So that l = 3.75 k = 32.9

Transcription

1 Sheet No. The hemical Reactions ============================================================ 1- Propane ( 3 8 ) is burned with 75 percent excess during a combustion process. ssuming complete combustion, determine the ratio. O 3. N ao bo lo kn balance: 3 = a balance: 8 = b so that b = O balance: = a + b + l so that = 10 + l N balance: 3.76 = k m actual theoritical 8 actual 3 So that l = 3.75 k = n 1 m n cetylene ( ) is burned with stoichiometric amount of during a combustion process. ssuming complete combustion, determine the ratio on a mass and on a mole basis. O 3. N ao bo kn 76 balance: = a balance: = b so that b = 1 O balance: = a + b so that =.5 N balance: 3.76 = k = n 1 m

2 kmol kmol 3- One kmol of ethane ( 6 ) is burned with an unknown amount of during a combustion process. n analysis of the combustion products reveals that the combustion is complete, and there are 3 kmol of free O in the products. Determine (a) the ratio and (b) the percentage of theoretical used during this process. O 3.76N ao bo O kn 6 3 balance: = a balance: 6 = b so that b = 3 O balance: = a + b + 6 so that = 6.5 N balance: 3.76 = k =. act. the n 1 m 1 n 1 m m n act. the So that: the amount of excess used is 85.7% of theoretical used. - Propylene ( 3 6 ) is burned with 50 percent excess during a combustion process. ssuming complete combustion and a total pressure of 105 kpa, determine (a) the ratio and (b) the temperature at which the water vapor in the products will start condensing. O 3. N ao bo lo kn balance: 3 = a balance: 6 = b so that b = 3 O balance: = a + b + l so that = 9 + l N balance: 3.76 = k

3 actual theoritical m n 6 actual So that l =.5 k = n 1 m or the amount of water formed n 3 O O ntotal P O O Ptotal kpa rom steam table at P O = kpa the T sat. = mixture of 0 percent by mass methane ( ) and 80 percent by mass ethanol ( 6 O), is burned completely with theoretical. If the total flow rate of the is 31 /s, determine the required flow rate of O 76 O 3. N ao b O kn balance: * = a = balance: * *6 = b so that b =.393 O balance: = a + b so that =.393 N balance: 3.76 = k = So that the amount of used be m s

4 6- Octane ( 8 18 ) is burned with 50 percent theoretical, which enters the combustion chamber at 5.ssuming complete combustion and a total pressure of 1 atm, determine (a) the ratio and (b) the dew-point temperature of the products. O 3. N ao b O lo kn balance: 8 = a balance: 18 = b so that b = 9 O balance: = a + b + l so that = 5 + l N balance: 3.76 = k m actual theoritical n 18 actual So that l = k = n 1 m or the amount of water formed no 9 O n total P P atm O O total rom steam table at P O = kpa the T sat. = In a combustion chamber, ethane ( 6 ) is burned at a rate of 8 /h with that enters the combustion chamber at a rate of 176 /h. Determine the percentage of excess used during this process. or actual combustion m m or theoretical combustion

5 O 3. N ao b O kn 76 6 balance: = a balance: 6 = b so that b = 3 O balance: = a + b so that = 3.5 N balance: 3.76 = k = n 1 m act. the So that: the amount of excess used is 37.% of theoretical used. 8- certain coal has the following analysis on a mass basis: 8 percent, 5 percent O, percent, 1 percent O, and 10 percent ash. The coal is burned with 50 percent excess. Determine the ratio. or actual combustion: O O 76 O 3. N ao bo lo kn balance: = a balance: = b so that b = O balance: = a + b + l so that = l N balance: 3.76 = k actual theoritica l or theoretical combustion: O O 76 balance: = a O 3. N ao bo kn

6 balance: = b so that b = O balance: = a + b so that = N balance: 3.76 = k actual theoritical So that l = k = Methane ( ) is burned with dry. The volumetric analysis of the products on a dry basis is 5.0 percent O, 0.33 percent O, 11. percent O, and 83.3 percent N. Determine (a) the ratio and (b) the percentage of theoretical used. or actual combustion: a O 3.76N 5.O 0.33O bo 11.O 83. 3N balance: a = = 5.53 balance: a = b so that b = O balance: = * *11. so that =.135 N balance: 3.76*.135 = 83.3 a n 1 m or theoretical combustion: O 3. N ao b O kn 76 balance: 1 = a balance: = b so that b = O balance: = *a + b so that = N balance: 3.76* = n 1 m The percent of theoretical used is 17.16

7 percent of theoretical act the