# Electronic Fuel Injection System

### Electronic Fuel Injection System

Electronic Fuel Injection system is divided into three subsystems: fuel delivery system, air induction system, and the electronic control system.

### The fuel delivery system incorporates the following components:

1. Fuel Tank

2. Fuel pump

3. Fuel pipe and in line filter

4. Fuel delivery pipe

5. Pulsation damper

6. Fuel injectors

7. Cold Start injector

8. Fuel pressure regulator

9. Fuel return pipe

In this article design of few of these components will be discussed.

### Fuel Pumps

In recent cars two types of fuel pumps are used namely, externally mounted in-line pump (roller cell pump) and in-tank pumps.

In roller cell pump, the fuel enters the pump and being compressed by rotating cells which force it through the pump at a high pressure. The pump can produce a pressure of 8 bar (120 psi) with a delivery rate of approximately 4 to 5 litres per minute. Within the pump is a pressure relief valve that lifts off its seat at 8 bar to arrest the pressure if a blockage in the filter or fuel lines or elsewhere causes it to become obstructed. The other end of the pump (output) is home to a non-return valve which, when the voltage to the pump is removed, closes the return to the tank and maintains pressure within the system. The normal operating pressure within this system is approximately 2 bar (30 psi), at which the current draw on the pump is 3 to 5 amps.

### Fuel Pressure Regulator

The fuel pressure regulator is a diaphragm operated pressure relief valve. To maintain precise fuel metering, the fuel pressure regulator maintains a constant pressure differential across the fuel injector.

The specified pressure differential is either 36 PSI or 41 PSI depending on engine application.

Below is the calculation for the appropriate fuel supply from the fuel pump and the pressure regulator. The following fuel supply calculation can be worked backward or forward.

Suppose we have a 300hp engine

Flow = Horsepower x BSFC (Brake Specific Fuel Consumption)

Assuming this is a turbocharged engine having BSFC of 0.55. Therefore,

Flow = 300 x 0.55

Flow = 165 lb/hr

= 27.5 gallons/hr

Below is the schematic diagram of fuel pressure regulator.

Mathematical model description of fuel pressure regulator

Motion equation: Forces acting on the diaphragm during a transient process are mainly due to the pressure difference across the diaphragm, the spring and its preload force and the inertia forces required to accelerate the diaphragm, Damping forces as compared to the spring and pressure forces. The motion equation for the diaphragm is given by:

### Flow equation

Fuel delivery to the regulator:

The fuel flow delivered from the fuel pump to the regulator is given by:

By pass flow:

The excess flow returned to the tank is given by:

Where, Y is the diaphragm displacement.

Air flow to air chamber:

The term sgn (P3-P4) is used to indicate the air flow in both directions during the transient process.

### Continuity Equations:

Pressure P of a fuel rail volume is a function of the volume included between the pump outlet and regulator inlet (fuel rail volume), the effective bulk modulus of elasticity of the fuel (β) and the net influx of fuel to the volume:

Diaphragm Fuel Chamber Volume:

Variation of fuel pressure in bottom side of diaphragm, has an important role in regular operation. This variation is given by

where, Ad is the diaphragm effective area.

Diaphragm air chamber volume:

The pressure in the upper volume of diaphragm is described by:

### Pulsation Damper

Although fuel pressure is maintained at a constant value by the pressure regulator, the pulsing of the injectors causes minor fluctuations in rail pressure. The pulsation damper acts as an accumulator to smooth out these pulsations, ensuring accurate fuel metering. It is a vessel with gas inside, normally Nitrogen.

### Flotronic Pulsation Dampers

When a pulsation damper has been installed the volume supplied by the pump during a complete rotation or work cycle is divided in two parts; one is going to the circuit needs and the other part goes into the pulsation dampener. This volume stored into the dampener is returned immediately to the circuit while the pump is in its suction cycle.

To the amount of liquid going into and out of the dampener in each cycle or pumprevolution we will call “dv”.

When “dv” is introduced into the dampener the gas filled inside will reduce its volume and increase its pressure, the final gas volume plus the volume of liquid introduced will be equal to the initial gas volume.
The initial gas volume is the total dampener volume or the dampener size. The dampener size is the unknown value to calculate and that will depend in all cases on the pump performances. To the dampener size we will call “Vo”

We can establishthat: V2 + dv= Vo, (V2 is the final gas volume)

The initial gas volume is the total dampener volume or the dampener size. The dampener size is the unknown value to calculate and that will depend in all cases on the pump performances. To the dampener size we will call“Vo”

We can establishthat: V2 + dv= Vo, (V2 is the final gas volume)

Each dampener has a constant which depends on the charging gas value and its size; Po x Vo = constant.

When the dampeners are working is not convenient that all the liquid stored goes out in each cycle keeping the dampener empty of liquid, this will damage prematurelythe bladder or the membrane when the insert fixed on itis hammered against the dampener internal bottom.

We will have a newformula: V2 + dv + v =Vo

Where,“v” is a non used volume of liquid inside the dampener; as a norm this volume is the 10% of the total dampener volume the former formula will change to:

V2 + dv + 0,1Vo = Vo;and from this Vo = (V2 + dv) / 0.9

The following graph and the representing the three states of gas volume inside the dampeners willmake clearer everything exposed above.

At charging gas value “Po” there is no liquid inside the dampener. The curve cuts the ordinate axis in the point where corresponds a zero value in the abscissa axis.

The pressure “P1” is the gas pressure when the volume “v” has been introduced into the dampener; the pressure “P2” is the value reached by the gas when the additional volume “dv” is into the dampener.

From this curve we can see that for a fixed dampener size if the value “dv” increases then the pressure value “P2” will also increase or if we increase the dampener size keeping constant the value “dv” the final pressure gas value “P2” will be lower.

### Pressures changes in pulsation dampers

The data needed to calculate the dampener size are :

“dv” = volume of liquid that the dampener must store ( in the different types of pumps described below we will see the relation between “dv” and the capacity per revolution of each type of more used pumps )
“P1” y “P2” are the mini. and maxi. pressure values that are accepted in the circuit.
Lets see an example: If the theoretical or work pressure is “Pt” and the residual pulsation admitted is

+, -5% of this pressure, the P1 y P2 values will be:

P1 = Pt - (5/100) x Pt, andP2 = Pt + (5/100) x Pt

Note: The “Pt” pressure is that created at the outlet port of the pump

Knowing all this data dv ,P1 and P2 we can already calculate the dampener size “Vo”

Po x Vo = P1 x V1= P2 x V2= Constant. (1)

If V1 = Vo - v, and v = 0.1 x Vo

We have, V1 = 0.9 x Vo (2)

And also, V2 =V1- dv (3)

From (1) and (2) we obtain

Po=0.9 x P1 (4)

Finallyfrom; (1) (2)(3)and(4) we will obtain

Po x Vo=P2 x V2; 0.9P1 x Vo=P2 x (V1-dv) =P2 (0.9Vo-dv)

from the underlined equalities we have the final formula

P2 x dv
Vo= ------------------
0.9 (P2-P1)

This is the theoretical simplified formula to calculate the dampener volume in function of; dv, P1 and P2.

As we have already said, that the charging gas “Po” = 0.9P1, this relation between Po and P1 has been taken to avoid the complete liquid empty from the dampener in each work cycle. Having this extra quantity of liquid “v” into the dampener it will compensate the possible variations of gas pressures due to external temperatures variations, and consequently the theoretical “dv” calculated could not be introduced or stored into the dampener.

The former formula (1) Po x Vo=P1 x V1= constant does not comply in the practice ,because when a volume of gas is compressed (in a short time) the temperature rises making an extra increase of the pressure and when the gas expands its pressure drops an extra value because the temperature is reduced-effect refrigerator-
This effect is produced in all the majority of gases, Nitrogen and air included which are the more common used for charging the dampeners ( the atmospheric air can be used for pressures less than 10 Bar ,and always when there is not any chemical reaction between the oxygen in the air and the liquid pumped )

The formula (1) will be transformed;

g g
Po x Vo=P1 x V1= constant

g= specific heat relation of the gas at constant pressure and volume . For the majority of gases, g = 1.41 .This constant is also theoretical. In the practice the value that can be taken isg= 1.25

But in order not to complicate the calculation formula of dampener size we will use a new constant (0.8) that will give the same result.

P2 x dv

Vo= ------------------------
(0.8) x 0.9x (P2-P1)

This formula can be used in practice in all applications needed in the industry. The volume calculated with this formula many times will not be those of one standard manufacturer dampener; except in very exigent applications we can recommend to use the manufacturer standard lower volume, for cost reasons obviously

Note: we have not considered a possible temperature variation of the fluid or environment. This will change the charging gas value at 20º (take note that for a 10ºC of temperature variation the gas pressure will change approx. a 3%).

### Fuel Injectors

Fuel injectors are the valves that regulate the amount of fuel that enters the engine of a vehicle. The amount of time that the fuel injector stays open is referred to as the injector pulse width (IPW). Fuel injector duty cycle is a term used to describe the length of time each individual fuel injector remains open relative to the amount of time that it is closed. For example if, during each of the fuel injector's pulses, the injector is open for 75 milliseconds and closed for 25 milliseconds, the injector duty cycle (IDC) would be 75%. This is because the injector remains open for 75% percent of the time that it takes to complete one pulse. Knowing a fuel injector's duty cycle is important because it helps to determine if the injector is still functioning correctly and if injector is of the appropriate size.

### Selecting the proper injector size

In order to select the correct size of the injector for your application, the following formulas are to be used.

Fuel injector size = Horsepower x B.S.F.C

No.of Injectors x IDC

Assuming a 200 hp four cylinder engine, having BSFC of 0.5 lbs/hr and IDC of 80% we get,

Fuel injector size = 200 x 0.5 = 31.25 lbs/hr (per injector) = 328 cc/min4 x .8

Thus we get that for the above specifications the appropriate fuel injector size is 31.24 lbs/hr or 328 cc/min.

We can also calculate the fuel injector pulse width. It is the amount of time, measured in milliseconds (ms), a fuel injector stays open (delivers fuel) during a cylinder intake cycle. Typical injector pulse width for an idling engine at normal operating temperature is between 2.5 and 3.5 ms.

Pulse width can be calculated from the air flow as follows:

First the ECU determines the air mass flow rate from the sensors - Massair / Minute.

\frac{Mass_{air}}{Minute} * \frac{Minutes}{Revolution}* \frac{Revolutions}{Cycle}*\frac{1}{Strokes/Cycle}=\frac{Mass_{air}}{Stroke}

- Minutes / Revolution is the reciprocal of engine speed (RPM).

- A four stroke engine has Revolutions / Cycle = 2, and 1 / (Strokes / Cycle) = 1 / 4.

\frac{Mass_{air}}{Stroke_{intake}}*\frac{Mass_{Fuel}}{Mass_{Air}}=\frac{Mass_{Fuel}}{Stroke_{intake}}

- MassFuel / MassAir is the desired mixture ratio, usually stoichiometric, but often different depending on operating conditions.

\frac{Mass_{Fuel}}{Stroke_{intake}}*\frac{1}{Mass_{Fuel}/Minute} =\frac{Minutes}{Stroke_{intake}}=Pulsewidth

- 1 / (MassFuel / Minute) is the flow capacity of the injector, or its size.

\frac{Mass_{air}}{Minute} * \frac{Minutes}{Revolution}* \frac{Revolutions}{Cycle}*\frac{1}{Strokes/Cycle}*\frac{Mass_{Fuel}}{Mass_{Air}}*\frac{1}{Mass_{Fuel}/Minute} = Pulsewidth Combining the above three terms . . .

Substituting real variables for the 5.0 L engine at idle.

0.55 \frac{lb}{min} * \frac{1 min}{700 rev}* 2 \frac{rev}{cycle}*\frac{1 cycle}{4 strokes}*\frac{1}{14.64}*2.5\frac{min}{lb} = 6.7*10^{-5} min = 4ms

Substituting real variables for the 5.0L engine at maximum power.

28 \frac{lb}{min} * \frac{1 min}{5500 rev}* 2 \frac{rev}{cycle}*\frac{1 cycle}{4 strokes}*\frac{1}{11.00}*2.5\frac{min}{lb} = 57.9*10^{-5} min = 35ms

Injector pulsewidth typically ranges from 4 ms/engine-cycle at idle, to 35 ms per engine-cycle at wide-open throttle. The pulsewidth accuracy is approximately 0.01 ms.
Calculate fuel-flow rate from pulsewidth

* (Fuel flow rate) ≈ (pulsewidth) × (engine speed) × (number of fuel injectors)

Looking at it another way:

* (Fuel flow rate) ≈ (throttle position) × (rpm) × (cylinders)

Looking at it another way:

* (Fuel flow rate) ≈ (air-charge) × (fuel/air) × (rpm) × (cylinders)

Substituting real variables for the 5.0L engine at idle.

* (Fuel flow rate) = (2.0 ms/intake-stroke) × (hour/3,600,000 ms) × (24 lb-fuel/hour) × (4-intake-stroke/rev) × (700 rev/min) × (60 min/h) = (2.24 lb/h)

Substituting real variables for the 5.0L engine at maximum power.

* (Fuel flow rate) = (17.3 ms/intake-stroke) × (hour/3,600,000-ms) × (24 lb-fuel/hour) × (4-intake-stroke/rev) × (5500-rev/min) × (60-min/hour)

= (152 lb/h)

The fuel consumption rate is 68 times greater at maximum engine output than at idle.

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