Q19Engineering Chemistry

Part A: Bomb Calorimeter - Principle, Construction, and Working

Principle of Operation: The Bomb Calorimeter is the universally accepted standard apparatus used for determining the Gross Calorific Value (GCV) of solid fuels (like coal, coke) and heavy liquid fuels (like heavy fuel oil). The fundamental principle relies on burning a precisely weighed quantity of the fuel in an environment of highly pressurized, pure oxygen. Because the combustion occurs inside a sealed, rigid, heavy-steel vessel (the 'bomb'), the volume remains absolutely constant ($V=\text{constant}$, therefore $\Delta V=0$). Consequently, the heat released by the combustion is equal to the change in internal energy ($\Delta E$) of the system, not the enthalpy. The heat generated by the explosive combustion is completely absorbed by a known mass of water surrounding the bomb. By precisely measuring the temperature rise of this water, the calorific value is calculated using the principles of calorimetry (Heat lost by fuel = Heat gained by water and apparatus).

Detailed Construction: 1. The Bomb: The core is a strong, highly pressure-resistant cylindrical vessel made of corrosion-resistant austenitic stainless steel. It must withstand internal pressures exceeding $50-100\text{ atm}$ generated during the explosive combustion. The bomb has a heavy, gas-tight screw cap. 2. The Crucible: Inside the bomb, a small crucible made of non-reactive platinum, silica, or stainless steel is suspended. The finely powdered fuel sample is placed here. 3. Ignition System: A thin magnesium or platinum ignition wire is stretched across the crucible, firmly touching the fuel sample. This wire is connected to two sturdy metal electrodes that pass through the lid of the bomb to an external 6-volt battery. Passing a current through the wire causes it to glow red hot, instantaneously igniting the fuel. 4. Oxygen Valve: A high-pressure valve on the lid allows pure oxygen from a cylinder to be pumped into the bomb up to a pressure of $25-30\text{ atmospheres}$. This massive excess of oxygen guarantees absolutely complete, instantaneous, and smokeless combustion of the fuel. 5. The Calorimeter Vessel: The entire heavy steel bomb is carefully lowered into a highly polished copper calorimeter vessel containing a very precisely weighed mass of distilled water (usually around $2000-2500\text{ grams}$). 6. Stirrer and Thermometer: A mechanical, electrically driven stirrer is immersed in the water to ensure rapid and uniform heat distribution. A highly sensitive Beckmann thermometer (capable of reading temperature changes as small as $0.01^{\circ }\text{C}$ or even $0.001^{\circ }\text{C}$) is inserted into the water to record the temperature rise. 7. Insulating Jackets: To ensure the system is perfectly adiabatic (no heat enters or leaves from the surroundings), the copper calorimeter is surrounded by an air jacket, which is in turn surrounded by a larger water jacket. This prevents radiation and convection losses.

Working Procedure: 1. A perfectly dry, accurately weighed sample of the solid fuel (usually around $1.000\text{ gram}$, mass = $m$) is placed securely in the crucible. 2. The magnesium ignition wire is attached to the electrodes and arranged so it lies directly on top of the powdered fuel. 3. The heavy bomb lid is screwed down tightly to ensure a perfect high-pressure seal. 4. Pure oxygen gas is slowly pumped into the bomb until the pressure gauge registers exactly $25\text{ atm}$. The valve is then closed tight. 5. The loaded bomb is carefully lowered into the copper calorimeter, which has been pre-filled with a precisely weighed mass of distilled water (mass = $W$). 6. The motorized stirrer is turned on. It is allowed to run for several minutes to ensure the water temperature is perfectly uniform throughout the vessel. 7. The initial temperature of the water ($t_1$) is read from the Beckmann thermometer with extreme precision. 8. The external electrical circuit is closed. A 6-volt current surges through the ignition wire, instantly melting it and generating a shower of sparks that ignites the fuel sample. 9. In the extremely oxygen-rich, high-pressure environment, the fuel burns explosively and completely within fractions of a second. Massive amounts of heat are liberated instantly. 10. This heat conducts through the thick steel walls of the bomb and is rapidly transferred to the surrounding water. The stirrer ensures this heat is distributed evenly. 11. The temperature of the water begins to rise rapidly. The operator carefully monitors the thermometer and records the absolute maximum peak temperature reached ($t_2$). 12. After the experiment, the bomb is carefully removed, the high-pressure gases are slowly vented, and the interior of the bomb is meticulously washed with distilled water. This washings are collected and titrated to determine the amounts of sulfuric acid and nitric acid formed during combustion (for precise corrections).

Derivation and Calculation of GCV: Let: - $m$ = Mass of the fuel sample burned (in kg) - $W$ = Mass of the water in the calorimeter (in kg) - $w$ = Water equivalent of the calorimeter apparatus (bomb, stirrer, thermometer, copper vessel). This represents the mass of water that would absorb the exact same amount of heat as the metal apparatus for a $1^{\circ }\text{C}$ rise. (in kg) - $t_1$ = Initial temperature of water (${}^{\circ }\text{C}$) - $t_2$ = Maximum final temperature of water (${}^{\circ }\text{C}$) - $GCV$ = Gross Calorific Value of the fuel (in kcal/kg or kJ/kg) - Specific heat of water = $1{\text{ kcal/kg}}^{\circ }\text{C}$ (or $4.184{\text{ kJ/kg}}^{\circ }\text{C}$)

According to the principle of conservation of energy (assuming a perfectly adiabatic system): Total Heat Liberated by the Fuel = Total Heat Absorbed by Water and Apparatus

Heat liberated by burning $m$ kg of fuel = $m\times GCV$ Heat absorbed by the water = $W\times \text{Specific Heat}\times (t_2-t_1)$ Heat absorbed by the apparatus = $w\times \text{Specific Heat}\times (t_2-t_1)$

Equating them: $m\times GCV=(W+w)\times (t_2-t_1)$ Therefore, the uncorrected formula is: $GCV=\frac{(W+w)(t_2-t_1)}{m}$

Precise Corrections: To achieve highly accurate industrial results, several corrections must be applied to the basic formula: 1. Cooling Correction ($T_c$): Even with perfect insulation, the calorimeter slowly loses some heat to the surroundings once it becomes hotter than the room. This cooling correction is added to the temperature difference. 2. Acid Correction ($A_c$): At high pressures and temperatures inside the bomb, the nitrogen ($N_2$) and sulphur ($S$) in the coal react with oxygen and moisture to form Nitric Acid ($HN{O}_3$) and Sulfuric Acid ($H_{2}S{O}_4$). These formations are highly exothermic (they release heat). Since this heat is not part of the normal fuel combustion under atmospheric conditions, it must be subtracted. 3. Fuse Wire Correction ($F_c$): The electrical burning of the magnesium ignition wire releases a small but measurable amount of heat. This must be subtracted. 4. Cotton Thread Correction ($C_c$): If a small piece of cotton thread is used to aid ignition, the heat from its combustion must be subtracted.

The final, highly accurate formula for Gross Calorific Value is:

Part B: Numerical Problem - Minimum Air Required for Complete Combustion

The fundamental goal is to calculate exactly how much atmospheric air must be supplied to a furnace to completely burn $1\text{ kg}$ of a specific coal sample without leaving any unburnt carbon.

Step 1: Analyze the Coal Composition (per 1 kg of fuel) The problem provides the percentage composition by mass. We convert this directly to mass fractions per kg: - Carbon ($C$) = $88\%=0.88\text{ kg}$ - Hydrogen ($H$) = $4\%=0.04\text{ kg}$ - Oxygen ($O$) = $4\%=0.04\text{ kg}$ (Note: This oxygen is chemically bound within the coal and aids combustion) - Nitrogen ($N$) = $2\%=0.02\text{ kg}$ (Inert gas, does not burn, requires no oxygen) - Sulphur ($S$) = $2\%=0.02\text{ kg}$ (Burns to form $S{O}_2$)

Step 2: Calculate Theoretical Oxygen Demand for Each Combustible Element We use the stoichiometric chemical equations of combustion to find the exact mass of $O_2$ required for each element.

1. For Carbon (C): The combustion reaction is: $C+O_2\to C{O}_2$ Molar masses: $12\text{ kg}$ of C reacts completely with $32\text{ kg}$ of $O_2$. Therefore, $1\text{ kg}$ of C requires $\frac{32}{12}$ or $\frac{8}{3}\text{ kg}$ of $O_2$. Oxygen required to burn $0.88\text{ kg}$ of Carbon: $=0.88\times \frac{32}{12}=2.3467\text{ kg of }{O}_2$

2. For Hydrogen (H): Before calculating, we must account for the oxygen already present inside the coal structure. It is assumed that all intrinsic oxygen is bound to hydrogen in the form of water ($H_{2}O$). Since water is $H_{2}O$ (Mass ratio of H:O is 2:16 or 1:8), every $8\text{ parts}$ of oxygen chemically bind and neutralize $1\text{ part}$ of hydrogen. The mass of hydrogen already neutralized = $\frac{\text{Mass of Oxygen present}}{8}=\frac{0.04}{8}=0.005\text{ kg}$. The 'Available Hydrogen' that actually needs external oxygen from the air to burn: Available $H=\text{Total }H-\text{Neutralized }H=0.04-0.005=0.035\text{ kg}$. The combustion reaction is: $2H_2+O_2\to 2H_{2}O$ Molar masses: $4\text{ kg}$ of H reacts with $32\text{ kg}$ of $O_2$. Therefore, $1\text{ kg}$ of H requires $\frac{32}{4}=8\text{ kg}$ of $O_2$. Oxygen required to burn the available $0.035\text{ kg}$ of Hydrogen: $=0.035\times 8=0.2800\text{ kg of }{O}_2$

3. For Sulphur (S): The combustion reaction is: $S+O_2\to S{O}_2$ Molar masses: $32\text{ kg}$ of S reacts completely with $32\text{ kg}$ of $O_2$. Therefore, $1\text{ kg}$ of S requires $1\text{ kg}$ of $O_2$. Oxygen required to burn $0.02\text{ kg}$ of Sulphur: $=0.02\times 1=0.0200\text{ kg of }{O}_2$

Step 3: Calculate Total Theoretical Oxygen Required Total $O_2$ Required = $O_2$ for Carbon + $O_2$ for Available Hydrogen + $O_2$ for Sulphur Total $O_2$ Required = $2.3467+0.2800+0.0200=2.6467\text{ kg}$ of pure $O_2$ per kg of coal.

Step 4: Convert Oxygen Demand to Atmospheric Air Demand In reality, furnaces are fed atmospheric air, not pure oxygen. By mass, standard atmospheric air consists of approximately $23\%$ Oxygen and $77\%$ Nitrogen. This means that to supply exactly $23\text{ kg}$ of pure oxygen to a furnace, you must pump in $100\text{ kg}$ of atmospheric air. Therefore, the mass of Air required = Total $O_2$ required $\times \frac{100}{23}$

Final Answer Summary: To ensure absolutely complete combustion of $1\text{ kg}$ of this specific coal sample, the furnace must be supplied with a theoretical absolute minimum of $11.51\text{ kg}$ of atmospheric air. In industrial practice, a large percentage of 'excess air' (e.g., $20-50\%$) would be added on top of this theoretical minimum to guarantee that every single carbon atom finds an oxygen atom in the turbulent furnace environment.