Question Set Up In homework 7 problem one you established th

Question Set Up: In homework 7 problem one you established the thermal efficiency of a natural gas-fired industrial boiler operating with excess air such that the O2 concentration in the flue gases was 2% (volume or molar), measured after the removal of water vapor from the combustion products. The temperature of the gases leaving the boiler was 700 K. You assumed the natural gas was essentially methane, CH4. The air and methane entered the combustion chamber at 298 K

Chemical Eq: CH4 +2.188 (O2+3.76 N2)-->CO2 + 2 H2O + 0.188 O2 + 8.22 N2

Answer needed: What is the efficiency if the incoming air is preheated with a preheater that reduces the flue gas temperature to 500 K? Where efficiency=Qout/ LHV (methane) Hint you need to be clever with the control volume describing your system.

The Correct answer is 91%. I need the steps to get there.

Solution

Introduction - Up to the present purpose the warmth alphabetic character all told issues and examples was either a given price or was obtained from the primary Law relation. but in varied heat engines, gas turbines, and steam power plants the warmth is obtained from combustion processes, exploitation either solid fuel (e.g. coal or wood). liquid fuel (e.g. gasolene, kerosine, or diesel fuel), or vaporific fuel (e.g. fossil fuel or propane).

In this chapter we have a tendency to introduce the chemistry and natural philosophy of combustion of genericorganic compound fuels - (CxHy), within which the oxydizer is that the chemical element contained inatmospherical air. Note that we are going to not cowl the combustion of solid fuels or the advanced blends and mixtures of the hydrocarbons that frame gasoline, kerosene, or diesel fuels.

Atmospheric Air contains just about twenty first chemical element (O2) by volume. the opposite seventy nine of \"other gases\" is generally element (N2), therefore we are going to assume air to be composed of twenty firstchemical element and seventy nine element by volume. therefore every mole of chemical element required to oxidize the organic compound is among 79/21 = three.76 moles of element. exploitation this mix the molecular mass of air becomes twenty nine [kg/kmol]. Note that it\'s assumed that the element won\'t unremarkably bear anyreaction.

The Combustion method - the fundamental combustion method may be delineate by the fuel (the hydrocarbon)and oxydizer (air or oxygen) known as the Reactants, that bear a action whereas emotional heat to create themerchandise of combustion specified mass is preserved. within the simplest combustion method, called ratioCombustion, all the carbon within the fuel forms carbonic acid gas (CO2) and every one the chemical elementforms water (H2O) within the merchandise, therefore we will write the reaction as follows:


where z is thought because the ratio constant for the oxidant (air)

Note that this reaction yields 5 unknowns: z, a, b, c, d, therefore we want 5 equations to resolve. ratio combustion assumes that no excess chemical element exists within the merchandise, thus d = 0. we have a tendency toacquire the opposite four equations from equalization the amount of atoms of every component within thereactants (carbon, hydrogen, chemical element and nitrogen) with the amount of atoms of these componentswithin the merchandise. this suggests that no atoms ar destroyed or lost in a very combustion reaction.