2NOTE: you will need the table of enthalpy values given in lesson MAEB-4-3 to enable you to complete...

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2NOTE: you will need the table of enthalpy values given in lesson MAEB-4-3 to enable you to complete this TMA.Attempt ALL questions. 	A fuel gas has the following volumetric composition:   	 		 			CO2CO 			49c6% 		 		 			CH4 			309r 		 		 			H 			50 0 		 		 			O2 			2% 		 		 			N2 			8% 		 	  It is burned to produce a flue gas which contains on a dry basis by volume:CO;, 8%0 2 5 49cN2 86.69•Calculate: 	the theoretical air supply 	the actual air supply 	the No excess air.     3 	A fuel consists of carbon and hydrogen only. When completely burned in air the flue gas on a dry basis has the following analysis:  CO2 129r0 2 5%N2 83%The temperature of the flue gases produced is 1600°C and these are fed to a boiler used to raise steam. 15% of the heat available at 1600°C is lost to the surroundings and the flue gases leave the boiler at 300°C.Calculate: 	the amount of air supplied     	the composition of the fuel  (in) the amount of steam raised per hour if the enthalpy change from water feed to steam product is 2490 kJ kg* and the flow of wet flue gas = 100 kmol h‘*. 	A power plant burns 100 to methane (CH4), supplied at 25°C and1 atmosphere pressure. 209a excess air is supplied at the same conditions.     	Write a balanced equation for the reaction.     	Calculate the heat of combustion of methane at 25°C and 1 atmosphere from the data given below.     	Assuming complete combustion of the methane to carbon dioxide  , and water and no heat loss to the surroundings, confirm that the theoretical flame temperature will be approximately 2070 K. Thank you for using www.freepdfconvert.com service!Only two pages are converted. Please Sign Up to convert all pages. https://www.freepdfconvert.com/membership

David · Accepted Answer

1) Theoretical air is the quantity of air that contains the theoretical oxygen. It is the minimum air 
requires to burn fuel completely so that all the carbon are combusted. 
In the given Volume composition, the organic substance methane is present as 30%. 
To combust this amount of methane completely, we require the amount of oxygen which is 
stoichiometric ratio with respect to balanced equation. 
Therefore – 
The balanced equation for the combustion reaction is – 
       4 2 2 2CH 2O CO 2H Og g g g    
So every 1 mol of methane requires 2 mols oxygen theoretically, here mixture contains 30% 
methane, thus it requires 60% of oxygen to complete the combustion process. 
a)  Let there are 100  mL of total volume – 
Therefore number of mols of methane present = (30 mL)(1 mol /22400 mL) = 0.001339 mol 
To get the complete combustion, it requires 2*0.001339 mol = 0.002678 mol 
O2 in fuel = (2 mL /22400 mL)1 mol = 8.92*10-5  mol 
O2 theoretical = (0.002678 mol – 8.92*10-5 mol) = 0.

2 NOTE: you will need the table of enthalpy values given in lesson MAEB-4-3 to enable you to complete this TMA. Attempt ALL questions. A fuel gas has the following volumetric composition: CO2 CO 49c...

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