Join Date: Dec 2008
Location: Tyler, TX
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Due to the design of nearly every electric fuel pump made (and many mechanical pumps), if you increase the fuel pressure, you DECRESE the pump's FLOW CAPACITY. Fuel pumps are rated in GPH, LPH or lb/hr at a given pressure. There is a formula that gives the new flow at the new pressure.
If the pump will flow 255L/hr at 60 PSI, and you increase the pressure to 120 PSI it flows less. The amount less (or more if at lower pressure) goes as: new = rated (flow) times the square root of the pressure at rated (flow) the rated flow divided by actual pressure. Rated pressure is 60 psi ,flow is (assumed) 255l/hr.
Doing the math on it:
New flow => 60/(60+60) = 0.5 (ratio of old to new), we take the Sq Rt of this = 0.707 and multiply by the rated flow (255l/hr) to get 180.28l/hr.
So, we increased pressure, but DECREASED the peak flow capacity of the pump by nearly 30%.
To increase total fuel flow, especially with larger injectors (high power applications) you need a pump RATED for more flow, or more pumps.
If you do the math again, with to pressure pumps in series (one after the other) we find that the pressure is doubled, and the flow remains the same because the pressure drop (change actually) accross each pump is still at or below the rated pressure. 0->60 = 60 psi increase, 60-> 120 = 60 psi increase. What we did not accomplish is to increase the flow capacity of the pump(s), but we did increase the flow capacity of the supply system somewhat.
The REAL trick is to keep the stock pressure but increase flow capacity with the power levels. To give an example, let's say the stock power is 250 WHP, and we now make 500 WHP. We need to FLOW twice as much fuel. Most likely, the stock pump CAN keep up with demand, if the rest of the system can as well. If the stock fuel lines are 3/8" (0.375") ID, the cross-sectional area is pi*r^2 = 3.14 * (0.375/2)^2 = 3.14 * (3/16)^2 = 3.14 * 0.035" = 0.110 sq-in. To maintain flow at twice the power, the line now NEEDS to be 2 * 0.110 sq-in, or 0.220i sq-in.
What ID? 0.220/pi = r^2 = 0.0703. Remember is was r^2, so we need to find the sq-rt ( x-^2), and 0.0703-^2 = r = 0.264", d = 0.528" ~(17/32").
Most turbo-diesels have a BSFC of something close to 0.45. That's how many POUNDS of diesel it takes to make on HP (per hour). To figure out if the pump can support this, more math. 255l/hr =~ 67G/hr. Diesel weighs ~7.2lb/G, so 7.2 * 67 = 482.4 lbs/hr which if we divide bt 0.45 gies us HP/hr = 482.4/0.44 = ~1071 HP/hr.
Fuel supply size is one of the most often overlooked factors in making power.
'93 F-250 HD, 7.3L IDI, 5spd - FARM TRUCK
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