others) They have a good help line. call, or E-mail them your question. Good luck.

When I switch to ethanol, I will be swapping in larger needle and seats. Its dual inlet. Single carb.

Aeromotive's site says a single -10an line is good for 2500hp. A 3/8 (-6an) line is generally good for 600hp. So maybe another way would be to calculate the area differences and compute.

Figuring area of the lines with the ID being 12% smaller than its rated size-

3/8"-1.04

1/2"-1.38

5/8"-1.73

It would appear that the area is linear with diameter? I don't remember that being right. But when they state good for 2500hp, it may actually be a good bit higher than that just to cover their butts. By how I've calculated, the 1/2" line should be good for 1500hp on gasoline naturally aspirated.

Aeromotive's site states the pump is good for 1500hp nat asp. and 1200 forced induction with a carb. Due to the higher fuel pressure to combat air pressure. So the line would be good for 1200hp. And about 900hp with ethanol.

That number is based on free flow no restrictions. I think that it's based on water also. Gas S.G. is 0.74, 45.9 lbs/ft^3 . 1 gal (US)=0.13368 ft^3. 231 in^3= 1 gal (US). 1hr= 60 mins :)

Cheers

First, thanks for the replies. I greatly appreciate it. I love to learn.

900 lbs/hr at about 15 psi. Its an Aeromotive A1000. Supports 1100hp for forced induction. The pump flows enough. It has -10an fittings and I'm using an -8an hose.

I have seen fuel systems using -8an line that support 1000hp. But they are for EFI which needs more pressure and less volume. I'm concerned the 1/2" line won't support the volume.

I know it will support my power on gasoline, but I'm switching to E85 (ethanol) and need 30% more supply. Worst case scenario, I try it and keep an eye on fuel pressure.

My peak pressure is 15psi (6 base and 9 added) blow through carburetor. Most people that I talk to say I will be ok but close to its limit. I'd like to know where the limit is.

The nitrous system will most definitely have to have its own dedicated system.

How do I get ft/min from lbs/hr and lbs/sq.in?

I just started my first year for my MET BA degree, its been 7 years since I've had any math courses.....

First of all flow and pressure are independent. You can have flow and no psi, you can have psi and no flow. What the pump can deliver will depend on what flow will be available at x psi. 15 psi could stop it cold or it could develop. 1 GPM at 15 + X. Psi is amount of restriction of the outlet. So where to start?

The fuel consumed is going to be based on the power of the engine, and the efficiency of the engine.Aeromotive should give you the guidance you ask for. To get 1000 hp out of a low compression flat head design, will take a huge engine, and a truckload of fuel. But a small high revving engine with a turbo, or blower that is pumping 15 pounds of boost, will consume less fuel per horsepower. See how the answer you are looking for is harder than it sounds. Westech runs a dynomometer that sees hot rod motors on it all the time. If you can find a # for them, they should be able to match your engine combination to a similar one. From that they should be able to tell you the fuel consumption. Westech is in California I believe. Good luck.

"I have seen fuel systems using -8an line that support 1000hp. But they are for EFI which needs more pressure and less volume. I'm concerned the 1/2" line won't support the volume."

EFI needs more pressure, but the volume should still hold true. My belief anyway.

How do I get ft/min from lbs/hr and lbs/sq.in?

Simple. Just to get you thinking properly, I'll ask a few of questions, and I'll bet you can come up with an answer yourself.

What do you have to multiply ft/min by to get ft/hr? Answer: min/hr

What do you have to multiply ft/hr by to get lbs/hr? Answer: lb/ft

How do you get quantity for lb/ft? Answer: Density of gasoline/ethanol, diameter of pipe (or volume of 1 ft of pipe).

Volume of ft of line-.152 sq.ft

I'm having trouble finding the density of gasoline or E85. I found .78 on wikipedia, but I half doubt just about everything I read on there. A gallon weighs 6.5 lbs. Is the density 6.5lbs/gallon? What unit does that need to be in?

The formula I have so far is (900lb/hr)*(.152sq.ft/density)=ft/hr

I'm confused about the square, obviously the velocity can't be 2 dimensional so it stays right?

Hmmm . . . volume is cubic feet, not square feet. Density of gasoline is 49.5 lbs/cu ft, and the density of ethanol is essentially the same at 49.22 lbs/cu ft.

There are 7.48 gal per cu ft or 0.13368 cu ft per gal.

The formula you "have so far" doesn't make any sense whatsoever. To start with, lb/hr*ft/density = lb X ft/hr X density, not ft/hr. Density is not a unit of measure.

The cross sectional area of a 0.44" ID tube is calculated using the formula: Pi X r², meaning that the cross sectional area of a 0.44" ID tube is 0.152 sq in. The volume of a foot of tubing is 12 X 0.152 = 1.825 cu in. There's 1728 cu in per cu ft, so 1.825 cu in is 0.001056 cu ft, and that volume of gasohol weighs 0.052 lbs.

If you need 900 lb per hr, then you need 0.25 lb per sec, or by dividing 0.25 lb/sec by 0.052 lb/ft, you get a flow rate (velocity) of 4.8 ft/sec.

4.8 ft/sec is 288 ft/min. Putting that into the formula mayt2u gave me, I get 2.28 gallons per minute or 820 lbs per hour. I found two different calcs for fuel/hp. The one mayt2u gave me is .66 lbs/hp, another one I found is .5/hp per hour. Going with the less efficient number, I get 1242hp. In a perfect world, with no restrictions, no friction losses.

I will be making about 750hp. Since the E85 requires 30% more fuel, I get 975hp. If I were to use the same amount of gasoline. Seems like it will be ok, but close. I'll just keep an eye out for pressure droppage at high rpms.

Thanks for help and info!

Not sure what I was thinking with the sq. ft.

Putting it in a gallon jug and timing it is actually probably the best way........

You're wearing me out . . .

You should have come up with 900 lb/hr, which is the number I started with to get 4.8 ft/sec. Did you use the density of water to get 820 lb/hr?

Hehe, sorry. I plugged 4.8 ft/sec as velocity into mayt2u's formula. But thats the velocity of the gas coming out of the pump with a 5/8" outlet. Once it hits the 1/2" line, velocity is going to increase though right?

Eh, its not a big deal, I don't think even if I get the right answer its going to be exact. I'm sure you have more interesting or important things to do. Thanks though, I think my chances are good. E85 is very detonation resistant, so if it does go a little lean before I get out of the throttle, I shouldn't do any damage.

Now that it is all worked out, can you convert that into gallons per cubic hour.

Sure can't, but I found a website that will do it.

http://www.youtube.com/watch?v=fbGkxcY7YFU

Run it into a bucket and weigh it. See how many seconds it takes to fill it about halfway up then calculate how many seconds it takes to stuff a liter out of the hose end. In the arrangement you stated there are too many potential variables. But from what I know of diesel engines you should not be dealing with less than .75" short hoses. The fuel pump flow is what you should be most interested in. Size the fuel line to no more than 7 fps for transmission from a fuel farm to 10 fps within a few feet of the fuel pump.

It is an <...easy calc...> and to do it one needs the following:

• Viscosity (temperature-dependent variable)
• Density (temperature-dependent variable)
• Average velocity (volumetric flowrate divided by cross-sectional area of the pipe)
• Intended pressure drop along the pipe
• Length of the pipe.
• Diameter of the pipe

One feeds these into the Darcy-Weisbach equation. The calculation may need a couple of iterations to get close to what is intended.

<...How much...> depends on how hard it is being pumped, which is the effect of pressure drop along the line, which is why the calculation becomes attractive.