FIRST LAW ANALYSIS OF COMBUSTION SYSTEMS P M V Subbarao Professor Mechanical Engineering Department Efficient Use of Resource thruSufficient Supply of Oxygen…..

Phenomenological Modeling of Combustion Engineering Objective of Combustion: To Create Maximum Possible Temperature through conversion of microscopic potential energy into microscopic kinetic energy. Thermodynamic Strategy for conversion: Constant volume combustion Constant pressure combustion

Classification of Combustion Systems External Combustion Systems: Transfer the thermal energy liberated due to combustion to surroundings. Process Heat Utilization Surroundings. Power generating Surroundings. Air is the source of oxygen. Internal Combustion Systems: Thermal energy liberated due to combustion is sued generate Mechanical Power. Air is a working fluid and sources of oxygen.

Steam Power Plant: An External Combustion system in SSSF Wout Qin Qout Win

Furnace in A Steam Power Plant

First Law Analysis of External Combustion System: SSSF Thermal power plants running on Ranke Cycles use an external combustions system known as Coal (fuel) Fired Steam generator. First Law Analysis of a Combustion System (SSSF) in molar form :

First Law Analysis of A Furnace First Law Analysis of a Furnace (SSSF) in molar form :

Reciprocating IC Engine : An Internal Combust System : Control Mass

Turbojet Engine: : An Internal Combust System : Control Volume

A Rocket on the Way to Orbit

First Law Analysis of An Internal Combustion System First Law Analysis of a Furnace (SSSF) in molar form :

How to use the Calorific Value of Fuel in Design of Combustion Systems???

Enthalpy of Combustion Combustion is an exothermic reaction which releases a large amount of energy. This energy change is redefined as the enthalpy of combustion. The enthalpy change associated with a combustion reaction when the reactants and products are in their respective standard states is called the standard enthalpy of combustion. The standard heats of combustion of several compounds are the amount of energy that can be transferred as heat from the reactor to surroundings and is equal to -∆Hc.

Change in Enthalpy standard states? solid Na? liquid gas Hg? N2? There are many forms (or expressions) of change in enthalpy used in engineering. Hvaporization, Hfusion, Hcombination & Hformation. The most useful is standard enthalpy of formation (H°f): the enthalpy change that accompanies the formation of 1 mole of a compound from elements in their standard states. standard states? solid liquid gas Na? Hg? N2?

Enthalpy of Formation Enthalpies of formation can be used to calculate enthalpy of combustion. Enthalpy of formation: enthalpy to form 1 mole of any substance. Standard Enthalpy of Formation: Enthalpy change when 1 kmol of species is formed in its Standard State at a Specified Temperature from the most stable forms of its constituent elements in their standard forms (at the same temperature).

State of Zero Enthalpy The enthalpy of formation of any elemental substance in its most stable form is always zero. The Form favored in Equilibrium at 1 Atmosphere and specified temp. (usually 298.15 K) Carbon as graphite; oxygen as O2 gas; copper as an elemental solid (metal).

Standard Enthalpy of Formation For CO2 ∆Hf0 CO2(gas) = Standard Enthalpy of combustion/reaction. C (s, graphite) + O2 (g) → CO2(g) Nomenclature: ∆ Hf0 at 25 0C & 1 atm

Standard Enthalpy of Combustion Hess's Law: states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state. If we know the Hf of all reactants & products, we can calculate enthalpy of formation of any compound. C3H8 + 5O2  3CO2 + 4H2O (gas) (gas) (gas) (liquid) Expressed via Hess’s Law: C3H8 --> 3C + 4H2 H1 = -Hf [C3H8(g)] 3C + 3O2 --> 3CO2 H2 = 3Hf [CO2(g)] 4H2 + 2O2 --> 4H2O H3 = 4Hf [H2O(l)] C3H8 + 5O2 --> 3CO2 + 4H2O H°c = H1 + H2 + H3

Enthalpies of combustion of fuels Industrial fuels: fuel value (MJ/kg) Coal 31 - 32 Oil 45 Natural gas 49 Gasoline 48 H2 142

First Law Analysis of A General Combustion System First Law Analysis of a Furnace (SSSF) in molar form :

Variation of Specific Heat of Ideal Gases Air 1.05 -0.365 0.85 -0.39 Methane 1.2 3.25 0.75 -0.71 CO2 0.45 1.67 -1.27 0.39 Steam 1.79 0.107 0.586 -0.20 O2 0.88 -0.0001 0.54 -0.33 N2 1.11 -0.48 0.96 -0.42

g cp cv

Properties of Fuels C0 C1 C2 C3 C4 Fuel Methane -0.29149 26.327 -10.610 1.5656 0.16573 Propane -1.4867 74.339 -39.065 8.0543 0.01219 Isooctane -0.55313 181.62 -97.787 20.402 -0.03095 Gasoline -24.078 256.63 -201.68 64.750 0.5808 Diesel -9.1063 246.97 -143.74 32.329 0.0518

First Order Models for Variable Specific Heats For general ideal gases: ap = 0.9718 – 1.1 kJ/kg.K k1 = 1.32610-4 – 3.39510-4 kJ/kg.K2