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 2 1 1 2

To calculate the equivalence ratio, we need to first write out the stoichiometric reaction of ethane and oxygen Based on Mass: Based on Mole: (F/O)actual 0.938 1 (F/O)stoichiometric 30/112 = 0.268 1/(3.5) = 0.286 Equivalence Ratio 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 Theoretical Air (%) = X 100 Stoichiometric Air Actual Air – Stoichiometric Air Excess Air (%) = X 100 Stoichiometric Air

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: . . - (Based on Mass) S mph S - mrh p p r r . . S nph S (Based on Mole) - nrh p p r r Hp Hr

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

Specific Molar Enthalpy (kJ/kmol) 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 The standard enthalpy of reaction (hR) is -890,330 kJ/kmol of CH4 at 298K. o Step #1: Do Energy Balance = - = H

Step #2: Calculate H Reactants H Products H HR (To) T To = 298K T2 = 1500K H = HR(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: - nr and np are the stoichiometric coefficients for reactants and products of a chemical reaction h p ( r ( p r 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 ln 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