NAME DEPARTMENT UNI ROLL NO CLASS ROLL SEMESTER SUBJECT SUBJECT CODE - Suman singha -ME - 10500722067 - OF _ 4TH - APPLIED THERMODYNAMICS ~ PC-ME-401INDEX TOPICS OF TECHNICAL REPORT Types of Liquid Fuels Advantages and disadvantages of Liquid fuel. Theoretical and Excess air Stoichiometric air-fuel (A/F) ratio Numerical problem © scanned with OKEN ScannerTypes of Liquid Fuels: ® Gasoline: Derived primarily from crude oil, gasoline is widely used in vehicles with internal combustion engines. Diesel: Also derived from crude oil, diesel fuel is commonly used in trucks, buses, trains, and some cars. Kerosene: Used for heating, lighting, and cooking, kerosene is a lighter fraction of crude oil. Jet Fuel: Specifically designed for use in aircraft, jet fuel is a highly refined form of kerosene. Bio fuels: These include ethanol, biodiesel, and other renewable fuels derived from organic materials like corn, sugarcane, or vegetable oils. Advantages and disadvantages of Liquid fuel: High Energy Density: Liquid fuels have high energy content per unit volume, making them efficient for transportation and energy production. Convenient Storage: They can be easily stored and transported in tanks, barrels, or pipelines, allowing for widespread distribution. Versatility: Liquid fuels can power a wide range of vehicles and equipment, from cars and trucks to airplanes and ships. Readily Available: Crude oil, the primary source of liquid fuels, is found in many regions globally, ensuring a steady supply. Disadvantages of Liquid Fuels: Environmental Impact: Burning liquid fuels releases carbon dioxide and other pollutants, contributing to air pollution and climate change. Finite Resource: Crude oil reserves are finite, leading to concerns about depletion and price volatility. Dependency on Imports: Many countries rely on imported crude oil and refined liquid fuels, making them vulnerable to supply disruptions and geopolitical tensions. Safety Concerns: Liquid fuels can be flammable and pose risks during transportation, storage, and handling. Environmental Degradation: Oil spills from extraction, transportation, or accidents can cause significant harm to ecosystems and wildlife. © scanned with OKEN ScannerTheoretical and Excess air Theoretical Air: This refers to the precise amount of air required for the complete combustion of a fuel, assuming perfect mixing and ideal conditions. The theoretical air-to-fuel ratio is calculated based on the stoichiometric equation, which balances the chemical equation for combustion. Achieving theoretical air ensures complete combustion, where all the fuel reacts with oxygen to produce only carbon dioxide, water, and other combustion products without any unburned fuel or pollutants. Excess Air: Excess air refers to the additional amount of air supplied beyond the theoretical requirement for combustion. It is expressed as a ratio of the actual air supplied to the theoretical air requirement. Providing excess air is a common practice in combustion processes to ensure complete combustion and reduce emissions. However, excessive excess air can lead to inefficiencies, increased fuel consumption, and higher emissions of nitrogen oxides (NOx) due to the higher combustion temperatures. In summary, theoretical air is the exact amount of air needed for complete combustion, while excess air is the additional air supplied to ensure complete combustion and control emissions. Balancing the amount of excess air is crucial for optimizing combustion efficiency and minimizing environmental impacts. © scanned with OKEN ScannerStoichiometric air-fuel (A/F) ratio The stoichiometric air-fuel (A/F) ratio is the ideal ratio of air to fuel required for complete combustion in a combustion process. It represents the precise amount of air needed to completely burn a unit mass or volume of fuel, producing only carbon dioxide and water as combustion products, without any excess fuel or oxygen remaining. For hydrocarbon fuels such as gasoline or diesel, the stoichiometric A/F ratio is determined by the chemical composition of the fuel. It is calculated based on the stoichiometric equation, which balances the chemical reaction between the fuel and oxygen during combustion. Achieving the stoichiometric A/F ratio ensures optimal combustion efficiency and minimal emissions. However, in practice, combustion processes often operate with either excess air or excess fuel to account for variations in operating conditions, equipment design, and combustion stability. The stoichiometric A/F ratio serves as a reference point for tuning combustion systems, designing emission control strategies, and calculating fuel economy. It is a fundamental concept in combustion engineering and is essential for understanding and optimizing combustion processes in various applications, including automotive engines, industrial boilers, and power generation systems. © scanned with OKEN ScannerTo calculate the cycle efficiency, work ratio, and specific steam consumption of the Rankine cycle, we can use the following formulas: Cycle Efficiency (n): _ Net Work Output — Whet Heat Supplied Qin >. Work Ratio (WR): Actual Work Done by Turbine Wrurbine WR= = Maximum Possible Work Done by Turbine — Wiurbinemax 3. Specific Steam Consumption (SSC): Mass flow rate of steam m 85¢ = —______.___ = Net Work Output Waet First, we need to find the states of the Rankine cycle using the given pressures: 1. State1: P7= 30 bar (high pressure) dry saturated steam 2. State2: P2= 0.07 bar (low pressure) vapor Then, we calculate the enthalpies at these states using steam tables or other relevant data sources. After obtaining the enthalpies, we calculate the net work output ( W net) ,the actual work done by the turbine ( W turbine), and the mass flow rate of steam(m). Finally, we use the formulas above to find the cycle efficiency, work ratio, and specific steam consumption. Let's proceed with the calculations. © scanned with OKEN ScannerTo find the enthalpies at states 1 and 2, we can use steam tables or software. Let's assume: At state 1 (high pressure, dry saturated steam): * Enthalpy (h;) = 2800 kJ/kg (approximately) At state 2 (low pressure, saturated vapor): * Enthalpy (ha) = 2585 kJ/kg (approximately) Now, we can calculate the net work output (IV,.:) using the formula: Waet = hy = he Woyet = 2800 — 2585 = 215 kJ/kg Next, we calculate the actual work done by the turbine (Witurbine), Which is the difference between the enthalpies at states 1 and 2 (assuming ideal conditions): Wiurbine = hi — hg = 2800 — 2585 = 215 kJ/kg Now, we calculate the mass flow rate of steam (7m), which is usually not given directly but can be found using the given data and assumptions about the cycle: Wret 215 m= _ = = —*? ___ 0.0742 ke/ke of stez hi —hy 2800 — 2585 ayegcf eben Finally, we can calculate the cycle efficiency (77), work ratio (WR), and specific steam consumption (SSC): = Wret _ Wret = 215 ~ Qin hy —hy 2800 — 2585 Wrurbine _ Wiurbine Pa 215 me Wturbine,max Wret 215 m 0.0742 Wret 215 n = 0.0786 or 7.86% WR= SSC = ~ 0.000345 kg/kJ So, for the given Rankine cycle: * Efficiency: 7.86% * Work Ratio: 1 * Specific Steam Consumption: 0.000345 kg/kJ © scanned with OKEN Scanner