Explanation Write the formula to calculate the mass of the substance ( m ) from mole number. m = N M (I) Here, number of moles of substance is N and the molar mass of the substance is M . Write the formula to calculate the total mass of the mixture ( m m ) . m m = m 1 + m 2 + m 3 + .......... + m n (II) Here, suffices represents the number of substances. Write the expression to calculate the mass fraction ( mf ) . mf = m m m (III) Write the formula to calculate the constant pressure specific heat of mixture ( c p ) . c p = mf 1 c p , 1 + mf 2 c p , 2 + ............... + mf n c p , n (IV) Here, suffices represents the number of substances. Write the formula to calculate the individual gas constant ( R ) . R = R u M (V) Here, universal gas constant is R u . Write the relation between specific heats and gas constant. c v = c p − R (VI) Write the expression to determine the ratio of specific heats ( k ) . k = c p c v (VII) Here, constant volume specific heat is c v . Write the relation between temperatures and pressures for an isentropic process. T 2 = T 1 ( P 2 P 1 ) k − 1 k (VIII) Here, final temperature is T 2 , initial temperature is T 1 , final pressure is P 2 , and the initial pressure is P 1 . Write the expression to calculate the heat transfer ( q ) for a process. q = c p ( T 2 − T 1 ) (IX) Write the formula to calculate the thermal efficiency of the cycle. η th = 1 − q out q in (X) Here, heat rejected is q out and the heat addition is q in . Conclusion: Substitute 20 lbmol for N N 2 and 28 lbm / lbmol for M N 2 in Equation (I). m N 2 = ( 20 lbmol ) ( 28 lbm / lbmol ) = 560 lbm Substitute 5 lbmol for N O 2 and 32 lbm / lbmol for M O 2 in Equation (I). m O 2 = ( 5 lbmol ) ( 32 lbm / lbmol ) = 160 lbm Substitute 35 lbmol for N H 2 O and 18 lbm / lbmol for M H 2 O in Equation (I). m H 2 O = ( 35 lbmol ) ( 18 lbm / lbmol ) = 630 lbm Substitute 40 lbmol for N CO 2 and 44 lbm / lbmol for M CO 2 in Equation (I). m CO 2 = ( 40 lbmol ) ( 44 lbm / lbmol ) = 1760 lbm Substitute 560 lbm for m N 2 , 160 lbm for m O 2 , 630 lbm for m H 2 O , and 1760 lbm for m CO 2 in Equation (II). m m = 560 lbm + 160 lbm + 630 lbm + 1760 lbm = 3110 lbm Substitute 560 lbm for m N 2 and 3110 lbm for m m in Equation (III). mf N 2 = 560 lbm 3110 lbm = 0.1801 Substitute 160 lbm for m O 2 and 3110 lbm for m m in Equation (III). mf O 2 = 160 lbm 3110 lbm = 0.05145 Substitute 630 lbm for m H 2 O and 3110 lbm for m m in Equation (III). mf H 2 O = 630 lbm 3110 lbm = 0.2026 Substitute 1760 lbm for m CO 2 and 3110 lbm for m m in Equation (III). mf CO 2 = 1760 lbm 3110 lbm = 0.5659 From the Table A-2E of “Ideal gas specific heats of various common gases”, obtain the constant pressure specific heats as follows: c p , N 2 = 0.248 Btu / lbm ⋅ R c p , O 2 = 0.219 Btu / lbm ⋅ R c p , H 2 O = 0.445 Btu / lbm ⋅ R c p , CO 2 = 0.203 Btu / lbm ⋅ R Substitute 0.1801 for mf N 2 , 0

8th Edition

ISBN: 8220100257056

Author: CENGEL

Publisher: YUZU

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Chapter 13.3, Problem 61P

To determine

Thermal efficiency of the ideal Baryton cycle.

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Chapter 13.3, Problem 60P

Chapter 13.3, Problem 62P

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Chapter 13 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 13.3 - Consider a mixture of several gases of identical...Ch. 13.3 - Somebody claims that the mass and mole fractions...Ch. 13.3 - The sum of the mole fractions for an ideal-gas...Ch. 13.3 - Consider a mixture of two gases. Can the apparent...Ch. 13.3 - What is the apparent molar mass for a gas mixture?...Ch. 13.3 - Prob. 6P Ch. 13.3 - Consider a mixture of two gases A and B. Show that...Ch. 13.3 - The composition of moist air is given on a molar...Ch. 13.3 - Prob. 9P Ch. 13.3 - Prob. 10P

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