prsrizdgt,
Right from an ex-Ford engineer and now SCT...This is taken from the Modular Depot
"Before I can go into fuel systems and losing fuel pressure, I think the first thing to do is give you some information on fuel injectors and fuel pumps. This one will cover fuel injectors and the next one will cover fuel pumps.
I have been told before that this is just math and math has nothing to do with cars (this was on another site, not here) so if this is your attitude, then please just press the back button now and stop reading, this is not for you.
Fuel injectors have a flow rate. This flow rate of the injector is rated at a certain pressure drop across the injector. Meaning the injector flow is say 30#/hr at 42.5 PSI. This means that that injector will flow 30#/hr as long as the pressure at the supply side, minus the pressure in the manifold is 42.5 psi. A fuel pressure regulator is designed to keep the fuel injectors at a certain pressure drop. In Ford systems this pressure drop is 39.5 psi, in the rest of the industry it’s 42.5 PSI (don’t ask, I don’t know why).
This is why fuel pressure is lower under vacuum. The vacuum is a pressure less than atmospheric so it lowers the rail pressure to keep the pressure drop to 39.5 psi. At WOT; when the vacuum should be zero, the rail pressure should be pretty close to 39.5 psi. If your motor has 20” of vacuum, this is actually –10 psi (vacuum in inches of Hg, is double what the pressure is in psi). So with 20” Hg vacuum, you should have a fuel rail pressure of 39.5 psi –10 or 29.5 psi. If the engine has 10” of vacuum, that’s –5 psi, so your rail pressure should be 39.5 – 5 or 34.5 psi.
Now, on a blower car it’s the same thing. The fuel pressure regulators on the car have a gain of 1:1. This means that for every 1 psi of boost, it will raise rail pressure by 1 psi, to keep the pressure drop across the injector to 39.5 psi. So, if you are running 10 pounds of boost, your fuel pressure should be 39.5 + 10 or 49.5 psi.
This is why you should always set fuel pressure with the vacuum line off the regulator; this gives you the true pressure drop across the injector. Now, if you set your fuel pressure to 45 psi, all the above stuff is still true, but your base pressure is now 45 rather than 39.5.
Now, you can make a smaller injector flow more by raising fuel pressure, I think we all know that. There is a mathematical equation that tells you pretty much exactly how much the flow rate will change based on this new pressure drop.
To determine the new flow rate of your injectors you take the square root of the new pressure divided by the pressure at the rated flow times the rated flow. Let’s go through this once. If you have an injector that flows 30#/hr at 39.5 psi, and you raise fuel pressure to 55 psi, what is the new injector flow rate? You would take 55/39.5 which is 1.392, and then take the square root of this, which is 1.18 and then multiply this by 30, which is 35.4 #/hr. So, a 30# injector at 55 psi is now a 35.4# injector.
The same thing applies for losing pressure. If you have those same 30# injectors but your fuel pressure drops to 35 psi, your new injector flow rate is 28.2#/hr.
Now you can roughly calculate how much HP a set of injectors will support with some math as well. On an engine dyno there is one value that is measured called BSFC or Brake Specific Fuel Consumption. This is the pounds of fuel that it takes to make one Hp for one hour. Typically a Naturally Aspirated 4.6L engine will have a BSFC of about .5 and a blown car will be about .6, the actual numbers may be slightly different, but these are close.
If you have 8 injectors and each injector is rated at 30#/hr, then you have a total fuel flow of 8 time 30 or 240 #/hr of fuel. If your car is blown with a BSFC of .6, then you can take this 240#/hr of fuel and divided by the BSFC of .6, then these injectors would support 400 HP at the crankshaft. There are cars out there making more than this, I have worked on some, but in those cases fuel pressure was raised up higher. In some true race only cars, you can run them a little leaner and get a little lower BSFC as well. In the above example if you raised fuel pressure to 55 psi with these 30# injectors, you would be able to support 35.4 times 8 divided by .6 to get a HP of about 472.
Now the beauty of a returnless system is this. It measures the pressure drop across the injector so it constantly adjusts the injector flow rate based on the actual pressure across the injector. In addition, you can electronically raise fuel pressure, only at higher flows, to increase injector flow rate.
Like fuel injectors, fuel pumps have a rated flow at a certain PSI. For example, a 255L/hr pump is rated at 255 L/hr with a pressure drop of 40 PSI between the inlet and the outlet. Now, in most cases the inlet pressure is 0 psi, or atmospheric pressure, but in some cases it’s higher. Like when you add a T-Rex inline pump. Now your stock pump pumps to the T-Rex and the T-Rex pumps to the motor. The pressure at the inlet of the T-Rex should be higher than atmospheric since you have your intake pump pushing into it.
Like fuel injectors the flow rate of a pump changes based on the pressure drop across the pump. It even follows the same equation, the rated flow times the square root of the pressure at rated the rated flow divided by actual pressure.
But, unlike a fuel injector, whose flow increases when you raise the pressure drop across it, a fuel pump loses flow when you run more pressure. An easy way to understand this is the following example;
The faucet in your sink is turned off. There is pressure at the other side, but now no water comes out, there is no flow. As you turn the faucet on slowing, the flow increases, but the pressure drops across the valve in the faucet drops. At full on, there is probably the same pressure on both sides of the valve in the faucet, meaning no pressure drop, but it’s flowing a bunch of water.
Now, remember how the fuel pressure regulator keeps the pressure drop across the injector constant to 39.5 psi? (Even in a returnless car the same thing happens, it’s just done electronically). Well, if you have 10# of boost in the manifold, and a 40-psi pressure drop across the injector the rail pressure is 40+10 or 50 PSI. This is where rail pressure is important. That means the fuel pump is now pumping into 50 PSI rather than 40 psi. And since the pressure rise across the pump is greater, it’s flow rate drops. Follow the math;
You have a 255 L/hr pump that’s rated at 40 psi. You have a blower with 10# of boost so rail pressure is 50 psi. What is the new fuel pump flow rate? You take the pressure at which the flow is rated, 40 psi; divide this by your base fuel pressure, 39.5 psi, plus the amount of boost you are running, 10#. This is 40/(39.5+10) or value of .81. You then take the square root of this and multiply it by the rated flow. So the square root of .81 is .9 times 255 for a new fuel pump flow of 255 times .9 or 229 L/hr. So, with 10# of boost your 255 L/hr pump is really only a 229 L/hr pump.
Let’s go a step further, remember in the fuel injector example about running 30# injectors up to 55 psi to make them 35.4 # injectors? Let’s say that car is running 10# of boost also. Now you take 40/(55+10) which equals .615, now take the square root of this to get .78 and then multiply this by 255 to get a new fuel pump flow rate of 200 L/hr.
Now lets translate this fuel pump flow into HP. To get from L/hr to # of fuel per hour, assuming a density of gasoline of .76g/cc, you would multiply L/hr by 1.2733 to get #/hr of fuel. While I could show the math as to how I got the 1.2733, I don’t know if anyone is interested.
So, the same equation of fuel flow to HP is true as in the fuel injector example. You take #/hr of fuel delivered and divide it by the BSFC (Brake Specific Fuel Consumption) of the engine.
Going to the first example of a 255 L/hr pump blowing into 10# of boost reducing it’s flow to 229 L/hr. How much HP will this pump support? You’d take the 229 times the 1.273 to get a total fuel delivered of 292#/hr. You take this number and divide it by the BSFC of .6 to get a HP of 486 HP that this setup will support.
Here is another example; Let’s say you have a ’03 Cobra running 20 psi of boost. The factory pumps are rated at 155 L/hr each and there are two of them. So, total pump flow of 310 L/hr. If you take 40/(39.5+20) you get .6722, then take the square root of this to get .82 and multiply this by pump flow rate of 310 L/hr to get a new pump flow rate of 254 L/hr. This is about 324#/hr of fuel and with a BSFC of .6 it will support a HP of 539 at the crankshaft. I will post data in a post late tonight or tomorrow, (data, not opinion, or my cousins neighbors car did this) that show a stock ’03 Cobra pump and tank (no boost a pump) losing fuel pressure above 500 RWHP. I will also show this same car with a boost a pump and how that impacts fuel pressure.
Next example. You have a T-Rex inline pump. This pump is actually only a 190L/hr pump. And you have your stock intake pump of 110 L/hr. With 10# of boost how much flow is in your fuel system…
For the most part, the answer is, you don’t know unless you measure the pressure between the two fuel pumps. I have measured this pressure, and you’ll have to trust me on what that pressure is.
In a 350 RWHP car, with a stock intake pump and a T-Rex, there was 20 psi between the two pumps. So, now you take 40/(20+10) to get 1.333. Now I used 20 instead of 39.5 psi because the actual pressure drop across the pump is 20 psi, not 39.5. The square root of 1.333 is 1.155 and then this multiplied by the flow of 190L/hr is 219 L/hr. This is a total of 279#/hr of fuel delivered and with a BSFC of .6, it would support a HP of about 466 at the crankshaft. " - Jerry of SCT
2008 Acura TSX (5AT)
2012 Honda Ridgeline RTL (w/Navi)
Ex toy: 1995 T-Bird LX - ALLEN supercharged, 2000 Mustang GT 4.6L PI engine, lot's of goodies...
12.74 @ 109.45 mph (BEST E.T. BEST MPH)
325 RWHP/380 RWTQ (SAE) on a dynojet
Tuned by Jerry W. from SCT