11/17/10
Class 36 Wednesday, November
17th
Energy Balances on Reactive
Processes
(Heat of Formation Method)
A gas stream consisting of n-hexane in methane is fed to a
condenser at 60 °C and 1.2 atm. The dew point of the gas is 55
°C. The gas is cooled to 5°C in the condenser, recovering pure
hexane as a liquid. The effluent gas leaves the condenser
saturated with hexane at 5 °C and 1.1 atm and is fed to a boiler
furnace at a rate of 207.4 L/s, where it is burned with 100 %
excess air that enters the furnace at 200 °C. The stack gas
emerges at 400 °C and 1 atm and contains no carbon
monoxide or unburned hydrocarbons. The heat transferred
from the furnace is used to generate saturated steam at 10 bar
from liquid water at 25 °C. Find:
a) The mole fractions of hexane in the condenser feed and
product gas streams and the rate of hexane condensation
(liters condensate/s)
b) The rate at which heat must be transferred from the
condenser (kW) and the rate of generation of steam in the
boiler (kg/s).
H2O (l)
T = 25 °C
H2O (l)
T = 25 °C
Feed Gas
n-hexane (g)
methane (g)
T = 60 °C
Tdp = 55 °C
P = 1.2 atm
Condenser
Effluent sat’d gas
n-hexane (v)
methane (g)
T = 5 °C
P = 1.1 atm
100% n-hexane (l)
Sat’d steam
P = 10 bar
Sat’d steam
P = 10 bar
Q (kW)
Reactor
100% Excess Air
Stack Gas
CO2 (g)
O2 (g)
N2 (g)
H2O (v)
T = 400 °C
P = 1 atm
10 mol/s sat’d gas
0.7 mol/s C6H14(v)
9.3 mol/s CH4(g)
T = 5 °C
P = 1.1 atm
Plan A:
Q (kW)
Reactor
Stack Gas
CO2 (g)
O2 (g)
N2 (g)
H2O (v)
T = 400 °C
P = 1 atm
100% Excess Air
a)  Do atomic species balance around reactor
b)  Put together enthalpy table (choose reference states for all
substances)
c)  Calculate enthalpies, including heat of formation
d)  Calculate Q
e)  Calculate mass of water needed
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11/17/10
10 mol sat’d gas
0.7 mol C6H14 (v)
9.3 mol CH4 (g)
Reactor
Stack Gas
CO2 (g)
O2 (g)
N2 (g)
H2O (v)
100% Excess Air
The effluent gas leaves the condenser saturated with hexane at
5o C and 1.1 atm and is fed to a boiler furnace at a rate of 207.4
L/s, where it is burned with 100 % excess air that enters the
furnace at 200o C. The stack gas emerges at 400o C and 1 atm
and contains no carbon monoxide or unburned hydrocarbons.
The effluent gas leaves the condenser saturated with hexane at 5o C and 1.1
atm and is fed to a boiler furnace at a rate of 207.4 L/s, where it is burned
with 100 % excess air that enters the furnace at 200o C. The stack gas
emerges at 400o C and 1 atm and contains no carbon monoxide or unburned
hydrocarbons.
Balanced Reactions:
CH4
+
2 O2 
CO2 + 2 H2O
C6H14 + 19/2 O2  6 CO2 + 7 H2O
Incoming Air (mol/s):
Theoretical O2 =
mol O2
Excess O2 = 2.0 x 25.25 mol = 50.5 mol/s O2
Incoming N2 = 50.5 x 3.76 = 190 mol/s N2
Unreactive Species Balance: N2, IN = N2, OUT = 190 mol/s
The effluent gas leaves the condenser saturated with hexane at 5o C and 1.1
atm and is fed to a boiler furnace at a rate of 207.4 L/s, where it is burned
with 100 % excess air that enters the furnace at 200o C. The stack gas
emerges at 400o C and 1 atm and contains no carbon monoxide or unburned
hydrocarbons.
Carbon:
Subs
nin (mol/s) Ĥin (kJ/mol) nout (mol/s) Ĥout (kJ/mol)
CH4
9.3
Ĥ1
----
----
C6H14 (v)
0.7
Ĥ2
----
----
O2
50.5
Ĥ3
25.3
Ĥ5
N2
190
Ĥ4
190
Ĥ6
CO2
----
----
13.5
Ĥ7
H2O (v)
----
----
23.5
Ĥ8
Hydrogen:
Oxygen:
What’s our reference for CH4?
a) Vapor at 5oC
b) Vapor at 100oC
c) Vapor at 25oC
d) CH4 does not have a reference
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References: C (s), H2 (g), O2 (g), N2 (g) at 25 °C
Subs
References: C (s), H2 (g), O2 (g), N2 (g) at 25 °C
nin (mol/s) Ĥin (kJ/mol) nout (mol/s) Ĥout (kJ/mol)
Subs
nin (mol/s) Ĥin (kJ/mol) nout (mol/s) Ĥout (kJ/mol)
CH4
9.3
Ĥ1
----
----
CH4
9.3
Ĥ1
----
----
C6H14 (v)
0.7
Ĥ2
----
----
C6H14 (v)
0.7
Ĥ2
----
----
O2
50.5
Ĥ3
25.3
Ĥ5
O2
50.5
Ĥ3
25.3
Ĥ5
N2
190
Ĥ4
190
Ĥ6
N2
190
Ĥ4
190
Ĥ6
CO2
----
----
13.5
Ĥ7
CO2
----
----
13.5
Ĥ7
H2O (v)
----
----
23.5
Ĥ8
H2O (v)
----
----
23.5
Ĥ8
Plan A:
Find Ĥ’s of Elemental Species:
a)  Do atomic species balance around reactor
b)  Put together enthalpy table (choose reference states for
all substances)
c)  Calculate enthalpies, including heat of formation
d)  Calculate Q
e)  Calculate mass of water needed
References: C (s), H2 (g), O2 (g), N2 (g) at 25 °C
Subs
Ĥ3 = Ĥ(O2, 200°C) = 5.31 kJ/mol
(from Table B.8)
Ĥ4 = Ĥ(N2, 200°C) = 5.13 kJ/mol
Ĥ5 = Ĥ(O2, 400°C) = 11.72 kJ/mol
Ĥ6 = Ĥ(N2, 400°C) = 11.15 kJ/mol
References: C (s), H2 (g), O2 (g), N2 (g) at 25 °C
nin (mol/s) Ĥin (kJ/mol) nout (mol/s) Ĥout (kJ/mol)
Subs
nin (mol/s) Ĥin (kJ/mol) nout (mol/s) Ĥout (kJ/mol)
CH4
9.3
Ĥ1
----
----
CH4
9.3
-75.6
----
C6H14 (v)
0.7
Ĥ2
----
----
C6H14 (v)
0.7
-170.1
----
----
O2
50.5
5.31
25.3
11.72
O2
50.5
5.31
25.3
11.72
----
N2
190
5.13
190
11.15
N2
190
5.13
190
11.15
CO2
----
----
13.5
Ĥ7
CO2
----
----
13.5
Ĥ7
H2O (v)
----
----
23.5
Ĥ8
H2O (v)
----
----
23.5
Ĥ8
Calculate Ĥ of entering hydrocarbons:
Calculate Ĥ’s of Products:
Ĥ1a: C(s)(25°C, 1atm) + 2 H2(g)(25°C, 1atm)  CH4(g)(25°C, 1atm)
Ĥ2a: 6 C(s)(25°C, 1atm) + 7 H2(g)(25°C, 1atm)  C6H14(g)(25°C, 1atm)
Ĥ1a = (ΔĤf°)methane (g) = -74.85 kJ/mol
Ĥ2a = (ΔĤf°)hexane (v) = -167.2 kJ/mol
Ĥ = (ΔĤf°) +
(from Table B.1)
Ĥ1 = -75.6 kJ/mol, Ĥ2 = -170.1 kJ/mol
Using Tables B.1 and B.8,
Ĥ7 = (ΔĤf°)CO2(g) + Ĥ(CO2, 400°C) = -393.5 kJ/mol + 16.35 kJ/mol
= -377.15 kJ/mol
Ĥ8 = (ΔĤf°)H2O(v) + Ĥ(H2O, 400°C) = -241.83 kJ/mol + 13.23 kJ/mol
= -228.6 kJ/mol
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References: C (s), H2 (g), O2 (g), N2 (g) at 25 °C
Subs
nin (mol/s) Ĥin (kJ/mol) nout (mol/s) Ĥout (kJ/mol)
CH4
9.3
-75.6
----
C6H14 (v)
0.7
-170.1
----
----
O2
50.5
5.31
25.3
11.72
N2
190
5.13
190
11.15
CO2
----
----
13.5
-377.2
H2O (v)
----
----
23.5
----
-228.6
Specific Energy Balance:
Q = ΔH
Q = ΣnoutĤout – ΣninĤin = -8049 kJ – 421 kJ
Q = -8470 kW
Class 36 Wednesday, November 17th
Adiabatic Flame Temperature
Thermochemistry
The heat coming out of the reactor equals -8470 kJ. Solve
for the mass of water at 25 °C and 10 bar that can be
turned into saturated steam at 10 bar.
Subs
min (kg/s)
Ĥin (kJ/kg)
mout (kg/s)
Ĥout (kJ/kg)
H 2O
mw
104.8
mw
2776.2
Energy Balance:
Q = ΔH = mwΔĤ
Ĥ1(H2O(l), 25°C) = 104.8 kJ/kg
(from Table B.5)
Ĥ2(H2O(v), 10 bar) = 2776.2 kJ/kg
(from Table B.6)
8470 kJ = mw(2776.2 – 104.8)
mw = 3.17 kg/s
Adiabatic Flame Temperature (Tad) - the highest
achievable temperature that can be reached – for an
adiabatic system (generally in combustion)
ΔH = nfΔĤc° + ∑out niĤi(Tad) - ∑in niĤi(Tfeed) = 0
∑out niĤi(Tad) = -nfΔĤc° + ∑in niĤi(Tfeed)
where ΔĤc° = heat of combustion of the fuel at 25 °C and
1 atm
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Methane is burned with 25 % excess air in a continuous
adiabatic reactor. The methane enters the reactor at 25 °C
and 1.1 atm at a rate of 5.50 L/s, and the entering air is at 150
°C and 1.1 atm. Combustion in the reactor is complete, and
the reactor effluent emerges at 1.05 atm. Calculate the
temperature and degrees of superheat of the reactor effluent.
Incoming CH4 (mol/s):
CH4 + 2 O2  CO2 + 2 H2O
CH4, 5.50 L/s
T = 25 °C
P = 1.1 atm
Methane is burned with 25 % excess air in a
continuous adiabatic reactor. The methane enters
the reactor at 25 °C and 1.1 atm at a rate of 5.50
L/s, and the entering air is at 150 °C and 1.1 atm.
Do the proper material balances.
Ideal Gas Law: nCH4 =
REACTOR
25% Excess Air
T = 150 °C
P = 1.1 atm
(
)
)
(1.1atm) 5.50 L s
= 0.247 mol s
0.08206 L ⋅ atm mol ⋅ K (298K )
(
Effluent gas
Incoming Air:
T = Tad
P = 1.05 atm
Theoretical O2 = 0.247 (mol/s) x (2 mol O2/mol) = 0.495 mol/s
€
Excess O2 = 1.25 x 0.495 mol/s = 0.619 mol/s O2
Incoming N2 = 0.619 x 3.76 = 2.33 mol/s N2
Q=0
CH4 + 2 O2  CO2 + 2 H2O
Methane is burned with 25 % excess air in a
continuous adiabatic reactor. The methane enters
the reactor at 25 °C and 1.1 atm at a rate of 5.50
L/s, and the entering air is at 150 °C and 1.1 atm.
Do the proper material balances.
C: nCH4 = nCO2
nCO2 = 0.247 mol/s
0.247 mol/s CH4
0.619 mol/s O2
2.33 mol/s N2
H: 4 nCH4 = 2 nH2O
nH2O = 0.495 mol/s
O2
N2
CO2
H 2O
Methane is burned with 25 % excess air in a
continuous adiabatic reactor. The methane enters the
reactor at 25 °Cand 1.1 atm at a rate of 5.50 L/s, and
the entering air is at 150 °C and 1.1 atm. Combustion
in the reactor is complete, and the reactor effluent
emerges at 1.05 atm. What is the energy balance?
a) Q = ΔH
b) Q = ΔU
c) Q - W= ΔH
d) Q = 0
ΔH˙ = Q˙ = 0
ΔH˙ = n ΔHˆ  + ∑ n Hˆ (T
f
c
i
i
ad
out
) − ∑ n i Hˆ i (Tfeed )
in
n f ΔHˆ c + ∑ n i Hˆ i (Tad ) = ∑ n i Hˆ i (Tfeed )
out
in
O: 2 nO2 = nH2O + 2 nO2 + 2 nCO2
nO2 = 0.124 mol/s
€
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11/17/10
References: CH4, O2, N2, CO2, H2O at 25oC
References: CH4, O2, N2, CO2, H2O at 25oC
Subs
CH4
O2
nin
0.247
0.619
Ĥin
0
Ĥ1
nout
---0.124
Ĥout
---Ĥ3
Subs
CH4
O2
nin
0.247
0.619
Ĥin
0
3.78
nout
---0.124
Ĥout
---Ĥ3
N2
CO2
H2O (v)
2.33
-------
Ĥ2
-------
2.33
0.247
0.495
Ĥ4
Ĥ5
Ĥ6
N2
CO2
H2O (v)
2.33
-------
3.66
-------
2.32
0.247
0.495
Ĥ4
Ĥ5
Ĥ6
Which enthalpies will we will be able to look up?
From Table B.8:
a)  All of them b) none of them
Ĥ1(O2, 150°C) = 3.78 kJ/mol Ĥ2 (N2, 150°C) = 3.66 kJ/mol
c) Ĥ1 and Ĥ2
d) Ĥ1, Ĥ2, Ĥ3 and Ĥ4
n f ΔHˆ c + ∑ n i Hˆ i (Tad ) = ∑ n i Hˆ i (Tfeed )
out
€
Substance 10-3
in
Find ΔĤc° for CH4 at 25 °C:
Use Cp integration to solve for Tad
ΔĤc°(CH4, 25°C) = -890.36 kJ/mol
nf ΔĤc° = 0.247 x (-890.36) = -220 kW
Rest of the terms:
n f ΔHˆ c + ∑ n i Hˆ i (Tad ) = −220 + 0.619 × 3.78 + 2.33 × 3.66
out
n f ΔHˆ c + ∑ n i Hˆ i (Tad ) = −209kW
out
O2
H2O (v)
CO2
N2
29.10
33.46
36.11
29
10-5
1.158
0.6880
4.233
0.2199
10-8
-0.6076
0.7604
-2.887
0.5723
10-12
1.311
-3.593
7.464
-2.871
1.  List the heat capacity terms under each order of
magnitude (10-3, 10-5, etc.)
2.  Multiply each term by its corresponding number
of moles
For example O2: 0.124*29.10 = 3.608
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11/17/10
Substance
10-3
10-5
10-8
10-12
O2 (0.124)
H2O (0.494)
CO2 (0.247)
N2 (2.32)
3.608
16.53
8.919
67.28
96.34
0.1436
0.3399
1.045
0.5102
2.036
‐0.07534
0.3756
‐0.7131
1.328
0.9149
0.1626
‐1.775
1.844
‐6.661
‐6.429
Calculate Heat of Vaporization for H2O:
 T − 25  0.38
ΔHˆ v (25  C) = ΔHˆ v (Tbp ) c
 Using Watson’s correlation, p382
 Tc − Tbp 

ΔHˆ v (25 C) = 44.6 kJ mol
Putting it all together:
€
out
total
−1.61×10−12 (Tad4 − 25 4 ) + 22kW + 209kW = 0
+ btotal T + c total T 2 + dtotal T 3 dT
Tref
= atotal (Tad − 25) +
in
9.6 ×10−2 (Tad − 25) + 1.02 ×10−5 (Tad2 − 25 2 ) + 3.05 ×10−9 (Tad3 − 25 3 )
Tad
∫a
n f ΔHˆ c + ∑ n i Hˆ i (Tad ) = ∑ n i Hˆ i (Tfeed )
btotal 2
(Tad − 252 ) + c total
(Tad3 − 253 ) + dtotal
(Tad4 − 254 )
2
3
4
Solve for T using solver or goal seek: Tad = 1833 °C
€
€
Methane is burned with 25 % excess air in a
continuous adiabatic reactor. The methane enters the
reactor at 25o C and 1.1 atm at a rate of 5.50 L/s, and
the entering air is at 150o C and 1.1 atm. Combustion
in the reactor is complete, and the reactor effluent
emerges at 1.05 atm. What are the degrees of
superheat in the effluent?
a) 
b) 
c) 
d) 
0
56
1624
1777
y H 2O P = p*H 2O (Tdp )
Tdp = 56  C

From Table B.3
SH = 1833 − 56 = 1777  C
€
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