Transcript Slides

ME 475/675 Introduction to
Combustion
Lecture 6
Constant pressure and volume adiabatic flame temperature, Chemical
Equilibrium, Mixture entropy evaluation
Announcements
• Monday, Labor Day Holiday
• HW 2 due Wednesday, Ch 2 (25, 30, 31, 32, 35)
• Due Now (before lecture starts)
• Extra Credit example
Adiabatic (𝑄 = 0) Flame Temperature
Complete Combustion Products
CCO2 HH2O
PP = PR, T = TAd
Stoichiometric
Reactants
TR PR
𝑄𝐼𝑁 = 0
𝑊𝑂𝑈𝑇 = 0
• 1st Law, Steady Flow Reactor
• 𝑄𝐼𝑁 − 𝑊𝑂𝑈𝑇 = 0 = 𝐻𝑃 − 𝐻𝑅 = 𝑚 ℎ𝑃 − ℎ𝑅
• All chemical energy goes into heating the
products
• To find adiabatic flame temperature use
• PP = PR and ℎ𝑃 = ℎ𝑅
Constant pressure Evaluation method 𝑇𝐴𝑑,𝑃
• 𝐻𝑅 𝑇𝑅 = 𝐻𝑃 𝑇𝐴𝑑,𝑃
•
𝑜
𝑜
𝑁
ℎ
+
Δ
ℎ
𝑠
𝑅 𝑖 𝑓
•
• For 𝑇𝑅 = 𝑇𝑟𝑒𝑓
𝑜
𝑁
ℎ
𝑅 𝑖 𝑓 𝑖−
𝑖
= 298.15 𝐾
• 𝐻𝑐 =
•
𝑜
𝑜
𝑁
ℎ
+
Δ
ℎ
𝑠
𝑃 𝑖 𝑓
𝑜
𝑁
ℎ
𝑃 𝑖 𝑓
𝑜
𝑁
ℎ
𝑅 𝑖 𝑓
−
𝑖
=
𝑜
𝑁
ℎ
𝑃 𝑖 𝑓
𝑖
𝑜
𝑁
Δ
ℎ
𝑠
𝑃 𝑖
𝑖
𝑖
• Molecular constant-pressure heat of combustion
• 𝐻𝑐 = 𝑃 𝑁𝑖 Δℎ𝑠𝑜 𝑖 Could iterate Problem 2.32
𝑇
Δℎ𝑠𝑜 𝑖 = 𝑇 𝐴𝑑,𝑃
𝑐𝑝,𝑖 𝑇 𝑑𝑇 ≈ 𝑐𝑝,𝑖,𝑎𝑣𝑔 𝑇𝐴𝑑,𝑃
=𝑇
𝑅
𝑟𝑒𝑓
• 𝐻𝑐 =
•
=
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 𝑇𝐴𝑑,𝑃
𝐻𝑐
𝑇𝐴𝑑,𝑃 − 𝑇𝑅 =
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔
− 𝑇𝑅
𝑖
− 𝑇𝑅
Constant Volume Adiabatic Flame Temperature
𝑄=0
V, m
𝑊=0
• 𝑉𝑃 = 𝑉𝑅 = 𝑉
• 𝑄 − 𝑊 = 0 = 𝑈2 − 𝑈1 = 𝑈𝑃 − 𝑈𝑅 so 𝑈𝑅 = 𝑈𝑃
• Use definition: 𝑈 = 𝐻 − 𝑃𝑉 (since standard internal energy U is not tabulated)
• 𝐻𝑅 − 𝑃𝑅 𝑉 = 𝐻𝑃 − 𝑃𝑃 𝑉
• Idea gas: 𝑃𝑅 𝑉 = 𝑁𝑅 𝑅𝑢 𝑇𝑅 ; 𝑃𝑃 𝑉 = 𝑁𝑃 𝑅𝑢 𝑇𝑃 = 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑣 ,
• 𝐻𝑅 𝑇𝑅 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 = 𝐻𝑃 𝑇𝐴𝑑,𝑣 − 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑.𝑣
• Only 𝑇𝐴𝑑,𝑣 and 𝐻𝑅 𝑇𝐴𝑑,𝑣 are unknown
Constant volume evaluation methods
• 𝐻𝑅 𝑇𝑅
− 𝑁𝑅 𝑅𝑢 𝑇𝑅 = 𝐻𝑃 𝑇𝐴𝑑,𝑉
•
𝑜
𝑜
𝑁
ℎ
+
Δ
ℎ
𝑠
𝑅 𝑖 𝑓
•
𝑜
𝑁
ℎ
𝑅 𝑖 𝑓
• 𝐻𝑐 =
𝑖
−
𝑖
𝑜
𝑁
ℎ
𝑅 𝑖 𝑓 𝑖
• 𝐻𝑐 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 =
𝑜
𝑜
𝑁
ℎ
+
Δ
ℎ
𝑠
𝑃 𝑖 𝑓
− 𝑁𝑅 𝑅𝑢 𝑇𝑅 =
𝑜
𝑁
ℎ
𝑃 𝑖 𝑓
−
𝑖
− 𝑁𝑅 𝑅𝑢 𝑇𝑅 =
𝑜
𝑁
ℎ
𝑃 𝑖 𝑓 𝑖;
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔
• 𝑇𝐴𝑑,𝑉 − 𝑇𝑅 =
− 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉
𝑇 𝑑𝑇 ≈ 𝑐𝑝,𝑖,𝑎𝑣𝑔 𝑇𝐴𝑑,𝑉 − 𝑇𝑅
𝑇𝐴𝑑,𝑉 − 𝑇𝑅 − 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 = 𝑇𝐴𝑑,𝑉
𝐻𝑐 −𝑁𝑅 𝑅𝑢 𝑇𝑅 +𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔
• 𝑇𝐴𝑑,𝑉 − 𝑇𝑅 =
− 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉 (𝑇𝑅 = 𝑇𝑟𝑒𝑓 )
𝑖
𝑜
𝑁
Δ
ℎ
𝑠 𝑖
𝑃 𝑖
𝑇𝐴𝑑,𝑉
𝑜
Δℎ𝑠 𝑖 = 𝑇 =𝑇 𝑐𝑝,𝑖
𝑅
𝑟𝑒𝑓
• 𝐻𝑐 − 𝑁𝑅 𝑅𝑢 𝑇𝑅 + 𝑇𝑅
• 𝑇𝐴𝑑,𝑉 =
− 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔
− 𝑁𝑃 𝑅𝑢 𝑇𝐴𝑑,𝑉
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑁𝑃 𝑅𝑢
𝐻𝑐 −𝑁𝑅 𝑅𝑢 𝑇𝑅 +𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑇𝑅 𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑁𝑃 𝑅𝑢
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑁𝑃 𝑅𝑢
𝐻𝑐 + 𝑁𝑃 −𝑁𝑅 𝑅𝑢 𝑇𝑅
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 −𝑁𝑃 𝑅𝑢
=
𝐻𝑐 + 𝑁𝑃 −𝑁𝑅 𝑅𝑢 𝑇𝑅
𝑃 𝑁𝑖 𝑐𝑝,𝑖,𝑎𝑣𝑔 − 𝑃 𝑁𝑖 𝑅𝑢
=
𝐻𝑐 + 𝑁𝑃 −𝑁𝑅 𝑅𝑢 𝑇𝑅
𝑃 𝑁𝑖 𝑐𝑣,𝑖,𝑎𝑣𝑔
Chemical Equilibrium
𝑄=0
V, m
𝑊=0
𝑄𝐼𝑁 = 0
𝑊𝑂𝑈𝑇 = 0
• “Ideal” combustion assumes CCO2 HH2O
• CxHy + a(O2+3.76N2)  _b_CO2 + _c_ H2O + _d_ N2 + _e_ CXHY + _f_ O2
• Specifies product mole fractions in the products
• To find Tad (1st law)
• Constant Volume: 𝑈𝑅 = 𝑈𝑃
• Constant Pressure: 𝐻𝑃 = 𝐻𝑅
• In real high temperature combustion the product species may dissociate (break apart)
yielding: H2, OH, CO, H, O, N, NO, …. (no CxHy?)
• How to find actual final molar fractions of each component for different temperatures
and pressures?
• Chemical Equilibrium (2nd Law)
Constant Volume Reaction Constraints
• Conservation of atoms (C, H, O, N)
• For Constant Volume and no interaction with surroundings
• Q=W=0
• Time-dependent Ideal Gas Equation of State: P(t)V = N(t)RuT(t)
• Initially: T1 P1 and 𝜒𝑖 1 (initial mole fraction of each specie)
• Find: T2 P2 and 𝜒𝑖
2
• Before we specified final mole fractions 𝜒𝑖
• Now we find them.
2
• Chemical Equilibrium
• 1st and 2nd laws of Thermodynamics
• Reaction stops when entropy reaches a maximum
V, m
𝑄=𝑊=0
Example: Carbon Dioxide Formation/Dissociation
• 𝐶𝑂
1
+ 𝑂2
2
→ 𝐶𝑂2
• At high temperature some of the 𝐶𝑂2 will
1
dissociate and form 𝐶𝑂 + 𝑂2
𝛼
2
2
• −𝛼𝐶𝑂2 + 𝛼𝐶𝑂 + 𝑂2
• 𝐶𝑂
1
+ 𝑂2
2
→ 1 − 𝛼 𝐶𝑂2 + 𝛼𝐶𝑂 +
𝛼
𝑂
2 2
• Were 𝛼 =? is the fraction of 𝐶𝑂2 that dissociates
• 0≤𝛼≤1
• How do we find 𝛼?
• First step: Find 𝑇𝐴𝑑 = 𝑓𝑛(𝛼) [as a function of 𝛼]
• Find 𝑁𝑖 = 𝑓𝑛(𝛼) and 𝜒𝑖 𝑃 = 𝑓𝑛(𝛼)
• Use 1st Law (𝑈𝑃 = 𝑈𝑅 ) to find 𝑇𝐴𝑑 = 𝑓𝑛(𝛼)
• Use ideal gas law to find 𝑃𝑃 = 𝑃𝑅
𝑇𝐴𝑑
𝑇𝑅
= 𝑓𝑛(𝛼)
• How much dissociation (𝛼) will actually take place?
1−𝛼
Find Entropy of Product Mixture
• Entropy is a function of composition, temperature, and pressure
• Product Mixture
• 𝐶𝑂
1
+ 𝑂2
2
→ 1 − 𝛼 𝐶𝑂2 + 𝛼𝐶𝑂
• Extensive Property
𝛼
+ 𝑂2
2
• 𝑆𝑀𝑖𝑥 𝑇𝑓 , 𝑃 = 𝑛𝑖=1 𝑁𝑖 𝑠𝑖 𝑇𝑓 , 𝑃𝑖
𝛼
= 1 − 𝛼 𝑠𝐶𝑂2 𝑇𝑓 , 𝑃𝐶𝑂2 + 𝛼 𝑠𝐶𝑂 𝑇𝑓 , 𝑃𝐶𝑂 + 𝑠𝑂2 𝑇𝑓 , 𝑃𝑂2
2
• Where partial pressure of species 𝑖 is: 𝑃𝑖 = 𝜒𝑖 𝑃𝑓
• 𝑃𝐶𝑂2 =
• 𝑃𝐶𝑂 =
• 𝑃𝑂2 =
1−𝛼
𝛼 𝑃𝑓 =
1−𝛼 +𝛼+ 2
𝛼
1−𝛼
𝛼
1+ 2
𝑃𝑓
𝛼𝑃
1+ 2 𝑓
𝛼
2
𝛼
1+ 2
𝑃𝑓
• How does 𝑆𝑀𝑖𝑥 𝑇𝑓 , 𝑃 𝑇𝑓
vary with 𝑇𝑓 ?
Need 𝑠𝑖 𝑇𝑓 , 𝑃𝑖
Gibb’s
nd
2
equation
• 𝑇𝑑𝑠 = 𝑑 ℎ − 𝑣𝑑𝑃
• 𝑑𝑠 =
𝑑ℎ
𝑇
𝑣
𝑇
− 𝑑𝑃 =
𝑐𝑃 (𝑇)
𝑑𝑇
𝑇
−
𝑑𝑃
𝑅𝑢
𝑃
• For Ideal Gases: 𝑑ℎ = 𝑐𝑃 (𝑇)𝑑𝑇; 𝑃𝑣 = 𝑅𝑢 𝑇
•
𝑇,𝑃
𝑑𝑠
𝑇𝑅𝑒𝑓, 𝑃𝑜
=
𝑇
𝑐𝑃 (𝑇)
𝑑𝑇
𝑇𝑅𝑒𝑓 𝑇
• 𝑠 𝑇, 𝑃 = 𝑠 𝑇𝑅𝑒𝑓, 𝑃
𝑜
+
−
𝑃 𝑑𝑃
𝑅𝑢 𝑃𝑜
𝑃
𝑇
𝑐𝑃 (𝑇)
𝑑𝑇
𝑇𝑅𝑒𝑓 𝑇
• Standard entropy
• 𝑠 𝑜 (𝑇) = 𝑠 𝑇𝑅𝑒𝑓, 𝑃𝑜 +
• Appendix A, pp 687-699
𝑇
𝑐𝑃 (𝑇)
𝑑𝑇
𝑇𝑅𝑒𝑓 𝑇
− 𝑅𝑢 ln
𝑃
𝑃𝑜
𝑜
= 𝑠 (𝑇) − 𝑅𝑢 ln
𝑃
𝑃𝑜
Equilibrium
• Plot 𝑆𝑀𝑖𝑥 𝑇𝑓 versus 1 − 𝛼
• Maximum at 1 − 𝛼 = 0.51
• For Q = W = 0 (constant volume) and constant mass, 2nd law requires 𝑑𝑠 > 0
• Once a system gets to 𝑆𝑀𝑖𝑥 𝑚𝑎𝑥 it can no longer change spontaneously (on its
own)
• At equilibrium for given U, V and m: 𝑑𝑠 𝑈,𝑉,𝑚 = 0
End 2015
Find equilibrium composition
(reactants and products) for a given
Temperature, Pressure & Mass
𝑊=
𝑃𝑑𝑉
G
Q
T,P
• aA + bB + …  eE + fF + …
• Use Q and boundary work 𝑊 = 𝑃𝑑𝑉 to achieve P and T
• Gibbs Function: 𝐺 𝑇, 𝑃 = 𝐻 𝑇 − 𝑇𝑆 𝑇, 𝑃
• 𝐺𝑀𝑖𝑥 = 𝑁𝑖 𝑔𝑖
• 𝑔𝑖 = ℎ𝑖 𝑇 + 𝑇𝑠𝑖 𝑇, 𝑃 = ℎ𝑖 𝑇 + 𝑇 𝑠𝑖𝑜 (𝑇) − 𝑅𝑢 ln
𝑃
𝑃𝑜
a
= 𝑔𝑖𝑜 𝑇 + 𝑇𝑅𝑢 ln
𝑃
𝑃0
0
• 𝑔𝑖𝑜 𝑇 is Gibbs function at P = Po = 1 atm (what is difference between 𝑔𝑖𝑜 𝑇 and 𝑔𝑖,𝑇
?)
• For a constant temperature, pressure, and mass, and only boundary work (PdV)
• It can be shown that second Law requires 𝑑𝐺 𝑇,𝑃,𝑚 ≤ 0 for spontaneous changes (no
outside influence) (don’t spent too much time explaining this)
• 𝑑𝐺
• 𝑑𝐺𝑀𝑖𝑥 =
𝑇,𝑃,𝑚
=0
0
𝑑𝑁𝑖 𝑔𝑖,𝑇
It can be shown…
+
𝑃𝑖
𝑅𝑢 𝑇 ln 𝑜
𝑃
+
𝑁𝑖 𝑑
0
𝑔𝑖,𝑇
+
𝑃𝑖
𝑅𝑢 𝑇 ln 𝑜
𝑃
=0
For General Reactions
• aA + bB + …  eE + fF + …
• dNA = -a(dk), dNB = -b(dk)
• dNE = +e(dk), dNF = +f(dk)
•
𝑃𝑖
+ 𝑅𝑢 𝑇 ln 𝑜 = 0
𝑃
𝑃𝐴
𝑃𝐵
0
0
𝑔𝐴,𝑇 + 𝑅𝑢 𝑇 ln 𝑜 −b(dk) 𝑔𝐵,𝑇 + 𝑅𝑢 𝑇 ln 𝑜 − ⋯
𝑃
𝑃
𝑃𝐸
𝑃𝐹
𝑜
𝑜
+e(dk) 𝑔𝐸,𝑇 + 𝑅𝑢 𝑇 ln 𝑜 +f(dk) 𝑔𝐹,𝑇 + 𝑅𝑢 𝑇 ln 𝑜
0
𝑑𝑁𝑖 𝑔𝑖,𝑇
• −a(dk)
• 𝑅𝑢 𝑇 ln
𝑃𝐸 𝑒 𝑃𝐹 𝑓
…
𝑃𝑜
𝑃𝑜
𝑃𝐴 𝑎 𝑃𝐵 𝑏
…
𝑃𝑜
𝑃𝑜
𝑃
𝑃
=0
𝑜
𝑜
𝑜
𝑜
= − 𝑒𝑔𝐸,𝑇
+ 𝑓𝑔𝐹,𝑇
+ ⋯ − 𝑎𝑔𝐴,𝑇
− 𝑏𝑔𝐵,𝑇
−⋯
Solution Method
• 𝑅𝑢 𝑇 ln 𝐾𝑃 = − Δ𝐺𝑇𝑜
• 𝐾𝑃 = exp
−Δ𝐺𝑇𝑜
𝑅𝑢 𝑇
• Equilibrium Constant
• 𝐾𝑃 =
𝑃𝐸 𝑒 𝑃𝐹 𝑓
…
𝑃𝑜
𝑃𝑜
𝑃𝐴 𝑎 𝑃𝐵 𝑏
…
𝑃𝑜
𝑃𝑜
=
𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
• Standard State Gibbs Function Change
𝑜
𝑜
𝑜
𝑜
• Δ𝐺𝑇𝑜 = 𝑒𝑔𝐸,𝑇
+ 𝑓 𝑔𝐹,𝑇
+ ⋯ − 𝑎𝑔𝐴,𝑇
− 𝑏𝑔𝐵,𝑇
− ⋯ = 𝑓𝑛(𝑇)
𝑜
𝑜
𝑜
𝑜
• Δ𝐺𝑇𝑜 = 𝑒𝑔𝑓,𝐸
+ 𝑓𝑔𝑓,𝐹
+ ⋯ − 𝑎𝑔𝑓,𝐴
− 𝑏𝑔𝑓,𝐵
− ⋯ = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
𝑜
• In terms of Gibbs functions of formation 𝑔𝑓,𝑖
= 𝑓𝑛(𝑇) (tabulated in App. A and B)
• Products are “favored” as 𝐾𝑃 increases, which happens when Δ𝐺𝑇𝑜 decreases
Example 2.7 (p 43) Turn in next time for EC
• Consider the dissociation of CO2 as a function of temperature and pressure:
1
𝐶𝑂2 ↔ 𝐶𝑂 + 𝑂2 .
2
• Find the composition of the mixture, i.e. the mole fractions of 𝐶𝑂2 , 𝐶𝑂 and 𝑂2 ,
that results from subjecting the originally pure 𝐶𝑂2 to various temperatures, and
pressures
• T = 1500, 2000, 2500 and 3000K
• P = 0.1, 1, 20 and 100 atm
• Find final values of: 𝜒𝐶𝑂2 , 𝜒𝐶𝑂 , 𝜒𝑂2
• Initially 𝜒𝐶𝑂2 =___?
• MathCAD solution
• http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/index.htm
Pressure and Temperature
Dependence
• For this reaction
1
• 𝐶𝑂2 ↔ 𝐶𝑂 + 𝑂2
2
𝑜
• Δ𝐺𝑇𝑜 = Δ𝐻 − 𝑇Δ𝑆 𝑜 > 0 (Δ𝐻 𝑜 > 0 exothermic) and Nproducts > Nreactants
• 𝜒𝐶𝑂 (product) increases as T increases and P decreases
Pressure and Temperature
Dependence
• 𝐾𝑃 =
𝑃𝐸 𝑒 𝑃𝐹 𝑓
…
𝑃𝑜
𝑃𝑜
𝑃𝐴 𝑎 𝑃𝐵 𝑏
…
𝑃𝑜
𝑃𝑜
=
𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
= exp
−Δ𝐺𝑇𝑜
𝑅𝑢 𝑇
;
• 𝐾𝑃 Products are “favored” as KP increases and Δ𝐺𝑇𝑜 = Δ𝐻 𝑜 − 𝑇Δ𝑆 𝑜 decreases
• Standard State Gibbs Function Change
𝑜
𝑜
𝑜
𝑜
• Δ𝐺𝑇𝑜 = 𝑒𝑔𝑓,𝐸
+ 𝑓𝑔𝑓,𝐹
+ ⋯ − 𝑎𝑔𝑓,𝐴
− 𝑏𝑔𝑓,𝐵
− ⋯ = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
Gibbs Function
• Helps find equilibrium composition for given initial 𝑇1 , 𝑃1 and 𝜒𝑖
1
• 𝐺 = 𝐻 − 𝑇𝑆 = 𝑁𝑔 = 𝑁 ℎ − 𝑇𝑠
• For a fixed temperature, pressure and mass, all spontaneous changes in the
absence of all work except boundary work, must cause changes to G such that
𝑑𝐺 𝑇,𝑃,𝑚 ≤ 0 (G must decrease)
• At equilibrium 𝑑𝐺
𝑇,𝑃,𝑚
=0
• For mixtures
• 𝐺𝑀𝑖𝑥 =
𝑁𝑖 𝑔𝑖 (𝑇, 𝑃𝑖 )
• 𝑔𝑖 𝑇, 𝑃𝑖 = ℎ𝑖 𝑇 − 𝑇𝑠𝑖 𝑇, 𝑃 = ℎ𝑖 𝑇 − 𝑇 𝑠𝑖 𝑇𝑅𝑒𝑓, 𝑃𝑜 +
𝑜
= 𝑔𝑖,𝑇
− 𝑇𝑅𝑢 ln
𝑃𝑖
𝑃𝑜
𝑐𝑃,𝑖 𝑇
𝑇
𝑇𝑅𝑒𝑓 𝑇
𝑑𝑇 − 𝑇𝑅𝑢 ln
𝑃𝑖
𝑃𝑜
Gibbs Function of pure species i as standard pressure
•
𝑜
𝑔𝑖,𝑇
= ℎ𝑖 𝑇 − 𝑇 𝑠𝑖 𝑇𝑅𝑒𝑓, 𝑃
𝑜
+
𝑐𝑃,𝑖 (𝑇)
𝑇
𝑑𝑇
𝑇𝑅𝑒𝑓 𝑇