Note that equivalence ratio is a normalized quantity

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: stoichiometric condition. f <1 is a lean mixture, f ¼ 1 is a stoichiometric mixture, and f >1 is a rich mixture. Similar to f, the range of f is bounded by zero and 1 corresponding to the limits of pure air and fuel respectively. Note that equivalence ratio is a normalized quantity that provides the information regarding the content of the combustion mixture. An alternative 20 2 Thermodynamics of Combustion variable based on AFR is frequently used by combustion engineers and is called lambda (l). Lambda is the ratio of the actual air-fuel ratio to the stoichiometric air-fuel ratio defined as AFR 1=f 1 1 ¼ ¼ ¼ AFRs 1=fs f =fs f l¼ (2.16) Lambda of stoichiometric mixtures is 1.0. For rich mixtures, lambda is less than 1.0; for lean mixtures, lambda is greater than 1.0. Percent Excess Air: The amount of air in excess of the stoichiometric amount is called excess air. The percent excess air, %EA, is defined as   ma À mas ma %EA ¼ 100 ¼ 100 À1 mas mas (2.17) For example, a mixture with %EA ¼ 50 contains 150% of the theoretical (stoichiometric) amount of air. Converting between quantification methods: Given one of the three variables (f, f, and %EA), the other two can be deduced as summarized in Table 2.1 with their graphic relations. In general, the products of combustion include many different % of excess air Equivalence ratio, φ 10 8 6 4 2 0 −100 0 100 200 300 400 % of excess air 400 300 200 100 0 −100 % of excess air 0.5 f = 0.05 0.4 s 0.3 0.2 0.1 0 0 200 400 % of excess air 10 fs = 0.05 8 6 4 2 0 0 0.1 0.2 0.3 0.4 0.5 Fuel air ratio (mass) Equivalence ratio, φ 0.5 fs = 0.05 0.4 0.3 0.2 0.1 0 0 2 4 6 8 10 Equivalence ratio, φ fuel air ratio (mass) fuel air ratio (mass) Table 2.1 Relations among the three variables for describing reacting mixtures f (fuel air ratio by mass) f (equivalence ratio) %EA (% of excess air) f 1Àf f ¼ fs Á f f¼ %EA ¼ 100 fs f 100 Á fs f¼ 100 1 À f = fs %EA þ 100 f¼ %EA ¼ 100 %EA þ 100 f = fs 0.5 1.0 1.5 2.0 Equivalence ratio, φ...
View Full Document

Ask a homework question - tutors are online