Explanation Write the expression for balanced reaction equation for the complete combustion with the theoretical amount of air. C 8 H 18 + 12.5 ( O 2 + 3.76 N 2 ) → 8 CO 2 + 9 H 2 O + 47 N 2 Here, oxygen is O 2 , nitrogen is N 2 , carbon dioxide is CO 2 , carbon monoxide is CO and water vapor is H 2 O . Write the expression for balanced reaction equation for the combustion with the 90 % of theoretical amount of air. C 8 H 18 + 11.25 O 2 + 42.3 N 2 → [ 5.5 CO 2 + 2.5 CO + 9 H 2 O + 42.3 N 2 ] (I) Write the expression for energy rate balance equation. [ 5.5 ( h ¯ f 0 + Δ h ¯ ) CO 2 + 2.5 ( h ¯ f 0 + Δ h ¯ ) CO + 9 ( h ¯ f 0 + Δ h ¯ ) CO + 42.3 ( h ¯ f 0 + Δ h ¯ ) N 2 ] = ( h ¯ f 0 ) C 8 H 18 [ 5.5 [ Δ h ¯ ] CO 2 + 2.5 [ Δ h ¯ ] CO + 9 [ Δ h ¯ ] H 2 O + 42.3 [ Δ h ¯ ] N 2 ] = [ [ h ¯ f 0 ] C 8 H 18 − 5.5 [ h ¯ f 0 ] CO 2 − 2.5 [ h ¯ f 0 ] CO − 9 [ h ¯ f 0 ] H 2 O ] (II) Here, the specific enthalpy formation for the C 8 H 18 is ( h ¯ f 0 ) C 8 H 18 , the specific enthalpy formation for the nitrogen is ( h ¯ f 0 ) N 2 , the specific enthalpy formation for the carbon-dioxide is ( h ¯ f 0 ) CO 2 , the specific enthalpy formation for the carbon mono-oxide is ( h ¯ f 0 ) CO and the specific enthalpy formation for the water is ( h ¯ f 0 ) H 2 O . Write the expression for the change in enthalpy. Δ h ¯ = h ( T P ) − h ( T ) (III) Here, the enthalpy is h , final temperature is T P and initial temperature is T . Write the expression for the adiabatic flame temperature using the trail and error method. z n = 5.5 h ( T P ) CO 2 + 2.5 h ( T P ) CO + 9 h ( T P ) H 2 O + 42.3 h ( T P ) N 2 (IV) Here, energy of the adiabatic flame in kJ / kmol ⋅ K at temperature T n is z n . Write the expression for adiabatic flame temperature using interpolation method. T 1 − T P T P − T 2 = z 1 − z z − z 2 (V) Here, first assumed temperature for interpolation is T 1 and energy of the adiabatic flame in kJ / kmol ⋅ K is z 1 , second assumed temperature for interpolation is T 2 and energy of the adiabatic flame in kJ / kmol ⋅ K is z 2 . Conclusion: Refer to Table A-23, “Ideal gas properties of selected gases” to obtain the value of ( h ¯ f 0 ) CO 2 as − 393520 kJ / kmol , ( h ¯ f 0 ) H 2 O as − 241820 kJ / kmol , ( h ¯ f 0 ) CO as − 110530 kJ / kmol and ( h ¯ f 0 ) C 8 H 18 as − 249910 kJ / kmol . Substitute − 393520 kJ / kmol for ( h ¯ f 0 ) CO 2 , − 241820 kJ / kmol for ( h ¯ f 0 ) H 2 O , − 110530 kJ / kmol for ( h ¯ f 0 ) CO , and − 249910 kJ / kmol for ( h ¯ f 0 ) C 8 H 18 in Equation (II). [ 5.5 [ Δ h ¯ ] CO 2 + 2.5 [ Δ h ¯ ] CO + 9 [ Δ h ¯ ] H 2 O + 42.3 [ Δ h ¯ ] N 2 ] = [ ( − 249910 kJ / kmol ) − 5.5 ( − 393520 kJ / kmol ) − 2.5 ( − 110530 kJ / kmol ) − 9 ( − 241820 kJ / kmol ) ] [ 5.5 [ Δ h ¯ ] CO 2 + 2.5 [ Δ h ¯ ] CO + 9 [ Δ h ¯ ] H 2 O + 42.3 [ Δ h ¯ ] N 2 ] = [ 4367155 kJ / kmol ] (VI) Substitute h ( T P ) − h ( T ) for Δ h ¯ in Equation (VI). [ 5.5 [ h ( T P ) − h ( T ) ] CO 2 + 2.5 [ h ( T P ) − h ( T ) ] CO + 9 [ h ( T P ) − h ( T ) ] H 2 O + 42.3 [ h ( T P ) − h ( T ) ] N 2 ] = [ 4367155 kJ / kmol ] (VII) Substitute 298 K for T in above Equation (VII). [ 5.5 [ h ( T P ) − h ( 298 K ) ] CO 2 + 2.5 [ h ( T P ) − h ( 298 K ) ] CO + 9 [ h ( T P ) − h ( 298 K ) ] H 2 O + 42

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Chapter 13.8, Problem 65P

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The temperature of the exit products.

The comparison of the temperature of the exit products to the results of the Example 13.8.

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Chapter 13.8, Problem 64P

Chapter 13.8, Problem 66P

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FUNDAMENTALS OF ENGINEERING THERMODYNAM

Ch. 13.8 - Prob. 1E Ch. 13.8 - Prob. 2E Ch. 13.8 - Prob. 3E Ch. 13.8 - Prob. 4E Ch. 13.8 - Prob. 5E Ch. 13.8 - Prob. 6E Ch. 13.8 - Prob. 7E Ch. 13.8 - Prob. 8E Ch. 13.8 - Prob. 9E Ch. 13.8 - Prob. 10E

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The temperature of the exit products. The comparison of the temperature of the exit products to the results of the Example 13.8.

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The temperature of the exit products. The comparison of the temperature of the exit products to the results of the Example 13.8.

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The temperature of the exit products.

The comparison of the temperature of the exit products to the results of the Example 13.8.