Solution Found!
n-Pentane is burned with excess air in a continuous
Chapter 4, Problem 4.70(choose chapter or problem)
n-Pentane is burned with excess air in a continuous combustion chamber. (a) A technician runs an analysis and reports that the product gas contains 0.270 mole% pentane, 5.3% oxygen, 9.1 % carbon dioxide, and the balance nitrogen on a dry basis. Assume 100 mol of dry product gas as a basis of calculation, draw and label a flowchart, perform a degree-offreedom analysis based on atomic species balances, and show that the system has -1 degrees of freedom. Interpret this result. (b) Use balances to prove that the reported percentages could not possibly be correct. (c) The technician reruns the analysis and reports new values of 0.304 mole% pentane, 5.9% oxygen, 10.2% carbon dioxide, and the balance nitrogen. Verify that this result could be correct and, assuming that it is, calculate the percent excess air fed to the reactor and the fractional conversion of pentane
Questions & Answers
QUESTION:
n-Pentane is burned with excess air in a continuous combustion chamber. (a) A technician runs an analysis and reports that the product gas contains 0.270 mole% pentane, 5.3% oxygen, 9.1 % carbon dioxide, and the balance nitrogen on a dry basis. Assume 100 mol of dry product gas as a basis of calculation, draw and label a flowchart, perform a degree-offreedom analysis based on atomic species balances, and show that the system has -1 degrees of freedom. Interpret this result. (b) Use balances to prove that the reported percentages could not possibly be correct. (c) The technician reruns the analysis and reports new values of 0.304 mole% pentane, 5.9% oxygen, 10.2% carbon dioxide, and the balance nitrogen. Verify that this result could be correct and, assuming that it is, calculate the percent excess air fed to the reactor and the fractional conversion of pentane
ANSWER:Problem 4.70n-Pentane is burned with excess air in a continuous combustion chamber.(a) A technician runs an analysis and reports that the product gas contains 0.270 mole%pentane, 5.3% oxygen, 9.1 % carbon dioxide, and the balance nitrogen on a dry basis. Assume100 mol of dry product gas as a basis of calculation, draw and label a flowchart, perform adegree-of freedom analysis based on atomic species balances, and show that the system has-1 degrees of freedom. Interpret this result.(b) Use balances to prove that the reported percentages could not possibly be correct.(c) The technician reruns the analysis and reports new values of 0.304% mole pentane, 5.9%oxygen, 10.2% carbon dioxide, and the balance nitrogen. Verify that this result could be correctand, assuming that it is, calculate the percent excess air fed to the reactor and the fractionalconversion of pentane Step-by-step solution Step 1 of 3 ^(a)The equation for the combustion of butane is given below: C512 O 2O + 6H 2 2Assume 100 moles of dry product gas, then the mole fraction of dry product gases are: C5120.0027 O 20.053 CO : 0.091 2 N2 0.853Sketch and label the diagram for this process as shown below:\nHere, n i1moles of pentane in the feed gas, n is moles of oxy2n in the feed air, and n is 3moles of water produced. The moles of nitrogen in the feed air is 3.76 (79/21) times the molesof oxygen in the feed air. The mole fraction of N in the product 2s is one minus the molefraction of other dry gases.The number of degrees of freedom is given by the equation, ndf n unknowns nindep eqnsHere, n unknowns is number of unknowns and n indep equations number of independent equations thatcan be written for the process.In this process, the unknowns are molar flow rates (n , n , n ) so the 1mb2 of3nknowns is 4.The atomic balanced equations can be written for C, H, O each and the mole balance fornitrogen molecule can be written.So, the number of degrees of freedom is, n = n - n df unknowns indep equations = 3 - 3 - 1 = -1Thus, the number of degrees of freedom is .