2nd Workshop: Dual-Fuel Combustion FACULTY OF MECHANICAL ...
Department of Mechanical Engineering ME 322 …...ME 322 – Mechanical Engineering Thermodynamics...
Transcript of Department of Mechanical Engineering ME 322 …...ME 322 – Mechanical Engineering Thermodynamics...
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Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 37
Heat of Reaction
1st Law Analysis of Combustion Systems
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Combustion System Analysis
2
Consider the complete combustion of octane in 150%
theoretical air,
Combustion
Chamber
C8H18
PTA = 150%
Products (p)
Q
In the previous lecture, we found the balanced reaction,
8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N
The First Law applied to the system identified above is,
a a f f p p
p
Q n h n h n h
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Combustion System Analysis
3
Combustion
Chamber
C8H18
PTA = 150%
Products (p)
Q
reactants
(air and fuel)
combustion
products
Potential issue: There are no Dh values (except for O2 and
N2). This has the potential to cause a datum state problem.
8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N
a a f f p p
p
Q n h n h n h
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Resolving the Datum State Problem
4
In combustion calculations, the enthalpy of all stable* elements
is defined as zero at the standard reference state (SRS),
25 C 298.15 K at 0.1 MPa 1 atm
77 F 536.67 R at 14.5 psia 1 atm
*‘Stable’ means chemically stable at the SRS. For example,
diatomic oxygen (O2) is stable at the SRS. Monatomic
oxygen (O) is not stable at the SRS.
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Enthalpy of Formation
0
fh Heat released in an exothermic reaction (or absorbed
in an endothermic reaction) when a compound is
formed from its elements. (Elements and compound at the SRS)
The enthalpy of a compound at the standard reference state
4 4 2 2
2
4 2
4 4 4
4 4
4
CH CH C C H H
HCCH C H
CH CH CH
0
CH ,CH
CH
f
Q n h n h n h
nnQh h h
n n n
Qh h
n
Example – Methane
C
2H 2
CH 4
25 ° C
1 atm
25 ° C
1 atm
Q
5
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Enthalpy of Formation Values
Using EES* …
*Unit setting = molar
7
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Enthalpy of Formation Values
Conclusion: EES uses the SRS as the datum state for enthalpy for
the ideal gases! Therefore, enthalpy of formation values can be
calculated from EES using the ideal gas substances (except AIR)
8
Results ...
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Enthalpy Values in Combustion
9
What do we know so far?
1. The enthalpy of a stable element at the SRS is 0
2. The SRS is 25°C, 0.1 MPa
3. The enthalpy of a compound at the SRS is the
enthalpy of formation (Table 15.1 or from EES)
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Enthalpy Values at Other States
10
0
,i f i ih T h h D
The enthalpy of a component at any temperature in a
combustion process can be evaluated by,
0
,i f i i i SRSh T h h T h T
Accounts for the
enthalpy difference
relative to the SRS
How is the enthalpy difference in brackets determined??
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1. If the heat capacity of the component can be assumed constant,
2. If the constant heat capacity assumption is not accurate enough,
then use the ideal gas tables (Table C.16c). In this case the datum
state for the table does not have to match the enthalpy of formation.
3. Use a set of property tables for all components that has all enthalpy
values referenced to the SRS. Does such a thing exist?
Enthalpy Values at Other States
11
0
, ,i f i p i SRSh T h c T T
0
,i f i i i SRSh T h h T h T
Three possibilities ...
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Enthalpy Values at Other States
12
0
,i f i i i SRSh T h h T h T
Exploring Option 3 from the previous slide ...
If a thermodynamically consistent set of tables exists, then
0
,f i i SRSh h T
Therefore the enthalpy of the component could simply be
looked up in a table at the given temperature,
i ih T h T
If something like this were available ... combustion
calculations would be EESy!
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Enthalpy Values at Other States
13
ALL of the ideal gas enthalpy reference states (except for
the ideal gas ‘AIR’) in EES are referenced to the SRS!
This is from the EES Help Menu for the ideal gas CO2 ...
All other ideal gases in EES (except AIR) say the same thing!
Significance: Combustion calculations just became EESy!
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Heat of Reaction
14
Combustion
Chamber
Fuel
Air
Products (P)
Q
Consider an aergonic combustion process as shown below
i i i i
P R
Q n h n h
The First Law applied to this system results in,
Reactants (R)
i ii i
P Rfuel fuel fuel
n nQq h h
n n n
Dividing by the molar flow rate of the fuel,
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Heat of Reaction
15
i ii i
P Rfuel fuel fuel
n nQq h h
n n n
The molar flow rate ratios are the molar coefficients from the
balanced combustion reaction! Therefore,
i i i i
P Rfuel
Qq h h
n
This is known as the molar heat of reaction.
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Heating Values of Fuels
16
Given: Gaseous octane (C8H18) is burned completely in
100% theoretical air. The reactants and the products are at
the SRS.
Find: The heat released during this combustion process per
mole of fuel for the following cases,
(a) the water in the products is all vapor
(b) the water in the products is all liquid
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
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The system boundary is drawn
around the combustion chamber. Applying the First Law
results in,
Heating Values
17
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
i i i i
P R
Q n h n h
Dividing both sides of this equation by the molar flow rate of
the fuel,
i ii i
P Rfuel fuel fuel
n nQh h
n n n
Number of moles of
reactant per mole of fuel
Number of moles of product
species per mole of fuel
How are these found?
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The molar flow rate ratios on the
previous slide are the molar coefficients from the balanced
combustion reaction (for one mole of fuel)! Therefore,
Heating Values
18
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
i ii i
P Rfuel fuel fuel
n nQh h
n n n
i i i i
P Rfuel
Qq h h
n
Observations ...
1. The importance of being able to balance the
combustion reaction is evident!
2. As long as the combustion process is aergonic, the
First Law will be as written above, independent of the
conditions in and out of the combustion chamber!
Notice: PTA = 100% means
stoichiometric combustion
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For the complete combustion of
normal octane in 100% theoretical air, we previously found,
Heating Values
19
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
i i i i
P Rfuel
Qq h h
n
Applying the First Law to the system,
2 2 2 2 8 18 2 21 CO 2 H O 4 O 5 N C H 0 O N3.67
fuel
Qq h h h h h h h
n
2 2 8 181 ,CO 2 ,H O ,C Hf f f
fuel
Qq h h h
n
8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N
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Heating Values
20
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
2 2 8 181 ,CO 2 ,H O ,C Hf f f
fuel
Qq h h h
n
Now we have an interesting problem. Is the water liquid or gas (or both)?
8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N
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Let’s consider both extremes
(1) the H2O is all vapor and (2) the H2O is all liquid.
Heating Values
21
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
All vapor water ...
2 2 8 181 ,CO 2 ,H O ,C Hf f f
fuel
Qq h h h
n
MJ MJ MJ MJ
8 393.522 9 241.827 208.447 5,116.172kmol kmol kmol kmolfuel
n
All liquid water ...
MJ MJ MJ MJ
8 393.522 9 285.838 208.447 5,512.271kmol kmol kmol kmolfuel
n
8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N
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Heating Values
22
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
All vapor water ...
MJ5,116.172
kmolfuel
n
All liquid water ...
MJ5,512.271
kmolfuel
n
Lower Heating Value (LHV)
Higher Heating Value (HHV)
Observations ...
1. The reactants and products are at the SRS
2. The reaction occurs with PTA = 100% (i = i )
3. The difference between the HHV and the LHV is the
enthalpy of vaporization of water!
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Heating Values
23
1 mol fuel
Stoichiometric air
Products (vapor H2O)
Products (liquid H2O)
,fgh2 2H O H O
,HHV LHV fgh 2 2H O H O
TSRS, PSRS
TSRS
PSRS
q
LHV
HHV
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Heating Values
24
The heating values represent the maximum possible heat
transfer that can occur per mole of fuel.
• The reactants and products are at the SRS
• The HHV represents fully condensed water vapor
• The LHV represents all water vapor
These values provide a basis for the combustion efficiency,
Actual heat transferred per mole of fuel
HHV or LHVcomb
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Example
25
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
Back to our problem ... Is there
liquid water in the products at the SRS? If so, how much?
8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N
90.1406
8 9 47
0.1406 0.1 MPa 0.01406 MPa 14.06 kPa
52.6 C
w
w w
dp
y
P y P
T
Will water condense?
Since TSRS < Tdp, water will condense
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Example
26
Q
8 18C H
PTA 100%
ProductsCombustionChamberR SRST T
P SRST T
How much water will condense?
At TSRS = 25°C, the mole fraction of water vapor in the
products is,
8 47
vv
v
ny
n
8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N
The mole fraction of the water vapor at 25°C can be found,
0.03142 bar0.03101
1.01325 bar
ww
Py
P