3/14/2016
Prob 8‐74b
Overall mass balance
• “Delivered air” flow rate stays the same, but 6/7 is recycled
1
m in , no recycle
7
1
 m out , no recycle
7
m in , recycle 
m out , recycle
Heat balance
What was the purpose of Problem 8.74?
1
Qrecycle  Qno recycle
7
Q recycle  H   m i Hˆ i

• Practice the following concepts:
 
out
  m i Hˆ i

in
Q recycle  m H 2O ,liq Hˆ H 2O ,liq  m delivered air Hˆ delivered air
 m inlet air Hˆ inlet air  m recycled air Hˆ recycled air
– Energy balances
– Mass balances
– Psychrometric chart
– Recycle
– Dry air as a constant quantity in balances
We could also get Heats of Reaction
1. Definition
Schedule
• Today – Heats of Reaction, Heat of Combustion
• Wed – Energy Balance with Reaction
– WARNING on Reading:
Skip Heat of reaction method on pages 450‐451, Skip Examples 9.5‐1 and 9.5‐3
•
•
•
•
Fri – Practice Energy Balance with Reaction
Mon – Adiabatic Flame Temperature Wed – Transient Balances
Fri – Review for Exam 3 (Fri, March 29 thru Tues, Apr 2)
• Hr = heat of reaction (kJ/mol or Btu/lbmol or cal/mol)
– If Hr < 0, exothermic (gives off heat)
– If Hr > 0, endothermic (needs heat)
• Hr0 = standard heat of reaction (i.e., at 25C)
• Heat of reaction always defined by complete reaction
(i.e., XA = 1)
– Heat absorbed per mole reacted
• Remember ,
,
,
• Therefore Hsystem =  Hr
·
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3/14/2016
Heats of Reaction
2. From Heats of Reaction
Heats of Reaction
3. Path Independent
• Hf0 = heat required to form species at 1 atm, 25C
• C(s) + O2  CO2
Hr0 = ‐393.51 kJ/mol
• Hr0 =  iHf,i0
• C(s) + ½O2  CO
• CO + ½O2  CO2
• C(s) + O2  CO2
Hr0 = ‐110.52 kJ/mol
Hr0 = ‐282.99 kJ/mol
Hr0 = ‐393.51 kJ/mol
– Tabulated in Table B.1, 2nd to last column
– Assumes all reactants and products at 25C
– Remember negative sign for reactants
– Like products minus reactants
• Example: Gasification of carbon
C(s) + H2O(g)  CO(g) + H2(g)
Species
C(s)
H2O
CO
H2
i
‐1
‐1
1
1
Hf,i0 (kJ/mol)
0.0
Hr0 = ‐110.52 + 0 – 0 – (‐241.83) = +131.31 kJ/mol
‐241.83
‐110.52
Endothermic!
0.0
• Hess’s Law: – Add or subtract reactions to get correct Hr
Heats of Reaction
4. Heat of Combustion
Heats of Reaction
6. Review
• Hc0 = heat of combustion at 1 atm, 25C
–
–
–
–
–
–
• How do you find at different temperatures?
Tabulated in Table B.1, 2nd to last column
Assumes all reactants and products at 25C
All C  CO2 (g)
All H  H2O (liq) (for “high heating value”)
All S  SO2 (g)
All N  N2 (g)
=∆
• Suppose and ∆
0.079
,
• Find • Heating value = ‐ Hc0
• Example: NH3 + ¾ O2  3/2 H2O(liq) + ½ N2
•
at 400
= 74.85
= 74.85
Hf,i0 (kJ/mol): ‐46.19 0.0 ‐285.84 0.0
74.85
,
0.079
0.079 ∗ 400
25
45.22
Hr = 3/2(‐285.84) – (‐46.19) = ‐382.57 kJ/mol (same as Hc0 in Table B.1)
Multiple Species
Multiple Species
• Calculate the energy required to raise a mixture from 25C to 400C.
Species
Gram‐moles
O2
0.05
CO2
1.0
HO
0.3
∑
• Now get∆
0.05 ∗
400
25
1.0 ∗
400
25
0.30 ∗
400
25
2
∆
400
400
400
25
3
25
25
400
400
400
25
4
25
25
400
25
400
400
25
25
∆
∆
This is a lot of programming or punching buttons on a calculator!
∆
,
400
4
400
25
2
400
25
3
400
25
25
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Shortcut
Two Methods to Find Hrxn
(Combine coefficients)
0.05 ∗
400
25
1.0 ∗
400
25
2
400
25
400
3
400
25
25
400
4
400
25
25
400
Hr at T
A
0.30 ∗
400
25
400
25
400
25
400
• Heat of Formation
• Path Method (AB)
25
B
Δ
Δ
Define new variables:
∑
∑
∑
A at 25C
∑
,
25
Δ
=Δ
Hr at 25C
Δ
,
,
,
,
B at 25C
∑
or
∆
,
,
Now H becomes:
∆
∆
400
25
2
400
25
3
400
25
4
400
25
′
′ ∑
, ′ ∑
, etc., where I is negative for reactants and positive for products
Δ
Δ
,
= Δ
,
∆
25
25
25
25
See Excel Sheet
Example
• CO + ½ O2  CO2
• Find the heat of reaction at 1200C
A. Cp approach (path method)
B. Hout ‐ Hin (Hf0 approach)
See spreadsheet
3