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calculation of air fuel ratio in a carburetor

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					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:

                                  qw        h2  h1   1 C22  C12                .(1)
                                                            2
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

                                      C2          2h1  h2                            (2)

       Assuming air to be a perfect gas, we get

                                              h  c p T then

                                         C2  2c p T1  T2 
                                                                                         (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,

                                                          1
                                            T2  p 2                                   (4)
                                  then         
                                            T1  p1 
                                                
                                                               1
                                                                  
                                                         p2   
                                                   1   
                                  T1  T2      T1
                                                     p1                              (5)
                                                      
                                                                  


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

                                                    1
                                                       
                                       1   p 2  
                                                     
                         C 2  2c p T1                           ( 6)
                                         p1  
                                          
                                                       

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

                              .
                             m a  1 A1C1   2 A2 C 2           (7 )

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.

                                                  
                                      p1v1  p 2 v 2        (8 A)

                                       p1        p2
                                         
                                                 
                                                           (8 B )
                                       1        2

                                                       1
                                                   p 
                                          2  1  2 
                                                  p 
                                                   1

                                                          1
                                      1
                                                             
                      .         p2               p2   
                                              1   
                     m a  1   A2 2c p T1
                               p                                   (9)
                                1             p1  
                                                 
                                                             

       For a perfect gas we have

                                                  p1
                                        1                (9 A)
                                                 RT1

       Thus

                                                     1
                             1
                                                        
                 .    p 2   p1              p2   
                                         1   
                 
                ma   
                      p  RT A2 2c p T1   p                      (10 A)
                      1        1
                                           1 
                                                        

and rearranging the above equation we have

                                                           1
                                                 2
                                                               
                          .   A2 p1         p2    p2   
                                              
                      
                     ma              2c p        p            (10 B)
                              R T1         p1     1 
                                          
                                                              

       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.
                                                                          1.43                  1.71
                          .         A p p                                        p 
                          
                         ma  0.1562 2 1  2                                      2
                                      T1  p1 
                                          
                                                                                   p 
                                                                                    1

                                              A2 p1
                                   0.1562                                               (11)
                                                T1

where

                                                              1.43                1.71
                                        p                            p 
                                       2
                                        p                            2
                                                                       p 
                                         1                            1

       Here, pressure p is in N/m2 , area A is in m2, and temperature T is in K. If we take
the ambient temperature T 1 = 300K and ambient pressure p1 = 105 N/m2 , then

                                      .
                                     ma  901 .8 A2
                                                                                (12 )

        Equation 11 gives the theoretical mass flow rate of air. The actual mass flow rate,
 .
m a , can be obtained by multiplying the equation by the coefficient of discharge for the
venturi, Cd,a. Thus

                              .                                A2 p1
                           ma  0.1562 C d ,a                                           (13)
                                                                     T1

where

                                                          .
                                                     ma
                                          C d ,a         .
                                                                              (14)
                                                      
                                                     ma

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

                                          .          p1
                                      ma                                    (15)
                                                     T1

        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

                                   p1       p2       C2
                                                               gz
                                                      f
                                                                               (16)
                                  f        f        2

where ρf is the density of the fuel in kg/m3, C f 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

                                        p  p2      
                                C f  2 1       gz                           (17)
                                        f
                                                    
                                                     

        Applying the continuity equation for the fuel, we can obtain the theoretical mass
            .
flow rate, mf , from

                            .
                          mf   f A f C f


                                 A f 2  f  p1  p 2   f gz                      (18)

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

                                                       .
                                                     mf
                                        Cd , f        .
                                                                       (19 )
                                                     m f

then


                        m f  Cd , f A f 2 f  p1  p2   f gz 
                                                            .
                                                                                        (20)

        Since

                                                                .
                                      Air  A m
                                            .a                          (21)
                                     Fuel F m
                                               f
                A          C d ,a A2                   p1
                   0.1562
                                          2  f T1  p1  p 2   f gz 
                                                                                    (22)
                F          Cd , f A f

        If we put p a  p1  p 2 , we get the following equation for the air-fuel ratio

                      A C d ,a A2        a        p a
                       
                      F Cd , f A f       f    pa   f gz              (23)


where

                                                                1
                                                     1 2
                                          2
                                                        
                                    p2    p2   
                                       
                             p1 
                                     
                                              p 
                                               1      
                       
                                 
                                                                           (24)
                            1            p2      
                                        1  
                                             p        
                                             1       
                                                       

                                                                           p a
        For the normal carburetor operating range, where                         0.1 , the effects of
                                                                            p1
compressibility which reduce Φ below 1.0 are small.

        The equivalence ratio, φ, where

                                         A
                                         
                                         F s
                                                           (25 )
                                           A
                                          F

is given by

                         A
                                                                   1

                         F  s C d ,a A2      a       gz  2
                                                  1  f 
                                                                            (26)
                           Cd , f A f         f       p a 
                                                              

        In Eq. 22, if we take T 1 = 300K and p1 = 105 N/m2 then

                  A         C d ,a A2                  
                     901.8
                                            2  f  p1  p 2   f gz 
                                                                               (27)
                  F         Cd , f A f
        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, and
    9. Fluid surface tension.

The use of the orifice Reynolds number

                                             VDo
                                   Re o                    (28 )
                                              

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

                                                       2
                                   p1       p2       C2
                                                           (29)
                                   a       a        2

       Thus

                                         p  p2 
                                 C 2  2 1                   (30 )
                                         a 

       Applying the continuity equation for the fuel, we can obtain the theoretical mass
            .
            
flow rate, ma , from

                             .
                            
                           ma   a A2 C 2


                                   A2 2  a  p1  p 2            (31)
where A2 is the venturi in m2. If Cd,a is the coefficient of discharge of the venturi given by

                                                 .
                                                ma
                                     C d ,a     .
                                                              (32)
                                                 
                                                ma

then

                                                     .

                          m a  C d ,a A2 2  a  p1  p 2 
                           .
                                                                            (33)

Since
                                                         .
                                  Air  A m
                                        .a                       (34 )
                                 Fuel F m
                                           f



                      A C d ,a A2              a  p1  p 2 
                       
                                         f  p1  p 2   f gz 
                                                                                   (35)
                      F Cd , f A f



                    A C d ,a A2       a              p1  p2 
                     
                    F Cd , f A f      f    p  1     p 2   f gz 
                                                                                   (35 A)


        If we assume z = 0, then

                               A C d ,a A2               a
                                                                    (36)
                               F Cd , f A f              f

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).

				
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