صفحه محصول - پاورپوینت واکنش های شیمیایی

پاورپوینت واکنش های شیمیایی (pptx) 45 اسلاید

دسته بندی : پاورپوینت

نوع فایل : PowerPoint (.pptx) ( قابل ویرایش و آماده پرینت )

تعداد اسلاید: 45 اسلاید

قسمتی از متن PowerPoint (.pptx) :

واکنش های شیمیایی 2 فرایند احتراق یک واکنش شیمیایی است که در آن سوخت، اکسید می شود و انرژی آزاد می گردد. سوخت ها معمولا از ترکیبات یا مخلوط هایی شامل کربن و هیدروژن تشکیل یافته اند. مثال هایی از سوخت های هیدروکربنی: CH4 Methane C8H18 Octane Coal Mixture of C, H2, S, O2, N2 and non-combustibles در این بحث ما تنها واکنش های کامل را در نظر می گیریم. اجزای شرکت کننده در فرایند را قبل از انجام فرایند، واکنشگر (reactants) و بعد از واکنش محصول احتراق (products) می نامند. 3 فرایند احتراق، کامل یا استویکیومتریک نامیده می شود وقتی همه کربن و هیدروژن موجود در سوخت بسوزند و به ترتیب به دی اکسید کربن و آب تبدیل گردند. این دو واکنش احتراق کامل به این صورت می باشند. Example 15-1 A complete combustion of octane in oxygen is represented by the balanced combustion equation. The balanced combustion equation is obtained by making sure we have the same number of atoms of each element on both sides of the equation. That is, we make sure the mass is conserved. Note we often can balance the C and H for complete combustion by inspection. 4 The amount of oxygen is found from the oxygen balance. It is better to conserve species on a monatomic basis as shown for the oxygen balance. Note: Mole numbers are not conserved, but we have conserved the mass on a total basis as well as a specie basis. The complete combustion process is also called the stoichiometric combustion, and all coefficients are called the stoichiometric coefficients. In most combustion processes, oxygen is supplied in the form of air rather than pure oxygen. Air is assumed to be 21 percent oxygen and 79 percent nitrogen on a volume basis. For ideal gas mixtures, percent by volume is equal to percent by moles. Thus, for each mole of oxygen in air, there exists 79/21 = 3.76 moles of nitrogen. Therefore, complete or theoretical combustion of octane with air can be written as 5 Air-Fuel Ratio Since the total moles of a mixture are equal to the sum of moles of each component, there are 12.5(1 + 3.76) = 59.5 moles of air required for each mole of fuel for the complete combustion process. Often complete combustion of the fuel will not occur unless there is an excess of air present greater than just the theoretical air required for complete combustion. To determine the amount of excess air supplied for a combustion process, let us define the air-fuel ratio AF as Thus, for the above example, the theoretical air-fuel ratio is 6 On a mass basis, the theoretical air-fuel ratio is Percent Theoretical and Percent Excess Air In most cases, more than theoretical air is supplied to ensure complete combustion and to reduce or eliminate carbon monoxide (CO) from the products of combustion. The amount of excess air is usually expressed as percent theoretical air and percent excess air. 7 Show that these results may be expressed in terms of the moles of oxygen only as Example 15-2 Write the combustion equation for complete combustion of octane with 120 percent theoretical air (20 percent excess air). Note that (1)(12.5)O2 is required for complete combustion to produce 8 kmol of carbon dioxide and 9 kmol of water; therefore, (0.2)(12.5)O2 is found as excess oxygen in the products. 8 Second method to balance the equation for excess air (see the explanation of this technique in the text) is: Incomplete Combustion with Known Percent Theoretical Air Example 15-3 Consider combustion of C8H18 with 120 % theoretical air where 80 % C in the fuel goes into CO2. 9 O balance gives Why is X > 2.5? Then the balanced equation is Combustion Equation When Product Gas Analysis Is Known Example 15-4 Propane gas C3H8 is reacted with air such that the dry product gases are 11.5 percent CO2, 2.7 percent O2, and 0.7 percent CO by volume. What percent theoretical air was supplied? What is the dew point temperature of the products if the product pressure is 100 kPa? We assume 100 kmol of dry product gases; then the percent by volume can be interpreted to be mole numbers. But we do not know how much fuel and air were supplied or water formed to get the 100 kmol of dry product gases.