Chapter 2. Fundamental
Concepts in Understanding
Bioenergy and Biobased
Products
Engineering Thermodynamics

Introduction
•Thermodynamics essential to designing processing systems for biorenewable resources ( Net energy output must be positive!!!
)
•Fundamental concepts include
–Mass balances
–Energy balances
•These lectures not a substitute for a course in engineering thermodynamics

Kinds of Systems
• Isolated system –neither mass nor energy enters the system
• Closed system –mass does not enter or leave the system (no restriction on energy flow)
• Open system –both mass and energy can flow through the system

Describing mass flow through an open system

Mass Balances for Combustion
Processes
• The fuel-oxygen ratio (F/O) = mass of fuel per mass of oxygen (Sometime, F/O ratio could be written based on mole rather than mass.)
• The equivalence ratio ( 
)
• (F/O) stoichiometric is the Fuel-Oxygen ratio at which exactly all the available oxygen is used to burn the fuel completely
• The advantage of using equivalence ratio over fuel– oxidizer ratio is that it does not depend on the units being used.

Mass Balances for Combustion
Processes (Continues)
Example
Consider a mixture of one mole of ethane (C
2
H
6
) and one mole of oxygen (O
2
).
F/O ratio of this mixture based on the mass of fuel and O
2 is:
F/O ratio of this mixture based on the number of moles of fuel and O
2 is:
0.5
2

• To calculate the equivalence ratio, we need to first write out the stoichiometric reaction of ethane and oxygen
(F/O) actual
(F/O) stoichiometric
Equivalence
Ratio
Based on Mass:
0.938
30/112 = 0.268
3.5
Based on Mole:
0.5
1/(7) = 0.143
3.5

Mass Balances for Combustion
Processes (Continues)
• The air in excess of the stoichiometric amount is called the excess air
Actual Air
• Theoretical Air (%) =
Stoichiometric Air
X 100
• Excess Air (%) =
Actual Air – Stoichiometric Air
Stoichiometric Air
X 100

Describing energy flow through an open system

Energy Balance for Open System
*h e
*h e
*H e or h i or h i or H i
= specific enthalpy (energy / mass)
= specific molar enthalpy (energy / mole)
= enthalpy (energy)

Energy Balance for Open System
• Within each inlet and outlet stream, we can have a multiple species. In this case, we need to add the enthalpy contribution from each and every species in the stream:
S p
.
m p h -
S p
.
n p h p p
-
r
S
S r
.
m
.
n r h r r h r
(Based on Mass)
(Based on Mole)
H p
H r

Energy Balance for Open System
Example:

Let us assume that 1 kmole/hr of biogas is produced by anaerobic digestion of animal waste consists of 60% of CH
4 reacts with 1.2 kmol/hr of O
2 and 40% of CO to form CO
2 and H
2
2
(molar basis). The biogas
O (no other products).
Biogas + O
2
T = T o
= 298K
Q
T = T
2
= 1500K
CO
2
+ H
2
O
0.6 CH
4
+ 0.4 CO
2
+ 1.2 O
2
 CO
2
+ 1.2 H
2
O

We want to calculate Q under steady state condition for this example with following additional info.
T (K)
298
1500
CH
-
-
4
Specific Molar Enthalpy (kJ/kmol)
O
2
8,682
-
CO
2
9,364
71,078
H
2
O
9,904
57,999
The standard enthalpy of reaction (
 h o
R
) is -890,330 kJ/kmol of CH
4 at 298K.
Step #1: Do Energy Balance
=
S p
.
n p h p
r
S
.
n r h r
H p
H r
=

H

Step #2: Calculate

H
H

H
R
(T o
)
Reactants
Products
S p

H
.
n p
[h (T
2
) –h (T o
)]
T o
= 298K

H =

H
R
(T o
) +
T
S p n
.
T p
2
= 1500K
[h (T
2
) –h (T o
)]

H = (-890,00)*(0.6) + [1*(71,078-9,364) + (1.2)*(57,999-9,904)] = -414,770 kJ/hr

Energy Balances
• For well-characterized fuels, standard enthalpies of reaction can be calculated from tabulations of specific enthalpies of formation, , of chemical compounds from their elements at a standard state: h
 p
(
 r
(
 r p are the stoichiometric coefficients for reactants and products of a chemical reaction
Example:
Calculate the standard heat of reaction for the dehydrogenation of ethane:
C
2
H
6
 C
2
H
4
+H
2

• Most biomass fuels are not well characterized in terms of their chemical constituents
–Often simpler to perform calorimetric tests on biomass fuels to determine enthalpy of reaction

Thermodynamic efficiency
• Every energy conversion process can be characterized by its thermodynamic efficiency

Chemical Equilibrium
A B
• At equilibrium condition, a rate of forward reaction equals to a rate of reverse reaction = No Net Changes!!!
• Gibbs Free Energy and Entropy are two important thermodynamic properties in understanding chemical equilibrium

Chemical Equilibrium
• At the equilibrium condition, the equilibrium constant (K) can be defined.
• For a reaction involving ideal gases, the equilibrium constant based on partial pressures (K p
) can be expressed as:

Chemical Equilibrium
If

G > 0, then the reaction is not spontaneous.
If

G < 0, then the reaction is spontaneous.
• Everything in nature moves toward the equilibrium condition .
• The relationship between the Gibbs Free Energy and equilibrium condition can be written as: ln