For enthalpy 2 going in the heater, the result is -0.15 kj/mol air. How was this obtained?
For enthalpy 2 going in the heater, the result is -0.15 kj/mol air. How was this obtained?
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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For enthalpy 2 going in the heater, the result is -0.15 kj/mol air. How was this obtained?

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500
(---)
T
1000
(-)
652 APPENDIX B Physical Property Tables
Q Search
A
0
← →
0.000734
Ĥ 1878
2293
3021
Û 1791
2169
2765
0.001726 0.002491
0.005112
1798
2051
2316
2594
1653
1888
2127
2369
0.001446 0.001628 0.001893 0.002246
0.002000
672 of 695
0.000000
2723
2529
0.003882
U.VIVZ
TABLE B.9 Specific Enthalpies of Selected Gases:
U.S. Customary Units
H
OP
TABLE B.8 Specific Enthalpies of Selected Gases: SI Units
Ĥ (kJ/mol)
Reference state: Gas, Pref= 1 atm, Tref = 25°C
CO
CO₂ H₂O
Air 0₂
-0.72 -0.73
0.00
N₂
-0.73
H₂
-0.72
0.00
0
-0.73
-0.92 -0.84
25
0.00 0.00
0.00
0.00
0.00
100
2.19
2.24
2.19
2.16
2.19
2.90
2.54
200
5.15
5.31
5.13
5.06
5.16
7.08
6.01
300
8.17
8.47
8.12
7.96
8.17 11.58
9.57
13.23
17.01
400 11.24 11.72 11.15 10.89 11.25 16.35
500 14.37 15.03 14.24 13.83 14.38 21.34
600 17.55 18.41 17.39 16.81 17.57 26.53 20.91
700 20.80 21.86 20.59 19.81 20.82 31.88 24.92
800 24.10 25.35 23.86 22.85 24.13 37.36 29.05
900 27.46 28.89 27.19 25.93 27.49 42.94 33.32
1000 30.86 32.47 30.56
29.04
30.91 48.60 37.69
1100 34.31 36.07 33.99 32.19 34.37 54.33 42.18
1200 37.81 39.70 37.46 35.39 37.87 60.14 46.78
1300 41.34 43.38 40.97 38.62 41.40 65.98 51.47
1400 44.89 47.07 44.51 41.90 44.95 71.89 56.25
1500 48.45 50.77 48.06 45.22 48.51 77.84 61.09
V.VIIT
3248
3439
2946
3091
0.006112 0.007000
2857
2591
0.002668
V.VIZU
3105
2795
0.003106
CVIJO
3610
3224
0.007722
3324
2971
0.003536
VUITT
3771
3350
0.008418
3526
3131
0.003953
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Example 8.3-5
Solution
Energy Balance on a Gas Preheater
A stream containing 10% CH4 and 90% air by volume is to be heated from 20°C to 300°C. Calculate
the required rate of heat input in kilowatts if the flow rate of the gas is 2.00 × 10³ liters (STP)/min.
Q Search
441 of 695
Basis: Given Flow Rate
Assume ideal-gas behavior.
2000 L (STP)/min, 20°C
n(mol/min)
0.100 mol CH4/mol
0.900 mol air/mol
CD
n =
2000 L (STP)
min
HEATER
(010)
Q(kW)
Recall that specifying the flow rate in liters (STP)/min does not imply that the feed gas is at standard
temperature and pressure; it is simply an alternative way of giving the molar flow rate.
8.3 Changes in Temperature 421
1 mol
22.4L (STP)
n(mol/min), 300°C
0.100 mol CH₂/mol
0.900 mol air/mol
= 89.3 mol/min
The energy balance with kinetic and potential energy changes and shaft work omitted is Q = AH. The
task is to evaluate AH = [n¡Â¡- ΣnĤ₁. Since each species has only one inlet condition and one outlet
out
in
condition in the process, two rows are sufficient for the enthalpy table.
References: CH4(g, 20°C, 1 atm), air(g, 25°C, 1 atm)
nin
Ĥin
nout
Substance (mol/min) (kJ/mol) (mol/min)
Ĥ out
(kJ/mol)
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