Chapter4_A - Chapter 4 - Section A - Mathcad Solutions 4.1...

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Ans. 4.2 (a) T 0 473.15 K := n 10 mol := Q 800 kJ := For ethylene: A 1.424 := B 14.394 10 3 K := C 4.392 10 6 K 2 := τ 2 := (guess) Given Qn R AT 0 ⋅τ 1 () B 2 T 0 2 2 1 + C 3 T 0 3 3 1 + = τ Find τ := τ 2.905 = T τ T 0 := T 1374.5K = Ans. (b) T 0 533.15 K := n 15 mol := Q 2500 kJ := For 1-butene: A 1.967 := B 31.630 10 3 K := C 9.873 10 6 K 2 := Chapter 4 - Section A - Mathcad Solutions 4.1 (a) T 0 473.15 K := T 1373.15 K := n 10 mol := For SO2: A 5.699 := B 0.801 10 3 := C 0.0 := D 1.015 10 5 := H R ICPH T 0 T , A , B , C , D , := H 47.007 kJ mol = H := Q 470.073kJ = Ans. (b) T 0 523.15 K := T 1473.15 K := n 12 mol := For propane: A 1.213 := B 28.785 10 3 := C 8.824 10 6 := D0 := H R ICPH T 0 T , A , B , C , 0.0 , := H 161.834 kJ mol = H := Q 1.942 10 3 × kJ = 76
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τ 2.256 = T τ T 0 := T 1202.8K = Ans. T 1705.4degF = 4.3 Assume air at the given conditions an ideal gas. Basis of calculation is 1 second. P 1 atm := T 0 122 degF := V 250 ft 3 := T 932 degF := Convert given values to SI units V 7.079m 3 = T T 32degF ( ) 273.15K + := T 0 T 0 32degF () 273.15K + := T 773.15K = T 0 323.15K = n PV RT 0 := n 266.985mol = For air: A 3.355 := B 0.575 10 3 := C 0.0 := D 0.016 10 5 := H R ICPH T 0 T , A , B , C , D , := τ 3 := (guess) Given Qn R AT 0 ⋅τ 1 B 2 T 0 2 2 1 + C 3 T 0 3 3 1 + = τ Find τ := τ 2.652 = T τ T 0 := T 1413.8K = Ans. (c) T 0 500 degF := n 40 lbmol := Q1 0 6 BTU := Values converted to SI units T 0 533.15K := n 1.814 10 4 × mol = Q 1.055 10 6 × kJ = For ethylene: A 1.424 := B 14.394 10 3 K := C 4.392 10 6 K 2 := τ 2 := (guess) Given R 0 1 B 2 T 0 2 2 1 + C 3 T 0 3 3 1 + = τ Find τ := 77
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P 2 101.3 kPa := P 3 104.0 kPa := T 2 T 3 P 2 P 3 := T 2 290.41K = C P 30 J mol K := (guess) Given T 2 T 1 P 2 P 1 R C P = C P Find C P () := C P 56.95 J mol K = Ans. 4.9 a) Acetone: T c 508.2K := P c 47.01bar := T n 329.4K := H n 29.10 kJ mol := T rn T n T c := T rn 0.648 = Use Eq. (4.12) to calculate H at T n ( H ncalc ) H ncalc RT n 1.092 ln P c bar 1.013 0.930 T rn := H ncalc 30.108 kJ mol = Ans. H 13.707 kJ mol = Qn H := Q 3.469 10 3 × BTU = Ans. 4.4 molwt 100.1 gm mol := T 0 323.15 K := T 1153.15 K := n 10000 kg molwt := n 9.99 10 4 × mol = For CaCO3: A 12.572 := B 2.637 10 3 := C 0.0 := D 3.120 10 5 := H R ICPH T 0 T , A , B , C , D , := H 9.441 10 4 × J mol = H := Q 9.4315 10 6 × kJ = Ans. 4.7 Let step 12 represent the initial reversible adiabatic expansion, and step 23 the final constant-volume heating. T 1 298.15 K := T 3 298.15 K := P 1 121.3 kPa := 78
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To compare with the value listed in Table B.2, calculate the % error. %error H ncalc H n H n := %error 3.464% = Values for other components in Table B.2 are given below. Except for acetic acid, acetonitrile. methanol and nitromethane, agreement is within 5% of the reported value. H n (kJ/mol) % error Acetone 30.1 3.4% Acetic Acid 40.1 69.4% Acetonitrile 33.0 9.3% Benzene 30.6 -0.5% iso-Butane 21.1 -0.7% n-Butane 22.5 0.3% 1-Butanol 41.7 -3.6% Carbon tetrachloride 29.6 -0.8% Chlorobenzene 35.5 0.8% Chloroform 29.6 1.1% Cyclohexane 29.7 -0.9% Cyclopentane 27.2 -0.2% n-Decane 40.1 3.6% Dichloromethane 27.8 -1.0% Diethyl ether 26.6 0.3% Ethanol 40.2 4.3% Ethylbenzene 35.8 0.7% Ethylene glycol 51.5 1.5% n-Heptane 32.0 0.7% n-Hexane 29.0 0.5% Methanol 38.3 8.7% Methyl acetate 30.6 1.1% Methyl ethyl ketone 32.0 2.3% Nitromethane 36.3 6.7% n-Nonane 37.2 0.8%
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This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter4_A - Chapter 4 - Section A - Mathcad Solutions 4.1...

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