# vehicle emission assignment (calculation)

Need to provide solution for all the attached questions..

Quite urgent!!

eg:

In areas with severe CO problems during the winter months, all the gasoline sold must contain at least 2.7 percent by weight oxygen. The gasoline companies are meeting this requirement by

blending into their ordinary summer gasoline enough methanol (CH3OH), ethanol (C2H5OH), or

methyl tert-butyl ether, MTBE (CH3OC4H9), to meet the required 2.7 percent by weight oxygen.

(i) If we assume that their ordinary summer gasoline is the equivalent of C8H17, and that

they plan to meet this oxygen requirement by blending enough Ethanol with their

ordinary summer gasoline, what weight percent Ethanol must there be in the final

blend?

(ii) What is the stoichiometric A/F ratio for this mixture?

C_8 H_17 + C_2 H_5OH(ETHANOL)

Now, mass of the C_8 H_17 in the mixture:

(8x12)+(17x1) = 113 g

And mass of C_8 H_5OH :

x{(2X12)+(5X1)+(1X16)+(1X1)}

=46x g

Mass of the oxygen in the mixture :

16x g

From question the mixture should contain 2.7% oxygen .

Hence,

16x/(113+46x) = 2.7/100

x=0.206

The mass of gasoline in the mixture :113g

The mass of ethanol in the mixture :

46X0.206=9.5g

Hence, the % of Ethanol in the mixture :

9.5/(9.5+113) X 100 = 7.76%

To find stoichiometric A/F ratio for this mixture:

The complete combustion equation:

C_8 H_17 + 0.206 C_2 H_5OH + z(o_2+3.76N_2 ) = a ?Co?_2 + bH_2o + 3.76N_2

To balance the C:

8 + (0.206x2)= a

a = 8.412

balancing for H:

17+(0.206x6)=b

b = 9.11

balancing for O:

0.206 + 2z = 2a+b

z = 12.867

hence the air fuel ratio:

{12.867 X (32+3.76X28)}/ 113+(0.206X46)

1765.42/122.47 = 14.42 kg/kg

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