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CE-HW4-Ch04soln

CE-HW4-Ch04soln - HW#4 Chap 4 Problems 7a,d 9 13a,b 17 18...

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HW #4 - Chap 4 Problems: 7a,d, 9, 13a,b, 17, 18, 21, 24 (4-7) a, d Stoichiometric combustion reaction CH 3 0H + 1.5 O 2 + (1.5)(3.76) N 2 CO 2 + 2 H 2 O+ (1.5*3.76) N 2 At equivalence ratio = 0.75 CH 3 0H + (1.5/0.75) O 2 + (1.5/0.75)(3.76) N 2 CO 2 + 2 H 2 O+ (1.5/0.75)*(3.76) N 2 + 0.5 O 2 This reduces to CH 3 0H + 2 O 2 + (2)(3.76) N 2 CO 2 + 2 H 2 O+ (2)*(3.76) N 2 + 0.5 O 2 (a) From Eqs. (2-55) and (4-1) AF = m a /m f = (N a M a /N f M f ) = [(2)(4.76)(29)]/[(1)(32)] = 8.63 (b) not assigned Mole fraction of water x = N water /N total = 2/[1 + 2 + 2(3.76) + 0.5] = 0.1815 Vapor pressure of water P v = xP total = (0.1815) (101 kPa) = 18.33 kPa Using steam tables from the reference cited above T DP = 58 o C (d) From Eq.(4-9) AKI = (MON + RON)/2 = (92 + 106)/2 = 99 (4-9) Stoichiometric combustion equation C 2 H 5 OH + 3 O 2 + 3(3.76) N 2 → 2 CO 2 + 3 H 2 0 + 3(3.76) N 2 (a) Eq. (2-57) and stoichiometric value from Table A-2 (AF) act = (AF) stoi = 9.0/1.10 = 8.18 (b) using Fig. 3-2 and Eqs. (3-4) and (3-5) for conditions at end of compression T 2 = T 1 (r c ) k-1 = (333 K)(10) 0.35 = 745 K P 2 = P 1 (r c ) k = (101 kPa)(10) 1.35 = 2261 kPa Eq. (3-11) Q HV C . = (AF + 1) C v (T 3 T 2 ) (26,950 KJ/kg)(0.97) = (8.18 + 1)(0.821 kJ/kg-K)( T 3 - 745 K) T 3 = T peak = 4214 K = 3941° C (c) at constant volume P 3 = P 2 (T 3 / T 2 ) = (2261 kPa) (4214/745) = 12.789 kPa

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CE-HW4-Ch04soln - HW#4 Chap 4 Problems 7a,d 9 13a,b 17 18...

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