Transcript Engineering Thermodynamics

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 (C2H6) and one mole
of oxygen (O2).
F/O ratio of this mixture based on the mass of fuel and O2 is:
F/O ratio of this mixture based on the number of moles of fuel
and O2 is:
2
0.5
• To calculate the equivalence ratio, we need to first write out
the stoichiometric reaction of ethane and oxygen
Based on Mass:
(F/O)actual
(F/O)stoichiometric
Equivalence
Ratio
0.938
Based on Mole:
0.5
30/112 = 0.268
1/(7) = 0.143
3.5
3.5
Mass Balances for Combustion
Processes (Continues)
• The air in excess of the stoichiometric amount is called the
excess air
Actual Air
X 100
• Theoretical Air (%) =
Stoichiometric Air
• Excess Air (%) =
Actual Air – Stoichiometric Air
Stoichiometric Air
X 100
Describing energy flow through an
open system
Energy Balance for Open System
*he or hi = specific enthalpy (energy / mass)
*he or hi = specific molar enthalpy (energy / mole)
*He or Hi = 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:
.
Sm h
.
- Smh
.
Sn h
.
- Snh
p
p
-
p p
p p
Hp
r r
r
r
r r
Hr
(Based on Mass)
(Based on Mole)
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 CH4 and 40% of CO2 (molar basis). The biogas
reacts with 1.2 kmol/hr of O2 to form CO2 and H2O (no other products).
Biogas + O2
T = To = 298K
Q
T = T2 = 1500K
CO2 + H2O
0.6 CH4 + 0.4 CO2 + 1.2 O2  CO2 + 1.2 H2O
We want to calculate Q under steady state condition for this example with
following additional info.
Specific Molar Enthalpy (kJ/kmol)
T (K)
CH4
O2
CO2
H2O
298
-
8,682
9,364
9,904
1500
-
-
71,078
57,999
o
The standard enthalpy of reaction (hR) is -890,330 kJ/kmol of CH4 at 298K.
Step #1: Do Energy Balance
=
.
Sn h
p
p p
Hp
.
- Snh
r
-
r r
Hr
=
H
Step #2: Calculate H
Reactants
H
Products
H
.
Sn
HR (To)
p
p [h p (T2)
–h p (To)]
T
To = 298K
T2 = 1500K
.
Sn
H = HR(To) + p
p [h p (T2)
–h p (To)]
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 (
- npr and nrp are the stoichiometric coefficients for
reactants and products of a chemical reaction
Example:
Calculate the standard heat of reaction for the dehydrogenation of ethane:
C2H6  C2H4 +H2
• 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 (Kp) 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