Explanation Express the volume of carbon monoxide. v CO = R T P (I) Here, gas constant of air is R , temperature is T and pressure is P . Express mass flow rate of carbon monoxide. m ˙ CO = v ˙ CO v CO (II) Here, volume flow rate of carbon monoxide is v ˙ CO . Express the molar air fuel ratio. AF ¯ = N air N fuel = m ˙ air / M air m ˙ CO / M CO (III) Here, number of moles of air is N air , number of moles of fuel is N fuel , mass flow rate of air is m ˙ air , molar mass of air is M air , and molar mass of carbon monoxide is M CO . Express the number of moles of O 2 per mole of carbon monoxide. N O 2 = AF ¯ N dry air (IV) Here, number of moles of dry air is N dry air . Express the combustion equation. CO + N O 2 ( O 2 + 3.76 N 2 ) → CO 2 + 0.137 O 2 + 2.40 N 2 (V) Here, carbon monoxide is CO , nitrogen is N 2 , carbon dioxide is CO 2 and oxygen is O 2 . Express the combustion equation by applying the energy balance under steady flow conditions. − Q ¯ out = ∑ N P ( h ¯ f o + h ¯ − h ¯ o ) P − ∑ N R ( h ¯ f o + h ¯ − h ¯ o ) R = ∑ N P ( h ¯ f o + h ¯ 900 K − h ¯ 298 K ) P − ∑ N R ( h ¯ f o + h ¯ 310 K − h ¯ 298 K ) R (VI) Here, number of moles of products is N P , number of moles of reactants is N R , enthalpy of formation is h ¯ f . Express the rate of heat transfer from the combustion chamber. Q out = [ m ˙ CO M CO ] Q out (VII) Conclusion: Refer Table A-1, “molar mass, gas constant and critical point properties”, and write the molecular mass of methane and oxygen. M CO = 28 kg / kmol M air = 29 kg / kmol Refer Table A-1, “molar mass, gas constant and critical point properties”, and write the universal gas constant of air. R = 0.2968 kPa ⋅ m 3 / kg ⋅ K Since each kmol of O 2 in air is accompanied by 3 .76 kmol of N 2 , thus N air = 1 kmol O 2 + 3.76 kmol N 2 = 4.76 kmol air Substitute 0.2968 kPa ⋅ m 3 / kg ⋅ K for R , 37°C for T and 110 kPa for P in equation (I). v CO = ( 0.2968 kPa ⋅ m 3 / kg ⋅ K ) ( 37°C ) 110 kPa = ( 0.2968 kPa ⋅ m 3 / kg ⋅ K ) [ ( 37 + 273 ) K ] 110 kPa = ( 0.2968 kPa ⋅ m 3 / kg ⋅ K ) ( 310 K ) 110 kPa = 0.836 m 3 / kg Substitute 0 .4 m 3 / min for v ˙ CO and 0.836 m 3 / kg for v CO in Equation (II)

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ISBN: 9781260917055

Author: CENGEL

Publisher: MCG

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Chapter 15.7, Problem 103RP

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The rate of heat transfer from the combustion chamber.

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Chapter 15.7, Problem 102RP

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Chapter 15 Solutions

CENGEL'S 9TH EDITION OF THERMODYNAMICS:

Ch. 15.7 - What are the approximate chemical compositions of...Ch. 15.7 - How does the presence of N2 in air affect the...Ch. 15.7 - Prob. 3P Ch. 15.7 - Prob. 4P Ch. 15.7 - Is the airfuel ratio expressed on a mole basis...Ch. 15.7 - How does the presence of moisture in air affect...Ch. 15.7 - Prob. 7P Ch. 15.7 - Prob. 8P Ch. 15.7 - Prob. 9P Ch. 15.7 - Are complete combustion and theoretical combustion...

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The rate of heat transfer from the combustion chamber.

Solution Summary: The author explains the rate of heat transfer from the combustion chamber and the molar air fuel ratio.

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