Fifteen kmol/min of air is cooled from 430°C to 100°C. Calculate the required heat removal rate using (1) heat capacity formulas from Table B.2, (2) specific enthalpies from Table B.8, and (3) the Enthalpy function of APEX.

Elements Of Electromagnetics
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I need help understanding the easy to solve this problem.  Explain everything, including the linear interpolation.  

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Example 8.3-3
Solution
←→
When enthalpies must be calculated frequently for a species, it is convenient to prepare a table
of H(T) for the species (as was done for water in the steam tables) to avoid having to integrate the
formula for C₂(T) again and again. Tables B.8 and B.9 in Appendix B list specific enthalpies of
species involved in combustion reactions-air, O₂, N2, H₂ (a fuel), CO, CO2, and H₂O(v). The
values in these tables were generated by integrating C₂(7) from the specified reference state (25°C
for Table B.8, 77°F for Table B 9) to the listed temperatures. The next example illustrates the use
of these tables.
Evaluation of AH Using Heat Capacities and Tabulated Enthalpies
Fifteen kmol/min of air is cooled from 430°C to 100°C. Calculate the required heat removal rate using (1) heat
capacity formulas from Table B.2, (2) specific enthalpies from Table B.8, and (3) the Enthalpy function of
APEX.
Q Search
ΔΗ
437
With AEK, AĖp, and W, deleted, the energy balance is
kJ
mol
Assume ideal-gas behavior, so that pressure changes (if there are any) do not affect AĤ.
1. The hard way. Integrate the heat capacity formula in Table B.2.
Cp(T) dT
[0.02894 +0.4147 × 10-³T +0.3191 × 10-87² — 1.965 × 10-¹²7³] AT
0.4147 x 10-5
2
=
=
of 695 3
= (0.497 mol)(-1.250 kJ/mol) = -0.621 KJ
Q = AH = nair Hair,out - nairĤair.in = nair AĤ
100°C
100°C
air(g, 430°C) → air(g, 100°C)
J430°C
= 0.02894(100 - 430) +
-(100² - 430²)
0.3191 x 10-8
1.965 × 10-12
-(100³ - 430³) -
3
4
= (-9.5502-0.3627 -0.0835 +0.0167) kJ/mol = -9.98 kJ/mol
H
(100¹ - 430¹) kJ/mol
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Transcribed Image Text:||! PDF *Elementary Principles of Chemic X PDF Elementary Principles of Chemica X + 158°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdf Draw D T Read aloud Example 8.3-3 Solution ←→ When enthalpies must be calculated frequently for a species, it is convenient to prepare a table of H(T) for the species (as was done for water in the steam tables) to avoid having to integrate the formula for C₂(T) again and again. Tables B.8 and B.9 in Appendix B list specific enthalpies of species involved in combustion reactions-air, O₂, N2, H₂ (a fuel), CO, CO2, and H₂O(v). The values in these tables were generated by integrating C₂(7) from the specified reference state (25°C for Table B.8, 77°F for Table B 9) to the listed temperatures. The next example illustrates the use of these tables. Evaluation of AH Using Heat Capacities and Tabulated Enthalpies Fifteen kmol/min of air is cooled from 430°C to 100°C. Calculate the required heat removal rate using (1) heat capacity formulas from Table B.2, (2) specific enthalpies from Table B.8, and (3) the Enthalpy function of APEX. Q Search ΔΗ 437 With AEK, AĖp, and W, deleted, the energy balance is kJ mol Assume ideal-gas behavior, so that pressure changes (if there are any) do not affect AĤ. 1. The hard way. Integrate the heat capacity formula in Table B.2. Cp(T) dT [0.02894 +0.4147 × 10-³T +0.3191 × 10-87² — 1.965 × 10-¹²7³] AT 0.4147 x 10-5 2 = = of 695 3 = (0.497 mol)(-1.250 kJ/mol) = -0.621 KJ Q = AH = nair Hair,out - nairĤair.in = nair AĤ 100°C 100°C air(g, 430°C) → air(g, 100°C) J430°C = 0.02894(100 - 430) + -(100² - 430²) 0.3191 x 10-8 1.965 × 10-12 -(100³ - 430³) - 3 4 = (-9.5502-0.3627 -0.0835 +0.0167) kJ/mol = -9.98 kJ/mol H (100¹ - 430¹) kJ/mol { J I D ENG Sign in 10:22 AM 5/6/2023 • la + O
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418 CHAPTER 8
Balances on Nonreactive Processes
Cofa
OD
(100¹ - 430¹³) -
438 of 695
(2) T
The easy way. Use tabulated enthalpies from Table B.8.
Ĥ for air at 100°C can be read directly from Table B.8 and Ĥ at 430°C can be estimated by linear
interpolation from the values at 400°C (11.24 kJ/mol) and 500°C (14.37 kJ/mol).
Q Search
ΔΗ = ἡΔῇ =
3
4
= (-9.5502-0.3627 -0.0835+ 0.0167) kJ/mol = -9.98 kJ/mol
3. The easiest way. In a spreadsheet cell, insert =Enthalpy("air",430,100). The value -9.98 (in kJ/mol)
will be returned. In using the spreadsheet, it is absolutely essential to keep track of units by adding
explicit notation where appropriate; otherwise the probability of getting units wrong increases
substantially.
Whichever way AĤ is determined
15.0 kmol
min
Ĥ(100°C) = 2.19 kJ/mol
Ĥ(430°C) = [11.24 +0.30(14.37 - 11.24)] kJ/mol = 12.17 kJ/mol
||
AĤ = (2.19 - 12.17) kJ/mol = -9.98 kJ/mol
(100* — 430*)] kJ/mol
10³ mol
1kmol
- 9.98 kJ
mol
1 min
60 s
1 kW
1 kJ/s
-2500 kW
Reminder: (The enthalpies listed in Tables B.8 and B.9(and for that matter, the heat capacity
formulas of Table B.2 and the Enthalpy function of APEx apply strictly to heating and cooling at
a constant pressure of 1 atm. The tabulated enthalpies and heat capacities may also be used for
nonisobaric heating and cooling of ideal or nearly ideal gases; however, at pressures high enough
for temperatures low enough) for the gases to be far from ideal, enthalpy tables on more accurate
heat capacity formulas should be used.
<
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Transcribed Image Text:||! PDF *Elementary Principles of Chemic X PDF Elementary Principles of Chemica X + 158°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdf Draw D T Read aloud 418 CHAPTER 8 Balances on Nonreactive Processes Cofa OD (100¹ - 430¹³) - 438 of 695 (2) T The easy way. Use tabulated enthalpies from Table B.8. Ĥ for air at 100°C can be read directly from Table B.8 and Ĥ at 430°C can be estimated by linear interpolation from the values at 400°C (11.24 kJ/mol) and 500°C (14.37 kJ/mol). Q Search ΔΗ = ἡΔῇ = 3 4 = (-9.5502-0.3627 -0.0835+ 0.0167) kJ/mol = -9.98 kJ/mol 3. The easiest way. In a spreadsheet cell, insert =Enthalpy("air",430,100). The value -9.98 (in kJ/mol) will be returned. In using the spreadsheet, it is absolutely essential to keep track of units by adding explicit notation where appropriate; otherwise the probability of getting units wrong increases substantially. Whichever way AĤ is determined 15.0 kmol min Ĥ(100°C) = 2.19 kJ/mol Ĥ(430°C) = [11.24 +0.30(14.37 - 11.24)] kJ/mol = 12.17 kJ/mol || AĤ = (2.19 - 12.17) kJ/mol = -9.98 kJ/mol (100* — 430*)] kJ/mol 10³ mol 1kmol - 9.98 kJ mol 1 min 60 s 1 kW 1 kJ/s -2500 kW Reminder: (The enthalpies listed in Tables B.8 and B.9(and for that matter, the heat capacity formulas of Table B.2 and the Enthalpy function of APEx apply strictly to heating and cooling at a constant pressure of 1 atm. The tabulated enthalpies and heat capacities may also be used for nonisobaric heating and cooling of ideal or nearly ideal gases; however, at pressures high enough for temperatures low enough) for the gases to be far from ideal, enthalpy tables on more accurate heat capacity formulas should be used. < J D ENG Sign in 10:22 AM 5/6/2023 • la + O
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