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What is the air to fuel ratio AF (mass basis) of the complete combustion of propane? C3H8 + 5 (02 + 3.76 N2) -> 3 CO2 + 4 H20 + 18.8 N2 Take: M(C3Ha) = 44.1 g/mol M(air) = 28.91 g/mol a. 7.63 kg(air)/kg(fuel) Ob. 36.31 kg(air)/kg(fuel) 3.28 kg(air)/kg(fuel) C. d. 15.6 kg(air)/kg(fuel)

What is the air to fuel ratio AF (mass basis) of the complete combustion of propane? C3H8 + 5 (02 + 3.76 N2) -> 3 CO2 + 4 H20 + 18.8 N2 Take: M(C3Ha) = 44.1 g/mol M(air) = 28.91 g/mol a. 7.63 kg(air)/kg(fuel) Ob. 36.31 kg(air)/kg(fuel) 3.28 kg(air)/kg(fuel) C. d. 15.6 kg(air)/kg(fuel)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Question 8 What is the air to fuel ratio AF (mass basis) of the complete combustion of propane? C3Hg + 5 (02 + 3.76 N2) -> 3 CO2 + 4 H20 + 18.8 N2 Take: M(C3Hg) = 44.1 g/mol M(air) = 28.91 g/mol %3D 7.63 kg(air)/kg(fuel) a. b. 36.31 kg(air)/kg(fuel) 3.28 kg(air)/kg(fuel) C. 15.6 kg(air)/kg(fuel) d. Moving to the next question prevents changes to this answer.
Transcribed Image Text:Question 8 What is the air to fuel ratio AF (mass basis) of the complete combustion of propane? C3Hg + 5 (02 + 3.76 N2) -> 3 CO2 + 4 H20 + 18.8 N2 Take: M(C3Hg) = 44.1 g/mol M(air) = 28.91 g/mol %3D 7.63 kg(air)/kg(fuel) a. b. 36.31 kg(air)/kg(fuel) 3.28 kg(air)/kg(fuel) C. 15.6 kg(air)/kg(fuel) d. Moving to the next question prevents changes to this answer.
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