Calculation of Air-fuel ratio in a carburetor

Fig. 1 Simple Carburetor

Applying the steady flow energy equation to sections A-A and B-B per unit mass flow of air:

.(1)

Here, q and w are the heat and work transfers from the entrance to the throat and h and C stand for enthalpy and velocity respectively.

If we assume reversible adiabatic conditions, and there is no work transfer, q=0, w=0, and if approach velocity C1≈0 we get

(2)

Assuming air to be a perfect gas, we get

then

(3)

If we assume that the distance from the inlet to the venture throat is short, we can consider it to be isentropic in the ideal case,

(4)

(5)

Substituting for T1 – T2 from Eq. 5 in Eq. 3, we get

By the continuity equation we can write down the theoretical mass flow rate of air

where A1 and A2 are the cross-sectional areas at the air inlet (point 1) and venturi throat (point 2).

To calculate the mass flow rate of air at the throat, we have assumed the flow to be isentropic till the throat so the equation relating p and v (or ρ) can be used.

1

1

212

p

p

For a perfect gas we have

Thus

and rearranging the above equation we have

Since the fluid flowing in the intake is air, we can put in the approximate values of R = 287 J/kgK, cp = 1005 J/kgK at 300K, and γ = 1.4 at 300K.

where

Here, pressure p is in N/m2, area A is in m2, and temperature T is in K. If we take the ambient temperature T1 = 300K and ambient pressure p1 = 105 N/m2, then

Equation 11 gives the theoretical mass flow rate of air. The actual mass flow rate,

, can be obtained by multiplying the equation by the coefficient of discharge for the

venturi, Cd,a. Thus

where

The coefficient of discharge and area are both constant for a given venturi, thus

Since we have to determine the air-fuel ratio, we have to now calculate the fuel flow rate. Since the fuel is a liquid before mixing with the air, it can be taken to be incompressible. We can apply Bernoulli’s equation between the atmospheric conditions prevailing at the top of the fuel surface in the float bowl, which corresponds to point 1 and the point where the fuel will flow out, at the venturi, which corresponds to point 2. Fuel flow will take place because of the drop in pressure at point 1 due to the venturi effect. Thus

where ρf is the density of the fuel in kg/m3, Cf is the velocity of the fuel at the exit of the fuel nozzle (fuel jet), and z is the depth of the jet exit below the level of fuel in the float

bowl. This quantity must always be above zero otherwise fuel will flow out of the jet at all times. The value of z is usually of the order of 10 mm.

From Eq. 16 we can obtain an expression for the fuel velocity at the jet exit as

Applying the continuity equation for the fuel, we can obtain the theoretical mass

flow rate, , from

where Af is the exit area of the fuel jet in m2. If Cd,f is the coefficient of discharge of the fuel nozzle (jet) given by

then

Since

If we put , we get the following equation for the air-fuel ratio

where

For the normal carburetor operating range, where , the effects of

compressibility which reduce Φ below 1.0 are small.

The equivalence ratio, φ, where

is given by

In Eq. 22, if we take T1 = 300K and p1 = 105 N/m2 then

The coefficient of discharge defined in Eq 19 represents the effect of all deviations from the ideal one-dimensional isentropic flow. It is influenced by many factors of which the most important are:

1. Fluid mass flow rate,2. Orifice length-to-diameter ratio,3. Orifice area-to-approach area ratio,4. Orifice surface area,5. Orifice surface roughness,6. Orifice inlet and exit chamfers,7. Fluid specific gravity,8. Fluid viscosity, and9. Fluid surface tension.

The use of the orifice Reynolds number

as a correlating parameter for the coefficient of discharge accounts for the effects of mass flow rate, fluid density and viscosity, and length scale to a good approximation. The discharge coefficient of a typical carburetor main fuel-metering system orifice increases smoothly with increasing orifice Reynolds number, Reo.

Air-fuel ratio neglecting compressibility of air

If we assume air to be incompressible, then we can apply Bernoulli’s equation to air flow also. Since initial velocity is assumed zero, we have

Thus

Applying the continuity equation for the fuel, we can obtain the theoretical mass

flow rate, , from

where A2 is the venturi in m2. If Cd,a is the coefficient of discharge of the venturi given by

then

Since

If we assume z = 0, then

Carburetor Performance

In Eq. 26, the terms A1, A2, ρa, and ρf are all constant for a given carburetor, fuel, and ambient conditions. Also, for very low flows, Δpa » ρfgz. However, the discharge coefficients Cd,a and Cd,f and Φ, all vary with flow rate. Hence, the equivalence ratio delivered by an elementary carburetor is not constant.

Figure 2 shows the performance of an elementary carburetor. The top graph shows the variation of Cd,a and Cd,f and Φ with the venturi pressure drop. For Δpa ≤ ρfgz, there is no fuel flow. Once fuel starts to flow, the fuel flow rate increases more rapidly than the air flow rate. The carburetor delivers a mixture of increasing equivalence ratio as the flow rate increases.

Fig. 2 Performance of Elementary Carburetor

Suppose the venturi and fuel orifice (jet) are sized to give a stoichiometric mixture at an air flow rate corresponding to 1 kN/m2 venturi pressure drop (middle graph of Fig 2). At higher flow rates, the carburetor will deliver a fuel-rich mixture. At very high flow rates the carburetor will deliver an essentially constant equivalence ratio. At lower air flow rates, the mixture delivered leans out rapidly.

Thus, the elementary carburetor cannot provide the variation in mixture ratio which the engine requires over the complete load range at any given speed.

Summary of the Deficiencies of the Elementary Carburetor

1. At low loads, the mixture becomes leaner; the engine requires the mixture to be enriched at low loads. The mixture is richest at idle.

2. At intermediate loads, the equivalence ratio increases slightly as the air flow rate increases; the engine requires an almost constant equivalence ratio.

3. As the air flow approaches the maximum (WOT) value, the equivalence ratio remains essentially constant; the engine requires an equivalence ratio of about 1.1 at maximum engine power.

4. The elementary carburetor cannot compensate for transient phenomena in the intake manifold. It also cannot provide a rich mixture during engine starting and warm-up.

5. It cannot adjust to changes in ambient air density due to changes in altitude.

Modern Carburetor Design

The changes required in the elementary carburetor so that it provides the equivalence ratio required at various air flow rates are as follows.

1. The main metering system must be compensated to provide a constant lean or stoichiometric mixture over 20 to 80% of the air flow range.

2. An idle system must be added to meter the fuel flow at idle and light loads to provide a rich mixture.

3. An enrichment system must be provided so that the engine can get a rich mixture as WOT conditions is approached and maximum power can be obtained.

4. An accelerator pump must be provided so that additional fuel can be introduced into the engine only when the throttle is suddenly opened.

5. A choke must be added to enrich the mixture during cold starting and warm-up to ensure that a combustible mixture is provided to each cylinder at the time of ignition.

6. Altitude compensation is necessary to adjust the fuel flow which makes the mixture rich when air density is lowered.

7. Increase in the magnitude of the pressure drop available for controlling the fuel flow is provided by introducing boost venturis (Venturis in series) or Multiple-barrel carburetors (Venturis in parallel).