(a) Explanation Write the expression for the combustion reaction of C 3 H 8 with theoretical air. C 3 H 8 +5 ( O 2 +3 .76N 2 ) → 3CO 2 +4H 2 O+18 .8N 2 Write the expression for the combustion reaction of C 3 H 8 with 200 % of theoretical air. C 3 H 8 +2×5 ( O 2 +3 .76N 2 ) → 3CO 2 +4H 2 O+5O 2 +2×18 .8N 2 C 3 H 8 +10 ( O 2 +3 .76N 2 ) → 3CO 2 +4H 2 O+5O 2 +37 .6N 2 (I) Write the expression for the molar flow rate of fuel. n ˙ f u e l = p v R ¯ T o (III) Here, the pressure is p ,the volume is v ,the reference temperature is T o and the universal gas constant is R ¯ . Write the expression for the energy rate balance for controlled volume. 0 = Q ˙ c v − W ˙ c v + n ˙ f u e l [ ( h ¯ C 3 H 8 + 10 h ¯ O 2 + 37.6 h ¯ N 2 ) − ( 3 h ¯ C O 2 + 4 h ¯ H 2 O + 5 h ¯ O 2 + 37.6 h ¯ N 2 ) ] (IV) Here, the heat transfer is Q ˙ c v , the rate of work is W ˙ c v and the enthalpy per unit mass is h ¯ . Since the state is steady. Substitute 0 for W ˙ c v in Equation (IV). 0 = Q ˙ c v − 0 + n ˙ f u e l [ ( h ¯ C 3 H 8 + 10 h ¯ O 2 + 37.6 h ¯ N 2 ) − ( 3 h ¯ C O 2 + 4 h ¯ H 2 O + 5 h ¯ O 2 + 37.6 h ¯ N 2 ) ] Q ˙ c v = n ˙ f u e l [ ( 3 h ¯ C O 2 + 4 h ¯ H 2 O + 5 h ¯ O 2 + 37.6 h ¯ N 2 ) − ( h ¯ C 3 H 8 + 10 h ¯ O 2 + 37.6 h ¯ N 2 ) ] (V) Write the expression for the enthalpy of carbon dioxide. h ¯ C O 2 = h ¯ f C O 2 o + ( h ¯ 500 − h ¯ 298 ) (VI) Here, The enthalpy of formation of carbon dioxide is h ¯ f C O 2 o ,the enthalpy at 500 K is h ¯ 500 and the enthalpy at 298 K is h ¯ 298 . Write the expression for the enthalpy of oxygen. h ¯ O 2 = h ¯ f O 2 o + ( h ¯ 500 − h ¯ 298 ) (VII) Here, The enthalpy of formation of oxygen is h ¯ f O 2 o ,the enthalpy at 500 K is h ¯ 500 and the enthalpy at 298 K is h ¯ 298 . Write the expression for the enthalpy of water. h ¯ H 2 O = h ¯ f H 2 O o + ( h ¯ 500 − h ¯ 298 ) (VIII) Here, The enthalpy of formation of water is h ¯ f H 2 O o ,the enthalpy at 500 K is h ¯ 500 and the enthalpy at 298 K is h ¯ 298 . Write the expression for the enthalpy of nitrogen. h ¯ N 2 = h ¯ f N 2 o + ( h ¯ 500 − h ¯ 298 ) (IX) Here, The enthalpy of formation of nitrogen is h ¯ f N 2 o ,the enthalpy at 500 K is h ¯ 500 and the enthalpy at 298 K is h ¯ 298 . Conclusion: Substitute 101.3 N / m 2 for p , 0.03 m 3 / min for v 25 ° C for T o and 8.314 kJ / kmol ⋅ K for R ¯ in Equation (III). n ˙ f u e l = ( 101.3 N / m 2 ) × ( 0.03 m 3 / min ) ( 8.314 kJ / kmol ⋅ K ) × ( 25 ° C ) = ( 101.3 N / m 2 ) × ( 0.03 m 3 / min ) ( 8.314 kJ / kmol ⋅ K ) × ( 25 ° C+273 ) K = ( 101.3 N / m 2 ) × ( 0.03 m 3 / min ) ( 8.314 kJ / kmol ⋅ K ) × ( 298 K ) = 0.00126604 kmol / min Refer Table A-23, “Ideal gas properties of selected gases” to obtain the values of enthalpy h ¯ f C O 2 o as − 393520 kJ / kmol , h ¯ 500 o as 17678 kJ / kmol and h ¯ 298 o as 9364 kJ / kmol with respect to temperature 500 K and 298 K Substitute − 393520 kJ / kmol for h ¯ f C O 2 o , 17678 kJ / kmol for h ¯ 500 o and 9364 kJ / kmol for h ¯ 298 o in Equation (VI) (b) To determine The rate of entropy production. (c) To determine The exergitic efficiency.

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 13.8, Problem 106P

(a)

To determine

The rate of heat transfer in kJ/min

(b)

To determine

The rate of entropy production.

(c)

To determine

The exergitic efficiency.

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Chapter 13.8, Problem 105P

Chapter 13.8, Problem 107P

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Can you solve it analytically using laplace transforms and with Matlab code as well please. Thank You

Can you solve it analytically using laplace transforms and with Matlab code as well please. Thank You.

Chapter 13 Solutions

Fundamentals of Engineering Thermodynamics, Binder Ready Version

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 rate of heat transfer in kJ / min

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The rate of heat transfer in kJ/min

The rate of entropy production.

The exergitic efficiency.

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Chapter 13 Solutions